2. For any $[g] \in [(M, \partial M), (SO(3), 1)]$, we have $[g]^2 \in \langle [\rho_M(B^3)] \rangle$.

3. The group $[(M, \partial M), (SO(3), 1)]$ is abelian.

PROOF: Let $g \in [(M, \partial M), (SO(3), 1)]_m$. Assume that $H^1(g; \mathbb{Z}/2\mathbb{Z})$ is trivial. Choose a cell decomposition of $M$ relative to its boundary, with only one three-cell, no zero-cell if $\partial M \neq \emptyset$, one zero-cell if $\partial M = \emptyset$, one-cells, and two-cells. See [Hir94, Chapter 6, Section 3]. Then after a homotopy relative to $\partial M$, we may assume that $g$ maps the one-skeleton of $M$ to 1. Next, since $\pi_2(SO(3)) = 0$, we may assume that $g$ maps the two-skeleton of $M$ to 1, and therefore that $g$ maps the exterior of some 3-ball to 1. Now $g$ becomes a map from $B^3/\partial B^3 = S^3$ to $SO(3)$, and its homotopy class is $k[\tilde{\rho}]$ in $\pi_3(SO(3)) = \mathbb{Z}[\tilde{\rho}]$. Therefore, the map $g$ is homotopic to $\rho_M(B^3)^k$. This proves the first assertion.

Since $H^1(g^2; \mathbb{Z}/2\mathbb{Z}) = 2H^1(g; \mathbb{Z}/2\mathbb{Z})$ is trivial, the second assertion follows.

For the third assertion, first note that $[\rho_M(B^3)]$ belongs to the center of $[(M, \partial M), (SO(3), 1)]$ because it can be supported in a small ball disjoint from the support (preimage of $SO(3) \setminus \{1\}$) of a representative of any other element. Thus, according to the second assertion, any square is in the center. In particular, if $f$ and $g$ are elements of $[(M, \partial M), (SO(3), 1)]$, we have

$$(gf)^2 = (fg)^2 = (f^{-1}f^2g^2f)(f^{-1}g^{-1}fg),$$

where the first factor is equal to $f^2g^2 = g^2f^2$. Exchanging $f$ and $g$ yields $f^{-1}g^{-1}fg = g^{-1}f^{-1}gf$. Therefore, the commutator, which is a power of $[\rho_M(B^3)]$, thanks to Lemma 5.1 and to the first assertion, has a vanishing square, and thus a vanishing degree. So it is trivial. $\square$

## 5.2 Any oriented 3-manifold is parallelizable.

In this subsection, we prove the following standard theorem. The spirit of our proof is the same as the Kirby proof in [Kir89, p.46]. But instead of assuming familiarity with the obstruction theory described by Norman Steenrod in [Ste51, Part III], we use this proof to introduce this theory.

**Theorem 5.3 (Stiefel).** Any orientable 3-manifold is parallelizable.

**Lemma 5.4.** Let $M$ be an oriented 3-manifold. Let $S$ be a closed surface, orientable or not, immersed in $M$. Then the restriction to $S$ of the tangent bundle $TM$ to $M$ is trivializable.
PROOF: Let us first prove that this bundle is independent of the immersion. It is the direct sum of the tangent bundle to the surface and of its normal one-dimensional bundle. This normal bundle is trivial when $S$ is orientable. Otherwise, its unit bundle is the 2-fold orientation cover of the surface. (The orientation cover of $S$ is its 2-fold orientable cover that is trivial over annuli embedded in the surface). Since any surface $S$ can be immersed in $\mathbb{R}^3$, the restriction $TM|_S$ is the pull-back of the trivial bundle of $\mathbb{R}^3$ by such an immersion, and it is trivial. □

Using Stiefel-Whitney classes, the proof of Theorem 5.3 goes as follows. QUICK PROOF OF THEOREM 5.3: Let $M$ be an orientable smooth 3-manifold, equipped with a smooth triangulation.<fnref n="1" /> By definition, the first Stiefel-Whitney class $w_1(TM) \in H^1(M; \mathbb{Z}/2\mathbb{Z} = \pi_0(GL(\mathbb{R}^3)))$ viewed as a map from $\pi_1(M)$ to $\mathbb{Z}/2\mathbb{Z}$ maps the class of a loop $c$ embedded in $M$ to 0 if $TM|_c$ is orientable, and to 1 otherwise. It is the obstruction to the existence of a trivialization of $TM$ over the one-skeleton of $M$. Since $M$ is orientable, the first Stiefel-Whitney class $w_1(TM)$ vanishes. So $TM$ can be trivialized over the one-skeleton of $M$. The second Stiefel-Whitney class $w_2(TM) \in H^2(M; \mathbb{Z}/2\mathbb{Z} = \pi_1(GL^+(\mathbb{R}^3)))$ viewed as a map from $H_2(M; \mathbb{Z}/2\mathbb{Z})$ to $\mathbb{Z}/2\mathbb{Z}$ maps the class of a connected closed surface $S$ to 0 if $TM|_S$ is trivializable, and to 1 otherwise. The second Stiefel-Whitney class $w_2(TM)$ is the obstruction to the existence of a trivialization of $TM$ over the two-skeleton of $M$ when $w_1(TM) = 0$. According to the above lemma, we have $w_2(TM) = 0$, and $TM$ can be trivialized over the two-skeleton of $M$. Then since $\pi_2(GL^+(\mathbb{R}^3)) = 0$, any parallelization over the two-skeleton of $M$ can be extended as a parallelization of $M$. □

We detail the involved arguments without mentioning Stiefel-Whitney classes, (in fact, by almost defining $w_2(TM)$) below. The elementary proof below can serve as an introduction to the obstruction theory used above. ELEMENTARY PROOF OF THEOREM 5.3: Let $M$ be an oriented 3-manifold. Choose a triangulation of $M$. For any cell $c$ of the triangulation, define an arbitrary trivialization $\tau_c: c \times \mathbb{R}^3 \to TM|_c$ such that $\tau_c$ induces the orientation of $M$. This defines a trivialization $\tau^{(0)}: M^{(0)} \times \mathbb{R}^3 \to TM|_{M^{(0)}}$ of $M$ over the 0-skeleton $M^{(0)}$ of $M$, which is the set of 0-dimensional cells of the triangulation. Let $C_k(M)$ be the set of $k$-cells of the triangulation. Equip every cell with an arbitrary orientation. Let $e \in C_1(M)$ be an edge of the triangulation. On $\partial e$, the trivialization $\tau^{(0)}$ may be written as $\tau^{(0)} = \tau_e \circ \psi_{\mathbb{R}}(g_e)$, for a map $g_e: \partial e \to GL^+(\mathbb{R}^3)$. Since $GL^+(\mathbb{R}^3)$ is connected, the map $g_e$ extends to $e$, and $\tau^{(1)} = \tau_e \circ \psi_{\mathbb{R}}(g_e)$ extends $\tau^{(0)}$ to $e$. Doing so for all the edges extends

<footnote n="1">¹A theorem of Henry Whitehead proved in the Munkres book [Mun66] ensures the existence of such a triangulation.</footnote>
$\tau^{(0)}$ to a trivialization $\tau^{(1)}$ of the one-skeleton $M^{(1)}$ of $M$, which is the union of the edges of the triangulation.

Let $t$ be a triangle of the triangulation. There is a map $g_t: \partial t \to GL^+(\mathbb{R}^3)$ such that $\tau^{(1)} = \tau_t \circ \psi_{\mathbb{R}}(g_t)$ on $\partial t$. Let $E(t, \tau^{(1)})$ be the homotopy class of $g_t$ in $(\pi_1(GL^+(\mathbb{R}^3)) = \pi_1(SO(3)) = \mathbb{Z}/2\mathbb{Z})$. Then $E(., \tau^{(1)}): C_2(M) \to \mathbb{Z}/2\mathbb{Z}$ extends linearly to a cochain, which is independent of the $\tau_t$. When $E(., \tau^{(1)}) = 0$, $\tau^{(1)}$ extends to a trivialization $\tau^{(2)}$ over the two-skeleton of $M$, as before. Since $\pi_2(GL^+(\mathbb{R}^3)) = 0$, the trivialization $\tau^{(2)}$ can next be extended over the three-skeleton of $M$. So it extends over $M$.

Let us now study the obstruction cochain $E(., \tau^{(1)})$ whose vanishing guarantees the existence of a parallelization of $M$.

Choose a map $d(e): (e, \partial e) \to (GL^+(\mathbb{R}^3), 1)$ for every edge $e$ of the triangulation. Let $\tau^{(1)'}$ be the trivialization of $TM|_{M^{(1)}}$ obtained as above by changing the map $g_e$ associated to $e$ to $d(e)g_e$. Define the cochain

$$D(\tau^{(1)}, \tau^{(1)'}) : (\mathbb{Z}/2\mathbb{Z})^{C_1(M)} \to \mathbb{Z}/2\mathbb{Z}$$

that maps $e$ to the homotopy class of $d(e)$. Then $(E(., \tau^{(1)'}) - E(., \tau^{(1)}))$ is the coboundary of $D(\tau^{(1)}, \tau^{(1)'})$. (See Section A.1, before Theorem A.9.)

Let us prove that $E(., \tau^{(1)})$ is a cocycle. Consider a 3-simplex $T$. The trivialization $\tau^{(0)}$ extends to $T$. Assume that $\tau_T$ is an extension of $\tau^{(0)}$ to $T$, that $\tau_t$ is the restriction of $\tau_T$ to $t$ for any face $t$ of $T$, and that the above $\tau^{(1)'}$ coincides with $\tau_T$ on the edges of $\partial T$. Then we have $E(., \tau^{(1)'})(\partial T) = 0$. Since any coboundary maps $\partial T$ to 0, we also have $E(., \tau^{(1)})(\partial T) = 0$.

Now, it suffices to prove that the cohomology class of $E(., \tau^{(1)})$ (which is equal to $w_2(TM)$) vanishes in order to prove the existence of an extension $\tau^{(1)'}$ of $\tau^{(0)}$ on $M^{(1)}$ that extends on $M$.

Since $H^2(M; \mathbb{Z}/2\mathbb{Z}) = \operatorname{Hom}(H_2(M; \mathbb{Z}/2\mathbb{Z}); \mathbb{Z}/2\mathbb{Z})$, it suffices to prove that $E(., \tau^{(1)})$ maps any 2-dimensional $\mathbb{Z}/2\mathbb{Z}$-cycle $C$ to 0.

We represent the class of such a cycle $C$ by a closed surface $S$, orientable or not, as follows.

Let $N(M^{(0)})$ be a disjoint union of balls around the vertices and let $N(M^{(1)})$ be a small regular neighborhood of $M^{(1)}$ in $M$ as in Figure 5.1. We assume that $N(M^{(1)}) \cap (M \setminus N(M^{(0)}))$ is a disjoint union, over the edges $e$, of solid cylinders $B_e$ identified with $]0, 1[ \times D^2$. The core $]0, 1[ \times \{0\}$ of $B_e = ]0, 1[ \times D^2$ is a connected part of the interior of the edge $e$. (As in Figure 5.1, the neighborhood $N(M^{(1)})$ is thinner than $N(M^{(0)})$.)

Construct $S$ in the complement of $N(M^{(0)}) \cup N(M^{(1)})$ as the intersection of the support of $C$ with this complement. Then the closure of $S$ meets the part $[0, 1] \times S^1$ of every $\bar{B}_e$ as an even number of parallel intervals from $\{0\} \times S^1$ to $\{1\} \times S^1$. Complete $S$ in $M \setminus N(M^{(0)})$ by connecting the intervals
<figure><img src="image_45.png" /><figcaption>Figure 5.1: The neighborhoods $N(M^{(1)})$ and $N(M^{(0)})$</figcaption></figure>

pairwise in $\bar{B}_e$ by disjoint bands. After this operation, the boundary of the closure of $S$ is a disjoint union of circles in the boundary of $N(M^{(0)})$. Glue disjoint disks of $N(M^{(0)})$ along these circles to finish the construction of $S$.

Extend $\tau^{(0)}$ to $N(M^{(0)})$, assume that $\tau^{(1)}$ coincides with this extension over $M^{(1)} \cap N(M^{(0)})$, and extend $\tau^{(1)}$ to $N(M^{(1)})$. The bundle $TM|_S$ is trivial, and we may choose a trivialization $\tau_S$ of $TM$ over $S$ that coincides with our extension of $\tau^{(0)}$ over $N(M^{(0)})$, over $S \cap N(M^{(0)})$. We have a cell decomposition of $(S, S \cap N(M^{(0)}))$ with only 1-cells and 2-cells, for which the 2-cells of $S$ are in one-to-one canonical correspondence with the 2-cells of $C$, and one-cells correspond bijectively to bands connecting two-cells in the cylinders $B_e$. These one-cells are equipped with the trivialization of $TM$ induced by $\tau^{(1)}$. Then we can define 2-dimensional cochains $E_S(., \tau^{(1)})$ and $E_S(., \tau_S)$ as before, with respect to this cellular decomposition of $S$. The cochain $(E_S(., \tau^{(1)}) - E_S(., \tau_S))$ is again a coboundary, and $E_S(., \tau_S) = 0$. So $E_S(C, \tau^{(1)}) = 0$, and we also have $E(C, \tau^{(1)}) = 0$ since $E(C, \tau^{(1)}) = E_S(C, \tau^{(1)})$. $\square$

Theorem 5.3 has the following immediate corollary.

**Proposition 5.5.** Any punctured oriented 3-manifold $\tilde{R}$ as in Definition 3.6 can be equipped with an asymptotically standard parallelization.

**PROOF:** The oriented manifold $R$ admits a parallelization $\tau_0: R \times \mathbb{R}^3 \to TR$. Over $\tilde{B}_{1,\infty} \setminus \{\infty\}$, $\tau_s = \tau_0 \circ \psi_R(g)$ for a map $g: \tilde{B}_{1,\infty} \setminus \{\infty\} \to GL^+(\mathbb{R}^3)$. For $r \in [1, 2]$, let $\tilde{B}_{r,\infty}$ (resp. $B_{r,\infty}$) be the complement in $S^3$ of the closed (resp. open) ball $B(r)$ of radius $r$ in $\mathbb{R}^3$. Since $\pi_2(GL^+(\mathbb{R}^3)) = \{0\}$, the restriction of $g$ to $B_{7/4,\infty} \setminus \tilde{B}_{2,\infty}$ extends to a map of $\tilde{B}_{1,\infty} \setminus \tilde{B}_{2,\infty}$ that maps $\tilde{B}_{1,\infty} \setminus \tilde{B}_{5/3,\infty}$ to 1. After smoothing, we get a smooth map $\tilde{g}: \tilde{B}_{1,\infty} \setminus \{\infty\} \to GL^+(\mathbb{R}^3)$ that coincides with $g$ on $\tilde{B}_{2,\infty}$ and that maps $\tilde{B}_{1,\infty} \setminus \tilde{B}_{3/2,\infty}$ to 1. Extend $\tilde{g}$ to $\tilde{R}$ so that it maps $\tilde{R} \setminus \tilde{B}_{3/2,\infty}$ to 1. So $\tau_0 \circ \psi_R(\tilde{g})$ is an asymptotically standard parallelization as wanted. $\square$
## 5.3 The homomorphism induced by the degree

In this section, $M$ is a compact connected oriented 3-manifold, with or without boundary. Let $S$ be a closed surface, orientable or not, embedded in the interior of our manifold $M$. Let $\tau$ be a parallelization of our 3-manifold $M$. We define a twist $g(S, \tau) \in [(M, \partial M), (SO(3), 1)]_m$ below.

The surface $S$ has a tubular neighborhood $N(S)$, which is a $[-1, 1]$-bundle over $S$. This bundle admits (orientation-preserving) bundle charts $\phi: [-1, 1] \times D \to N(S)$ for disks $D$ of $S$, such that the changes of coordinates restrict to the fibers as $\pm 1$, where $1$ denotes the identity map. Let $\phi(t, s)$ be in the image of such a chart. Let $N_s = T_{\phi(0,s)}\phi([-1, 1] \times s)$ be the tangent vector to the fiber $\phi([-1, 1] \times s)$ at $\phi(0, s)$. Define $g(S, \tau)(\phi(t, s))$ to be the rotation with angle $\pi(t+1)$ and with axis

$$p_2(\tau^{-1}(N_s) = (\phi(0, s), p_2(\tau^{-1}(N_s))) ).$$

Since this rotation coincides with the rotation with opposite axis and with opposite angle $\pi(1-t)$, this provides a consistent definition of $g(S, \tau)|_{N(S)}$. Extend $g(S, \tau)|_{N(S)}$ to $M$ so that

$$g(S, \tau): (M, \partial M) \to (SO(3), 1)$$

maps $M \setminus N(S)$ to 1.

The homotopy class of $g(S, \tau)$ depends only on the homotopy class of $\tau$ and on the isotopy class of $S$. When $M = B^3$, when $\tau$ is the standard parallelization of $\mathbb{R}^3$, and when $\frac{1}{2}S^2$ denotes the sphere $\frac{1}{2}\partial B^3$ inside $B^3$, the homotopy class of $g(\frac{1}{2}S^2, \tau)$ coincides with the homotopy class of $\rho$.

We will see later (Proposition 5.33) that the homotopy class of $g(S, \tau)$ depends only on the Euler characteristic $\chi(S)$ of $S$ and on the class of $S$ in $H_2(M; \mathbb{Z}/2\mathbb{Z})$. Thus, we will simply denote $g(S, \tau)$ by $g(S)$. We will also see (Corollary 5.34) that any element of $[(M, \partial M), (SO(3), 1)]$ can be represented by some $g(S)$.

**Lemma 5.6.** The morphism $H^1(g(S, \tau); \mathbb{Z}/2\mathbb{Z})$ maps the generator of

$$H^1(SO(3); \mathbb{Z}/2\mathbb{Z})$$

to the mod 2 intersection with $S$ in

$$\mathrm{Hom}(H_1(M, \partial M; \mathbb{Z}/2\mathbb{Z}), \mathbb{Z}/2\mathbb{Z}) = H^1(M, \partial M; \mathbb{Z}/2\mathbb{Z}).$$

Thus, the morphism $H^1(\cdot; \mathbb{Z}/2\mathbb{Z}) : [(M, \partial M), (SO(3), 1)] \to H^1(M, \partial M; \mathbb{Z}/2\mathbb{Z})$ is onto.
**PROOF:** The first assertion is obvious.

The second one follows since $H^1(M, \partial M; \mathbb{Z}/2\mathbb{Z})$ is the Poincaré dual of $H_2(M; \mathbb{Z}/2\mathbb{Z})$ and since any element of $H_2(M; \mathbb{Z}/2\mathbb{Z})$ is the class of a closed surface. $\square$

**Lemma 5.7.** *The degree is a group homomorphism*

$$ \deg: [(M, \partial M), (SO(3), 1)] \to \mathbb{Z} $$

$$ \textit{and } \deg(\rho_M(B^3)^k) = 2k. $$

**PROOF:** It is easy to see that $\deg(fg) = \deg(f) + \deg(g)$ when $f$ or $g$ is a power of $[\rho_M(B^3)]$.

Let us prove that $\deg(f^2) = 2\deg(f)$ for any $f$. According to Lemma 5.6, there is an unoriented embedded surface $S_f$ of the interior of $C$ such that $H^1(f; \mathbb{Z}/2\mathbb{Z}) = H^1(g(S_f, \tau); \mathbb{Z}/2\mathbb{Z})$ for some trivialization $\tau$ of $TM$. According to Lemmas 5.1 and 5.2, $fg(S_f, \tau)^{-1}$ is homotopic to some power of $\rho_M(B^3)$. So, it suffices to prove that the degree of $g^2$ is $2\deg(g)$ for $g = g(S_f, \tau)$. This can be done easily, by noticing that $g^2$ is homotopic to $g(S_f^{(2)}, \tau)$, where $S_f^{(2)}$ is the boundary of the tubular neighborhood of $S_f$. In general, we have

$$ \deg(fg) = \frac{1}{2} \deg((fg)^2) = \frac{1}{2} \deg(f^2 g^2) = \frac{1}{2} (\deg(f^2) + \deg(g^2)), $$

and the lemma is proved. $\square$

Lemmas 5.2 and 5.7 imply the following lemma.

**Lemma 5.8.** *The degree induces an isomorphism*

$$ \deg: [(M, \partial M), (SO(3), 1)] \otimes_{\mathbb{Z}} \mathbb{Q} \to \mathbb{Q}. $$

Any group homomorphism $\psi: [(M, \partial M), (SO(3), 1)] \to \mathbb{Q}$ may be expressed as

$$ \frac{1}{2}\psi(\rho_M(B^3)) \deg. $$

<figure><img src="image_46.png" /></figure>

Recall that $M$ is a compact connected oriented 3-manifold in this section. So $H^3(M, \partial M; \pi_3(SO(3)))$ is canonically isomorphic to $\mathbb{Z}$.

**Proposition 5.9.** Let $P: H^1(M, \partial M; \mathbb{Z}/2\mathbb{Z}) \to H_2(M; \mathbb{Z}/2\mathbb{Z})$ be the Poincaré duality isomorphism.
Let $i: H^3(M, \partial M; \pi_3(SO(3))) \to [(M, \partial M), (SO(3), 1)]$ be the group morphism that sends the canonical generator $[M, \partial M]$ of $H^3(M, \partial M; \pi_3(SO(3)))$ to $[\rho_M(B^3)]$. Then the sequence

$$0 \to H^3(M, \partial M; \pi_3(SO(3))) \xrightarrow{i} [(M, \partial M), (SO(3), 1)]$$

$$\xrightarrow{P \circ H^1([\cdot]; \mathbb{Z}/2\mathbb{Z})} H_2(M; \mathbb{Z}/2\mathbb{Z}) \to 0,$$

is a canonical exact sequence.

PROOF: The proposition is a consequence of Lemmas 5.2, 5.6, and 5.7. $\square$

## 5.4 On the groups $SU(n)$

Let $K = \mathbb{R}$ or $\mathbb{C}$. Let $n \in \mathbb{N}$. The stabilization maps induced by the inclusions

$$i: \begin{array}{ccc} GL(K^n) & \to & GL(K \oplus K^n) \\ g & \mapsto & i(g) : (x, y) \mapsto (x, g(y)) \end{array}$$

are denoted by $i$. We represent elements of $GL(K^n)$ by matrices whose columns contain the coordinates of the images of the basis elements with respect to the standard basis of $K^n$. View $S^3$ as the unit sphere of $\mathbb{C}^2$. So, its elements are the pairs $(z_1, z_2)$ of complex numbers such that $|z_1|^2 + |z_2|^2 = 1$. The group $SU(2)$ may be identified with $S^3$ by the homeomorphisms

$$m_r^{\mathbb{C}}: \begin{array}{ccc} S^3 & \to & SU(2) \\ (z_1, z_2) & \mapsto & \begin{bmatrix} z_1 & -\bar{z}_2 \\ z_2 & \bar{z}_1 \end{bmatrix} \end{array} \quad \text{and} \quad \bar{m}_r^{\mathbb{C}}: \begin{array}{ccc} S^3 & \to & SU(2) \\ (z_1, z_2) & \mapsto & \begin{bmatrix} \bar{z}_1 & \bar{z}_2 \\ -z_2 & z_1 \end{bmatrix}. \end{array}$$

So the first nontrivial homotopy group of $SU(2)$ is $\pi_3(SU(2)) = \mathbb{Z}[\overline{m_r^{\mathbb{C}}}]$, where $[\overline{m_r^{\mathbb{C}}}] = -[m_r^{\mathbb{C}}]$ and $\overline{m_r^{\mathbb{C}}}$ is a group homomorphism (it induces the group structure of $S^3$). The long exact sequence associated to the fibration

$$SU(n-1) \xrightarrow{i} SU(n) \to S^{2n-1},$$

described in Theorem A.14, shows that $i_*^n: \pi_j(SU(2)) \to \pi_j(SU(n+2))$ is an isomorphism for $j \le 4$ and $n \ge 0$. In particular, the group $\pi_j(SU(4))$ is trivial for $j \le 2$ and we have

$$\pi_3(SU(4)) = \mathbb{Z}[i_*^2(\overline{m_r^{\mathbb{C}}})],$$
where $i_*^2(\overline{m_r^C})$ is the following map

$$ i_*^2(\overline{m_r^C}): \begin{array}{ccc} (S^3 \subset \mathbb{C}^2) & \to & SU(4) \\ (z_1, z_2) & \mapsto & \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & \overline{z_1} & \overline{z_2} \\ 0 & 0 & -z_2 & z_1 \end{bmatrix} \end{array} $$

## 5.5 Definition of relative Pontrjagin numbers

Let $M_0$ and $M_1$ be two compact connected oriented 3-manifolds whose boundaries have collars identified by a diffeomorphism. Let $\tau_0: M_0 \times \mathbb{R}^3 \to TM_0$ and $\tau_1: M_1 \times \mathbb{R}^3 \to TM_1$ be two parallelizations (which respect the orientations) that agree on the collar neighborhoods of $\partial M_0 = \partial M_1$. Then the relative Pontrjagin number $p_1(\tau_0, \tau_1)$ is the Pontrjagin obstruction to extending the trivialization of $TW \otimes \mathbb{C}$ induced by $\tau_0$ and $\tau_1$ across the interior of a signature 0 cobordism $W$ from $M_0$ to $M_1$. Details follow.

Let $M$ be a compact connected oriented 3-manifold. A special complex trivialization of $TM$ is a trivialization of $TM \otimes \mathbb{C}$ that is obtained from a trivialization $\tau_M: M \times \mathbb{R}^3 \to TM$ by composing $(\tau_M^C = \tau_M \otimes \mathbb{R} \mathbb{C}): M \times \mathbb{C}^3 \to TM \otimes \mathbb{C}$ by

$$ \psi(G): \begin{array}{ccc} M \times \mathbb{C}^3 & \to & M \times \mathbb{C}^3 \\ (x, y) & \mapsto & (x, G(x)(y)) \end{array} $$

for a map $G: (M, \partial M) \to (SL(3, \mathbb{C}), 1)$, which is a map $G: M \to SL(3, \mathbb{C})$ that maps $\partial M$ to 1. The definition and properties of relative Pontrjagin numbers $p_1(\tau_0, \tau_1)$, given with more details below, are valid for pairs $(\tau_0, \tau_1)$ of special complex trivializations.

The signature of a 4-manifold is the signature of the intersection form on its $H_2(\cdot; \mathbb{R})$ (i.e., the number of positive entries minus the number of negative entries in a diagonalized version of this form). It is well known that any closed oriented three-manifold bounds a compact oriented 4-dimensional manifold. See [Rou85] for an elegant elementary proof. The signature of such a bounded 4-manifold may be changed arbitrarily by connected sums with copies of $\mathbb{C}P^2$ or $-\mathbb{C}P^2$. A cobordism from $M_0$ to $M_1$ is a compact oriented 4-dimensional manifold $W$ with ridges such that

$$ \partial W = -M_0 \cup_{\partial M_0 \sim 0 \times \partial M_0} ([-0, 1] \times \partial M_0) \cup_{\partial M_1 \sim 1 \times \partial M_0} M_1, $$

where $W$ is identified with an open subspace of one of the products $[0, 1] \times M_0$ or $]0, 1] \times M_1$ near $\partial W$, as the following picture suggests.
$$ \{0\} \times M_0 = M_0 \quad \left[ \begin{array}{ccc} & & W^4 \\ & & \vec{N} \\ \hline & & \end{array} \right] \quad \{1\} \times M_1 = M_1 \\ [0, 1] \times (-\partial M_0)' \end{array} $$

Let $W = W^4$ be such a cobordism from $M_0$ to $M_1$, with signature 0. Consider the complex 4-bundle $TW \otimes \mathbb{C}$ over $W$. Let $\vec{N}$ be the tangent vector to $[0, 1] \times \{\text{pt}\}$ over $\partial W$ (under the above identifications). Let $\tau(\tau_0, \tau_1)$ denote the trivialization of $TW \otimes \mathbb{C}$ over $\partial W$ obtained by stabilizing either $\tau_0$ or $\tau_1$ into $\vec{N} \oplus \tau_0$ or $\vec{N} \oplus \tau_1$. Then the obstruction to extending this trivialization to $W$ is the relative first Pontrjagin class

$$ p_1(W; \tau(\tau_0, \tau_1)) [W, \partial W] \in (H^4(W, \partial W; \mathbb{Z} = \pi_3(SU(4))) = \mathbb{Z}[W, \partial W] $$

of the trivialization.

Now, we specify our sign conventions for this Pontrjagin class. They are the same as in [MS74]. In particular, $p_1$ is the opposite of the complexified tangent bundle's second Chern class $c_2$. See [MS74, p. 174]. Let us describe these conventions. The determinant bundle of $TW$ is trivial because $W$ is oriented, and $\det(TW \otimes \mathbb{C})$ is also trivial. Our parallelization $\tau(\tau_0, \tau_1)$ over $\partial W$ is special with respect to the trivialization of $\det(TW \otimes \mathbb{C})$. Equip $M_0$ and $M_1$ with Riemannian metrics that coincide near $\partial M_0$. Equip $W$ with a Riemannian metric that coincides with the orthogonal product metric of one of the products $[0, 1] \times M_0$ or $[0, 1] \times M_1$ near $\partial W$. Equip $TW \otimes \mathbb{C}$ with the associated Hermitian structure. Up to homotopy, assume that $\tau(\tau_0, \tau_1)$ is unitary with respect to the Hermitian structure of $TW \otimes \mathbb{C}$ and the standard Hermitian form of $\mathbb{C}^4$. Since $\pi_i(SU(4)) = \{0\}$ when $i < 3$, the trivialization $\tau(\tau_0, \tau_1)$ extends to a special unitary trivialization $\tau$ outside the interior of a 4-ball $B^4$ and defines

$$ \tau: S^3 \times \mathbb{C}^4 \to (TW \otimes \mathbb{C})|_{S^3} $$

over the boundary $S^3 = \partial B^4$ of this 4-ball $B^4$. Over this 4-ball $B^4$, the bundle $TW \otimes \mathbb{C}$ admits a special unitary trivialization

$$ \tau_B: B^4 \times \mathbb{C}^4 \to (TW \otimes \mathbb{C})|_{B^4}. $$

Then $\tau_B^{-1} \circ \tau(v \in S^3, w \in \mathbb{C}^4) = (v, \phi(v)(w))$ for a map $\phi: S^3 \to SU(4)$ whose homotopy class may be written as

$$ [\phi] = p_1(W; \tau(\tau_0, \tau_1)) [i_*^2(\overline{m_r^C})] \in \pi_3(SU(4)), $$

where $i_*^2(\overline{m_r^C})$ was defined at the end of Section 5.4.

$$ \text{Define } p_1(\tau_0, \tau_1) = p_1(W; \tau(\tau_0, \tau_1)). $$
**Proposition 5.10.** Let $M_0$ and $M_1$ be two compact connected oriented 3-manifolds whose boundaries have collars identified by a diffeomorphism. Let $\tau_0: M_0 \times \mathbb{C}^3 \to TM_0 \otimes \mathbb{C}$ and $\tau_1: M_1 \times \mathbb{C}^3 \to TM_1 \otimes \mathbb{C}$ be two special complex trivializations (which respect the orientations) that agree on the collar neighborhoods of $\partial M_0 = \partial M_1$. The (first) Pontrjagin number $p_1(\tau_0, \tau_1)$ is well defined by the above conditions.

**PROOF:** According to the Nokivov additivity theorem, if a closed 4-manifold $Y$ is decomposed as $Y = Y^+ \cup_X Y^-$, where $Y^+$ and $Y^-$ are two 4-manifolds with boundary, embedded in $Y$, which intersect along a closed 3-manifold $X$ (their common boundary, up to orientation), then we have

$$ \mathrm{signature}(Y) = \mathrm{signature}(Y^+) + \mathrm{signature}(Y^-). $$

According to a theorem of Vladimir Rohlin (see [Roh52] or [GM86, p. 18]), when $Y$ is a compact oriented 4-manifold without boundary, we have

$$ p_1(Y) = 3 \mathrm{signature}(Y). $$

We only need to prove that $p_1(\tau_0, \tau_1)$ is independent of the signature 0 cobordism $W$. Let $W_E$ be a 4-manifold of signature 0 bounded by $(-\partial W)$. Then $W \cup_{\partial W} W_E$ is a 4-dimensional manifold without boundary whose signature is

$$ \mathrm{signature}(W_E) + \mathrm{signature}(W) = 0 $$

by the Novikov additivity theorem. According to the Rohlin theorem, the first Pontrjagin class of $W \cup_{\partial W} W_E$ is also zero. On the other hand, this first Pontrjagin class is the sum of the relative first Pontrjagin classes of $W$ and $W_E$ with respect to $\tau(\tau_0, \tau_1)$. These two relative Pontrjagin classes are opposite. Therefore, the relative first Pontrjagin class of $W$ with respect to $\tau(\tau_0, \tau_1)$ does not depend on $W$. $\square$

Similarly, it is easy to prove the following proposition.

**Proposition 5.11.** Under the above assumptions, except for the assumption on the signature of the cobordism $W$, we have

$$ p_1(\tau_0, \tau_1) = p_1(W; \tau(\tau_0, \tau_1)) - 3 \mathrm{signature}(W). $$

<figure><img src="image_47.png" /></figure>

**Remark 5.12.** When $\partial M_1 = \emptyset$ and $M_0 = \emptyset$, the map $p_1(= p_1(\tau(\emptyset), .))$ coincides with the map $h$ studied by Friedrich Hirzebruch in [Hir73, §3.1], and by Robion Kirby and Paul Melvin in [KM99] under the name of *Hirzebruch defect*.
**Definition 5.13.** When $(\tilde{R}, \tau)$ is an asymptotic rational homology $\mathbb{R}^3$, set

$$p_1(\tau) = p_1((\tau_s)|_{B^3}, \tau|_{B_R}),$$

with the notation of Proposition 5.10.

**Lemma 5.14.** Let $(\tilde{R}, \tau)$ be an asymptotic rational homology $\mathbb{R}^3$ as in Definition 3.8. The parallelization $\tau: \tilde{R} \times \mathbb{R}^3 \to T\tilde{R}$ induces the parallelization $\bar{\tau}: (-\tilde{R}) \times \mathbb{R}^3 \to T(-\tilde{R})$ such that $\bar{\tau}(x, v) = -\tau(x, v)$.

Compose the orientation-preserving identification of a neighborhood of $\infty$ in $R$ with $\tilde{B}_{1,\infty}$ by the (restriction of the) multiplication by $(-1)$ in $\mathbb{R}^3 \cup \{\infty\}$ in order to get an orientation-preserving identification of a neighborhood of $\infty$ in $(-R)$ with $\tilde{B}_{1,\infty}$. Then

$$\left(-\tilde{R} = \tilde{\overbrace{(-R), \bar{\tau}}}\right)$$

is an asymptotic rational homology $\mathbb{R}^3$, and we have $p_1(\bar{\tau}) = -p_1(\tau)$.

**PROOF:** Use a signature 0 cobordism $W$ from $\{0\} \times B^3$ to $\{1\} \times B_R$ to compute $p_1(\tau)$. Extend the trivialization of $TW \otimes \mathbb{C}$ on $\partial W$, which may be expressed as $\vec{N} \oplus \tau_s$ on $\{0\} \times B^3 \cup (-[0, 1] \times \partial B^3)$, and $\vec{N} \oplus \tau$ on $\{1\} \times B_R$, to a special trivialization on the complement of an open ball $\tilde{B}^4$ in $W$. Let $\overline{W}$ be the cobordism obtained from $W$ by reversing the orientation of $W$. Equip $W \setminus \tilde{B}^4$ with the trivialization obtained from the above trivialization by a composition by $\mathbf{1}_{\mathbb{R}} \times (-\mathbf{1})_{\mathbb{R}^3}$. Then the changes of trivializations $\phi: \partial B^4 \to SU(4)$ and $\bar{\phi}: \partial \overline{B^4} \to SU(4)$ are obtained from one another by the orientation-preserving conjugation by $\mathbf{1}_{\mathbb{R}} \times (-\mathbf{1})_{\mathbb{R}^3}$. Since $\partial \overline{B^4}$ and $\partial B^4$ have opposite orientations, we get the result. $\square$

Back to the invariant $\Theta$ defined in Corollary 4.8, we can now prove the following proposition.

**Proposition 5.15.** For any $\mathbb{Q}$-sphere $R$, we have $\Theta(-R) = -\Theta(R)$.

**PROOF:** Note that $C_2(-R)$ is naturally identified with $C_2(R)$, with the same orientation. Recall the orientation-reversing diffeomorphism $\iota$ of $C_2(R)$ that exchanges the two coordinates in $\tilde{R}^2 \setminus \Delta(\tilde{R}^2)$. If $\omega$ is a propagating form of $(C_2(R), \tau)$, then $\iota^*(\omega)$ is a propagating form of $(C_2(-R), \bar{\tau})$. So we have $\Theta(-R, \bar{\tau}) = \int_{C_2(-R)} \iota^*(\omega^3)$. This proves $\Theta(-R, \bar{\tau}) = -\Theta(R, \tau)$. Corollary 4.8 and Lemma 5.14 yield the conclusion. $\square$
## 5.6 On the groups $SO(3)$ and $SO(4)$

Let $\mathbb{H}$ denote the vector space $\mathbb{C} \oplus \mathbb{C}j$. Set $k = ij$. The *conjugate* of an element $(z_1 + z_2j)$ of $\mathbb{H}$ is

$$ \overline{z_1 + z_2j} = \overline{z_1} - z_2j. $$

**Lemma 5.16.** The bilinear map that maps $(z_1+z_2j, z'_1+z'_2j)$ to $(z_1z'_1-z_2z'_2)+ (z_2z'_1+z_1z'_2)$ maps $(i, j)$ to $k$, $(j, k)$ to $i$, $(k, i)$ to $j$, $(j, i)$ to $(-k)$, $(k, j)$ to $(-i)$, $(i, k)$ to $(-j)$, $(i, i)$, $(j, j)$, and $(k, k)$ to $(-1)$, and $(z_1+z_2j, \overline{z_1+z_2j})$ to $|z_1|^2 + |z_2|^2$. It defines an associative product on $\mathbb{H}$ such that this product and the addition make $\mathbb{H}$ a field.

**PROOF:** Exercise.

□

The noncommutative field $\mathbb{H}$, which contains $\mathbb{C}$ as a subfield, is the field of quaternions. It is equipped with the scalar product $\langle \cdot, \cdot \rangle$ that makes $(1, i, j, k)$ an orthonormal basis of $\mathbb{H}$. The associated norm maps $(z_1 + z_2j)$ to

$$ \sqrt{(z_1 + z_2j)\overline{z_1 + z_2j}}. $$

It is multiplicative. The unit sphere of $\mathbb{H}$ is the sphere $S^3$. It is equipped with the group structure induced by the product of $\mathbb{H}$. The elements of $\mathbb{H}$ are the quaternions. The real part of a quaternion $(z_1 + z_2j)$ is the real part of $z_1$. The pure quaternions are the quaternions with zero real part.

For $\mathbb{K} = \mathbb{R}$ or $\mathbb{C}$ and $n \in \mathbb{N}$, the $\mathbb{K}$ (Euclidean or Hermitian) oriented vector space with the direct orthonormal basis $(v_1, \dots, v_n)$ is denoted by $\mathbb{K}\langle v_1, \dots, v_n \rangle$. The cross product or vector product of two elements $v$ and $w$ of $\mathbb{R}^3 = \mathbb{R}\langle i, j, k \rangle$ is the element $v \times w$ of $\mathbb{R}^3$ such that for any $x \in \mathbb{R}^3$, $x \wedge v \wedge w = \langle x, v \times w \rangle i \wedge j \wedge k$ in $\Lambda^3 \mathbb{R}^3 = \mathbb{R}$.

**Lemma 5.17.** The product of two pure quaternions $v$ and $w$ is

$$ vw = -\langle v, w \rangle + v \times w. $$

Every element of $S^3$ may be expressed as $\cos(\theta) + \sin(\theta)v$ for a unique $\theta \in [0, \pi]$ and a pure quaternion $v$ of norm 1, which is unique when $\theta \notin \{0, \pi\}$. For such an element, the restriction to $\mathbb{R}\langle i, j, k \rangle$ of the conjugation

$$ R(\theta, v): w \mapsto (\cos(\theta) + \sin(\theta)v)w(\overline{\cos(\theta) + \sin(\theta)v}) $$

is the rotation with axis directed by $v$ and with angle $2\theta$.
PROOF: It is easy to check the first assertion. The conjugation $R(\theta, v)$ preserves the scalar product of $\mathbb{H}$ and fixes $\mathbb{R} \oplus \mathbb{R}v$, pointwise. Therefore, it restricts to $\mathbb{R}\langle i, j, k \rangle$ as an orthonormal transformation that fixes $v$. Let $w$ be a pure quaternion orthogonal to $v$.

$$R(\theta, v)(w) = (\cos(\theta) + \sin(\theta)v)w(\cos(\theta) - \sin(\theta)v)$$

is equal to

$$ \begin{aligned} R(\theta, v)(w) &= \cos^2(\theta)w - \sin^2(\theta)vw + \cos(\theta)\sin(\theta)(vw - wv) \\ &= \cos(2\theta)w + \sin(2\theta)v \times w. \end{aligned} $$

□

**Lemma 5.18.** *The group morphism*

$$ \begin{array}{rcl} \tilde{\rho}: & S^3 & \to SO(\mathbb{R}\langle i, j, k \rangle) = SO(3) \\ & x & \mapsto (w \mapsto x.w.\bar{x}) \end{array} $$

is surjective, and its kernel is $\{-1, +1\}$. The morphism $\tilde{\rho}$ is a two-fold covering map, and this definition of $\tilde{\rho}$ coincides with the previous one (after Definition 4.4), up to homotopy.

PROOF: According to Lemma 5.17, the map $\tilde{\rho}$ is surjective. Its kernel is the center of the group of unit quaternions, which is $\{-1, +1\}$. Thus, the map $\tilde{\rho}$ is a two-fold covering map.

This two-fold covering map clearly coincides with the previous one, up to homotopy and orientation, since both classes generate $\pi_3(SO(3)) = \mathbb{Z}$. We take care of the orientation using the outward normal first convention to orient boundaries, as usual. Consider the diffeomorphism

$$ \psi: \begin{array}{rcl} ]0, \pi[ & \times S^2 & \to S^3 \setminus \{-1, 1\} \\ (\theta, v) & \mapsto & \cos(\theta) + \sin(\theta)v. \end{array} $$

We study $\psi$ at $(\pi/2, i)$. At $(\psi(\pi/2, i) = i)$, the space $\mathbb{R}\langle i, j, k \rangle$ is oriented by the outward normal $o_i$ to $S^2$ followed by the orientation of $S^2$. The field $\mathbb{H}$ is oriented as $\mathbb{R} \oplus \mathbb{R}\langle i, j, k \rangle$. Since $o_i$ coincides with the outward normal to $S^3$ in $\mathbb{H}$, the sphere $S^3$ is oriented by the opposite of the real part followed by the orientation of $S^2$. Since cos is an orientation-reversing diffeomorphism at $\pi/2$, $\psi$ preserves the orientation near $(\pi/2, i)$. So the diffeomorphism $\psi$ preserves the orientation everywhere, and the two maps $\tilde{\rho}$ are homotopic. □
The following two group morphisms from $S^3$ to $SO(4)$ induced by the multiplication in $\mathbb{H}$

$$ m_{\ell}: \begin{array}{rcl} S^3 & \to & (SO(\mathbb{H}) = SO(4)) \\ x & \mapsto & (m_{\ell}(x) : v \mapsto x.v) \end{array} $$

$$ \overline{m_r}: \begin{array}{rcl} S^3 & \to & SO(\mathbb{H}) \\ y & \mapsto & (\overline{m_r}(y) : v \mapsto v.\overline{y}) \end{array} $$

together induce the surjective group morphism

$$ \begin{array}{rcl} S^3 \times S^3 & \to & SO(4) \\ (x, y) & \mapsto & (v \mapsto x.v.\overline{y}). \end{array} $$

The kernel of this group morphism is $\{(-1, -1), (1, 1)\}$. So this morphism is a two-fold covering map. In particular, we have $\pi_3(SO(4)) = \mathbb{Z}[m_{\ell}] \oplus \mathbb{Z}[\overline{m_r}]$. Define

$$ m_r: \begin{array}{rcl} S^3 & \to & (SO(\mathbb{H}) = SO(4)) \\ y & \mapsto & (m_r(y) : v \mapsto v.y). \end{array} $$

**Lemma 5.19.** In $\pi_3(SO(4)) = \mathbb{Z}[m_{\ell}] \oplus \mathbb{Z}[\overline{m_r}]$, we have

$$ i_*([\tilde{\rho}]) = [m_{\ell}] + [\overline{m_r}] = [m_{\ell}] - [m_r]. $$

**PROOF:** The $\pi_3$-product in $\pi_3(SO(4))$ coincides with the product induced by the group structure of $SO(4)$. $\square$

**Lemma 5.20.** Recall that $m_r$ denotes the map from the unit sphere $S^3$ of $\mathbb{H}$ to $SO(\mathbb{H})$ induced by the right-multiplication. Denote the inclusions $SO(n) \subset SU(n)$ by $c$. Then we have

$$ c_*([m_r]) = 2 [i_*^2(m_r^C)] $$

in $\pi_3(SU(4))$.

**PROOF:** Let $\mathbb{H} + I\mathbb{H}$ denote the complexification of $\mathbb{R}^4 = \mathbb{H} = \mathbb{R}\langle 1, i, j, k \rangle$. Here, we have $\mathbb{C} = \mathbb{R} \oplus I\mathbb{R}$ and $I^2 = -1$. When $x \in \mathbb{H}$ and $v \in S^3$, we have $c(m_r(v))(Ix) = Ixv$. Let $\varepsilon = \pm 1$. Define

$$ \mathbb{C}^2(\varepsilon) = \mathbb{C} \left\langle \frac{\sqrt{2}}{2}(1 + \varepsilon Ii), \frac{\sqrt{2}}{2}(j + \varepsilon Ik) \right\rangle. $$

Consider the quotient $\mathbb{C}^4/\mathbb{C}^2(\varepsilon)$. In this quotient, we have $Ii = -\varepsilon 1$, $Ik = -\varepsilon j$, and, since $I^2 = -1$, $I1 = \varepsilon i$ and $Ij = \varepsilon k$. Therefore, this quotient is
isomorphic to $\mathbb{H}$ as a real vector space with its complex structure $I = \varepsilon i$. Then it is easy to see that $c(m_r(v))$ maps $\mathbb{C}^2(\varepsilon)$ to $0$ in this quotient, for any $v \in S^3$. We get $c(m_r(v))(\mathbb{C}^2(\varepsilon)) = \mathbb{C}^2(\varepsilon)$. Now, observe that $\mathbb{H} + I\mathbb{H}$ is the orthogonal sum of $\mathbb{C}^2(-1)$ and $\mathbb{C}^2(1)$. In particular, $\mathbb{C}^2(\varepsilon)$ is isomorphic to the quotient $\mathbb{C}^4/\mathbb{C}^2(-\varepsilon)$, which is isomorphic to $(\mathbb{H}; I = -\varepsilon i)$, and $c(m_r)$ acts on it by the right multiplication. Therefore, with respect to the orthonormal basis $\frac{\sqrt{2}}{2}(1 - Ii, j - Ik, 1 + Ii, j + Ik)$, $c(m_r(z_1 + z_2j = x_1 + y_1i + x_2j + y_2k))$ may be written as

$$c(m_r(x_1+y_1i+x_2j+y_2k)) = \begin{bmatrix} x_1+y_1I & -x_2+y_2I & 0 & 0 \\ x_2+y_2I & x_1-y_1I & 0 & 0 \\ 0 & 0 & x_1-y_1I & -x_2-y_2I \\ 0 & 0 & x_2-y_2I & x_1+y_1I \end{bmatrix}$$

Therefore, the homotopy class of $c(m_r)$ is the sum of the homotopy classes of

$$(z_1 + z_2j) \mapsto \begin{bmatrix} m_r^{\mathbb{C}}(z_1, z_2) & 0 \\ 0 & 1 \end{bmatrix} \quad \text{and} \quad (z_1 + z_2j) \mapsto \begin{bmatrix} 1 & 0 \\ 0 & m_r^{\mathbb{C}} \circ \iota(z_1, z_2) \end{bmatrix},$$

where $\iota(z_1, z_2) = (\overline{z_1}, \overline{z_2})$. Since the first map is conjugate to $i_*^2(m_r^{\mathbb{C}})$ by a fixed element of $SU(4)$, it is homotopic to $i_*^2(m_r^{\mathbb{C}})$. Since $\iota$ induces the identity on $\pi_3(S^3)$, the second map is homotopic to $i_*^2(m_r^{\mathbb{C}})$, too. $\square$

The following lemma finishes to determine the maps

$$c_*: \pi_3(SO(4)) \to \pi_3(SU(4))$$

$$\text{and } c_*i_*: \pi_3(SO(3)) \to \pi_3(SU(4)).$$

**Lemma 5.21.** We have

$$c_*([\overline{m_r}]) = c_*([m_\ell]) = -2 [i_*^2(m_r^{\mathbb{C}})] = 2 [i_*^2(\overline{m_r^{\mathbb{C}}})],$$

and

$$c_*\left(i_*([\tilde{\rho}])\right) = 4 [i_*^2(\overline{m_r^{\mathbb{C}}})].$$

**PROOF:** According to Lemma 5.19, $i_*([\tilde{\rho}]) = [m_\ell] + [\overline{m_r}]$. Using the conjugacy of quaternions, we have $m_\ell(x)(v) = x.v = \overline{v.x} = \overline{m_r}(x)(\overline{v})$. Therefore, $m_\ell$ is conjugated to $\overline{m_r}$ via the conjugacy of quaternions, which lies in $(O(4) \subset U(4))$.

Since $U(4)$ is connected, the conjugacy by an element of $U(4)$ induces the identity on $\pi_3(SU(4))$. Thus, we get

$$c_*([m_\ell]) = c_*([\overline{m_r}]) = -c_*([m_r]) = -2 [i_*^2(m_r^{\mathbb{C}})] = 2 [i_*^2(\overline{m_r^{\mathbb{C}}})],$$

$$\text{and } c_*\left(i_*([\tilde{\rho}])\right) = c_*([m_\ell]) + c_*([\overline{m_r}]) = 4 [i_*^2(\overline{m_r^{\mathbb{C}}})]. \quad \square$$
## 5.7 Relating the Pontrjagin number to the degree

We finish proving Theorem 4.5 by proving the following proposition. See Lemmas 5.2, 5.7, and 5.8.

**Proposition 5.22.** Let $M_0$ and $M$ be two compact connected oriented 3-manifolds whose boundaries have collars identified by a diffeomorphism. Let $\tau_0: M_0 \times \mathbb{C}^3 \to TM_0 \otimes \mathbb{C}$ and $\tau: M \times \mathbb{C}^3 \to TM \otimes \mathbb{C}$ be two special complex trivializations (which respect the orientations) that coincide on the collar neighborhoods of $\partial M_0 = \partial M$. Let $[(M, \partial M), (SU(3), 1)]$ denote the group of homotopy classes of maps from $M$ to $SU(3)$ that map $\partial M$ to 1. For any

$$g: (M, \partial M) \to (SU(3), 1),$$

define

$$\psi(g): \begin{array}{ccc} M \times \mathbb{C}^3 & \to & M \times \mathbb{C}^3 \\ (x, y) & \mapsto & (x, g(x)(y)). \end{array}$$

Then $(p_1(\tau_0, \tau \circ \psi(g)) - p_1(\tau_0, \tau))$ is independent of $\tau_0$ and $\tau$. Set

$$p'_1(g) = p_1(\tau_0, \tau \circ \psi(g)) - p_1(\tau_0, \tau).$$

The map $p'_1$ induces an isomorphism from the group $[(M, \partial M), (SU(3), 1)]$ to $\mathbb{Z}$, and, if $g$ is valued in $SO(3)$, then we have

$$p'_1(g) = 2 \deg(g).$$

To prove Proposition 5.22, we first prove the following lemma.

**Lemma 5.23.** Under the hypotheses of Proposition 5.22,

$$p_1(\tau_0, \tau \circ \psi(g)) - p_1(\tau_0, \tau) = p_1(\tau, \tau \circ \psi(g)) = -p_1(\tau \circ \psi(g), \tau)$$

is independent of $\tau_0$ and $\tau$.

**PROOF:** The equalities of the statement are easy to observe. Let us prove that $p_1(\tau, \tau \circ \psi(g))$ is independent of $\tau$. Let $\tau_W$ be the trivialization of the restriction to $\partial([0, 1] \times M)$ of the complexified tangent bundle $T[0, 1] \times M \otimes_{\mathbb{R}} \mathbb{C}$ to $[0, 1] \times M$ given by $T[0, 1] \oplus \tau$ on $(\{0\} \times M) \cup ([0, 1] \times \partial M)$, and $T[0, 1] \oplus \tau \circ \psi(g)$ on $\{1\} \times M$. Let $\tilde{g}: \partial([0, 1] \times M) \to SU(4)$ map $(\{0\} \times M) \cup ([0, 1] \times \partial M)$ to 1 and coincide with $i \circ g$ on $\{1\} \times M$. Then the obstruction $p_1(\tau, \tau \circ \psi(g))$ to extending $\tau_W$ to $[0, 1] \times M$ is the obstruction to extending the map $\tilde{g}$ to $[0, 1] \times M$. It lies in $\pi_3(SU(4))$ since $\pi_i(SU(4)) = 0$ for $i < 3$. It is independent of $\tau$. $\square$

Lemma 5.23 guarantees that $p'_1$ defines two group homomorphisms to $\mathbb{Z}$ from $[(M, \partial M), (SU(3), 1)]$ and from $[(M, \partial M), (SO(3), 1)]$.
**Lemma 5.24.** Under the hypotheses of Proposition 5.22

$$p'_1: [(M, \partial M), (SO(3), 1)] \to \mathbb{Z}$$

is a group homomorphism and

$$p'_1: [(M, \partial M), (SU(3), 1)] \to \mathbb{Z}$$

is a group isomorphism.

**PROOF:** Since $\pi_i(SU(3))$ is trivial for $i < 3$ and since $\pi_3(SU(3)) = \mathbb{Z}$, the group of homotopy classes $[(M, \partial M), (SU(3), 1)]$ is generated by the class of a map that maps the complement of a 3-ball $B$ to 1 and that factors through a map whose homotopy class generates $\pi_3(SU(3))$ on $B$. By definition of the Pontrjagin classes, $p'_1$ sends such a generator to $\pm 1$, and it induces an isomorphism from the group $[(M, \partial M), (SU(3), 1)]$ to $\mathbb{Z}$. $\square$

**Lemma 5.25.** We have

$$p'_1(\rho_M(B^3)) = 4.$$

**PROOF:** Let $g = \rho_M(B^3)$. Extend the map $\tilde{g}$ of the proof of Lemma 5.23 by the constant map with value 1 outside $[\varepsilon, 1] \times B^3 \cong B^4$, for some $\varepsilon \in ]0, 1[$. We have

$$[\tilde{g}|_{\partial B^4}] = p_1(\tau, \tau \circ \psi(g)) [i_*^2(\overline{m_r^C})]$$

in $\pi_3(SU(4))$. Since $\tilde{g}|_{\partial B^4}$ is homotopic to $c \circ i \circ \tilde{\rho}$, Lemma 5.21 allows us to conclude. $\square$

**PROOF OF PROPOSITION 5.22:** According to Lemmas 5.8 and 5.24, the restriction of $p'_1$ to $[(M, \partial M), (SO(3), 1)]$ is equal to $p'_1(\rho_M(B^3))^{\frac{\deg}{2}}$. Conclude with Lemma 5.25. $\square$

# 5.8 Properties of Pontrjagin numbers

**Proposition 5.26.** Let $M_0$ and $M_1$ be two compact connected oriented 3-manifolds whose boundaries have collars identified by a diffeomorphism. Let $\tau_0: M_0 \times \mathbb{C}^3 \to TM_0 \otimes \mathbb{C}$ and $\tau_1: M_1 \times \mathbb{C}^3 \to TM_1 \otimes \mathbb{C}$ be two special complex trivializations (which respect the orientations) that agree on the collar neighborhoods of $\partial M_0 = \partial M_1$.

Then the first Pontrjagin number $p_1(\tau_0, \tau_1)$ satisfies the following properties.
1. Let $M_2$ be a compact 3-manifold whose boundary has a collar neighborhood identified with a collar neighborhood of $\partial M_0$. Let $\tau_2$ be a special complex trivialization of $TM_2$ that agrees with $\tau_0$ near $\partial M_2$. If two of the Lagrangians of $M_0$, $M_1$, and $M_2$ coincide in $H_1(\partial M_0; \mathbb{Q})$, then we have

$$p_1(\tau_0, \tau_2) = p_1(\tau_0, \tau_1) + p_1(\tau_1, \tau_2).$$

In particular, we also have $p_1(\tau_1, \tau_0) = -p_1(\tau_0, \tau_1)$ since $p_1(\tau_0, \tau_0) = 0$.

2. Let $D$ be a connected compact 3-manifold that contains $M_0$ in its interior. Let $\tau_D$ be a special complex trivialization of $TD$ that restricts as the special complex trivialization $\tau_0$ on $TM_0$. Let $D_1$ be obtained from $D$ by replacing $M_0$ by $M_1$. Let $\tau_{D_1}$ be the trivialization of $TD_1$ that agrees with $\tau_1$ on $TM_1$, and with $\tau_D$ on $T(D \setminus M_0)$. If the Lagrangians

<figure><img src="image_48.png" /><figcaption>Figure 5.2: The manifolds $D$ and $D_1$ of Proposition 5.26</figcaption></figure>

of $M_0$ and $M_1$ coincide, then we have

$$p_1(\tau_D, \tau_{D_1}) = p_1(\tau_0, \tau_1).$$

The proof uses a weak form of the Wall Non-Additivity theorem. We quote the weak form we need.

**Theorem 5.27 ([Wal69]).** Let $Y$ be a compact oriented 4-manifold (with possible boundary). Let $X$ be a 3-manifold properly embedded in $Y$ that separates $Y$ and that induces the splitting $Y = Y^+ \cup_X Y^-$, for two 4-manifolds $Y^+$ and $Y^-$ in $Y$, as in the following figure of $Y$:

$$X^- \boxed{Y^- \mid X Y^+} X^+$$

The manifold $X$ is the intersection $Y^+ \cap Y^-$. It is oriented as a part of the boundary of $Y^-$. Set

$$X^+ = \overline{\partial Y^+ \setminus (-X)} \quad \text{and} \quad X^- = -\overline{\partial Y^- \setminus X}.$$
Let $\mathcal{L}$, $\mathcal{L}^{-}$, and $\mathcal{L}^{+}$ denote the Lagrangians of $X$, $X^{-}$, and $X^{+}$, respectively. They are Lagrangian subspaces of $H_{1}(\partial X, \mathbb{Q})$. Then

$$ (\mathrm{signature}(Y) - \mathrm{signature}(Y^{+}) - \mathrm{signature}(Y^{-})) $$

is the signature of an explicit quadratic form on

$$ \frac{\mathcal{L} \cap (\mathcal{L}^{-} + \mathcal{L}^{+})}{(\mathcal{L} \cap \mathcal{L}^{-}) + (\mathcal{L} \cap \mathcal{L}^{+})}. $$

Furthermore, this space is isomorphic to $\frac{\mathcal{L}^{+} \cap (\mathcal{L} + \mathcal{L}^{-})}{(\mathcal{L}^{+} \cap \mathcal{L}) + (\mathcal{L}^{+} \cap \mathcal{L}^{-})}$ and $\frac{\mathcal{L}^{-} \cap (\mathcal{L} + \mathcal{L}^{+})}{(\mathcal{L}^{-} \cap \mathcal{L}) + (\mathcal{L}^{-} \cap \mathcal{L}^{+})}$.

We do not describe the involved quadratic form because we use this theorem only when the above space is trivial.

PROOF OF PROPOSITION 5.26: Let us prove the first property. Let $Y^{-} = W$ be a signature 0 cobordism from $X^{-} = M_{0}$ to $X = M_{1}$. Let $Y^{+}$ be a signature 0 cobordism from $M_{1}$ to $X^{+} = M_{2}$. Then it is enough to prove that the signature of $Y = Y^{+} \cup_{X} Y^{-}$ is zero. With the notation of Theorem 5.27, under our assumptions, the space $\frac{\mathcal{L} \cap (\mathcal{L}^{-} + \mathcal{L}^{+})}{(\mathcal{L} \cap \mathcal{L}^{-}) + (\mathcal{L} \cap \mathcal{L}^{+})}$ is trivial. Therefore, according to the Wall theorem, the signature of $Y$ is zero. The first property follows.

We now prove that under the assumptions of the second property, we have

$$ p_{1}(\tau_{D}, \tau_{D_{1}}) = p_{1}(\tau_{0}, \tau_{1}). $$

Set $Y^{+} = ([0, 1] \times (D \setminus \mathring{M}_{0}))$ and $X = -[0, 1] \times \partial M_{0}$. Let $Y^{-} = W$ be a signature 0 cobordism from $\mathring{M}_{0}$ to $M_{1}$. Note that the signature of $Y^{+}$ is zero. In order to prove the desired equality, it is enough to prove that the signature of $Y = Y^{+} \cup_{X} Y^{-}$ is zero. Here, we have $H_{1}(\partial X; \mathbb{Q}) = H_{1}(\partial M_{0}) \oplus H_{1}(\partial M_{0})$. Let $j: H_{1}(\partial M_{0}) \to H_{1}(D \setminus \mathring{M}_{0})$ and $j_{\partial D}: H_{1}(\partial D) \to H_{1}(D \setminus \mathring{M}_{0})$ be the maps induced by inclusions. With the notation of Theorem 5.27, we have

$$
\begin{align*}
\partial X &= -(\partial[0,1]) \times \partial M_0, \\
X^{-} &= -\{1\} \times M_1 \cup \{(0) \times M_0\}, \\
X^{+} &= -[0,1] \times \partial D \cup ((\partial[0,1]) \times (D \setminus \mathring{M}_0)), \\
\mathcal{L} &= \{(x,-x) : x \in H_1(\partial M_0)\}, \\
\mathcal{L}^{-} &= \{(x,y) : x \in \mathcal{L}_{M_0}, y \in \mathcal{L}_{M_1}\}, \\
\mathcal{L}^{+} &= \{(x,y) : (j(x),j(y)) = (j_{\partial D}(z \in H_1(\partial D)), -j_{\partial D}(z))\} \\
&= \{(y,-y) : j(y) \in \mathrm{Im}(j_{\partial D})\} \oplus \{(x,0) : j(x) = 0\}, \\
\mathcal{L} \cap \mathcal{L}^{-} &= \{(x,-x) : x \in (\mathcal{L}_{M_0} \cap \mathcal{L}_{M_1} = \mathcal{L}_{M_0})\}, \\
\mathcal{L} \cap \mathcal{L}^{+} &= \{(x,-x) : j(x) \in \mathrm{Im}(j_{\partial D})\}.
\end{align*}
$$
Let us prove $\mathcal{L} \cap (\mathcal{L}^{-} + \mathcal{L}^{+}) = (\mathcal{L} \cap \mathcal{L}^{-}) + (\mathcal{L} \cap \mathcal{L}^{+})$. For a subspace $K$ of $H_1(\partial M_0; \mathbb{Q})$, let $j_{MV}(K)$ be the subspace $\{(x, -x) : x \in K\}$ of $H_1(\partial X; \mathbb{Q})$. Then we have $\mathcal{L} = j_{MV}(H_1(\partial M_0))$, $\mathcal{L} \cap \mathcal{L}^{+} = j_{MV}(j^{-1}(\mathrm{Im}(j_{\partial D})))$, and

$$ \mathcal{L} \cap (\mathcal{L}^{-} + \mathcal{L}^{+}) = \mathcal{L} \cap \mathcal{L}^{+} + j_{MV} (\mathcal{L}_{M_1} \cap (\mathcal{L}_{M_0} + \mathrm{Ker}(j))). $$

Since $\mathcal{L}_{M_0} = \mathcal{L}_{M_1}$, we have $(\mathcal{L}_{M_1} \cap (\mathcal{L}_{M_0} + \mathrm{Ker}(j))) = \mathcal{L}_{M_0}$. So $\mathcal{L} \cap (\mathcal{L}^{-} + \mathcal{L}^{+}) = (\mathcal{L} \cap \mathcal{L}^{+}) + j_{MV}(\mathcal{L}_{M_0})$. Then the second property is proved because Wall's theorem guarantees the additivity of the signature in this case. $\square$

**The parallelizations of $S^3$.** As a Lie group, $S^3$ has two natural homotopy classes of parallelizations $\tau_\ell$ and $\tau_r$, which we describe below. Identify the tangent space $T_1S^3$ to $S^3$ at 1 with $\mathbb{R}^3$ (arbitrarily, with respect to the orientation). For $g \in S^3$, the multiplication induces two diffeomorphisms $m_\ell(g)$ and $m_r(g)$ of $S^3$, $m_\ell(g)(h) = gh$ and $m_r(g)(h) = hg$. Let $T(m_\ell(g))$ and $T(m_r(g))$ denote their respective tangent maps at 1. Then we have

$$ \tau_{\ell}(h \in S^3, v \in (\mathbb{R}^3 = T_1S^3)) = (h, T_1(m_{\ell}(h))(v)) $$

$$ \text{and } \tau_r(h, v) = (h, T_1(m_r(h))(v)). $$

**Proposition 5.28.** We have $p_1(\tau_\ell) = 2$ and $p_1(\tau_r) = -2$.

**PROOF:** Regard $S^3$ as the unit sphere of $\mathbb{H}$. So $T_1S^3 = \mathbb{R}\langle i, j, k \rangle$. The unit ball $B(\mathbb{H})$ of $\mathbb{H}$ has the standard parallelization of a real vector space equipped with a basis. The trivialization $\tau(\tau_\ell)$ induced by $\tau_\ell$ on $\partial B(\mathbb{H})$ is such that $\tau_\ell(h \in S^3, v \in \mathbb{H}) = (h, hv) \in (S^3 \times \mathbb{H} = T\mathbb{H}|_{S^3})$. So, we have $c_*([m_\ell]) = p_1(\tau_\ell) [i_*^2(\overline{m_r^C})]$ in $\pi_3(SU(4))$ by Definition 5.13 of $p_1$. According to Lemma 5.21, we get $p_1(\tau_\ell) = 2$. We similarly get $p_1(\tau_r) = -2$. $\square$

**On the image of $p_1$.** For $n \ge 3$, a spin structure of a smooth $n$-manifold is a homotopy class of parallelizations over a 2-skeleton of $M$ (or, equivalently, over the complement of a point, if $n = 3$ and if $M$ is connected).

The class of the covering map $\tilde{\rho}$ described after Definition 4.4 is the standard generator of $\pi_3(SO(3)) = \mathbb{Z}[\tilde{\rho}]$. Recall the map $\rho_M(B^3)$ of Lemma 4.6. Set $[\tilde{\rho}][\tau] = [\tau\psi_{\mathbb{R}}(\rho_M(B^3))]$. The homotopy classes of parallelizations that induce a given spin structure constitute an affine space with translation group $\pi_3(SO(3))$. According to Theorem 4.5 and Lemma 4.6, $p_1([\tilde{\rho}][\tau]) = p_1(\tau)+4$.

**Definition 5.29.** The Rohlin invariant $\mu(M, \sigma)$ of a smooth closed 3-manifold $M$, equipped with a spin structure $\sigma$, is the mod 16 signature of a compact 4-manifold $W$, bounded by $M$, equipped with a spin structure that restricts to $M$ as a stabilization of $\sigma$.
The first Betti number of M is the dimension of $H_1(M; \mathbb{Q})$. It is denoted by $\beta_1(M)$. Robion Kirby and Paul Melvin proved the following theorem [KM99, Theorem 2.6].

**Theorem 5.30.** For any closed oriented 3-manifold M, for any parallelization $\tau$ of M, we have

$$ (p_1(\tau) - \text{dimension}(H_1(M; \mathbb{Z}/2\mathbb{Z})) - \beta_1(M)) \in 2\mathbb{Z}. $$

Let M be a closed 3-manifold equipped with a given spin structure $\sigma$. Then $p_1$ is a bijection from the set of homotopy classes of parallelizations of M that induce $\sigma$ to

$$ 2(\text{dimension}(H_1(M; \mathbb{Z}/2\mathbb{Z})) + 1) + \mu(M, \sigma) + 4\mathbb{Z}. $$

When M is a $\mathbb{Z}$-sphere, $p_1$ is a bijection from the set of homotopy classes of parallelizations of M to $(2 + 4\mathbb{Z})$.

Thanks to Proposition 5.26(2), Theorem 5.30 implies Proposition 5.31 below.

**Proposition 5.31.** Let $M_0$ be the unit ball of $\mathbb{R}^3$. Let $\tau_s$ be the standard parallelization of $\mathbb{R}^3$,

* for any given $\mathbb{Z}$-ball M, $p_1(\cdot) = p_1((\tau_s)|_{B^3}, \cdot)$ defines a bijection from the set of homotopy classes of parallelizations of M that are standard near $\partial M = S^2$ to $4\mathbb{Z}$.

* For any $\mathbb{Q}$-ball M, for any trivialization $\tau$ of M that is standard near $\partial M = S^2$, we have

$$ (p_1(\tau) - \text{dimension}(H_1(M; \mathbb{Z}/2\mathbb{Z}))) \in 2\mathbb{Z}. $$

## 5.9 More on $[(M, \partial M), (SO(3), 1)]$

This section is a complement to the study of $[(M, \partial M), (SO(3), 1)]$ started in Sections 5.1 and 5.3. It is not used later in this book. We show how to describe all the elements of $[(M, \partial M), (SO(3), 1)]$ as twists across surfaces, and we describe the structure of $[(M, \partial M), (SO(3), 1)]$ precisely, by proving the following theorem.
**Theorem 5.32.** Let $M$ be a compact oriented 3-manifold.

If all the closed surfaces embedded in $M$ have an even Euler characteristic, then $[(M, \partial M), (SO(3), 1)]$ is canonically isomorphic to $H^3(M, \partial M; \mathbb{Z}) \oplus H_2(M; \mathbb{Z}/2\mathbb{Z})$, and the degree maps $[(M, \partial M), (SO(3), 1)]$ onto $2\mathbb{Z}$.

If $H_1(M; \mathbb{Z})$ has no 2-torsion, then all the closed surfaces embedded in $M$ have an even Euler characteristic.

If $M$ is connected, and if there exists a closed surface $S$ of $M$ with odd Euler characteristic, then the degree maps $[(M, \partial M), (SO(3), 1)]$ onto $\mathbb{Z}$, and $[(M, \partial M), (SO(3), 1)]$ is isomorphic to $\mathbb{Z} \oplus \text{Ker}(e\partial_B)$, where

$$e\partial_B: H_2(M; \mathbb{Z}/2\mathbb{Z}) \to \mathbb{Z}/2\mathbb{Z}$$

maps the class of a surface to its Euler characteristic modulo 2, and the kernel of $e\partial_B$ has a canonical image in $[(M, \partial M), (SO(3), 1)]$.

**Representing the elements of $[(M, \partial M), (SO(3), 1)]$ by surfaces.**

**Proposition 5.33.** Let $S$ be a surface, orientable or not, embedded in a 3-manifold $M$ equipped with a parallelization $\tau$. Recall the map $g(S, \tau)$ from the beginning of Section 5.3. If $M$ is connected, then $g(S, \tau)^2$ is homotopic to $\rho_M(B^3)^{\chi(S)}$. In particular, the homotopy class of $g(S, \tau)$ depends only on $\chi(S)$ and on the class of $S$ in $H_2(M; \mathbb{Z}/2\mathbb{Z})$.

**PROOF:** Assume that $S$ is connected and oriented. Perform a homotopy of $\tau$ so that $p_2 \circ \tau^{-1}$ maps the positive normal $N^+(S) = T_{(u,s)}([-1, 1] \times s)$ to $u \times S$ to a fixed vector $v$ of $S^2$, on $[-1, 1] \times (S \setminus D)$, for a disk $D$ of $S$. Then there is a homotopy from $[0, 1] \times [-1, 1] \times (S \setminus D)$ to $SO(3)$,

* which factors through the projection onto $[0, 1] \times [-1, 1]$,

* which maps $(1, u, s)$ to the rotation $g(S, \tau)^2(u, s)$ with axis $v$ and angle $2\pi(u + 1)$, for any $(u, s) \in [-1, 1] \times (S \setminus D)$, and

* which maps $(\partial([0, 1] \times [-1, 1]) \setminus (\{1\} \times [-1, 1])) \times (S \setminus D)$ to 1.

This homotopy extends to a homotopy $h: [0, 1] \times [-1, 1] \times S \to SO(3)$ from $h_0 = \rho_{[-1,1]\times S}(B^3)^k$ for some $k \in \mathbb{Z}$, to $h_1 = g(S, \tau)^2|_{[-1,1]\times S}$, such that $h([0, 1] \times \{-1, 1\} \times D) = 1$. Thus, the map $g(S, \tau)^2$ is homotopic to $\rho_M(B^3)^k$.

Let us show that $k$ depends only on the degree of the Gauss map $G$ from $(D, \partial D)$ to $(S^2, v)$ that maps $s \in (D = \{0\} \times D)$ to $p_2(\tau^{-1}(N^+(s)))$. Endow $\mathbb{R}^3$ with an orthonormal basis beginning with $v$. Identify the tangent space to $D$ with $(D \times)\mathbb{R}^2$ and $T([-1, 1] \times D)|_D$ with $([-1, 1] \times D) \times (\mathbb{R}N^+(D) \oplus \mathbb{R}^2)$. These identifications allow us to regard $p_2 \circ \tau^{-1}$ as a map $\Psi$ from $[-1, 1] \times D$
to $SO(3)$. The map $\Psi$ determines $\tau$. It sends $[-1, 1] \times \partial D$ to $(i(SO(2)) = SO(2))$. Since $\pi_2(SO(3))$ is trivial, the homotopy class of $\Psi|_D$ is determined by its restriction $\Psi|_{\partial D}$ to $\partial D$. The map from $\pi_2(S^2)$ to $\pi_1(SO(2) = S^1)$ in the long exact sequence associated to the fibration

$$SO(2) \xrightarrow{i} SO(3) \to S^2,$$

described in Theorem A.14 sends the class of $G$ to the class of $\Psi|_{\partial D}$. So the degree of $G$ determines the homotopy class of the restriction of $\tau$ to $D$ relative to $\partial D$. This homotopy class determines the homotopy class relative to $[-1, 1] \times \partial D$ of the restriction of $\tau$ to $[-1, 1] \times D$, which determines the homotopy class relative to $\partial([-1, 1] \times D)$ of the restriction of $g(S, \tau)^2$ to $[-1, 1] \times D$. Therefore, the degree $d$ of $G$ determines $k$. Cutting $D$ into smaller disks shows that $k$ depends linearly on $d$. Note that $d$ is the degree of the Gauss map from $S$ to $S^2$ before the homotopy of $\tau$. For a standard sphere $S^2$, we have $d = 1$ and $g(S^2, \tau) = \rho_M(B^3)$. So, we get $k = 2d$. It remains to see that the degree $d$ of the Gauss map is $\frac{\chi(S^2)}{2}$. This is easy to observe for a standard embedding of $S$ into $\mathbb{R}^3$ equipped with its standard trivialization. Up to homotopy, the trivializations of $TM|_S$ are obtained from the standard one by compositions by rotations with fixed axis $v$ supported in neighborhoods of curves outside the preimage of $v$. So the degree (at $v$) is independent of the trivialization. Thus, the proposition is proved when $S$ is orientable and connected. When $S$ is not orientable, the map $g(S, \tau)^2$ is homotopic to $\rho_M(B^3)^k$, for some $k$, according to Lemma 5.2. Furthermore, it is homotopic to $g(S^{(2)}, \tau)$, where $S^{(2)}$ is the orientable boundary of the tubular neighborhood of $S$. The Euler characteristic of $S^{(2)}$ is $2\chi(S)$. So $g(S, \tau)^4$ is homotopic to $\rho_M(B^3)^{2k}$ and to $\rho_M(B^3)^{2\chi(S)}$. Since the arguments are local, they extend to the disconnected case and prove that $g(S, \tau)^2$ is homotopic to $\rho_M(B^3)^{\chi(S)}$ for any $S$. Then Proposition 5.9 and Lemma 5.6 allow us to conclude that the homotopy class of $g(S, \tau)$ depends only on $\chi(S)$ and on the class of $S$ in $H_2(M; \mathbb{Z}/2\mathbb{Z})$. $\square$

Hence, $g(S, \tau)$ will be denoted by $g(S)$. Lemma 5.6 and Propositions 5.9 and 5.33 easily imply the following corollary.

**Corollary 5.34.** All elements of $[(M, \partial M), (SO(3), 1)]$ can be represented by $g(S)$ for some embedded disjoint union of closed surfaces $S$ of $M$. $\square$

<figure><img src="image_49.png" /></figure>

**Structure of** $[(M, \partial M), (SO(3), 1)]$. Tensoring a free chain complex $C_*(M; \mathbb{Z})$ whose homology is $H_*(M; \mathbb{Z})$ by the short exact sequence

$$0 \to \mathbb{Z} \xrightarrow{\times 2} \mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} \to 0$$
yields the associated long exact homology sequence

$$ \dots \to H_*(M; \mathbb{Z}) \xrightarrow{\times 2} H_*(M; \mathbb{Z}) \to H_*(M; \mathbb{Z}/2\mathbb{Z}) \xrightarrow{\partial_B} H_{*-1}(M; \mathbb{Z}) \to \dots, $$

where $\partial_B$ is the Bockstein morphism.

**Definition 5.35.** The self-linking number of a torsion element $x$ of $H_1(M; \mathbb{Z})$ is the linking number $lk(c, c')$ of a curve $c$ that represents $x$ and a parallel $c'$ of $c$, modulo $\mathbb{Z}$.<fnref n="2" /> It belongs to $\mathbb{Q}/\mathbb{Z}$.

**Proposition 5.36.** There is a canonical group homomorphism

$$ e\partial_B: H_2(M; \mathbb{Z}/2\mathbb{Z}) \to \mathbb{Z}/2\mathbb{Z} $$

that admits the following two equivalent definitions:

1. For any embedded surface $S$, $e\partial_B$ maps the class of $S$ to the Euler characteristic of $S$ modulo 2.

2. The map $e\partial_B$ is the composition of the Bockstein morphism

$$ \partial_B: H_2(M; \mathbb{Z}/2\mathbb{Z}) \to \text{2-torsion of } H_1(M; \mathbb{Z}) $$

and the map

$$ e: \text{2-torsion of } H_1(M; \mathbb{Z}) \to \mathbb{Z}/2\mathbb{Z} $$

that maps the class of a curve $x$ to 1 if the self-linking number of $x$ is $\frac{1}{2}$ modulo $\mathbb{Z}$, and to 0 otherwise.

**PROOF:** The map $e\partial_B$ is well defined by the second definition. It is a group homomorphism. Let $S$ be a connected closed surface. If $S$ is orientable, then the Euler characteristic of $S$ is even, and the long exact sequence shows that $\partial_B([S]) = 0$. Otherwise, there is a curve $x$ (Poincaré dual to $w_1(S)$) such that $S \setminus x$ is orientable, and the boundary of the closure of the domain of the embedding $S \setminus x$ maps to $(\pm 2x)$, so $\partial_B([S]) = [x]$, by definition. The characteristic curve $x$ may be assumed to be connected. Then the tubular neighborhood of $x$ in $S$ is either a Möbius band or an annulus. In the first case, we have $e([x]) = 1$ and $\chi(S)$ is odd. Otherwise, we have $e([x]) = 0$ and $\chi(S)$ is even. $\square$

**Proposition 5.37.** Let $M$ be an oriented connected 3-manifold. Then we have

$$ [(M, \partial M), (SO(3), 1)] \cong \mathbb{Z} \oplus \text{Ker}(e\partial_B), $$

and the degree maps $[(M, \partial M), (SO(3), 1)]$ onto $2\mathbb{Z}$ when $e\partial_B = 0$, and onto $\mathbb{Z}$ otherwise.

<footnote n="2">²Equivalently, it is the linking number of two disjoint representatives of $x$, modulo $\mathbb{Z}$.</footnote>
PROOF OF PROPOSITION 5.37 AND THEOREM 5.32: The class of a surface with even Euler characteristic in $H_2(M; \mathbb{Z}/2\mathbb{Z})$ can be represented by a surface $S$ with null Euler characteristic (a disjoint union with trivial bounding surfaces). According to Proposition 5.33, for such an $S$, the class of $g(S)$ is a 2-torsion element of $[(M, \partial M), (SO(3), 1)]$ called $\sigma([S])$. This defines a canonical partial section

$$ \sigma: (\mathrm{Ker}(e\partial_B) \subset H_2(M; \mathbb{Z}/2\mathbb{Z})) \to \mathrm{Ker}(\mathrm{deg}: [(M, \partial M), (SO(3), 1)] \to \mathbb{Z}) $$

of the sequence of Proposition 5.9. Therefore, if $e\partial_B = 0$, we have

$$ [(M, \partial M), (SO(3), 1)] = \mathbb{Z}[\rho_M(B^3)] \oplus \sigma(\mathrm{Ker}(e\partial_B) = H_2(M; \mathbb{Z}/2\mathbb{Z})). $$

If $e\partial_B \neq 0$, there exists a closed surface $S_1$ with $\chi(S_1) = 1$ in $M$. Since the degree is a group homomorphism from $[(M, \partial M), (SO(3), 1)]$ to $\mathbb{Z}$, Proposition 5.33 implies $\mathrm{deg}(g(S_1)) = 1$ for such an $S_1$. Thus we get

$$ [(M, \partial M), (SO(3), 1)] = \mathbb{Z}[g(S_1)] \oplus \sigma(\mathrm{Ker}(e\partial_B)). $$

□
<figure><img src="image_50.png" /></figure>
# Part II

## The general invariants
<figure><img src="image_51.png" /></figure>
# Chapter 6

## Introduction to finite type invariants and Jacobi diagrams

This chapter introduces the target space of the invariant $\mathcal{Z}$ studied in this book. It is a space generated by uni-trivalent graphs called Jacobi diagrams. Dror Bar-Natan has studied this space in his fundamental article [BN95a], where most of the results of this chapter come from. In this chapter, the field $\mathbb{K}$ is $\mathbb{Q}$ or $\mathbb{R}$.

### 6.1 Definition of finite type invariants

A $\mathbb{K}$-valued *invariant* of oriented 3-manifolds is a function from the set of 3-manifolds, considered up to orientation-preserving diffeomorphism, to $\mathbb{K}$. Let $\cup_{i=1}^n S_i^1$ denote a disjoint union of $n$ circles, where each $S_i^1$ is a copy of $S^1$. Here, an *n-component link* in a 3-manifold $R$ is an equivalence class of smooth embeddings $L: \cup_{i=1}^n S_i^1 \to R$ under the equivalence relation that identifies two embeddings $L$ and $L'$ if and only if there is an orientation-preserving diffeomorphism $h$ of $R$ such that $h(L) = L'^1$. A *knot* is a one-component link. A *link invariant* (resp. a *knot invariant*) is a function of links (resp. knots). For example, $\Theta$ is an invariant of $\mathbb{Q}$-spheres, and the linking number is a rational invariant of two-component links in rational homology spheres.

In order to study a function, it is common to study its derivative, and the derivatives of its derivative. The derivative of a function is defined from its variations. For a function $f$ from $\mathbb{Z}^d = \bigoplus_{i=1}^d \mathbb{Z}e_i$ to $\mathbb{K}$, one can define its

<footnote n="1">¹This relation is equivalent to the usual equivalence relation defined by isotopies when $R$ is $\mathbb{R}^3$ or $S^3$. In general 3-manifolds, two equivalent links are not necessarily isotopic, but the link invariants described in this book are invariant under the above equivalence relation.</footnote>
first-order derivatives $\frac{\partial f}{\partial e_i}: \mathbb{Z}^d \to \mathbb{K}$ by

$$ \frac{\partial f}{\partial e_i}(z) = f(z + e_i) - f(z), $$

and check that all the first-order derivatives of $f$ vanish if and only if $f$ is constant. Inductively define an $n$-order derivative to be a first-order derivative of an $(n-1)$-order derivative for a positive integer $n$. Then it can be checked that all the $(n+1)$-order derivatives of a function $f: \mathbb{Z}^d \to \mathbb{K}$ vanish if and only if $f$ is a polynomial of degree not greater than $n$ in the coordinates. In order to study topological invariants, we can similarly study their variations under *simple operations*.

Below, $X$ denotes one of the following sets

• $\mathbb{Z}^d$,

• the set $\mathcal{K}$ of knots in $\mathbb{R}^3$, the set $\mathcal{K}_k$ of $k$-component links in $\mathbb{R}^3$,$^2$

• the set $\mathcal{M}$ of $\mathbb{Z}$-spheres, the set $\mathcal{M}_{\mathbb{Q}}$ of $\mathbb{Q}$-spheres (up to orientation-preserving diffeomorphism).

and $\mathcal{O}(X)$ denotes a set of *simple operations* acting on some elements of $X$. For $X = \mathbb{Z}^d$, $\mathcal{O}(X)$ consists of the operations $(z \to z \pm e_i)$.

For knots or links in $\mathbb{R}^3$, the *simple operations* are *crossing changes*. A *crossing change ball* of a link $L$ is a ball $B$ of the ambient space, where $L \cap B$ is a disjoint union of two arcs $\alpha_1$ and $\alpha_2$ properly embedded in $B$, and there exist two disjoint topological disks $D_1$ and $D_2$ embedded in $B$, such that, for $i \in \{1, 2\}$, the topological circle $\partial D_i$ is the union of $\alpha_i$ and an arc of $\partial B$ as in the following picture:

<figure><img src="image_52.png" /></figure>

After an isotopy, a projection of $(B, \alpha_1, \alpha_2)$ looks like $\overbrace{(\times)}$ or $\overbrace{(\times)}$. A *crossing change* is a change that does not change $L$ outside $\tilde{B}$ and that modifies it inside $B$ by a local move

$$ (\overbrace{(\times)} \to \overbrace{(\times)} ) \text{ or } (\overbrace{(\times)} \to \overbrace{(\times)} ). $$

For the left move, the crossing change is *positive*. It is *negative* for the move of the right-hand side.

<footnote n="2">²Recall that a knot is an isotopy class of knot embeddings.</footnote>
For integer (resp. rational) homology 3-spheres, the simple operations are integral (resp. rational) *LP-surgeries*, defined in Subsection 1.3.2, and $\mathcal{O}(M)$ (resp. $\mathcal{O}(M_{\mathbb{Q}})$) is denoted by $\mathcal{O}_{\mathcal{L}}^{\mathbb{Z}}$ (resp. $\mathcal{O}_{\mathcal{L}}^{\mathbb{Q}}$).

Say that crossing changes are *disjoint* if they sit inside disjoint 3-balls. Say that LP-surgeries $(A'/A)$ and $(B'/B)$ in a manifold $R$ are *disjoint* if $A$ and $B$ are disjoint in $R$. Two operations on $\mathbb{Z}^d$ are always *disjoint* (even if they look identical). In particular, disjoint operations *commute* (i.e., their result does not depend on which one is performed first). Set $\underline{n} = \{1, 2, \dots, n\}$. Consider the vector space $\mathcal{F}_0(X) = \mathcal{F}_0(X; \mathbb{K})$ freely generated by $X$ over $\mathbb{K}$. For an element $x$ of $X$ and $n$ pairwise disjoint operations $o_1, \dots, o_n$ acting on $x$, let $x((o_i)_{i \in I})$ denote the element of $X$ obtained by performing the operations $o_i$ on $x$ for $i \in I$. Define

$$[x; o_1, \dots, o_n] = \sum_{I \subseteq \underline{n}} (-1)^{|I|} x((o_i)_{i \in I}) \in \mathcal{F}_0(X).$$

Then define $\mathcal{F}_n(X) = \mathcal{F}_n(X; \mathbb{K})$ as the $\mathbb{K}$-subspace of $\mathcal{F}_0(X)$ generated by the $[x; o_1, \dots, o_n]$, for all $x \in X$ equipped with $n$ pairwise disjoint simple operations $o_1, \dots, o_n$ acting on $x$. Since we have

$$[x; o_1, \dots, o_n, o_{n+1}] = [x; o_1, \dots, o_n] - [x(o_{n+1}); o_1, \dots, o_n],$$

we get $\mathcal{F}_{n+1}(X) \subseteq \mathcal{F}_n(X)$ for all $n \in \mathbb{N}$.

**Definition 6.1.** A $\mathbb{K}$-valued function $f$ on $X$, extends uniquely as a $\mathbb{K}$-linear map on

$$\mathcal{F}_0(X)^* = \operatorname{Hom}(\mathcal{F}_0(X); \mathbb{K}),$$

which is still denoted by $f$. For an integer $n \in \mathbb{N}$, the invariant (or function) $f$ is of degree $\le n$ if and only if $f(\mathcal{F}_{n+1}(X)) = 0$. The degree of such an invariant is the smallest integer $n \in \mathbb{N}$ such that $f(\mathcal{F}_{n+1}(X)) = 0$. An invariant is of *finite type* if it is of degree $n$ for some $n \in \mathbb{N}$. This definition depends on the chosen set of operations $\mathcal{O}(X)$. We fixed our choices for our sets $X$, but other choices could lead to different notions. See [GGP01] for $X = \mathcal{M}$.

Let $\mathcal{I}_n(X) = (\mathcal{F}_0(X)/\mathcal{F}_{n+1}(X))^*$ be the space of invariants of degree at most $n$. Of course, we have $\mathcal{I}_n(X) \subseteq \mathcal{I}_{n+1}(X)$ for all $n \in \mathbb{N}$.

**Exercise 6.2.** Prove that $\mathcal{I}_n(\mathbb{Z}^d)$ is the space of polynomials of degree at most $n$ on $\mathbb{Z}^d$.

**Lemma 6.3.** If $f \in \mathcal{I}_m(X)$ and $g \in \mathcal{I}_n(X)$, then $fg \in \mathcal{I}_{m+n}(X)$.
PROOF: Let $[x; (o_i)_{i \in \underline{m+n+1}}] \in \mathcal{F}_{m+n+1}(X)$. The lemma is a direct consequence of the equality

$$fg([x; (o_i)_{i \in \underline{m+n+1}}]) = \sum_{J \subseteq \underline{m+n+1}} f([x; (o_j)_{j \in J}]) g([x((o_j)_{j \in J}); (o_i)_{i \in \underline{m+n+1} \setminus J}]),$$

which is proved as follows. The right-hand side is equal to

$$\sum_{J \subseteq \underline{m+n+1}} (-1)^{|J|} \left( \sum_{K: K \subseteq J} (-1)^{|K|} f(x((o_i)_{i \in K})) \right) \left( \sum_{L: J \subseteq L} (-1)^{|L|} g(x((o_i)_{i \in L})) \right) \\ = \sum_{(K,L): K \subseteq L \subseteq \underline{m+n+1}} (-1)^{|K|+|L|} f(x((o_i)_{i \in K})) g(x((o_i)_{i \in L})) \left( \sum_{J: K \subseteq J \subseteq L} (-1)^{|J|} \right),$$

$$\text{where } \sum_{J: K \subseteq J \subseteq L} (-1)^{|J|} = \begin{cases} 0 & \text{if } K \subsetneq L \\ (-1)^{|K|} & \text{if } K = L. \end{cases} \quad \square$$

**Lemma 6.4.** Any $n$-component link in $\mathbb{R}^3$ can be transformed to the trivial $n$-component link below by a finite number of disjoint crossing changes.

$$U_1 \circled{\bigcirc} U_2 \circled{\bigcirc} \dots \circled{\bigcirc} U_n \circled{\bigcirc}$$

PROOF: Let $L$ be an (embedding representing an) $n$-component link in $\mathbb{R}^3$. Since $\mathbb{R}^3$ is simply connected, there is a homotopy that carries $L$ to the trivial link. Such a homotopy $h: [0, 1] \times \cup_{i=1}^n S_i^1 \to \mathbb{R}^3$ can be chosen to be smooth and such that $h(t, \cdot)$ is an embedding, except for finitely many times $t_i$, $0 < t_1 < \dots < t_i < t_{i+1} < \dots < 1$, at which $h(t_i, \cdot)$ is an immersion with one double point and no other multiple points, and the link $h(t, \cdot)$ changes exactly by a crossing change when $t$ crosses a $t_i$. (For an alternative elementary proof of this fact, see [Les05, Subsection 7.1] before Definition 7.5, for example). $\square$

In particular, a degree 0 invariant of $n$-component links of $\mathbb{R}^3$ must be constant since it does not vary under a crossing change.

**Exercise 6.5.** 1. Check that $\mathcal{I}_1(\mathcal{K}) = \mathbb{K}c_0$, where $c_0$ is the constant map that maps any knot to 1.

2. Check that the linking number is a degree 1 invariant of 2-component links of $\mathbb{R}^3$.

3. Check that $\mathcal{I}_1(\mathcal{K}_2) = \mathbb{K}c_0 \oplus \mathbb{K}lk$, where $c_0$ is the constant map that maps any two-component link to 1.
## 6.2 Introduction to chord diagrams

A singular knot with $n$ double points in $\mathbb{R}^3$ is an immersion of a circle with $n$ transverse double points in $\mathbb{R}^3$. Such a double point $\mathcal{X}$ can be desingularized in two ways, the positive one $\mathcal{X}$ and the negative one $\mathcal{X}$. For example, desingularizing the double points of the singular knot $\mathcal{X}$ in the positive way produces the knot $\mathcal{X}$. Note that the sign of the desingularization is defined from the orientations of the knot and of the ambient space.

Let $o_i$ be disjoint negative crossing changes $\mathcal{X} \to \mathcal{X}$ to be performed on a knot $K$. We represent $[K; o_1, \dots, o_n]$ as a singular knot with $n$ double points such that $K((o_i)_{i \in I})$ is obtained from the singular knot by desingularizing the crossings of $I$ in the negative way, and the others in the positive way, for $I \subseteq \underline{n}$. Thus, singular knots represent elements of $\mathcal{F}_0(\mathcal{K})$. Three singular knots that coincide outside a ball, inside which they look as in the following skein relation

$$ \mathcal{X} = \mathcal{X} - \mathcal{X}, $$

satisfy this relation in $\mathcal{F}_0(\mathcal{K})$.

Define the chord diagram $\Gamma_C([K; o_1, \dots, o_n])$ associated to $[K; o_1, \dots, o_n]$ as follows. Draw the preimage of the associated singular knot with $n$ double points as an oriented dashed circle equipped with the $2n$ preimages of the double points, and join the pairs of preimages of a double point by a plain segment called a chord. For example, we have

$$ \Gamma_C(\mathcal{O}) = \mathcal{X}. $$

Formally, a chord diagram with $n$ chords (on a circle) is a cyclic order of the $2n$ ends of the $n$ chords, up to a permutation of the chords and up to exchanging the two ends of a chord.

**Lemma 6.6.** If $f$ is an invariant of knots in $\mathbb{R}^3$ of degree at most $n$, then $f([K; o_1, \dots, o_n])$ depends only on $\Gamma_C([K; o_1, \dots, o_n])$.

**PROOF:** Since $f$ is of degree $n$, $f([K; o_1, \dots, o_n])$ is invariant under a crossing change outside the balls of the $o_i$, that is outside the double points of the associated singular knot. Therefore, $f([K; o_1, \dots, o_n])$ depends only on the cyclic order of the $2n$ arcs involved in the $o_i$ on $K$. (A more detailed proof can be found in [Les05, Subsection 7.3].) $\square$

Let $\mathcal{D}_n$ be the $\mathbb{K}$-vector space freely generated by the $n$-chord diagrams
(i.e., the diagrams with $n$ chords) on $S^1$. For example, we have

$$ \mathcal{D}_0 = \mathbb{K}(\bigcirc), \mathcal{D}_1 = \mathbb{K}(\bigcirc), \mathcal{D}_2 = \mathbb{K}(\bigcirc) \oplus \mathbb{K}(\bigcirc), \text{ and} $$

$$ \mathcal{D}_3 = \mathbb{K}(\bigcirc) \oplus \mathbb{K}(\bigcirc) \oplus \mathbb{K}(\bigcirc) \oplus \mathbb{K}(\bigcirc) \oplus \mathbb{K}(\bigcirc) \oplus \mathbb{K}(\bigcirc). $$

**Lemma 6.7.** The map $\phi_n$ from $\mathcal{D}_n$ to $\mathcal{F}_n(\mathcal{K})/\mathcal{F}_{n+1}(\mathcal{K})$ that maps an $n$-chord diagram $\Gamma$ to some $[K; o_1, \dots, o_n]$ whose diagram is $\Gamma$ is well-defined and surjective.

PROOF: Use the arguments of the proof of Lemma 6.6. $\square$

For example, we have

$$ \phi_3(\bigcirc) = [\bigcirc]. $$

Lemma 6.7 implies that

$$ \phi_n^*: \left( \frac{\mathcal{F}_n(\mathcal{K})}{\mathcal{F}_{n+1}(\mathcal{K})} \right)^* \to \mathcal{D}_n^* $$

is injective. The kernel of the restriction below

$$ \left( \mathcal{I}_n(\mathcal{K}) = \left( \frac{\mathcal{F}_0(\mathcal{K})}{\mathcal{F}_{n+1}(\mathcal{K})} \right)^* \right) \to \left( \frac{\mathcal{F}_n(\mathcal{K})}{\mathcal{F}_{n+1}(\mathcal{K})} \right)^* $$

is $\mathcal{I}_{n-1}(\mathcal{K})$. Thus, $\mathcal{I}_n(\mathcal{K})/\mathcal{I}_{n-1}(\mathcal{K})$ injects into $\mathcal{D}_n^*$, and the dimension of $\mathcal{I}_n(\mathcal{K})$ is finite for all $n$. In particular, $\mathcal{F}_0(\mathcal{K})/\mathcal{F}_{n+1}(\mathcal{K})$ is finite-dimensional, and the above restriction is surjective. Therefore, we have

$$ \frac{\mathcal{I}_n(\mathcal{K})}{\mathcal{I}_{n-1}(\mathcal{K})} = \mathrm{Hom} \left( \frac{\mathcal{F}_n(\mathcal{K})}{\mathcal{F}_{n+1}(\mathcal{K})}; \mathbb{K} \right). $$

An isolated chord in a chord diagram is a chord between two points of $S^1$ that are consecutive on the circle $S^1$.

**Lemma 6.8.** Let $D$ be a diagram on $S^1$ that contains an isolated chord. Then $\phi_n(D) = 0$. Let $D^1, D^2, D^3$, and $D^4$ be four $n$-chord diagrams identical outside three portions of circles, inside which they look like

$$ D^1 = \text{\textcircled{3} with a circled point}, D^2 = \text{\textcircled{3} with a circled point}, D^3 = \text{\textcircled{3} with a circled point}, \text{ and } D^4 = \text{\textcircled{3} with a circled point}. $$

Then we have

$$ \phi_n(-D^1 + D^2 + D^3 - D^4) = 0. $$
PROOF: For the first assertion, observe

$$ \phi_n(\textcircled{2}) = [\infty] - [\infty]. $$

Let us prove the second one. We may represent

$$ D^1 = \textcircled{3} \textcircled{1} $$

by a singular knot $K^1$ with $n$ double points, which intersects a ball as

$$ K^1 = \textcircled{2} \textcircled{3} \textcircled{1} $$

Let $K^2, K^3, K^4$ be the singular knots with $n$ double points that coincide with $K^1$ outside this ball, and that intersect this ball as shown in the pictures:

$$ K^2 = \textcircled{2} \textcircled{3} \textcircled{1} , \quad K^3 = \textcircled{2} \textcircled{3} \textcircled{1} , \quad K^4 = \textcircled{2} \textcircled{3} \textcircled{1} $$

Then the chord diagram $D(K^2)$ associated to $K^2$ is $D^2$. Similarly, we have $D(K^3) = D^3$ and $D(K^4) = D^4$. Therefore, we have $\phi_n(-D^1 + D^2 + D^3 - D^4) = -[K^1] + [K^2] + [K^3] - [K^4]$.

Thus, it is enough to prove that we have

$$ -[K^1] + [K^2] + [K^3] - [K^4] = 0 $$

in $\mathcal{F}_n(\mathcal{K})$. Let us prove it.

Let $K_0$ be the singular knot with $(n-1)$ double points that intersects our ball as

$$ K_0 = \textcircled{2} \textcircled{3} \textcircled{1} $$

and that coincides with $K^1$ outside this ball.

The strands 1 and 2 in the pictured double point are in the horizontal plane. They orient it. The strand 3 is vertical. It intersects the horizontal plane in a positive way between the tails of 1 and 2. Now, make 3 turn around the double point counterclockwise so that it successively becomes the knots with $(n-1)$ double points:

$$ K_1 = \textcircled{1} \textcircled{3} \textcircled{2} , \quad K_2 = \textcircled{1} \textcircled{3} \textcircled{2} , \quad \text{and} \quad K_3 = \textcircled{1} \textcircled{3} \textcircled{2} $$
On its way, it goes successively through our four knots $K^1$, $K^2$, $K^3$, and $K^4$ with $n$ double points, which appear inside matching parentheses, in the following obvious identity in $\mathcal{F}_{n-1}(K)$

$$([K_1] - [K_0]) + ([K_2] - [K_1]) + ([K_3] - [K_2]) + ([K_0] - [K_3]) = 0.$$

Now, we have $[K^i] = \pm ([K_i] - [K_{i-1}])$, where the sign $\pm$ is $+$ when the vertical strand goes through an arrow from $K_{i-1}$ to $K_i$, and minus when it goes through a tail. Therefore, the above equality can be written as

$$-[K^1] + [K^2] + [K^3] - [K^4] = 0.$$

□

Let $\overline{\mathcal{A}}_n$ denote the quotient of $\mathcal{D}_n$ by the four-term relation (4T). It is the quotient of $\mathcal{D}_n$ by the vector space generated by the $(-D^1 + D^2 + D^3 - D^4)$ for all the 4-tuples $(D^1, D^2, D^3, D^4)$ as in Lemma 6.8. Call $(1T)$ the relation that identifies a diagram with an isolated chord with 0. So $\overline{\mathcal{A}}_n/(1T)$ is the quotient of $\overline{\mathcal{A}}_n$ by the vector space generated by diagrams with an isolated chord.

According to Lemma 6.8 above, the map $\phi_n$ induces a map

$$\bar{\phi}_n: \overline{\mathcal{A}}_n/(1T) \to \frac{\mathcal{F}_n(K)}{\mathcal{F}_{n+1}(K)}.$$

The fundamental theorem of Vassiliev invariants (which are finite type knot invariants) can now be stated.

**Theorem 6.9 (Bar-Natan, Kontsevich).** There exists a family

$$(\tilde{\mathcal{Z}}_n : \mathcal{F}_0(\mathcal{K}) \to \overline{\mathcal{A}}_n/(1T))_{n \in \mathbb{N}}$$

of linear maps satisfying:

* $\tilde{\mathcal{Z}}_n(\mathcal{F}_{n+1}(\mathcal{K})) = 0$.

* Let $\overline{\mathcal{Z}}_n$ be the map induced by $\tilde{\mathcal{Z}}_n$ from $\mathcal{F}_n(\mathcal{K})/\mathcal{F}_{n+1}(\mathcal{K})$ to $\overline{\mathcal{A}}_n/(1T)$. Then $\overline{\mathcal{Z}}_n \circ \bar{\phi}_n$ is the identity map of $\overline{\mathcal{A}}_n/(1T)$.

In particular, the inverse isomorphisms $\overline{\mathcal{Z}}_n$ and $\bar{\phi}_n$ identify $\mathcal{F}_n(\mathcal{K})/\mathcal{F}_{n+1}(\mathcal{K})$ with $\overline{\mathcal{A}}_n/(1T)$, and we have

$$\frac{\mathcal{I}_n(\mathcal{K})}{\mathcal{I}_{n-1}(\mathcal{K})} \cong (\overline{\mathcal{A}}_n/(1T))^*.$$
Maxim Kontsevich and Dror Bar-Natan proved this theorem by using the Kontsevich integral $Z^K = (Z_n^K)_{n \in \mathbb{N}}$ [BN95a], for $K = \mathbb{R}$. It is also true when $K = \mathbb{Q}$. It is reproved in Section 17.6 using the invariant $\mathcal{Z}$ studied in this book. An invariant $\tilde{\mathcal{Z}}$ as in the above statement has the following universality property.

**Corollary 6.10.** For any real-valued finite type invariant $f$ of knots in $\mathbb{R}^3$ of degree at most $n$, there exist linear forms $\psi_i: \overline{\mathcal{A}}_i/(1T) \to \mathbb{R}$, for $i = 0, \dots, n$, such that

$$f = \sum_{i=0}^{n} \psi_i \circ \tilde{\mathcal{Z}}_i.$$

**PROOF:** Let $\psi_n = f|_{\mathcal{F}_n} \circ \bar{\phi}_n$, then $(f - \psi_n \circ \tilde{\mathcal{Z}}_n)$ is an invariant of degree at most $n-1$. Conclude by induction. $\square$

By projection (or up to $(1T)$), the invariant $\tilde{\mathcal{Z}}$ defines a universal Vassiliev knot invariant with respect to the following definition.

An invariant $Y: \mathcal{F}_0(\mathcal{K}) \to \prod_{n \in \mathbb{N}} \overline{\mathcal{A}}_n/(1T)$ such that

* $Y_n(\mathcal{F}_{n+1}(\mathcal{K})) = 0$ and

* $Y_n$ induces a left inverse to $\bar{\phi}_n$ from $\mathcal{F}_n(\mathcal{K}; \mathbb{R})/\mathcal{F}_{n+1}(\mathcal{K}; \mathbb{R})$ to $\overline{\mathcal{A}}_n/(1T)$

is called a *universal Vassiliev knot invariant*.

The terminology is justified because such an invariant contains all the real-valued Vassiliev knot invariants as in Corollary 6.10.

As Daniel Altschüler and Laurent Freidel proved in [AF97], the restriction of the invariant $\mathcal{Z} = (\mathcal{Z}_n)_{n \in \mathbb{N}}$ to knots of $\mathbb{R}^3$ also satisfies the properties of Theorem 6.9. So it is also a *universal Vassiliev knot invariant*. We will give alternative proofs of generalizations of this Altschüler–Freidel theorem in Section 17.6.

Similar characterizations of the spaces of finite type invariants of links in $\mathbb{R}^3$, integer homology 3-spheres, and rational homology 3-spheres will be presented in Section 17.6 and Chapter 18, respectively. For integer homology 3-spheres and rational homology 3-spheres, the most difficult parts of the proofs are consequences of the splitting formulae satisfied by $\mathcal{Z}$, stated in Theorem 18.6.

We end this subsection with an example of a nontrivial linear form on $\overline{\mathcal{A}}_n/(1T)$.

**Example 6.11.** We first define a function $\tilde{w}_C$ of chord diagrams. Let $\Gamma$ be a chord diagram. Immerse $\Gamma$ in the unit disk $D_1$ of the plane so that the
chords of $\Gamma$ are embedded and attached to the left-hand side of the boundary $S^1$ of $D_1$, as in our former pictures like

$$ \text{[Chord diagram]} \quad \text{or} \quad \text{[Surgery diagram]} $$

Attach disjoint oriented bands $[0, 1]^2$, one “around each chord”, so that $(\partial[0, 1]) \times [0, 1]$ is a well-oriented neighborhood of the ends of the chord in the dashed circle $S^1$, as in the following figure. Perform the surgery on the dashed circle $S^1$ that replaces $(\partial[0, 1]) \times [0, 1]$ with $[0, 1] \times \partial[0, 1]$ as in the following figure.

$$ \text{[Surgery diagram]} \quad \text{[Surgery diagram]} \quad \text{[Surgery diagram]} $$

If the resulting naturally oriented one-manifold is connected, then $\tilde{w}_C(\Gamma) = 1$. Otherwise, $\tilde{w}_C(\Gamma) = 0$. For example, we have

$$ \tilde{w}_C(\text{[Chord diagram]}) = 1, \quad \tilde{w}_C(\text{[Surgery diagram]}) = 0, \quad \text{and} \quad \tilde{w}_C(\text{[Surgery diagram]}) = 0. $$

Since $\text{[Surgery diagram]}$ is connected, we have $\tilde{w}_C(\text{[Surgery diagram]}) = 1$. We also have

$$ \tilde{w}_C(\text{[Surgery diagram]}) = \tilde{w}_C(\text{[Surgery diagram]}) = 0. $$

More generally, the reader can check that one of our surgeries changes the mod 2 congruence class of the number of connected components of the surgere-d manifold. So $\tilde{w}_C(\Gamma) = 0$ for any chord diagram $\Gamma$ with an odd number of chords. Extend $\tilde{w}_C$ linearly over $D_n$. For 4-tuples $(D^1, D^2, D^3, D^4)$ as in Lemma 6.8, the reader can check that the extended $\tilde{w}_C$ maps $(-D^1 + D^2 + D^3 - D^4)$ to zero. It also maps diagrams with isolated chords to zero. Therefore, $\tilde{w}_C$ induces a linear map $w_C$ on $\overline{\mathcal{A}}_n$ (and on $\overline{\mathcal{A}}_n/(1T)$) for any $n$. For any chord diagram $\Gamma$, we have $w_C([\Gamma]) = \tilde{w}_C(\Gamma)$. Dror Bar-Natan and Stavros Garoufalidis introduced this linear map in [BNG96]. They called it the *Conway weight system*. It is zero when $n$ is odd. It is not zero when $n$ is even since we have

$$ w_C(\text{[Surgery diagram]}) \neq 0. $$

## 6.3 More spaces of diagrams

**Definition 6.12.** A *uni-trivalent graph* $\Gamma$ is a 6-tuple

$$ (H(\Gamma), E(\Gamma), U(\Gamma), T(\Gamma), p_E, p_V), $$

where
• $H(\Gamma)$, $E(\Gamma)$, $U(\Gamma)$, and $T(\Gamma)$ are finite sets, respectively called the *set of half-edges* of $\Gamma$, the *set of edges* of $\Gamma$, the *set of univalent vertices* of $\Gamma$, and the set of trivalent vertices of $\Gamma$,

• $p_E: H(\Gamma) \to E(\Gamma)$ is a two-to-one map (every element of $E(\Gamma)$ has two preimages under $p_E$) and

• $p_V: H(\Gamma) \to U(\Gamma) \sqcup T(\Gamma)$ is a map such that every element of $U(\Gamma)$ has one preimage under $p_V$ and every element of $T(\Gamma)$ has three preimages under $p_V$,

up to isomorphism. In other words, $\Gamma$ is a set $H(\Gamma)$ equipped with two partitions, a partition into pairs (induced by $p_E$), and a partition into singletons and triples (induced by $p_V$), up to the bijections that preserve the partitions. These bijections are the *automorphisms* of $\Gamma$.

**Definition 6.13.** Let $\mathcal{L}$ be a one-manifold, oriented or not. A *Jacobi diagram* $\Gamma$ with support $\mathcal{L}$, also called *Jacobi diagram on $\mathcal{L}$*, is a finite uni-trivalent graph $\Gamma$ equipped with an isotopy class $[i_\Gamma]$ of injections $i_\Gamma$ from the set $U(\Gamma)$ of univalent vertices of $\Gamma$ into the interior of $\mathcal{L}$. For such a $\Gamma$, a $\Gamma$-compatible injection is an injection in the class $[i_\Gamma]$. An *orientation* of a trivalent vertex of $\Gamma$ is a cyclic order on the set of the three half-edges that meet at this vertex. An *orientation* of a univalent vertex $u$ of $\Gamma$ is an orientation of the connected component $\mathcal{L}(u)$ of $i_\Gamma(u)$ in $\mathcal{L}$, for a choice of $\Gamma$-compatible $i_\Gamma$. This orientation associated to $u$ is also called (and thought of as) a *local orientation of $\mathcal{L}$ at $u$*.<fnref n="3" /> When $\mathcal{L}$ is oriented, the orientation of $\mathcal{L}$ orients the univalent vertices of $\Gamma$ naturally.

A *vertex-orientation* of a Jacobi diagram $\Gamma$ is an *orientation* of every vertex of $\Gamma$. A Jacobi diagram is *oriented* if it is equipped with a vertex-orientation.<fnref n="4" />

Unless otherwise mentioned, the supports of Jacobi diagrams are oriented, and we use the induced orientations of univalent vertices without mentioning them. Nevertheless, the above notion of local orientations will prove useful to state some properties of the invariant $Z$ studied in this book, such as the behavior under cablings in Theorem 13.12.

We represent an oriented Jacobi diagram $\Gamma$ by a planar immersion of $\Gamma \cup \mathcal{L} = \Gamma \cup U(\Gamma) \mathcal{L}$, with the following conventions. We represent the (oriented)

<footnote n="3">³A *local orientation* of $\mathcal{L}$ at $u$ is simply an orientation of $\mathcal{L}(u)$. However, since different vertices are allowed to induce different orientations, we think of these orientations as being *local*, i.e., defined in a neighborhood of $i_\Gamma(u)$ for a choice of $\Gamma$-compatible $i_\Gamma$.</footnote>

<footnote n="4">⁴When $\mathcal{L}$ is oriented, it suffices to specify the orientations of the trivalent vertices since $\mathcal{L}$ orients the univalent vertices.</footnote>
one-manifold $\mathcal{L}$ by dashed lines, the edges of the diagram $\Gamma$ by plain segments, and the vertices by big dots. The univalent vertices of $U(\Gamma)$ are located at their images under a $\Gamma$-compatible injection $i_{\Gamma}$. We represent the orientation of a trivalent vertex by the counterclockwise order of the three half-edges that meet at the vertex. Here is an example of a picture of a Jacobi diagram $\Gamma$ on the disjoint union $\mathcal{L} = S^1 \sqcup S^1$ of two (oriented) circles:

<figure><img src="image_53.png" /></figure>

The degree of such a diagram is half the number of all its vertices. Note that a chord diagram of $\mathcal{D}_n$ is a degree $n$ Jacobi diagram on $S^1$ without trivalent vertices. For an (oriented) one-manifold $\mathcal{L}$, $\mathcal{D}_n(\mathcal{L})$ denotes the $\mathbb{K}$-vector space freely generated by the degree $n$ oriented Jacobi diagrams on $\mathcal{L}$. For the (oriented) circle $S^1$, we have

$$ \mathcal{D}_1(S^1) = \mathbb{K}(\text{\Large $\bullet$}) \oplus \mathbb{K}(\text{\Large $\bullet\circ$}) \oplus \mathbb{K}(\text{\Large $\bullet\circ\circ$}) \oplus \mathbb{K}(\text{\Large $\bullet\circ\circ\circ$}) \oplus \mathbb{K}(\text{\Large $\bullet\circ\circ\circ\circ$}). $$

For an (oriented) one-manifold $\mathcal{L}$, $\mathcal{A}_n(\mathcal{L})$ denotes the quotient of $\mathcal{D}_n(\mathcal{L})$ by the following relations AS, Jacobi, and STU:

$$ \text{AS (or antisymmetry): } \text{\Large $\bullet$} + \text{\Large $\circ$} = 0 $$

$$ \text{Jacobi: } \text{\Large $\bullet$} + \text{\Large $\circ$} + \text{\Large $\circ$} = 0 $$

$$ \text{STU: } \text{\Large $\circ$} = \text{\Large $\bullet$} + \text{\Large $\circ$} - \text{\Large $\circ$} $$

Each of these relations relates oriented Jacobi diagrams which are identical outside the pictures. The quotient $\mathcal{A}_n(\mathcal{L})$ is the largest quotient of $\mathcal{D}_n(\mathcal{L})$ in which these relations hold. It is obtained by quotienting $\mathcal{D}_n(\mathcal{L})$ by the $\mathbb{K}$-vector space generated by elements of $\mathcal{D}_n(\mathcal{L})$ of the form

$$ (\text{\Large $\bullet$} + \text{\Large $\circ$}), (\text{\Large $\circ$} + \text{\Large $\circ$} + \text{\Large $\circ$}), \text{ or } (\text{\Large $\circ$} - \text{\Large $\circ$} + \text{\Large $\circ$}). $$

**Example 6.14.** We have

$$ \mathcal{A}_1(S^1) = \mathbb{K}(\text{\Large $\bullet$}) \oplus \mathbb{K}(\text{\Large $\circ$}). $$

**Note 6.15.** When $\partial\mathcal{L} = \emptyset$, any finite-dimensional Lie algebra equipped with a finite-dimensional representation and a nondegenerate bilinear symmetric invariant form provides a nontrivial linear map from $\mathcal{A}_n(\mathcal{L})$ to $\mathbb{K}$. Such a map is called a *weight system*. See [BN95a], [CDM12, Chapter 6] or [Les05,
**Lemma 6.18.** Let $\mathcal{L}$ be an oriented one-manifold. Let $\mathcal{L}^f$ denote the nonoriented manifold obtained from $\mathcal{L}$ by forgetting the orientation. Let $\Gamma^f$ be an oriented Jacobi diagram on $\mathcal{L}^f$. A univalent vertex $u$ of $\Gamma^f$ is $\mathcal{L}$-oriented if it induces the orientation of $\mathcal{L}$. Otherwise, it is $(-\mathcal{L})$-oriented. Let $\Gamma(\Gamma^f, \mathcal{L})$ be the oriented Jacobi diagram on $\mathcal{L}$ obtained from $\Gamma^f$ by reversing the local orientation of the $(-\mathcal{L})$-oriented univalent vertices. Let $k(\Gamma^f, \mathcal{L})$ be the number of $(-\mathcal{L})$-oriented univalent vertices of $\Gamma^f$. The linear map from $\mathcal{D}_n(\mathcal{L}^f)$ into $\mathcal{D}_n(\mathcal{L})$ that maps any oriented Jacobi diagram $\Gamma^f$ to $(-1)^{k(\Gamma^f, \mathcal{L})}\Gamma(\Gamma^f, \mathcal{L})$ and the linear canonical injection from $\mathcal{D}_n(\mathcal{L})$ into $\mathcal{D}_n(\mathcal{L}^f)$ induce canonical isomorphisms between $\mathcal{A}_n(\mathcal{L})$ and $\mathcal{A}_n(\mathcal{L}^f)$, which are inverse to each other, for any integer $n \in \mathbb{N}$.

**PROOF:** Exercise. $\square$

<figure><img src="image_54.png" /></figure>

The above lemma justifies using the same notation $\mathcal{A}_n(\cdot)$ for oriented and unoriented supports. We draw Jacobi diagrams on oriented supports by attaching the half-edges of univalent vertices to the left-hand side of the support, to avoid confusion and get rid of the support orientation more easily.

**Notation 6.19.** When $\mathcal{L} \neq \emptyset$, let $\check{\mathcal{A}}_n(\mathcal{L}) = \check{\mathcal{A}}_n(\mathcal{L}; \mathbb{K})$ denote the quotient of $\mathcal{A}_n(\mathcal{L}) = \mathcal{A}_n(\mathcal{L}; \mathbb{K})$ by the vector space generated by the diagrams that have at least one connected component without univalent vertices.

So $\check{\mathcal{A}}_n(\mathcal{L})$ is generated by the degree $n$ oriented Jacobi diagrams whose (plain) connected components contain at least one univalent vertex.

**Lemma 6.20.** The space $\check{\mathcal{A}}_n(S^1)$ is the quotient by the relations AS and STU of the vector space generated by the degree $n$ oriented Jacobi diagrams whose connected components contain at least one univalent vertex. In other words, the Jacobi relation is a consequence of the relations AS and STU in this vector space.

**PROOF:** We want to prove that the Jacobi relation holds in the quotient by AS and STU of the space of uni-trivalent diagrams on $S^1$ with at least one univalent vertex in each connected component. Consider three diagrams represented by three immersions which coincide outside a disk $D$, inside which they are as in the pictures involved in the Jacobi relation. Use STU as much as possible to remove all trivalent vertices that can be removed without changing the two vertices in $D$, on the three diagrams simultaneously. This transforms the Jacobi relation to be proved to a sum of similar relations, where one of the four entries of the disk is directly connected to $S^1$. Since the four entries play the same role in the Jacobi relation, we may assume that the Jacobi relation to be proved is
<figure><img src="image_55.png" /></figure>

Using STU twice and AS transforms the summands of the left-hand side to diagrams that can be represented by three straight lines from the entries 1, 2, 3 to three fixed points of the horizontal line numbered from left to right. When the entry $i \in \{1, 2, 3\}$ is connected to the point $\sigma(i)$ of the horizontal dashed line, where $\sigma$ is a permutation of $\{1, 2, 3\}$, the corresponding diagram is denoted by $(\sigma(1)\sigma(2)\sigma(3))$. Thus, the expansion of the left-hand side of the above equation is

$$ ((123) - (132) - (231) + (321)) -((213) - (231) - (132) + (312)) , -((123) - (213) - (312) + (321)) $$

which vanishes, and the lemma is proved. □

**Proposition 6.21.** The natural map from $\mathcal{D}_n$ to $\check{\mathcal{A}}_n(S^1)$ induces an isomorphism from the space $\overline{\mathcal{A}}_n$ of chord diagrams to $\check{\mathcal{A}}_n(S^1)$.

FIRST PART OF THE PROOF: According to STU, we have

<figure><img src="image_56.png" /></figure>

in $\check{\mathcal{A}}_n(S^1)$. So the natural map from $\mathcal{D}_n$ to $\check{\mathcal{A}}_n(S^1)$ factors through $4T$. Since STU allows us to inductively write any oriented Jacobi diagram whose connected components contain at least one univalent vertex as a combination of chord diagrams, the induced map from $\overline{\mathcal{A}}_n$ to $\check{\mathcal{A}}_n(S^1)$ is surjective. We will prove injectivity in Section 6.4 by constructing an inverse map. □

## 6.4 Multiplying diagrams

Set $\mathcal{A}(\mathcal{L}) = \prod_{n \in \mathbb{N}} \mathcal{A}_n(\mathcal{L})$, $\check{\mathcal{A}}(\mathcal{L}) = \prod_{n \in \mathbb{N}} \check{\mathcal{A}}_n(\mathcal{L})$, and $\overline{\mathcal{A}} = \prod_{n \in \mathbb{N}} \overline{\mathcal{A}}_n$.

Assume that a one-manifold $\mathcal{L}$ is decomposed as a union of two one-manifolds $\mathcal{L} = \mathcal{L}_1 \cup \mathcal{L}_2$ whose interiors in $\mathcal{L}$ do not intersect. Let $\Gamma_1$ be a Jacobi diagram with support $\mathcal{L}_1$ and let $\Gamma_2$ be a Jacobi diagram with support $\mathcal{L}_2$. Define the Jacobi diagram $\Gamma_1 \cup \Gamma_2$ on $\mathcal{L}$ to be the disjoint union of $\Gamma_1$ and $\Gamma_2$. (Formally, a $\Gamma_1 \cup \Gamma_2$-compatible injection restricts to $U(\Gamma_i)$ as a
$\Gamma_i$-compatible injection, for $i \in \{1, 2\}$.) Define the product associated to the decomposition $\mathcal{L} = \mathcal{L}_1 \cup \mathcal{L}_2$:

$$ \mathcal{A}(\mathcal{L}_1) \times \mathcal{A}(\mathcal{L}_2) \to \mathcal{A}(\mathcal{L}) $$

to be the continuous bilinear map that maps $([\Gamma_1], [\Gamma_2])$ to $[\Gamma_1 \cup \Gamma_2]$, for two diagrams $\Gamma_1$ and $\Gamma_2$ as above. In particular, the disjoint union of diagrams turns $\mathcal{A}(\emptyset)$ into a commutative algebra graded by the degree, and it turns $\mathcal{A}(\mathcal{L})$ into an $\mathcal{A}(\emptyset)$-module, for any 1-dimensional manifold $\mathcal{L}$.

An orientation-preserving diffeomorphism from a manifold $\mathcal{L}$ to another one $\mathcal{L}'$ induces a natural isomorphism from $\mathcal{A}_n(\mathcal{L})$ to $\mathcal{A}_n(\mathcal{L}')$ (and from $\tilde{\mathcal{A}}_n(\mathcal{L})$ to $\tilde{\mathcal{A}}_n(\mathcal{L}')$), for all $n$. Let $I = [0, 1]$ be the compact oriented interval. If $I = \mathcal{L}$, and if we identify $I$ with $\mathcal{L}_1 = [0, 1/2]$ and with $\mathcal{L}_2 = [1/2, 1]$ with respect to the orientation, then the above process turns $\mathcal{A}(I)$ and $\tilde{\mathcal{A}}(I)$ into algebras. The elements of $\mathcal{A}(I)$ with nonzero degree zero part admit an inverse.

**Proposition 6.22.** The algebra $\mathcal{A}([0, 1])$ is commutative. The projection from $[0, 1]$ to $S^1 = [0, 1]/(0 \sim 1)$ induces an isomorphism from $\mathcal{A}_n([0, 1])$ to $\mathcal{A}_n(S^1)$ for all $n$. So $\mathcal{A}(S^1)$ inherits a commutative algebra structure from this isomorphism. The choice of an oriented connected component $\mathcal{L}_j$ of $\mathcal{L}$ equips $\mathcal{A}(\mathcal{L})$ with an $\mathcal{A}([0, 1])-module$ structure $\#_j$, induced by the orientation-preserving inclusion from $[0, 1]$ to a small part of $\mathcal{L}_j$ outside the vertices, and the insertion of diagrams with support $[0, 1]$ there.

In order to prove this proposition, we present a useful trick in diagram spaces.

**Lemma 6.23.** Let $\mathcal{L}$ be a nonoriented one-manifold. Let $\Gamma_1$ be an oriented Jacobi diagram (resp. a chord diagram) with support $\mathcal{L}$ as in Definitions 6.13 and 6.16. Assume that $\Gamma_1 \cup \mathcal{L}$ is immersed in the plane so that $\Gamma_1 \cup \mathcal{L}$ meets an open annulus A embedded in the plane exactly along $n + 1$ embedded arcs $\alpha_1, \alpha_2, \dots, \alpha_n$, and $\beta$, and one vertex v on $\alpha_1$, as in the examples below, so that

1. the $\alpha_i$ are disjoint, they may be dashed or plain (they are dashed in the case of chord diagrams), they go from a boundary component of A to the other,

2. $\beta$ is a plain arc going from the boundary of A to v on $\alpha_1$,

3. the bounded component D of the complement of A does not contain a boundary point of $\mathcal{L}$,
4. the vertex-orientations are induced by the planar immersion by the local counterclockwise orders in the neighborhoods of vertices (as usual).

Let $\Gamma_i$ be the diagram obtained from $\Gamma_1$ by attaching the endpoint $v$ of $\beta$ to $\alpha_i$ instead of $\alpha_1$ on the same side, where the side of an arc is its side when going from the outside boundary component of $A$ to the inside one $\partial D$, as in the examples below. Then we have

$$ \sum_{i=1}^{n} \Gamma_i = 0 $$

in $\mathcal{A}(\mathcal{L})$ (resp. in the space $\overline{\mathcal{A}}$ of Section 6.2).

**Examples 6.24.** We have

<figure><img src="image_57.png" /></figure>

and

<figure><img src="image_58.png" /></figure>

**Remarks 6.25.** The second example shows that the STU relation is equivalent to the relation of the statement when the bounded component $D$ of $\mathbb{R}^2 \setminus A$ intersects $\Gamma_1$ in the neighborhood of a univalent vertex on $\mathcal{L}$. Similarly, the Jacobi relation is given by the statement’s relation when $D$ intersects $\Gamma_1$ in the neighborhood of a trivalent vertex. Also note that $AS$ corresponds to the case in which $D$ intersects $\Gamma_1$ along a dashed or plain arc.

**PROOF OF LEMMA 6.23:** Let us give the Bar-Natan [BN95a, Lemma 3.1] proof. See also [Vog11, Lemma 3.3]. Without loss of generality, assume that $v$ is always attached on the left-hand side of the $\alpha$'s.

We first treat the case of trivalent diagrams. Add to the sum the contribution of the sum of the diagrams obtained from $\Gamma_1$ by attaching $v$ to each of the three (dashed or plain) half-edges of each vertex $w$ of $\Gamma_1 \cup \mathcal{L}$ in $D$ on the right-hand side when the half-edges are oriented towards $w$ (i.e., by attaching $v$ to the hooks in $\overline{\gamma_w}$). The Jacobi and STU relations ensure that this contribution is trivial. Now, group the terms of the obtained sum
according to the edges of $\Gamma_1 \cup \mathcal{L}$, where $v$ is attached. Observe that the sum is zero, edge by edge, by AS.

For chord diagrams, similarly add to the sum the contribution of the sum of the diagrams obtained from $\Gamma_1$ by attaching $v$ to each of the four (dashed) half-edges adjacent to each chord $W$ of $\Gamma_1 \cup \mathcal{L}$ in $D$, on the right-hand side when the half-edges are oriented towards $W$ (i.e., by attaching $v$ to the hooks in $\frac{1}{2}W\frac{1}{2}$). Thanks to the 4T relation, this contribution is trivial. Again, group the terms of the obtained sum according to the dashed edges of $\Gamma_1 \cup \mathcal{L}$ where $v$ is attached. Again observe that the sum is zero, edge by edge, by AS. $\square$

END OF PROOF OF PROPOSITION 6.21:

As promised, we construct a map $f$ from $\tilde{\mathcal{A}}_n(S^1)$ to the space $\overline{\mathcal{A}}_n$ of chord diagrams up to 4T and AS, and we prove that it is an inverse of the natural surjective map $g$ from $\overline{\mathcal{A}}_n$ to $\tilde{\mathcal{A}}_n(S^1)$. Let $\mathcal{D}_{n,k}$ denote the vector space generated by the oriented uni-trivalent degree $n$ diagrams on $S^1$ that have at most $k$ trivalent vertices and at least one univalent vertex per connected component. Then $\mathcal{D}_{n,2n}$ is the vector space generated by the oriented uni-trivalent degree $n$ diagrams on $S^1$ with at least one univalent vertex per connected component.

We will define linear maps $\lambda_k$ from $\mathcal{D}_{n,k}$ to $\overline{\mathcal{A}}_n$ by induction on $k$ so that

1. $\lambda_0$ maps a chord diagram to its class in $\overline{\mathcal{A}}_n$,

2. the restriction of $\lambda_k$ to $\mathcal{D}_{n,k-1}$ is $\lambda_{k-1}$, and

3. $\lambda_k$ maps all the relations AS and STU that involve only elements of $\mathcal{D}_{n,k}$ to zero.

When we have succeeded in such a task, the linear map that maps a diagram $d$ of $\mathcal{D}_{n,2n}$ with $k$ trivalent vertices to $\lambda_k(d)$ will factor through STU and AS. Then the induced map $\bar{\lambda}$ will provide the desired inverse map and allow us to conclude the proof. Let us define our maps $\lambda_k$ with the announced properties.

Let $k \ge 1$, assume that $\lambda_{k-1}$ is defined on $\mathcal{D}_{n,k-1}$ and that $\lambda_{k-1}$ maps all the relations AS and STU that involve only elements of $\mathcal{D}_{n,k-1}$ to zero. We want to extend $\lambda_{k-1}$ on $\mathcal{D}_{n,k}$ to a linear map $\lambda_k$ that maps all the relations AS and STU that involve only elements of $\mathcal{D}_{n,k}$ to zero.

Let $d$ be a diagram with $k$ trivalent vertices, and let $e$ be an edge of $d$ that contains one univalent vertex and one trivalent vertex. Set

$$ \lambda \left( (d, e) = \begin{tikzpicture}[baseline={([yshift=-.5ex]current bounding box.center)}] \draw (-.5,0) -- (0,0) -- (0.5,0); \draw (-.5,0) -- (-.5,1); \draw (0,0) -- (0,1); \draw (0.5,0) -- (0.5,1); \node[above] at (-.5,1) {$e$}; \node[above] at (0,1) {$d$}; \node[right] at (0.5,1) {$\rightarrow$}; \node[left] at (-.5,0) {$\rightarrow$}; \end{tikzpicture} \right) = \lambda_{k-1} \left( \begin{tikzpicture}[baseline={([yshift=-.5ex]current bounding box.center)}] \draw (-.5,0) -- (0,0) -- (0.5,0); \draw (-.5,0) -- (-.5,1); \draw (0,0) -- (0,1); \draw (0.5,0) -- (0.5,1); \node[above] at (-.5,1) {$e$}; \node[above] at (0,1) {$d$}; \node[right] at (0.5,1) {$\rightarrow$}; \node[left] at (-.5,0) {$\rightarrow$}; \end{tikzpicture} - \begin{tikzpicture}[baseline={([yshift=-.5ex]current bounding box.center)}] \draw (-.5,0) -- (0,0) -- (0.5,0); \draw (-.5,0) -- (-.5,1); \draw (0,0) -- (0,1); \draw (0.5,0) -- (0.5,1); \node[above] at (-.5,1) {$e$}; \node[above] at (0,1) {$d$}; \node[right] at (0.5,1) {$\rightarrow$}; \node[left] at (-.5,0) {$\rightarrow$}; \end{tikzpicture} \right). $$
It suffices to prove that $\lambda(d, e)$ is independent of our chosen edge $e$ to conclude the proof by defining the linear map $\lambda_k$, which will obviously satisfy the desired properties, by

$$ \lambda_k(d) = \lambda(d, e). $$

Assume that two different edges $e$ and $f$ of $d$ connect a trivalent vertex to a univalent vertex. We prove that $\lambda(d, e) = \lambda(d, f)$. If $e$ and $f$ are disjoint, then the fact that $\lambda_{k-1}$ satisfies STU allows us to express both $\lambda(d, e)$ and $\lambda(d, f)$ as the same combination of four diagrams with $(k-2)$ vertices and conclude. Thus, we assume that $e$ and $f$ are two different edges that share a trivalent vertex $t$. If there exists another trivalent vertex that is connected to $S^1$ by an edge $g$, then $\lambda(d, e) = \lambda(d, g) = \lambda(d, f)$. Thus, we furthermore assume that $t$ is the unique trivalent vertex connected to $S^1$ by an edge. If $t$ is the unique trivalent vertex, then its component is necessarily

like the dotted circle and the fact that $\lambda(d, e) = \lambda(d, f)$ is a consequence of (4T).

Otherwise, the component of $t$ is of the form $e \to t \to f$ where the dotted circle represents a dashed diagram with only one pictured entry. So we have

$$ \lambda(d, e) = \lambda_{k-1} \left( \begin{array}{ccc} \text{圈} & \text{圈} & \text{圈} \\ \text{圈} & \text{圈} & \text{圈} \end{array} \right). $$

Then $\lambda(d, e)$ is zero because the expansion of $\text{圈}$ as a sum of chord diagrams commutes with any vertex in $\overline{\mathcal{A}}_n$, according to Lemma 6.23. We similarly have $\lambda(d, f) = 0$. So we get $\lambda(d, e) = \lambda(d, f)$ in this last case. $\square$

**Lemma 6.26.** For any one-manifold $\mathcal{L}$, the class of a Jacobi diagram with one univalent vertex vanishes in $\mathcal{A}_n(\mathcal{L})$.

**PROOF:** Exercise. (Use Lemma 6.23.) $\square$

**PROOF OF PROPOSITION 6.22:** Let $\Gamma'$ be a diagram with support $\mathcal{L}$ and let $\Gamma$ is a diagram with support $I$. Define $[\Gamma]_{\#_j}[\Gamma']$ to be the class in $\mathcal{A}(\mathcal{L})$ of the diagram obtained by inserting $\Gamma$ along $\mathcal{L}_j$ outside the vertices of $\Gamma$, according to the given orientation. For example, we have

$$ \left[ \begin{array}{c} \text{圈} \\ \text{点} \end{array} \right] \#_j \left[ \begin{array}{c} \text{圈} \\ \text{点} \end{array} \right] = \left[ \begin{array}{c} \text{圈} \\ \text{点} \end{array} \right] = \left[ \begin{array}{c} \text{圈} \\ \text{点} \end{array} \right]. $$
As shown in the first example that illustrates Lemma 6.23, the independence of the choice of the insertion locus is a consequence of Lemma 6.23, where $\Gamma_1$ is the disjoint union $\Gamma \sqcup \Gamma'$, and $\Gamma_1$ intersects $D$ along $\Gamma \cup I$. The continuous bilinear map

$$ \begin{array}{ccc} \mathcal{A}(I) \times \mathcal{A}(\mathcal{L}) & \to & \mathcal{A}(\mathcal{L}) \\ ([\Gamma], [\Gamma']) & \mapsto & [\Gamma] \#_j [\Gamma'] \end{array} $$

endows $\mathcal{A}(\mathcal{L})$ with the $\mathcal{A}(I)$-module structure $\#_j$.

Lemma 6.23 allows us to similarly prove that $\mathcal{A}(I)$ is a commutative algebra. Now, it suffices to prove that the morphism from $\mathcal{A}(I)$ to $\mathcal{A}(S^1)$ induced by the identification of the two endpoints of $I$ is an isomorphism. This is proved in the more general proposition below. $\square$

**Proposition 6.27.** Let $n \in \mathbb{N}$. Let $\mathcal{L}$ be a disjoint union of circles. The projection from $[0, 1]$ to $S^1 = [0, 1]/(0 \sim 1)$ induces an isomorphism from $\mathcal{A}_n([0, 1] \sqcup \mathcal{L})$ to $\mathcal{A}_n(S^1 \sqcup \mathcal{L})$.

**PROOF:** The morphism from $\mathcal{A}([0, 1] \sqcup \mathcal{L})$ to $\mathcal{A}(S^1 \sqcup \mathcal{L})$ induced by the identification of the two endpoints of $[0, 1]$ amounts to mod out $\mathcal{A}([0, 1] \sqcup \mathcal{L})$ by the relation that identifies two diagrams obtained from one another by moving the nearest univalent vertex to an endpoint of $[0, 1]$ near the other endpoint. Applying Lemma 6.23 (with $\beta$ coming from the inside boundary of the annulus) shows that this relation is a consequence of the relations in $\mathcal{A}([0, 1] \sqcup \mathcal{L})$. So this morphism is an isomorphism from $\mathcal{A}([0, 1] \sqcup \mathcal{L})$ to $\mathcal{A}(S^1 \sqcup \mathcal{L})$. $\square$

As the following exercise shows, Proposition 6.27 would not be satisfied if $\mathcal{L}$ were replaced by an interval.

**Exercise 6.28.** 1. Prove

$$ \begin{bmatrix} \begin{tikzpicture}[baseline={([yshift=-.5ex]current bounding box.center)}] \node (top) at (0, 0.5) {}; \node (bottom) at (0, -0.5) {}; \draw (top) -- (bottom); \node at (0, 0) {}; \node at (0, -0.5) {}; \draw (top) -- (0.5, 0.5) -- (0.5, -0.5) -- (bottom) -- (top); \end{tikzpicture} & \neq & \begin{bmatrix} \begin{tikzpicture}[baseline={([yshift=-.5ex]current bounding box.center)}] \node (top) at (0, 0.5) {}; \node (bottom) at (0, -0.5) {}; \draw (top) -- (bottom); \node at (0, 0) {}; \node at (0, -0.5) {}; \draw (top) -- (0.5, 0.5) -- (0.5, -0.5) -- (bottom) -- (top); \end{tikzpicture} \end{bmatrix}. $$

in $\mathcal{A}_2(\begin{tikzpicture}[baseline={([yshift=-.5ex]current bounding box.center)}] \node (top) at (0, 0.5) {}; \node (bottom) at (0, -0.5) {}; \draw (top) -- (bottom); \node at (0, 0) {}; \node at (0, -0.5) {}; \draw (top) -- (0.5, 0.5) -- (0.5, -0.5) -- (bottom) -- (top); \end{tikzpicture})$.

2. Prove that $\mathcal{A}_2(\begin{tikzpicture}[baseline={([yshift=-.5ex]current bounding box.center)}] \node (top) at (0, 0.5) {}; \node (bottom) at (0, -0.5) {}; \draw (top) -- (bottom); \node at (0, 0) {}; \node at (0, -0.5) {}; \draw (top) -- (0.5, 0.5) -- (0.5, -0.5) -- (bottom) -- (top); \end{tikzpicture}) \not\cong \mathcal{A}_2(\begin{tikzpicture}[baseline={([yshift=-.5ex]current bounding box.center)}] \node (top) at (0, 0.5) {}; \node (bottom) at (0, -0.5) {}; \draw (top) -- (bottom); \node at (0, 0) {}; \node at (0, -0.5) {}; \draw (top) -- (0.5, 0.5) -- (0.5, -0.5) -- (bottom) -- (top); \end{tikzpicture})$.

**Lemma 6.29.** Let $\pi : \mathcal{L}' \to \mathcal{L}$ be a smooth map between two unoriented compact one-manifolds $\mathcal{L}$ and $\mathcal{L}'$. Assume $\pi(\partial\mathcal{L}') \subset \partial\mathcal{L}$. Let $\Gamma$ be an oriented Jacobi diagram on $\mathcal{L}$ (as in Definitions 6.13 and 6.16) equipped with a $\Gamma$-compatible injection whose image avoids the critical values of $\pi$. Define $\pi^*(\Gamma)$ to be the class in $\mathcal{A}(\mathcal{L}')$ of the sum of all diagrams on $\mathcal{L}'$ obtained from $\Gamma$ by lifting each univalent vertex to one of its preimages under $\pi$. (These diagrams
have the same vertices and edges as $\Gamma$, and the local orientations at univalent vertices are naturally induced by the local orientations of the corresponding univalent vertices of $\Gamma$.) Then $\pi^*(\Gamma)$ depends only on the class of $\Gamma$ in $\mathcal{A}(\mathcal{L})$ and on the homotopy class of $\pi$. So one can unambiguously define the linear degree-preserving map

$$ \pi^*: \mathcal{A}(\mathcal{L}) \to \mathcal{A}(\mathcal{L}') $$

that maps the class of a diagram $\Gamma$ as above to $\pi^*(\Gamma)$. Furthermore, the map $\pi^*$ depends only on the homotopy class of $\pi$ relative to the boundary of $\mathcal{L}'$.

**PROOF:** It suffices to see that this operation is compatible with STU and that nothing bad happens when a univalent vertex of $\mathcal{L}$ moves across a critical value of $\pi$. $\square$

**Remark 6.30.** Propositions 6.22 and 6.27 and Lemma 6.29 remain valid if $\mathcal{A}$ is replaced by $\tilde{\mathcal{A}}$.

**Notation 6.31.** Let $\mathcal{L}$ be a one-manifold. Let $\mathcal{L}_0$ be a connected component of $\mathcal{L}$. Let

$$ \mathcal{L}(r \times \mathcal{L}_0) = (\mathcal{L} \setminus \mathcal{L}_0) \sqcup (\bigsqcup_{i=1}^r \mathcal{L}_0^{(i)}) $$

be the manifold obtained from $\mathcal{L}$ by duplicating $\mathcal{L}_0$ (r-1) times, that is by replacing $\mathcal{L}_0$ by r copies of $\mathcal{L}_0$. Let $\pi(r \times \mathcal{L}_0): \mathcal{L}(r \times \mathcal{L}_0) \to \mathcal{L}$ be the associated map, which is the identity on $(\mathcal{L} \setminus \mathcal{L}_0)$, and the trivial r-fold covering from $\bigsqcup_{i=1}^r \mathcal{L}_0^{(i)}$ to $\mathcal{L}_0$. The associated map is the duplication map:

$$ \pi(r \times \mathcal{L}_0)^*: \mathcal{A}(\mathcal{L}) \to \mathcal{A}(\mathcal{L}(r \times \mathcal{L}_0)). $$

**Example 6.32.** We have

$$ \pi(2 \times I)^* (\begin{smallmatrix} \ddot{\mathcal{C}} \\ \mathcal{C} \end{smallmatrix}) = \begin{smallmatrix} \ddot{\mathcal{C}} \\ \mathcal{C} \end{smallmatrix} + \begin{smallmatrix} \ddot{\mathcal{C}} \\ \mathcal{C} \end{smallmatrix} + \begin{smallmatrix} \ddot{\mathcal{C}} \\ \mathcal{C} \end{smallmatrix} + \begin{smallmatrix} \ddot{\mathcal{C}} \\ \mathcal{C} \end{smallmatrix}. $$

Note the following lemma.

**Lemma 6.33.** When $\mathcal{L}$ is a disjoint union of r intervals, an r-duplicated vertex commutes with an element of $\mathcal{A}(\mathcal{L})$. This sentence is explained by the pictures below. In the first picture

<figure><img src="image_59.png" /></figure>
there is a Jacobi diagram on $\mathcal{L}$ inside the rectangle, and the picture represents the sum of the diagrams obtained by attaching the free end of an edge (the end with the empty circle) of some other part of a Jacobi diagram to each of the hooks attached to the vertical strands. The second picture

<figure><img src="image_60.png" /></figure>

is similar, except that the edge with the free end is a part of a Jacobi diagram that is inside the box apart from this half-edge.

PROOF: This is a direct consequence of Lemma 6.23 as the pictures show.

<figure><img src="image_61.png" /></figure>

## 6.5 Coproduct on $\mathcal{A}(\mathcal{L})$

Recall that $\mathbb{K}$ is $\mathbb{Q}$ or $\mathbb{R}$ in this chapter. All the results are valid over a commutative field $\mathbb{K}$ of characteristic zero. Below, all tensor products are over the ground field $\mathbb{K}$. The canonical identification of $V \otimes \mathbb{K}$ with $V$ for a finite-dimensional vector space over $\mathbb{K}$ will always be implicit. In this section, $\mathcal{L}$ denotes a one-manifold.

For $n \in \mathbb{N}$, set

$$ (\mathcal{A}(\mathcal{L}) \otimes \mathcal{A}(\mathcal{L}))_n = \bigoplus_{i=0}^{n} \mathcal{A}_i(\mathcal{L}) \otimes \mathcal{A}_{n-i}(\mathcal{L}) $$

and

$$ \mathcal{A}(\mathcal{L}) \hat{\otimes} \mathcal{A}(\mathcal{L}) = \prod_{n \in \mathbb{N}} (\mathcal{A}(\mathcal{L}) \otimes \mathcal{A}(\mathcal{L}))_n. $$

The topological product $\mathcal{A}(\mathcal{L})$ is equipped with the following collection of linear maps

$$ \Delta_n: \mathcal{A}_n(\mathcal{L}) \to (\mathcal{A}(\mathcal{L}) \otimes \mathcal{A}(\mathcal{L}))_n. $$

The image of the class of a Jacobi diagram $\Gamma = \bigsqcup_{i \in I} \Gamma_i$ with $|I|$ nonempty connected components $\Gamma_i$, numbered arbitrarily in a set $I$, is

$$ \Delta_n ([\Gamma]) = \sum_{J \subseteq I} [\bigsqcup_{i \in J} \Gamma_i] \otimes [\bigsqcup_{i \in (I \setminus J)} \Gamma_i]. $$

Here, the connected components are not the connected components of $\Gamma \cup_{U(\Gamma)} \mathcal{L}$. They are the connected components of $\Gamma$, i.e., the connected components of the solid (i.e., nondashed) part of the figures. It is easy to check that $\Delta_n$ is
well-defined. The family $\Delta = (\Delta_n)_{n \in \mathbb{N}}$ defines a degree-preserving map from $\mathcal{A}(\mathcal{L})$ to $\mathcal{A}(\mathcal{L}) \hat{\otimes} \mathcal{A}(\mathcal{L})$.

There is a well-defined continuous linear map $\varepsilon: \mathcal{A}(\mathcal{L}) \to \mathbb{K}$ that maps the class of the empty diagram to 1 and $\mathcal{A}_i(\mathcal{L})$ to 0 for any $i > 0$. The ground field $\mathbb{K}$ is considered as a degree 0 vector space. So the map $\varepsilon$ is a degree-preserving homomorphism. Recall that **1** denotes the identity map.

In the following statement, we omit degrees. However, the following identities, which express the fact that $\Delta$ is a graded coproduct with associated counit $\varepsilon$, are collections of identities between collections of degree-preserving linear maps between finite-dimensional vector spaces. For example, the coas-sociativity identity

$$ (\Delta \otimes \mathbf{1}) \circ \Delta = (\mathbf{1} \otimes \Delta) \circ \Delta $$

means that

$$ (\Delta \otimes \mathbf{1})_n \circ \Delta_n = (\mathbf{1} \otimes \Delta)_n \circ \Delta_n $$

for any $n \in \mathbb{N}$, where both maps are valued in

$$ (\mathcal{A}(\mathcal{L}) \otimes \mathcal{A}(\mathcal{L}) \otimes \mathcal{A}(\mathcal{L}))_n = \bigoplus_{i,j,k: (i,j,k) \in \mathbb{N}^3, i+j+k=n} \mathcal{A}_i(\mathcal{L}) \otimes \mathcal{A}_j(\mathcal{L}) \otimes \mathcal{A}_k(\mathcal{L}). $$

**Lemma 6.34.** We have $(\varepsilon \otimes \mathbf{1}) \circ \Delta = (\mathbf{1} \otimes \varepsilon) \circ \Delta = \mathbf{1}$ and

$$ (\Delta \otimes \mathbf{1}) \circ \Delta = (\mathbf{1} \otimes \Delta) \circ \Delta. $$

**PROOF:** Exercise. $\square$

<figure><img src="image_62.png" /></figure>

Let

$$ \tau_n: \begin{array}{ccc} (\mathcal{A}(\mathcal{L}) \otimes \mathcal{A}(\mathcal{L}))_n & \to & (\mathcal{A}(\mathcal{L}) \otimes \mathcal{A}(\mathcal{L}))_n \\ x \otimes y & \mapsto & y \otimes x. \end{array} $$

Then we also immediately have the identity

$$ \tau \circ \Delta = \Delta, $$

which expresses the cocommutativity of $\Delta$.

## 6.6 Bialgebra structures

**Definition 6.35.** A connected, finite type, commutative, cocommutative, graded bialgebra over a field $\mathbb{K}$ is the topological product $\mathcal{H} = \prod_{n \in \mathbb{N}} \mathcal{H}_n$ of finite-dimensional vector spaces $\mathcal{H}_n$ over $\mathbb{K}$ equipped with

* a multiplication $m = (m_n: (\mathcal{H} \otimes \mathcal{H})_n \to \mathcal{H}_n)_{n \in \mathbb{N}}$, where $(\mathcal{H} \otimes \mathcal{H})_n = \bigoplus_{i=0}^n \mathcal{H}_i \otimes \mathcal{H}_{n-i}$,
• a coproduct $\Delta = (\Delta_n : \mathcal{H}_n \to (\mathcal{H} \otimes \mathcal{H})_n)_{n \in \mathbb{N}}$,

• a unit $v: K \to \mathcal{H}$, which maps $K$ to $\mathcal{H}_0$, and which is an isomorphism from $K$ to $\mathcal{H}_0$ (connectedness),

• a counit $\varepsilon: \mathcal{H} \to K$, where $K$ is again assumed to be of degree 0,

where $(m, \Delta, v, \varepsilon)$ are families of degree-preserving linear maps that satisfy

• the following identities, which express that $(m, v)$ is an associative and commutative product with unit $v(1)$:

$$m \circ (m \otimes 1) = m \circ (1 \otimes m)$$

$$m \circ (v \otimes 1) = m \circ (1 \otimes v) = 1$$

$$m \circ \tau = m,$$

where $\tau_n : (\mathcal{H} \otimes \mathcal{H})_n \to (\mathcal{H} \otimes \mathcal{H})_n$ maps $x \otimes y$ to $y \otimes x$.

• the following identities, which express that $(\Delta, \varepsilon)$ is a coassociative and cocommutative coproduct with counit $\varepsilon$:

$$(\Delta \otimes 1) \circ \Delta = (1 \otimes \Delta) \circ \Delta$$

$$(\varepsilon \otimes 1) \circ \Delta = (1 \otimes \varepsilon) \circ \Delta = 1$$

$$\tau \circ \Delta = \Delta.$$

• the following compatibility identity, which expresses the fact that $\Delta$ is an algebra morphism and that $m$ is a coalgebra morphism

$$\Delta \circ m = (m \otimes m) \circ (1 \otimes \tau \otimes 1) \circ (\Delta \otimes \Delta),$$

where the product on $\mathcal{H} \hat{\otimes} \mathcal{H} = \prod_{n \in \mathbb{N}} (\mathcal{H} \otimes \mathcal{H})_n$ is defined from $m$ so that it maps $(a \otimes b) \otimes (a' \otimes b')$ to $m(a \otimes a') \otimes m(b \otimes b')$.

**Lemma 6.36.** In a connected, finite type, commutative, cocommutative, graded bialgebra, we have $\varepsilon \circ v = 1$ and $\Delta(v(1)) = v(1) \otimes v(1)$. The element $v(1)$ is denoted by 1.

**PROOF:** Since $\Delta$ is degree-preserving, there exists $k \in K$ such that $\Delta(v(1)) = kv(1) \otimes v(1)$. Since $(\varepsilon \otimes 1) \circ \Delta = 1$, we have $k\varepsilon(v(1)) = 1$. Applying the compatibility identity to $v(1) \otimes x$ yields $\Delta(x) = k\Delta(x)$. So we get $k = 1 = \varepsilon(v(1))$. $\square$

In a connected, finite type, commutative, cocommutative, graded bialgebra, a *primitive element* is an element such that $\Delta(x) = 1 \otimes x + x \otimes 1$, and a *group-like element* is an element such that $\Delta(x) = x \otimes x$ and $\varepsilon(x) \neq 0$.

The proof of the following lemma is straightforward and left to the reader.
**Lemma 6.37.** Equipped with the product of Section 6.4, with the coproduct of Section 6.5, and with the counit that maps the class of the empty diagram to 1, the graded vector spaces $\mathcal{A}(S^1)$, $\mathcal{A}(S^1)$, and $\mathcal{A}(\emptyset)$ are connected, finite type, commutative, cocommutative, graded bialgebras.<fnref n="5" /> The unit $v(1)$ of these algebras is the class of the empty diagram. Furthermore, connected Jacobi diagrams are primitive elements in these algebras.

□

Note the elementary lemma.

**Lemma 6.38.** If $y$ is a primitive element of a connected, finite type, commutative, cocommutative, graded bialgebra, then $\exp(y)$ is group-like.

**PROOF:** It suffices to prove that $\Delta(y^n) = \sum_{k=0}^{n} \frac{n!}{k!(n-k)!} y^k \otimes y^{n-k}$. When $n=0$, this is Lemma 6.36. The compatibility identity implies $\Delta(y^n) = \Delta(y^{n-1})(y \otimes v(1) + v(1) \otimes y)$. □

□

We can now state a version of the Milnor-Moore theorem.

**Theorem 6.39.** Let $(\mathcal{H}; m, \Delta, v, \varepsilon)$ be a connected, finite type, commutative, cocommutative, graded bialgebra over a field $K$ of characteristic zero. Let $\mathcal{P}_n$ denote the set of primitive elements of $\mathcal{H}_n$. It is a finite-dimensional vector space. Pick a basis $b_n$ of each $\mathcal{P}_n$, for each $n$. For all $n \in \mathbb{N}$, the space $\mathcal{H}_n$ is the vector space freely generated by the degree $n$ monic monomials in the elements of $b_{\le n} = \cup_{k \in \mathbb{N}: k \le n} b_k$.<fnref n="6" />

**PROOF:** Since $\mathcal{P}_0 = \{0\}$ and $\mathcal{H}_0$ is freely generated by the monomial $v(1)$, the theorem holds for $n=0$. Let $n \ge 1$, let $d_n$ be the set of degree $n$ monic monomials in the elements of $b_{\le n-1}$. We want to prove that $\mathcal{H}_n$ is freely generated by $d_n \sqcup b_n$, by induction on $n$.

For $x \in \mathcal{H}_n$, set $\Delta'(x) = \Delta(x) - x \otimes v(1) - v(1) \otimes x$. According to Lemma 6.36, since $(\varepsilon \otimes \mathbf{1}) \circ \Delta = (\mathbf{1} \otimes \varepsilon) \circ \Delta = \mathbf{1}$, we have

$$ \Delta'(x) \in \bigoplus_{i=1}^{n-1} \mathcal{H}_i \otimes \mathcal{H}_{n-i} $$

and $\mathcal{P}_n$ is the kernel of $\Delta'$.

By induction, $\mathcal{H}_i$ (resp. $\mathcal{H}_{n-i}$) has a basis consisting of degree $i$ (resp. $(n-i)$) monic monomials in the elements of $b_{\le n-1}$. Thus, $\mathcal{H}_i \otimes \mathcal{H}_{n-i}$ has a

<footnote n="5">⁵These spaces can be furthermore equipped with the linear *antipode* that maps a product $\Pi$ of $p$ primitive elements to $(-1)^p \Pi$. So they are *Hopf algebras*.</footnote>

<footnote n="6">⁶By *monic monomials*, we mean monomials with coefficient one. So degree $n$ monic monomials in the elements of $b_{\le n}$ are of the form $\prod_{i \in I} p_i^{r_i(i)}$, for elements $p_i$ of $b_{\le n}$ of degree $d(i)$, and positive integers $r_i(i)$ such that $\sum_{i \in I} d(i)r_i(i) = n$.</footnote>
basis consisting of tensor products of these monomials. Multiplying two such monomials yields an element

$$ \prod_{i \in I} p_i^{r(i)} $$

of $d_n$, where the $p_i$ are distinct elements of $b_{\le n-1}$, and the $r(i)$ are positive integers. We have

$$ \Delta \left( \prod_{i \in I} p_i^{r(i)} \right) = \sum_{k: I \to \mathbb{N}: 0 \le k(i) \le r(i), \forall i} \left( \prod_{i \in I} \frac{r(i)!}{k(i)!(r(i)-k(i))!} \right) \left( \prod_{i \in I} p_i^{k(i)} \right) \otimes \left( \prod_{i \in I} p_i^{r(i)-k(i)} \right). $$

This formula proves that $\Delta'$ injects the vector space freely generated by the degree $n$ monic monomials in the elements of $b_{\le n-1}$ into $\oplus_{i=1}^{n-1} \mathcal{H}_i \otimes \mathcal{H}_{n-i}$. Thus, the degree $n$ monic monomials in the elements of $b_{\le n}$ form a free system of $\mathcal{H}_n$, and it suffices to prove that they generate $\mathcal{H}_n$. To do so, we only need to check that for every $x \in \mathcal{H}_n$, for every element $d \in d_n$, there exists a constant $a(x, d)$ such that

$$ \Delta'(x) = \sum_{d \in d_n} a(x, d) \Delta'(d). $$

Indeed, in this case, $(x - \sum_{d \in d_n} a(x, d)d)$ is primitive. Fix $x \in \mathcal{H}_n$.

Let $d = \prod_{i \in I} p_i^{r_d(i)} \in \mathcal{H}_n$, where $r_d(i) > 0$, for any $i \in I$. Let $E(d)$ be the set of maps $k: I \to \mathbb{N}$ such that $0 < \sum_{i \in I} k(i), k(i) \le r_d(i)$, and $d(k) \stackrel{\text{def}}{=} \prod_{i \in I} p_i^{k(i)} \in \mathcal{H}_j$ for some $j$ such that $j < n$. Then we have

$$ \Delta'(d) = \sum_{k \in E(d)} c_{k,r_d} d(k) \otimes d(r_d - k) $$

with

$$ c_{k,r_d} = \prod_{i \in I} \frac{r_d(i)!}{k(i)!(r_d(i)-k(i))!} \neq 0. $$

We also have

$$ \Delta'(x) = \sum_{d \in d_n} \left( \sum_{k \in E(d)} c_k(x, d) d(k) \otimes d(r_d - k) \right). $$

Now, it suffices to prove that, for any $d \in d_n$, there exists $a(x, d)$ such that $c_k(x, d) = a(x, d)c_{k,r_d}$ for any $k \in E(d)$. Fix $d \in d_n$ and set $r = r_d$.
Thanks to the coassociativity of $\Delta$, we have $(\Delta \otimes \mathbf{1}) \circ \Delta(x) = (\mathbf{1} \otimes \Delta) \circ \Delta(x)$. Therefore, if $h(i) \le k(i)$ for all $i \in I$ and $\sum_{i \in I} h(i) > 0$, then the coefficient

$$c_{h,k}c_k(x, d) = c_h(x, d)c_{k-h,r-h}$$

of $d(h) \otimes d(k-h) \otimes d(r-k)$ in $(\Delta \otimes \mathbf{1}) \circ \Delta(x)$ determines both the coefficient $c_h(x, d)$ of $d(h) \otimes d(r-h)$ and the coefficient $c_k(x, d)$ in $\Delta'(x)$. So, we have

$$c_h(x, d) = \frac{c_{h,k}}{c_{k-h,r-h}}c_k(x, d).$$

The coassociativity applied to $d$ similarly implies

$$c_h(d, d) = \frac{c_{h,k}}{c_{k-h,r-h}}c_k(d, d),$$

where $c_h(d, d) = c_{h,r}$. So, we get

$$c_h(x, d) = \frac{c_{h,r}}{c_{k,r}}c_k(x, d).$$

Choose $j \in I$. Let $\delta_j \in E(d)$ be such that $\delta_j(j) = 1$ and $\delta_j(i) = 0$ for all $i \in I \setminus \{j\}$. Set $a(x, d) = c_{\delta_j}(x, d)/c_{\delta_j,r}$. Then for all $k \in E(d)$ such that $k(j) \ne 0$, we have $c_k(x, d) = a(x, d)c_{k,r}$. If $\sum_{i \in I} r(i) > 2$, then for any $i \in I \setminus \{j\}$, the map $\delta_i + \delta_j$ is in $E(d)$. So we have $c_{\delta_i}(x, d) = a(x, d)c_{\delta_i,r}$ for any $i \in I$, and therefore $c_k(x, d) = a(x, d)c_{k,r}$ for any $k \in E(d)$. The only untreated case is $I = \{i, j\}$ with $r(i) = r(j) = 1$. In this case, the cocommutativity of $\Delta$ leads to the result. $\square$

**Corollary 6.40.** Under the assumptions of Theorem 6.39, there is a well-defined unique linear projection $p^c$ from $\mathcal{H}$ to $\mathcal{P} = \prod_{n \in \mathbb{N}} \mathcal{P}_n$ that maps the products of two homogeneous elements of positive degree to 0, and that maps $\mathcal{H}_0$ to 0.<fnref n="7" />

<figure><img src="image_63.png" /></figure>

**Theorem 6.41.** Let $\mathcal{H}$ be a connected, finite type, commutative, cocommutative, graded bialgebra. Let $\mathcal{P}$ be the space of its primitive elements, and let $p^c: \mathcal{H} \to \mathcal{P}$ be the projection of Corollary 6.40. Any group-like element $x$ of $\mathcal{H}$ is the exponential of a unique primitive element of $\mathcal{P}$. This element is $p^c(x)$.

<footnote n="7">⁷An element of $\mathcal{H}$ is homogeneous if it belongs to some $\mathcal{H}_j$ for some $j$.</footnote>
PROOF: First note that $p^c(\exp(y)) = y$ for any primitive element $y$ of $\mathcal{H}$. Therefore, if $\exp(y) = \exp(y')$ for two primitive elements $y$ and $y'$ of $\mathcal{H}$, then $y = y'$.

Set $y_n = p^c(x)_n$, and let us prove $x = \exp(y)$. Since $\varepsilon(x) \neq 0$, we have $x_0 = kv(1)$ with $k \neq 0$. Since $\Delta(x) = x \otimes x$, we get $k = 1$. So the equality $x = \exp(y)$ holds in degree 0. Assume that it holds until degree $(n-1)$. We have $\Delta_n(x_n) = \sum_{i=0}^n x_i \otimes x_{n-i}$. According to Lemma 6.38, we also have $\Delta_n(\exp(y)_n) = \sum_{i=0}^n \exp(y)_i \otimes \exp(y)_{n-i}$. Our induction hypothesis ensures

$$ \Delta_n(x_n - \exp(y)_n) = 1 \otimes (x_n - \exp(y)_n) + (x_n - \exp(y)_n) \otimes 1. $$

So $(x_n - \exp(y)_n)$ is primitive, and we get $(x_n - \exp(y)_n) = p^c(x_n - \exp(y)_n) = 0$. $\square$
# Chapter 7

## First definitions of Z

In this chapter, we introduce the invariant $\mathcal{Z}$ of links in $\mathbb{Q}$-spheres, which is the main object of this book. We illustrate the required definitions with many examples, skippable by a reader who only wants the definition.

### 7.1 Configuration spaces of Jacobi diagrams in 3-manifolds

Let $(\tilde{R}, \tau)$ be an asymptotic rational homology $\mathbb{R}^3$, as in Definition 3.8. Let $\mathcal{L}$ be a disjoint union of $k$ circles $S_i^1$, $i \in \underline{k}$. Let

$$L: \mathcal{L} \to \tilde{R}$$

denote a $C^\infty$ embedding from $\mathcal{L}$ to $\tilde{R}$. The embedding $L$ is a *link embedding*. Let $\Gamma$ be a Jacobi diagram with support $\mathcal{L}$ as in Definition 6.13. Let $U = U(\Gamma)$ denote the set of univalent vertices of $\Gamma$, and let $T = T(\Gamma)$ denote the set of trivalent vertices of $\Gamma$. A *configuration* of $\Gamma$ (with respect to $L$) is an embedding

$$c: U \cup T \hookrightarrow \tilde{R}$$

whose restriction $c|_U$ to $U$ may be written as $L \circ j$ for some $\Gamma$-compatible injection

$$j: U \hookrightarrow \mathcal{L}.$$

Denote the set of these configurations by $\tilde{C}(R, L; \Gamma)$, we have

$$\tilde{C}(R, L; \Gamma) = \{c: U \cup T \hookrightarrow \tilde{R} : (\exists j \in [i_{\Gamma}] : c|_U = L \circ j)\}.$$

In $\tilde{C}(R, L; \Gamma)$, the univalent vertices move along $L(\mathcal{L})$, while the trivalent vertices move in the ambient space $\tilde{R}$. The configuration space $\tilde{C}(R, L; \Gamma)$ is naturally an open submanifold of $\mathcal{L}^U \times \tilde{R}^T$.
An *orientation* of a set of cardinality at least 2 is a total order of its
elements up to an even permutation. When $L$ is oriented, such an orientation
of the set $V(\Gamma)$ of vertices of $\Gamma$ orients $\tilde{C}(R, L; \Gamma)$ naturally since it orders
the oriented odd-dimensional factors of $\mathcal{L}^U \times \tilde{R}^T$. Below, we associate an
orientation of $\tilde{C}(R, L; \Gamma)$ to a vertex-orientation of $\Gamma$ and an orientation of
the set $H(\Gamma)$ of half-edges of $\Gamma$.

Cut each edge of Γ into two half-edges. When an edge is oriented, define its *first* half-edge and its *second* one so that we meet the first half-edge first when following the orientation of the edge. When the edges of Γ are oriented, the orientations of the edges of Γ induce the following orientation of the set H(Γ). Order the set E(Γ) of edges of Γ arbitrarily, and order H(Γ) as (First half-edge of the first edge, second half-edge of the first edge, ..., second half-edge of the last edge). The induced orientation of H(Γ) is called the *edge-orientation* of H(Γ). Note that it does not depend on the order of E(Γ).

**Lemma 7.1.** *When Γ is equipped with a vertex-orientation, orientations of the manifold Č(R, L; Γ) are in canonical one-to-one correspondence with orientations of the set H(Γ).*

**PROOF:** Since $\tilde{C}(R, L; \Gamma)$ is naturally an open submanifold of $\mathcal{L}^U \times \tilde{R}^T$, it inherits $\mathbb{R}^{|U|+3|T|}$-valued charts from $\mathbb{R}$-valued orientation-preserving charts of $\mathcal{L}$, with respect to the (possibly local as in Definition 6.13) orientation(s) of $\mathcal{L}$, and $\mathbb{R}^3$-valued orientation-preserving charts of $\tilde{R}$. To define the orientation of $\mathbb{R}^{|U|+3|T|}$, it suffices to identify its factors and order them (up to even permutation). Each factor is labeled by an element of $H(\Gamma)$ as follows. The $\mathbb{R}$-valued local coordinate of an element of $\mathcal{L}$ corresponding to the image under $j$ of an element $u$ of $U$ sits in the factor labeled by the half-edge that contains $u$. The three $\mathbb{R}$-valued coordinates of the image under a configuration $c$ of an element $t$ of $T$, with respect to an arbitrary oriented local chart, belong to the factors labeled by the three half-edges containing $t$ so that the cyclic order of the three half-edges induced by the vertex-orientation of $\Gamma$ matches the order of the three factors. $\square$

We will use Lemma 7.1 to orient $\tilde{C}(R, L; \Gamma)$ as summarized in the follow-
ing immediate corollary.

**Corollary 7.2.** If $\Gamma$ is equipped with a vertex-orientation $o(\Gamma)$ and if the edges of $\Gamma$ are oriented, then the induced edge-orientation of $H(\Gamma)$ orients $\tilde{C}(R, L; \Gamma)$, via the canonical correspondence described in the proof of Lemma 7.1.
**Examples 7.3.** In Subsection 1.2.5, the orientations of the configuration spaces $\tilde{C}(K; \mathbb{X})$ and $\tilde{C}(K; \mathbb{X}^2)$ were induced by the order of the given co-ordinates. We can check that these orientations are also induced by the edge-orientations and by the orientation of the vertex $w$ in the following figures

<figure><img src="image_64.png" /></figure>

Recall that $\tilde{C}(K; \mathbb{X}^2)$, which may be described as

$$ \{ c: \{w, v_1, v_2, v_3\} \hookrightarrow \mathbb{R}^3 : \begin{array}{l} c(v_i) = K(z_i \in S^1), z_2 = \exp(2i\pi t_2)z_1, \\ z_3 = \exp(2i\pi t_3)z_1, 0 < t_2 < t_3 < 1 \end{array} \}, $$

was regarded as an open submanifold of $\mathbb{R}^3 \times (S^1)^3 = \{(X_1, X_2, X_3, z_1, z_2, z_3)\}$, where $(X_1, X_2, X_3) = c(w)$.

Let us check that the above coordinates orient $\tilde{C}(K; \mathbb{X}^2)$ as the orientation of edges does. For $i \in \underline{3}$, let $e(i)$ denote the edge from $v_i$ to $w$. This orders the three factors of $(S^2)^{E(\Gamma)}$. Distribute the coordinates $X_1, X_2, X_3$ so that $X_i$ is on the second half-edge of $e(i)$. Then the vertex-orientation and the edge-orientation of $\Gamma$ induce the orientation of $\tilde{C}(K; \mathbb{X}^2)$ represented by $(z_1, X_1, z_2, X_2, z_3, X_3)$, which is the same as the orientation represented by $(X_1, X_2, X_3, z_1, z_2, z_3)$. The case of $\tilde{C}(K; \mathbb{X}^2)$ is left to the reader.

**Example 7.4.** Equip the diagram $\mathcal{H}$ with its vertex-orientation induced by the picture. Orient its three edges so that they start from the same vertex. Then the orientation of $\tilde{C}(R, L; \mathcal{H})$ induced by this edge-orientation of $\mathcal{H}$ matches the orientation of $(\tilde{R} \times \tilde{R}) \setminus \Delta$ induced by the order of the two factors, where the first factor corresponds to the position of the vertex from which the three edges start, as shown in the following picture.

$$ \begin{tikzpicture}[baseline={([yshift=-.5ex]current bounding box.center)}] 
\node (1) at (0,0) {$\begin{tikzpicture}[baseline={([yshift=-.5ex]current bounding box.center)}] 
\draw (0,0) circle (1); 
\node at (0,0) {$1$};
\draw (0,1) circle (1); 
\node at (0,1) {$3$};
\draw (1,0) circle (1); 
\node at (1,0) {$4$};
\draw (1,1) circle (1); 
\node at (1,1) {$6$};
\end{tikzpicture}$};
\node at (1.5,0) {$\cong$};
\node at (3,0) {$\begin{tikzpicture}[baseline={([yshift=-.5ex]current bounding box.center)}] 
\draw (0,0) circle (1); 
\node at (0,0) {$1$};
\draw (0,1) circle (1); 
\node at (0,1) {$2$};
\draw (1,0) circle (1); 
\node at (1,0) {$4$};
\draw (1,1) circle (1); 
\node at (1,1) {$6$};
\end{tikzpicture}$};
\end{tikzpicture} $$

**Remark 7.5.** For a Jacobi diagram $\Gamma$ equipped with a vertex-orientation $o(\Gamma)$, an orientation of $V(\Gamma)$ induces the following orientation of $H(\Gamma)$. Fix a total order of $V(\Gamma)$ that induces its given orientation. Then the corresponding orientation of $H(\Gamma)$ is induced by a total order which starts with the half-edges adjacent to the first vertex, ordered with respect to $o(\Gamma)$ if the vertex is trivalent, and continues with the half-edges adjacent to the second vertex, to the third one, ... This orientation is called the *vertex-orientation* of $H(\Gamma)$ associated to $o(\Gamma)$ and to the orientation of $V(\Gamma)$. In particular, an orientation of $H(\Gamma)$ (such as the edge-orientation of $H(\Gamma)$ when the edges of $\Gamma$ are oriented) and a vertex-orientation $o(\Gamma)$ together induce an orientation of $V(\Gamma)$,
namely the orientation of $V(\Gamma)$ such that the induced vertex-orientation of $H(\Gamma)$ matches the given orientation of $H(\Gamma)$.

The dimension of $\tilde{C}(R, L; \Gamma)$ is

$$|U(\Gamma)| + 3|T(\Gamma)| = 2|E(\Gamma)|.$$

Since the degree of $\Gamma$ is $n = n(\Gamma) = \frac{1}{2}(|U(\Gamma)| + |T(\Gamma)|)$, we have

$$|E(\Gamma)| = 3n - |U(\Gamma)|.$$

## 7.2 Configuration space integrals

**Definition 7.6.** Let $A$ be a finite set. An $A$-numbered Jacobi diagram is a Jacobi diagram $\Gamma$ whose edges are oriented, equipped with an injection $j_E: E(\Gamma) \hookrightarrow A$. Such an injection numbers the edges when $A \subset \mathbb{N}$. Let $\mathcal{D}_n^e(\mathcal{L})$ denote the set of $3n$-numbered degree $n$ Jacobi diagrams with support $\mathcal{L}$ without looped edges like $\circled{0}$.

Note that the injection $j_E$ is a bijection for any diagram of $\mathcal{D}_n^e(\mathcal{L})$ without univalent vertices.

**Examples 7.7.** We have

$$\begin{align*}
\mathcal{D}_1^e(\emptyset) &= \left\{ \begin{smallmatrix} 1 \\ 2 \\ 3 \end{smallmatrix}, \begin{smallmatrix} 1 \\ 2 \\ 3 \end{smallmatrix}, \begin{smallmatrix} 1 \\ 2 \\ 3 \end{smallmatrix}, \begin{smallmatrix} 1 \\ 2 \\ 3 \end{smallmatrix} \right\}, \\
\mathcal{D}_1^e(S^1) &= \mathcal{D}_1^e(\emptyset) \sqcup \left\{ \begin{smallmatrix} 1 \\ 2 \\ 3 \end{smallmatrix}, S^1, \begin{smallmatrix} 1 \\ 2 \\ 3 \end{smallmatrix}, S^1 \right\}, \text{ and} \\
\mathcal{D}_1^e(S_1^1 \sqcup S_2^1) &= \mathcal{D}_1^e(\emptyset) \sqcup (\mathcal{D}_1^e(S_1^1) \setminus \mathcal{D}_1^e(\emptyset)) \sqcup (\mathcal{D}_1^e(S_2^1) \setminus \mathcal{D}_1^e(\emptyset)) \\
&\sqcup \left\{ \begin{smallmatrix} S_1^1 & \begin{smallmatrix} 1 \\ 2 \\ 3 \end{smallmatrix} & S_2^1 \\ S_1^1 & \begin{smallmatrix} 1 \\ 2 \\ 3 \end{smallmatrix} & S_2^1 \\ S_1^1 & \begin{smallmatrix} 1 \\ 2 \\ 3 \end{smallmatrix} & S_2^1 \end{smallmatrix}, \right. \\
&\qquad \left. S_1^1 \begin{smallmatrix} 1 \\ 2 \\ 3 \end{smallmatrix} S_2^1, S_1^1 \begin{smallmatrix} 1 \\ 2 \\ 3 \end{smallmatrix} S_2^1, S_1^1 \begin{smallmatrix} 1 \\ 2 \\ 3 \end{smallmatrix} S_2^1, \right. \\
&\qquad \left. S_1^1 \begin{smallmatrix} 1 \\ 2 \\ 3 \end{smallmatrix} S_2^1, S_1^1 \begin{smallmatrix} 1 \\ 2 \\ 3 \end{smallmatrix} S_2^1, S_1^1 \begin{smallmatrix} 1 \\ 2 \\ 3 \end{smallmatrix} S_2^1 \right\}.
\end{align*}$$

Let $\Gamma$ be a numbered degree $n$ Jacobi diagram with support $\mathcal{L}$. An edge $e$ oriented from a vertex $v_1$ to a vertex $v_2$ of $\Gamma$ induces the map

$$\begin{array}{rcl}
p_e: & \tilde{C}(R, L; \Gamma) & \to C_2(R) \\
& c & \mapsto (c(v_1), c(v_2)).
\end{array}$$

Let $o(\Gamma)$ be a vertex-orientation of $\Gamma$. For any $i \in \underline{3n}$, let $\omega(i)$ be a propagating form of $(C_2(R), \tau)$. Let $(\tilde{C}(R, L; \Gamma), o(\Gamma))$ denote the manifold $\tilde{C}(R, L; \Gamma)$, equipped with the orientation induced by $o(\Gamma)$ and by the edge-orientation of $\Gamma$, as in Corollary 7.2. Define

$$I(R, L, \Gamma, o(\Gamma), (\omega(i))_{i \in \underline{3n}}) = \int_{(\tilde{C}(R, L; \Gamma), o(\Gamma))} \bigwedge_{e \in E(\Gamma)} p_e^*(\omega(j_E(e))).$$
The convergence of this integral is a consequence of the following proposition, which will be proved in Chapter 8. (See the end of Section 8.2.)

**Proposition 7.8.** There exists a smooth compactification of $\tilde{C}(R, L; \Gamma)$, which will be denoted by $C(R, L; \Gamma)$, to which the maps $p_e$ extend smoothly.

According to this proposition, $\bigwedge_{e \in E(\Gamma)} p_e^*(\omega(j_E(e)))$ extends smoothly to $C(R, L; \Gamma)$, and we have

$$ \int_{(\tilde{C}(R,L;\Gamma),o(\Gamma))} \bigwedge_{e \in E(\Gamma)} p_e^*(\omega(j_E(e))) = \int_{(C(R,L;\Gamma),o(\Gamma))} \bigwedge_{e \in E(\Gamma)} p_e^*(\omega(j_E(e))). $$

**Examples 7.9.** For any three propagating forms $\omega(1)$, $\omega(2)$, and $\omega(3)$ of $(C_2(R), \tau)$, we have

$$ I(R, K_i \sqcup K_j : S_i^1 \sqcup S_j^1 \hookrightarrow \tilde{R}, S_i^1 \xrightarrow{\kappa} S_j^1, (\omega(i))_{i \in \bar{3}}) = lk(K_i, K_j) $$

and

$$ I(R, \emptyset, \mathcal{D}, (\omega(i))_{i \in \bar{3}}) = \Theta(R, \tau) $$

for any numbering of the (nondashed) diagrams (exercise).

**Examples 7.10.** For any oriented trivalent numbered degree $n$ Jacobi diagram $(\Gamma, o(\Gamma))$, we have

$$ I(\Gamma, o(\Gamma)) = I(S^3, \emptyset, \Gamma, o(\Gamma), (p_{S^2}^*(\omega_{S^2}))_{i \in \bar{3n}}) = 0. $$

Indeed, $I(\Gamma, o(\Gamma))$ is equal to

$$ \int_{(\tilde{C}(S^3, \emptyset; \Gamma), o(\Gamma))} \left( \prod_{e \in E(\Gamma)} p_{S^2} \circ p_e \right)^* \left( \bigwedge_{e \in E(\Gamma)} \omega_{S^2} \right), $$

where

* $\bigwedge_{e \in E(\Gamma)} \omega_{S^2}$ is a product volume form of $(S^2)^{E(\Gamma)}$ with total volume one,

* $\tilde{C}(S^3, \emptyset; \Gamma)$ is the space of injections of $\bar{3n}$ into $\mathbb{R}^3$,

* the degree of $\bigwedge_{e \in E(\Gamma)} \omega_{S^2}$ is equal to the dimension of $\tilde{C}(S^3, \emptyset; \Gamma)$, and

* the map $\prod_{e \in E(\Gamma)} p_{S^2} \circ p_e$ is never a local diffeomorphism since it is invariant under the action of global translations on $\tilde{C}(S^3, \emptyset; \Gamma)$.
**Examples 7.11.** Let $O$ denote the representative of the unknot of $S^3$, that is the image of the embedding of the unit circle $S^1$ of $\mathbb{C}$ regarded as $\mathbb{C} \times \{0\}$, into $\mathbb{R}^3$ regarded as $\mathbb{C} \times \mathbb{R}$. Let us compute

$$I(S^3, O, \Gamma, o(\Gamma), (p_{S^2}^*(\omega_{S^2}))_{i \in \underline{3n}}),$$

for the oriented Jacobi diagrams

$$\Gamma_1 = \text{循环}, \Gamma_2 = \text{对角线}, \Gamma_3 = \text{对角线}, \text{和 } \Gamma_4 = \text{对角线}.$$

Since all edges are equipped with the same standard propagating form $p_{S^2}^*(\omega_{S^2})$, we do not number the edges. For $i \in \underline{4}$, set

$$I(\Gamma_i) = I(S^3, O, \Gamma_i, o(\Gamma_i), (p_{S^2}^*(\omega_{S^2}))).$$

We are about to prove $I(\Gamma_1) = I(\Gamma_2) = I(\Gamma_3) = 0$ and $I(\Gamma_4) = \frac{1}{8}$. For $i \in \underline{4}$, set $\Gamma = \Gamma_i$. Recall that $I(\Gamma)$ is equal to

$$\int_{(\tilde{C}(S^3, O; \Gamma), o(\Gamma))} \left( \prod_{e \in E(\Gamma)} p_{S^2} \circ p_e \right)^* \left( \bigwedge_{e \in E(\Gamma)} \omega_{S^2} \right).$$

When $i \in \underline{2}$, the image of $\prod_{e \in E(\Gamma)} p_{S^2} \circ p_e$ lies in the subset of $(S^2)^2$ consisting of the pair of horizontal vectors. Since the interior of this subset is empty, the form $(\prod_{e \in E(\Gamma)} p_{S^2} \circ p_e)^* (\omega_{S^2}^{E(\Gamma)})$ vanishes identically. So we get $I(\Gamma_i) = 0$. When $i = 3$, the two edges with the same endpoints must have the same direction. So the image of $\prod_{e \in E(\Gamma)} p_{S^2} \circ p_e$ lies in the subset of $(S^2)^{E(\Gamma)}$, where two $S^2$-coordinates are identical (namely those in the $S^2$-factors corresponding to the above pair of edges), and we obtain $I(\Gamma_3) = 0$ as before.

**Lemma 7.12.** Let $\Gamma = \text{循环}$. Then we have

$$I(S^3, O, \Gamma, o(\Gamma), (p_{S^2}^*(\omega_{S^2}))) = \frac{1}{8}.$$

**PROOF:** Let $(S^1)_+^3$ be the subset of $(S^1)^3$ consisting of triples $(z_1, z_2, z_3)$ of pairwise distinct elements of $S^1$ such that the orientation of $S^1$ induces the cyclic order $(z_1, z_2, z_3)$. Then we write

$$\check{C} = \check{C}(S^3, O; \Gamma) = \{ (X, z_1, z_2, z_3) : X \in \mathbb{R}^3 \setminus \{z_1, z_2, z_3\}, (z_1, z_2, z_3) \in (S^1)_+^3 \},$$

where $X = (X_1, X_2, X_3)$. As explained in Example 7.3, these coordinates orient $\check{C}$ as does the orientation of edges. Let $\check{C}^+ = \{(X, z_1, z_2, z_3) \in \check{C} :$
$X_3 > 0\}$. The reflection $\sigma_h$ with respect to the horizontal plane acts on $\tilde{C}$ by an orientation-reversing diffeomorphism, which changes $X_3$ to $(-X_3)$ and leaves the other coordinates unchanged. It also acts on $S^2$ by an orientation-reversing diffeomorphism, which preserves the volume up to sign. Therefore, we have

$$I(\Gamma) = 2 \int_{\tilde{C}^+} \left( \prod_{e \in E(\Gamma)} p_{S^2} \circ p_e \right)^* \left( \bigwedge_{e \in E(\Gamma)} \omega_{S^2} \right).$$

Let $S_+^2$ denote the set of elements of $S^2$ with positive height (third coordinate), and let $(S^2)_+^3$ be the set of elements of $(S_+^2)^3$ which form a direct basis. Let us check that the volume of $(S^2)_+^3$ is $\frac{1}{16}$ (with respect to $\Lambda_{e \in E(\Gamma)} \omega_{S^2}$). The volume of $(S_+^2)^3$ is $\frac{1}{8}$. This is also the volume of the subset of $(S_+^2)^3$ consisting of triples of noncoplanar vectors. The involution that exchanges the last two vectors in the latter set sends the direct bases to the indirect ones, and it preserves the volume. Therefore, in order to prove $I(\Gamma_4) = \frac{1}{8}$, it suffices to prove that

$$\Psi: \quad \begin{array}{rcl} \tilde{C}^+ & \to & (S^2)^3 \\ c & \mapsto & \left( \prod_{e \in E(\Gamma)} p_{S^2} \circ p_e \right) (c) \end{array}$$

is an orientation-preserving diffeomorphism onto $(S^2)_+^3$. Let $c = (X, z_1, z_2, z_3)$ be a point of $\tilde{C}^+$. Let us first check that $\Psi(c) \in (S^2)_+^3$. For $j \in \underline{3}$, the vector $\overrightarrow{z_j X}$ may be written as $\lambda_j V_j$ for some $\lambda_j \in ]0, +\infty[$ and for some $V_j \in S_+^2$. Since $(V_1, \overrightarrow{z_1 z_2} = \lambda_1 V_1 - \lambda_2 V_2, \overrightarrow{z_1 z_3} = \lambda_1 V_1 - \lambda_3 V_3)$ is a direct basis of $\mathbb{R}^3$, so is $(V_1, V_2, V_3)$. Let us now compute the sign of the Jacobian of $\Psi$ at $c$. Let $T_c \Psi$ denote the tangent map to $\Psi$ at $c$. For $j \in \underline{3}$, let $Z_j$ denote the unit tangent vector to $S^1$ at $z_j$, and let $p_j: (\mathbb{R}^3)^3 \to \mathbb{R}^3/\mathbb{R}V_j$ be the projection onto the $j^{th}$ factor composed by the projection onto the tangent space $\mathbb{R}^3/\mathbb{R}V_j$ to $S^2$ at $V_j$. Then $p_j(T_c \Psi(Z_j)) = -Z_j$ in the tangent space to $S^2$ at $V_j$, which is generated by (the projections onto $\mathbb{R}^3/\mathbb{R}V_j$ of) $-Z_j$ and any vector $W_j$ such that $\det(V_j, -Z_j, W_j) = 1$. Reorder the oriented basis $(-Z_1, W_1, -Z_2, W_2, -Z_3, W_3)$ of $T(S^2)_+^3$ at $(V_1, V_2, V_3)$ by a positive permutation of the coordinates as $(W_1, W_2, W_3, -Z_1, -Z_2, -Z_3)$. Writing the matrix of $T_c \Psi$ with respect to this basis $(W_1, W_2, W_3, -Z_1, -Z_2, -Z_3)$ of the target space and the basis $(V_1, V_2, V_3)$ of $T_X \mathbb{R}^3$ followed by $(Z_1, Z_2, Z_3)$ for the domain $T_c \tilde{C}^+$ produces a matrix whose last three columns contain 1 on the diagonals as only nonzero entries. In the quotient

$$T(S^2)_+^3 / (\mathbb{R}(-Z_1, 0, 0) \oplus \mathbb{R}(0, -Z_2, 0) \oplus \mathbb{R}(0, 0, -Z_3)),$$

$T_c \Psi(V_1)$ may be expressed as

$$T_c \Psi(V_1) = \det(V_2, -Z_2, V_1)W_2 + \det(V_3, -Z_3, V_1)W_3,$$
where $\det(V_2, -Z_2, V_1) = \det(Z_2, V_2, V_1)$. Let us prove that $\det(Z_2, V_2, V_1) > 0$. When $z_2 = -1$, projecting $c$ on $(\mathbb{R} = \frac{\mathbb{C}}{i\mathbb{R}}) \times \mathbb{R}$ produces a picture as in Figure 7.1 that makes this result clear. The general result follows easily.

<figure><img src="image_65.png" /><figcaption>Figure 7.1: Partial projection of $(X, z_1, z_2, z_3)$ on $\frac{\mathbb{C}}{i\mathbb{R}} \times \mathbb{R}$ when $z_2 = -1$</figcaption></figure>

Finally, the Jacobian of $\Psi$ at $(X, z_1, z_2, z_3)$ is the determinant of

$$ \begin{bmatrix} 0 & \det(Z_1, V_1, V_2) & \det(Z_1, V_1, V_3) \\ \det(Z_2, V_2, V_1) & 0 & \det(Z_2, V_2, V_3) \\ \det(Z_3, V_3, V_1) & \det(Z_3, V_3, V_2) & 0 \end{bmatrix}, $$

which is

$$ \det(Z_1, V_1, V_2) \det(Z_2, V_2, V_3) \det(Z_3, V_3, V_1) + \det(Z_1, V_1, V_3) \det(Z_2, V_2, V_1) \det(Z_3, V_3, V_2). $$

It is positive since all the involved terms are. Let us finally check that every element $(V_1, V_2, V_3)$ of $(S^2)_+^3$ has a unique element in its preimage. Construct the three lines of $\mathbb{R}^3$ directed by $V_1, V_2$, and $V_3$ through the origin of $\mathbb{R}^3$. The line directed by $V_i$ intersects the horizontal plane at height $(-1)$ at a point $w_i$. There is a unique circle in this horizontal plane that contains $w_1, w_2$, and $w_3$. Let $\frac{1}{\lambda}$ be the radius of this circle, and let $w_0$ be its center. Then the unique element of $\Psi^{-1}(V_1, V_2, V_3)$ is $(-\lambda w_0, \lambda w_1 - \lambda w_0, \lambda w_2 - \lambda w_0, \lambda w_3 - \lambda w_0)$. $\square$

<figure><img src="image_66.png" /></figure>

## 7.3 Configuration space integrals associated to a chord

Let us now study the case of $I(\cdot, \cdot; s_j^1, (\omega(i))_{i \in \mathbb{S}})$. We will see that this integral depends on the chosen propagating forms and on the diagram numbering.

A *dilation* is a homothety with a positive ratio. Let $U^+K_j$ denote the fiber space over $K_j$ consisting of the tangent vectors to the knot $K_j$ of $\tilde{R}$ that orient $K_j$ up to dilation. The fiber of $U^+K_j$ consists of one point. So the total space of this *unit positive tangent bundle* to $K_j$ is canonically diffeomorphic to $K_j$. Set $U^-K_j = U^+(-K_j)$.
For a knot $K_j$ in $\tilde{R}$, we have

$$ \tilde{C}(K_j; \stackrel{\leftarrow}{\leftrightarrow} s_j^1) = \{ (K_j(z), K_j(z \exp(i\theta))) : (z, \theta) \in S^1 \times ]0, 2\pi[ \}. $$

Let $C_j = C(K_j; \stackrel{\leftarrow}{\leftrightarrow} s_j^1)$ be the closure of $\tilde{C}(K_j; \stackrel{\leftarrow}{\leftrightarrow} s_j^1)$ in $C_2(R)$. This closure is diffeomorphic to $S^1 \times [0, 2\pi]$, where $S^1 \times \{0\}$ is identified with $U^+K_j$, $S^1 \times \{2\pi\}$ is identified with $U^-K_j$, and $\partial C(K_j; \stackrel{\leftarrow}{\leftrightarrow} s_j^1) = U^+K_j - U^-K_j$.

**Lemma 7.13.** For any $i \in \underline{3}$, let $\omega(i)$ and $\omega'(i)$ be propagating forms of $(C_2(R), \tau)$, which restrict to $\partial C_2(R)$ as $p_\tau^*(\omega(i)_{S^2})$ and $p_\tau^*(\omega'(i)_{S^2})$, respectively. Then there exists a one-form $\eta(i)_{S^2}$ on $S^2$ such that $\omega'(i)_{S^2} = \omega(i)_{S^2} + d\eta(i)_{S^2}$, and we have

$$ I(\stackrel{\leftarrow}{\leftrightarrow} s_j^1, (\omega'(i))_{i \in \underline{3}}) - I(\stackrel{\leftarrow}{\leftrightarrow} s_j^1, (\omega(i))_{i \in \underline{3}}) = \int_{U^+K_j} p_\tau^*(\eta(k)_{S^2}) - \int_{U^-K_j} p_\tau^*(\eta(k)_{S^2}). $$

**Proof:** Such a form $\eta(i)_{S^2}$ exists for any $i$. According to Lemma 3.17, there exists a one-form $\eta(i)$ on $C_2(R)$ such that $\omega'(i) = \omega(i) + d\eta(i)$ and the restriction of $\eta(i)$ to $\partial C_2(R)$ is $p_\tau^*(\eta(i)_{S^2})$. Apply Stokes' theorem to $\int_{C_j} (\omega'(k) - \omega(k)) = \int_{C_j} d\eta(k)$. $\square$

**Exercise 7.14.** Find a knot $K_j$ of $\mathbb{R}^3$ and a form $\eta(k)$ of $C_2(\mathbb{R}^3)$ such that the right-hand side of Lemma 7.13 does not vanish. (Use Lemma 3.17. Hints can be found in Section 7.5.)

Recall that a propagating form $\omega$ of $(C_2(R), \tau)$ is homogeneous if its restriction to $\partial C_2(R)$ is $p_\tau^*(\omega_{S^2})$ for the homogeneous volume form $\omega_{S^2}$ of $S^2$ of total volume 1.

**Lemma 7.15.** For any $i \in \underline{3}$, let $\omega(i)$ be a homogeneous propagating form of $(C_2(R), \tau)$. Then

$$ I(\stackrel{\leftarrow}{\leftrightarrow} s_j^1, (\omega(i))_{i \in \underline{3}}) $$

does not depend on the choices of the $\omega(i)$, it is denoted by $I_\theta(K_j, \tau)$.

**Proof:** It is a corollary of Lemma 7.13. $\square$

<figure><img src="image_67.png" /></figure>

## 7.4 First definition of Z

From now on, the coefficients of our spaces of Jacobi diagrams are in $\mathbb{R}$. ($K = \mathbb{R}$.) Let $(\tilde{R}, \tau)$ be an asymptotic rational homology $\mathbb{R}^3$. Let $L: \mathcal{L} \hookrightarrow \tilde{R}$ be a link embedding.
Let $\Gamma$ be a numbered Jacobi diagram $\Gamma$ in the space $\mathcal{D}_n^e(\mathcal{L})$ of Definition 7.6. Let $[\Gamma, o(\Gamma)]$ denote the class in $\mathcal{A}_n(\mathcal{L})$ of $\Gamma$ equipped with a vertex-orientation $o(\Gamma)$. Then

$$I(R, L, \Gamma, o(\Gamma), (\omega(i))_{i \in \underline{3n}}) [\Gamma, o(\Gamma)] \in \mathcal{A}_n(\mathcal{L})$$

is independent of the vertex-orientation $o(\Gamma)$ of $\Gamma$. It is simply denoted by

$$I(R, L, \Gamma, (\omega(i))_{i \in \underline{3n}}) [\Gamma].$$

**Notation 7.16.** For $\Gamma \in \mathcal{D}_n^e(\mathcal{L})$, set

$$\zeta_{\Gamma} = \frac{(3n - |E(\Gamma)|)!}{(3n)!2^{|E(\Gamma)|}}.$$

Recall the definitions of propagating forms from Section 3.3. For any $i \in \underline{3n}$, let $\omega(i)$ be a propagating form of $C_2(R)$. For $n \in \mathbb{N}$, set

$$Z_n(\check{R}, L, (\omega(i))_{i \in \underline{3n}}) = \sum_{\Gamma \in \mathcal{D}_n^c(\mathcal{L})} \zeta_{\Gamma} I(R, L, \Gamma, (\omega(i))_{i \in \underline{3n}}) [\Gamma] \in \mathcal{A}_n(\mathcal{L}).$$

This $Z_n(\check{R}, L, (\omega(i))_{i \in \underline{3n}})$ is the hero of this book. Let us describe some of its variants.

Let $\mathcal{A}_n^c(\emptyset)$ denote the subspace of $\mathcal{A}_n(\emptyset)$ generated by connected trivalent Jacobi diagrams. Set $\mathcal{A}^c(\emptyset) = \prod_{n \in \mathbb{N}} \mathcal{A}_n^c(\emptyset)$, and let $p^c: \mathcal{A}(\emptyset) \to \mathcal{A}^c(\emptyset)$ be the projection that maps the empty diagram and diagrams with several connected components to 0. Let $\mathcal{D}_n^c$ denote the subset of $\mathcal{D}_n^e(\emptyset)$ that contains the connected diagrams of $\mathcal{D}_n^e(\emptyset)$. For $n \in \mathbb{N}$, set

$$z_n(\check{R}, (\omega(i))_{i \in \underline{3n}}) = p^c\left(Z_n(\check{R}, \emptyset, (\omega(i)))\right).$$

$$z_n(\check{R}, (\omega(i))) = \sum_{\Gamma \in \mathcal{D}_n^c} \zeta_{\Gamma} I(R, \emptyset, \Gamma, (\omega(i))_{i \in \underline{3n}}) [\Gamma] \in \mathcal{A}_n^c(\emptyset).$$

When all the forms $\omega(i)$ are equal to $\omega(1)$, $Z_n(\check{R}, L, \omega(1))$ and $z_n(\check{R}, \omega(1))$ respectively denote $Z_n(\check{R}, L, (\omega(i))_{i \in \underline{3n}})$ and $z_n(\check{R}, (\omega(i))_{i \in \underline{3n}})$.

We also use the projection $\check{p}: \mathcal{A}(\mathcal{L}) \to \check{\mathcal{A}}(\mathcal{L})$ that maps the diagrams with connected components without univalent vertices to zero and that maps the other diagrams to themselves. Set $\check{Z}_n = \check{p} \circ Z_n$. For example, we have

$$\check{Z}_n(\check{R}, L, \omega(1)) = \check{p}(Z_n(\check{R}, L, \omega(1))).$$
We also remove the subscript $n$ to denote the collection (or the sum) of the $Z_n$ for $n \in \mathbb{N}$. For example, we write

$$ \check{Z}(\check{R}, L, \omega(1)) = (\check{Z}_n(\check{R}, L, \omega(1)))_{n \in \mathbb{N}} = \sum_{n \in \mathbb{N}} \check{Z}_n(\check{R}, L, \omega(1)) \in \check{\mathcal{A}}(\mathcal{L}). $$

As a first example, let us prove the following proposition.

**Proposition 7.17.** Let $(\check{R}, \tau)$ be an asymptotic rational homology $\mathbb{R}^3$, then for any triple $(\omega(i))_{i \in \mathfrak{I}}$ of propagating forms of $(C_2(R), \tau)$, we have

$$ Z_1(\check{R}, \emptyset, (\omega(i))_{i \in \mathfrak{I}}) = z_1(\check{R}, (\omega(i))) = \frac{\Theta(R, \tau)}{12} [\Theta] $$

$$ \text{in } \mathcal{A}_1(\emptyset) = \mathcal{A}_1(\emptyset; \mathbb{R}) = \mathbb{R} [\Theta]. $$

**PROOF:** The diagram $\Theta$ is the only trivalent diagram with 2 vertices without looped edges, and it is easy to check that $\mathcal{A}_1(\emptyset) = \mathbb{R} [\Theta]$. Each of the three edges goes from one vertex to the other. There are 4 elements in $\mathcal{D}_1^e(\emptyset)$, depending on whether the orientations of Edge 2 and Edge 3 coincide with the orientation of Edge 1. See Example 7.7. When the three edges start from the same vertex, call the corresponding element $\theta_{++}$, and order the vertices so that the vertex from which the edges start is first. Recall from Example 7.4 that the vertex-orientation $o(\theta_{++})$ of $\Theta$ induced by the picture $\Theta$ and the edge-orientation of $H(\theta_{++})$ orient $C_2(R)$ as $\check{R}^2 \setminus \Delta(\check{R}^2)$, as in Corollary 7.2. Then, according to Theorem 4.1, for any triple $(\omega(i))_{i \in \mathfrak{I}}$ of propagating forms of $(C_2(R), \tau)$, we have

$$ I(\theta_{++}, o(\theta_{++}), (\omega(i))_{i \in \mathfrak{I}}) = \Theta(R, \tau). $$

Reversing the edge-orientation of Edge 2 transforms $\theta_{++}$ to $\theta_{-+}$ and $\omega(2)$ to $\iota^*(\omega(2))$. It changes the edge-orientation of $H(\theta_{\pm+})$. According to Lemma 3.16, the form $(-\iota^*(\omega(2)))$ is a propagating form of $(C_2(R), \tau)$. Therefore, we have

$$ I(\theta_{++}, (\omega(i))_{i \in \mathfrak{I}}) [\theta_{++}] = I(\theta_{-+}, (\omega(i))_{i \in \mathfrak{I}}) [\theta_{-+}]. $$

Similarly, the four graphs of $\mathcal{D}_1^e(\emptyset)$ contribute in the same way to

$$ Z_1(\check{R}, \emptyset, (\omega(i))_{i \in \mathfrak{I}}) = \frac{4}{3!2^3} \Theta(R, \tau) [\Theta]. $$

□
**Examples 7.18.** According to the computations in Example 7.10, we have

$$Z_n(\mathbb{R}^3, \emptyset, p_{S^2}^*(\omega_{S^2})) = z_n(\mathbb{R}^3, p_{S^2}^*(\omega_{S^2})) = 0$$

for $n > 0$, and $Z_0(\mathbb{R}^3, \emptyset, p_{S^2}^*(\omega_{S^2})) = [\emptyset]$, while $z_0(\mathbb{R}^3, p_{S^2}^*(\omega_{S^2})) = 0$.

For the embedding $O$ of the trivial knot in $\mathbb{R}^3$ of Example 7.11, we have

$$Z_0(\mathbb{R}^3, O, p_{S^2}^*(\omega_{S^2})) = 1 = [\emptyset].$$

Since $I_\theta(O, \tau_s) = 0$, we have $Z_1(\mathbb{R}^3, O, p_{S^2}^*(\omega_{S^2})) = 0$.

Let us now prove

$$Z_2(\mathbb{R}^3, O, p_{S^2}^*(\omega_{S^2})) = \frac{1}{24} [\begin{tikzpicture}[baseline={([yshift=-.5ex]current bounding box.center)}] \draw[->] (0,0) -- (1,0); \draw[->] (0,0) -- (0,1); \draw[->] (0,0) -- (-1,0); \draw[->] (0,0) -- (-1,1); \draw[->] (0,0) -- (0,2); \draw[->] (0,0) -- (1,2); \draw[->] (0,0) -- (0,3); \draw[->] (0,0) -- (-1,3); \draw[->] (0,0) -- (-1,3); \draw[->] (0,0) -- (0,4); \draw[->] (0,0) -- (1,4); \draw[->] (0,0) -- (0,5); \draw[->] (0,0) -- (1,5); \end{tikzpicture}].$$

Note $\iota_{S^2}^*(\omega_{S^2}) = -\omega_{S^2}$, where $\iota_{S^2}$ is the antipodal map of $S^2$. So, reversing the orientation of an edge does not change $I(S^3, O, \Gamma, p_{S^2}^*(\omega_{S^2})) [\Gamma]$, for a degree 2 numbered Jacobi diagram $\Gamma$, since it changes both the orientation of $\tilde{C}(S^3, 0; \Gamma)$ and the sign of the form to be integrated. Thus, $I(S^3, O, \Gamma, p_{S^2}^*(\omega_{S^2})) [\Gamma]$ depends only on the underlying Jacobi diagram. The degree 2 Jacobi diagrams all components of which have univalent vertices and without looped edges are

<figure><img src="image_68.png" /></figure>

As proved in Example 7.11, the diagrams $\begin{tikzpicture}[baseline={([yshift=-.5ex]current bounding box.center)}] \draw[->] (0,0) -- (1,0); \draw[->] (0,0) -- (0,1); \draw[->] (0,0) -- (-1,0); \draw[->] (0,0) -- (-1,1); \draw[->] (0,0) -- (0,2); \draw[->] (0,0) -- (1,2); \draw[->] (0,0) -- (0,3); \draw[->] (0,0) -- (1,3); \draw[->] (0,0) -- (0,4); \draw[->] (0,0) -- (1,4); \draw[->] (0,0) -- (0,5); \draw[->] (0,0) -- (1,5); \end{tikzpicture}$, $\begin{tikzpicture}[baseline={([yshift=-.5ex]current bounding box.center)}] \draw[->] (0,0) -- (1,0); \draw[->] (0,0) -- (0,1); \draw[->] (0,0) -- (-1,0); \draw[->] (0,0) -- (-1,1); \draw[->] (0,0) -- (0,2); \draw[->] (0,0) -- (1,2); \draw[->] (0,0) -- (0,3); \draw[->] (0,0) -- (1,3); \draw[->] (0,0) -- (0,4); \draw[->] (0,0) -- (1,4); \draw[->] (0,0) -- (0,5); \draw[->] (0,0) -- (1,5); \end{tikzpicture}$ do not contribute to $Z_2(\mathbb{R}^3, O, p_{S^2}^*(\omega_{S^2}))$.

Since Lemma 6.26 implies $[\begin{tikzpicture}[baseline={([yshift=-.5ex]current bounding box.center)}] \draw[->] (0,0) -- (1,0); \draw[->] (0,0) -- (0,1); \draw[->] (0,0) -- (-1,0); \draw[->] (0,0) -- (-1,1); \draw[->] (0,0) -- (0,2); \draw[->] (0,0) -- (1,2); \draw[->] (0,0) -- (0,3); \draw[->] (0,0) -- (1,3); \draw[->] (0,0) -- (0,4); \draw[->] (0,0) -- (1,4); \draw[->] (0,0) -- (0,5); \draw[->] (0,0) -- (1,5); \end{tikzpicture}] = 0$, the diagram $\begin{tikzpicture}[baseline={([yshift=-.5ex]current bounding box.center)}] \draw[->] (0,0) -- (1,0); \draw[->] (0,0) -- (0,1); \draw[->] (0,0) -- (-1,0); \draw[->] (0,0) -- (-1,1); \draw[->] (0,0) -- (0,2); \draw[->] (0,0) -- (1,2); \draw[->] (0,0) -- (0,3); \draw[->] (0,0) -- (1,3); \draw[->] (0,0) -- (0,4); \draw[->] (0,0) -- (1,4); \draw[->] (0,0) -- (0,5); \draw[->] (0,0) -- (1,5); \end{tikzpicture}$ does not contribute either. Lemma 7.12 implies

$$I(S^3, O, \begin{tikzpicture}[baseline={([yshift=-.5ex]current bounding box.center)}] \draw[->] (0,0) -- (1,0); \draw[->] (0,0) -- (0,1); \draw[->] (0,0) -- (-1,0); \draw[->] (0,0) -- (-1,1); \draw[->] (0,0) -- (0,2); \draw[->] (0,0) -- (1,2); \draw[->] (0,0) -- (0,3); \draw[->] (0,0) -- (1,3); \draw[->] (0,0) -- (0,4); \draw[->] (0,0) -- (1,4); \draw[->] (0,0) -- (0,5); \draw[->] (0,0) -- (1,5); \end{tikzpicture}, (p_{S^2}^*(\omega_{S^2}))) [\begin{tikzpicture}[baseline={([yshift=-.5ex]current bounding box.center)}] \draw[->] (0,0) -- (1,0); \draw[->] (0,0) -- (0,1); \draw[->] (0,0) -- (-1,0); \draw[->] (0,0) -- (-1,1); \draw[->] (0,0) -- (0,2); \draw[->] (0,0) -- (1,2); \draw[->] (0,0) -- (0,3); \draw[->] (0,0) -- (1,3); \draw[->] (0,0) -- (0,4); \draw[->] (0,0) -- (1,4); \draw[->] (0,0) -- (0,5); \draw[->] (0,0) -- (1,5); \end{tikzpicture}] = \frac{1}{8} [\begin{tikzpicture}[baseline={([yshift=-.5ex]current bounding box.center)}] \draw[->] (0,0) -- (1,0); \draw[->] (0,0) -- (0,1); \draw[->] (0,0) -- (-1,0); \draw[->] (0,0) -- (-1,1); \draw[->] (0,0) -- (0,2); \draw[->] (0,0) -- (1,2); \draw[->] (0,0) -- (0,3); \draw[->] (0,0) -- (1,3); \draw[->] (0,0) -- (0,4); \draw[->] (0,0) -- (1,4); \draw[->] (0,0) -- (0,5); \draw[->] (0,0) -- (1,5); \end{tikzpicture}].$$

When $\Gamma = \begin{tikzpicture}[baseline={([yshift=-.5ex]current bounding box.center)}] \draw[->] (0,0) -- (1,0); \draw[->] (0,0) -- (0,1); \draw[->] (0,0) -- (-1,0); \draw[->] (0,0) -- (-1,1); \draw[->] (0,0) -- (0,2); \draw[->] (0,0) -- (1,2); \draw[->] (0,0) -- (0,3); \draw[->] (0,0) -- (1,3); \draw[->] (0,0) -- (0,4); \draw[->] (0,0) -- (1,4); \draw[->] (0,0) -- (0,5); \draw[->] (0,0) -- (1,5); \end{tikzpicture}$, we have $\zeta_\Gamma = \frac{3!}{6!2^3}$, and there are $\frac{1}{3} \frac{6!2^3}{3!}$ numbered graphs of $\mathcal{D}_2^e(S^1)$ that are isomorphic to $\Gamma$ as a Jacobi diagram. This concludes the computation of $Z_2(\mathbb{R}^3, O, p_{S^2}^*(\omega_{S^2}))$.

See also Proposition 7.25 and Example 7.28. Enore Guadagnini, Maurizio Martellini, and Mihail Mintchev have performed alternative computations of similar quantities in [GMM90].

The following theorem is proved in Chapter 9. See Section 9.1, and Corollary 9.4 and Lemma 9.1, in particular.
**Theorem 7.19.** Let $(\tilde{R}, \tau)$ be an asymptotic rational homology $\mathbb{R}^3$. Let $n \in \mathbb{N}$. For any $i \in \underline{6n}$, let $\omega(i)$ be a propagating form of $C_2(R)$. Then $z_{2n}(\tilde{R}, (\omega(i))_{i \in \underline{6n}})$ is independent of the chosen $\omega(i)$, it depends only on the diffeomorphism class of $R$. It is denoted by $z_{2n}(R)$.

For odd $n$, $z_n(\tilde{R}, (\omega(i))_{i \in \underline{3n}})$ depends on the chosen $\omega(i)$. Theorem 7.20 explains how to deal with this dependence when the $\omega(i)$ are homogeneous propagating forms of $(C_2(R), \tau)$. Tadayuki Watanabe [Wat18a] and Tatsuro Shimizu [Shi16] have studied alternative compensations for this dependence.

We are going to prove the following theorem in the next chapters. The proof will be concluded in the end of Section 10.5.

**Theorem 7.20.** Let $(\tilde{R}, \tau)$ be an asymptotic rational homology $\mathbb{R}^3$. Let $\mathcal{L} = \bigsqcup_{j=1}^k S_j^1$ be a disjoint union of $k$ circles. Let $L: \mathcal{L} \hookrightarrow \tilde{R}$ be an embedding. Let $n \in \mathbb{N}$. For any $i \in \underline{3n}$, let $\omega(i)$ be a homogeneous propagating form of $(C_2(R), \tau)$.

Then $Z_n(\tilde{R}, L, (\omega(i))_{i \in \underline{3n}})$ is independent of the chosen $\omega(i)$. It depends only on the diffeomorphism class of $(R, L)$, on $p_1(\tau)$, and on the $I_\theta(K_j, \tau)$, for the components $K_j$ of $L$. It is denoted by $Z_n(\tilde{R}, L, \tau)$. More precisely, set

$$Z(\tilde{R}, L, \tau) = (Z_n(\tilde{R}, L, \tau))_{n \in \mathbb{N}} \in \mathcal{A}(\bigsqcup_{j=1}^k S_j^1).$$

There exist two constants $\alpha \in \check{\mathcal{A}}(S^1; \mathbb{R})$ and $\beta \in \mathcal{A}(\emptyset; \mathbb{R})$, called anomalies,<fnref n="1" /> such that $\alpha_{2n} = 0$ and $\beta_{2n} = 0$ for any $n \in \mathbb{N}$, and

$$\exp\left(-\frac{1}{4}p_1(\tau)\beta\right) \prod_{j=1}^{k} \left(\exp(-I_{\theta}(K_j, \tau)\alpha)\#_j\right) Z(\tilde{R}, L, \tau) = \mathcal{Z}(R, L)$$

depends only on the diffeomorphism class of $(R, L)$. Here $\exp(-I_\theta(K_j, \tau)\alpha)$ acts on $Z(\tilde{R}, L, \tau)$, on the copy $S_j^1$ of $S^1$ as indicated by the subscript $j$, as in Proposition 6.22.

Recall Notation 7.16. If $\tilde{R} = \mathbb{R}^3$, then the projection $\check{\mathcal{Z}}(S^3, L)$ of $Z(S^3, L)$ on $\check{\mathcal{A}}(\bigsqcup_{j=1}^k S_j^1)$ is a universal finite type invariant of links in $\mathbb{R}^3$, i.e., the projection of $\check{\mathcal{Z}}_n$ onto $\bar{\mathcal{A}}_n/(1T)$ satisfies the properties stated for $\bar{\mathcal{Z}}_n$ in Theorem 6.9. This result, due to Daniel Altschüler and Laurent Freidel [AF97], is proved in Section 17.6. See Theorem 17.30.

**Notation 7.21.** Let $\check{\mathcal{Z}}(R, L)$ denote the projection of $\mathcal{Z}(R, L)$ on $\check{\mathcal{A}}(\bigsqcup_{j=1}^k S_j^1)$, and set

$$\mathcal{Z}(R) = \mathcal{Z}(R, \emptyset).$$

<footnote n="1">¹The anomaly $\alpha$ is defined in Section 10.3, and $\beta$ is defined in Definition 10.5.</footnote>
The invariant $\mathcal{Z}$ of rational homology 3-spheres is the Kontsevich configuration space invariant studied by Greg Kuperberg and Dylan Thurston in [KT99], and described in [Les04a] as $Z_{KKT}$.

We will see in Chapter 11 that the anomalies $\alpha$ and $\beta$ are rational, i.e., that $\alpha \in \check{\mathcal{A}}(S^1; \mathbb{Q})$ (in Proposition 11.1) and $\beta \in \mathcal{A}(\emptyset; \mathbb{Q})$ (in Theorem 11.8).

**Examples 7.22.** According to Example 7.18, we have $\mathcal{Z}(\mathbb{R}^3, \emptyset, \tau_s) = 1 = [\emptyset]$. Since $p_1(\tau_s) = 0$, we also have $\mathcal{Z}(S^3) = 1$.

For the embedding $O$ of the trivial knot in $\mathbb{R}^3$ of Example 7.11, we have

$$I_{\theta}(O, \tau_s) = 0.$$

So, according to Example 7.18, we also have $\mathcal{Z}_0(S^3, O) = 1$, $\mathcal{Z}_1(S^3, O) = 0$, and

$$\mathcal{Z}_2(S^3, O) = \frac{1}{24} [\text{結点}].$$

**Definition 7.23.** Let $\Gamma$ be a Jacobi diagram on an oriented one-manifold $\mathcal{L}$ as in Definition 6.13. Let $j: U(\Gamma) \hookrightarrow \mathcal{L}$ be a $\Gamma$-compatible injection from its set $U(\Gamma)$ of univalent vertices to its support $\mathcal{L}$. An automorphism of $\Gamma$ is a permutation of the set $H(\Gamma)$ of half-edges of $\Gamma$ that maps a pair of half-edges of an edge to another such and a triple of half-edges that contain a vertex to another such, and such that, for the induced bijection $b$ of $U(\Gamma)$, $j \circ b$ is isotopic to $j$. (So the automorphisms preserve the components of the univalent vertices. They also preserve their linear order on intervals and their cyclic order on circles.) Let Aut($\Gamma$) denote the set of automorphisms of $\Gamma$.

**Examples 7.24.** There are six automorphisms of $\Theta$ that fix each vertex. They correspond to the permutations of the edges. The cardinality $|\text{Aut}(\Theta)|$ of Aut($\Theta$) is 12, and we have

$$|\text{Aut}(\hat{\zeta})| = 1 \text{ and } |\text{Aut}(\text{結点})| = 3$$

Recall from Notation 7.16 that $\zeta_{\Gamma} = \frac{(3n-|E(\Gamma)|)!}{(3n)!2^{|E(\Gamma)|}}$. Also recall that Lemma 3.16 ensures the existence of antisymmetric homogeneous propagating forms $\omega$ of $(C_2(R), \tau)$.

**Proposition 7.25.** Let $\mathcal{D}_n^u(\mathcal{L})$ denote the set of unnumbered, unoriented degree $n$ Jacobi diagrams on $\mathcal{L}$ without looped edges. Under the assumptions of Theorem 7.20, let $\omega$ be an antisymmetric homogeneous propagating form of $C_2(R)$. Then we have

$$Z_n(\check{R}, L, \omega) = \sum_{\Gamma \in \mathcal{D}_n^u(\mathcal{L})} \frac{1}{|\text{Aut}(\Gamma)|} I(R, L, \Gamma, (\omega)_{i \in \underline{3n}}) [\Gamma].$$
PROOF: Set $\omega(i) = \omega$ for any $i$. For a numbered graph $\Gamma$ (i.e., a graph equipped with the structure described in Definition 7.6), there are $\frac{1}{\zeta_{\Gamma}}$ ways of renumbering it (i.e., changing this structure), and $|\text{Aut}(\Gamma)|$ of them will produce the same numbered graph. Therefore, we have

$$ \sum_{\Gamma \in \mathcal{D}_n^c(\mathcal{L})} \zeta_{\Gamma} I(R, L, \Gamma, (\omega(i))_{i \in \underline{3n}}) [\Gamma] = \sum_{\Gamma \in \mathcal{D}_n^u(\mathcal{L})} \frac{1}{|\text{Aut}(\Gamma)|} I(R, L, \Gamma, (\omega)_{i \in \underline{3n}}) [\Gamma]. \quad \square $$

<figure><img src="image_69.png" /></figure>

**Remark 7.26.** Let $\omega$ be an antisymmetric homogeneous propagating form of $(C_2(R), \tau)$. The homogeneous definition of $Z_n(\tilde{R}, L, \tau) = Z_n(\tilde{R}, L, \omega)$ above makes clear that $Z_n(\tilde{R}, L, \tau)$ is a “measure” of graph configurations, where a graph configuration is an embedding of the set of vertices of a univalent graph into $\tilde{R}$, which maps univalent vertices to $L(\mathcal{L})$. The embedded vertices are connected by a set of abstract solid edges, which represent the measuring form. The factor $\frac{1}{|\text{Aut}(\Gamma)|}$ ensures that every such configuration of an unnumbered, unoriented graph is “measured” once.

**Lemma 7.27.** *Recall Notation 7.21. Under the assumptions of Theorem 7.20, we have*

$$ Z(\tilde{R}, L, \omega(1)) = Z(\tilde{R}, \emptyset, \omega(1)) \tilde{Z}(\tilde{R}, L, \omega(1)). $$

In particular, Theorem 7.20 will imply

$$ \mathcal{Z}(R, L) = \mathcal{Z}(R) \tilde{\mathcal{Z}}(R, L). $$

<figure><img src="image_70.png" /></figure>

PROOF: Remark 7.26 shows that the coefficients are correct.

**Example 7.28.** For the embedding $O$ of the trivial knot in $\mathbb{R}^3$ of Examples 7.11, according to the computations performed in the series of examples 7.18, and with the notation of Theorem 7.20 and Notation 7.16, as in Example 7.22, we have

$$ \begin{aligned} \mathcal{Z}_2(S^3, O) &= Z_2(\mathbb{R}^3, O, p_{S^2}(\omega_{S^2})) \\ &= \frac{1}{3} I(S^3, O, \stackrel{\leftarrow}{\bigcirc}_{\lambda}, p_{S^2}(\omega_{S^2})) [\stackrel{\rightarrow}{\bigcirc}_{\lambda}] \\ &= \frac{1}{24} [\stackrel{\rightarrow}{\bigcirc}_{\lambda}]. \end{aligned} $$

We end this section by stating Theorems 7.30 and 7.32 about the numerical invariants obtained from $\tilde{Z}_n$ by applying the Conway weight system $w_C$ of Example 6.11. For $n = 2$, we get the invariant $w_2$ discussed in Subsection 1.2.5.
Since $w_C$ is multiplicative, and since $w_C$ sends elements of odd degree to zero, $w_C$ sends $\exp(-I_\theta(K_j, \tau)\alpha)$ to the unit of $\check{\mathcal{A}}(S^1)$. So, with the notation of Theorem 7.20, we have $w_C(\check{Z}_n(R, L)) = w_C(\check{Z}_n(\check{R}, L, \tau))$ for any $n \in \mathbb{N}$, and we can forget the anomalies for $w_C \circ \check{Z}$. Theorem 7.30 will tell us that we can furthermore omit the homogeneity assumptions on the forms for $w_C \circ \check{Z}$ and reduce our averaging process. We will average only over some degree $n$ graphs whose edges are numbered in $3n-2$, or even in $3$ when $n=2$, in some cases.

(Since $\check{p}: \mathcal{A}(\mathcal{L}) \to \check{\mathcal{A}}(\mathcal{L})$ maps all graphs with less than 2 univalent vertices to zero (thanks to Lemma 6.26), the graphs that contribute to $\check{Z}_n(\check{R}, L, \omega(1))$ have at most $3n-2$ edges. So it is natural to average only over these graphs.)

**Notation 7.29.** For a finite set $A$, let $\mathcal{D}_{n,A}^e(\mathcal{L})$ denote the set of $A$-numbered degree $n$ Jacobi diagrams with support $\mathcal{L}$ without looped edges, as in Definition 7.6. (These diagrams have at most $|A|$ edges.) The coefficient $\zeta_\Gamma$ associated to a diagram $\Gamma \in \mathcal{D}_{n,A}^e(\mathcal{L})$ is

$$ \zeta_\Gamma = \frac{(|A| - |E(\Gamma)|)!}{|A|!2^{|E(\Gamma)|}}. $$

For any $i \in A$, let $\omega(i)$ be a propagating form of $C_2(R)$. Set

$$ Z_{n,A}(\check{R}, L, (\omega(i))_{i \in A}) = \sum_{\Gamma \in \mathcal{D}_{n,A}^e(\mathcal{L})} \zeta_\Gamma I(R, L, \Gamma, (\omega(i))_{i \in A}) [\Gamma] \in \mathcal{A}_n(\mathcal{L}). $$

For $m \in \mathbb{N}$, set $\mathcal{D}_{n,m}^e(\mathcal{L}) = \mathcal{D}_{n,m}^e(\mathcal{L})$ and

$$ Z_{n,m}(\check{R}, L, (\omega(i))_{i \in \bar{m}}) = Z_{n,m}(\check{R}, L, (\omega(i))_{i \in \bar{m}}). $$

Note that $\mathcal{D}_n^e = \mathcal{D}_{n,3n}^e$ and that $Z_{n,3n} = Z_n$. With the projection $\check{p}: \mathcal{A}_n(S^1) \to \check{\mathcal{A}}_n(S^1)$ of Notation 7.16, set

$$ \check{Z}_{n,m}(\check{R}, K, (\omega(i))_{i \in \bar{m}}) = \check{p}\left(Z_{n,m}(\check{R}, K, (\omega(i))_{i \in \bar{m}})\right). $$

**Theorem 7.30.** Let $(\check{R}, \tau)$ be an asymptotic rational homology $\mathbb{R}^3$. Let $K: S^1 \hookrightarrow \check{R}$ be an embedding. Let $n \in \mathbb{N}$. For any $i \in 3n-2$, let $\omega(i)$ be a propagating form of $C_2(R)$.

Let $p^c$ be the projection given by Corollary 6.40 from $\check{\mathcal{A}}(S^1)$ to the space $\check{\mathcal{A}}^c(S^1)$ of its primitive elements. Recall the linear form $w_C: (\check{\mathcal{A}}_n(S^1) = \bar{\mathcal{A}}_n) \to \mathbb{R}$ induced by the Conway weight system of Example 6.11.
$$ \text{Then } w_C(\check{Z}_{n,3n-2}(\check{R}, K, (\omega(i))_{i \in \underline{3n-2}})) \text{ and} \\ w_C \circ p^c(\check{Z}_{n,3n-2}(\check{R}, K, (\omega(i))_{i \in \underline{3n-2}})) $$

are independent of the chosen $\omega(i)$. They depend only on the diffeomorphism class of $(\check{R}, K)$. They are respectively denoted by $w_C\check{Z}_n(R, K)$ and

$$ w_C p^c \check{Z}_n(R, K). $$

At the end of Subsection 9.1, the proof of Theorem 7.30 will be reduced to the proof of Proposition 9.7, which is proved in Subsection 9.3.

**Remark 7.31.** Assuming both Theorems 7.30 and 7.20, there is no notation conflict between them. Indeed, with the notation of Theorem 7.20, we have

$$ w_C(\check{Z}_n(R, L)) = w_C(\check{Z}_n(\check{R}, L, \tau)) = w_C(\check{Z}_n(\check{R}, L, \omega)) \\ = \sum_{\Gamma \in \mathcal{D}_n^u(\mathcal{L})} \frac{1}{|\text{Aut}(\Gamma)|} I(R, L, \Gamma, (\omega)_{i \in \underline{3n}}) w_C([\Gamma]) $$

for any homogeneous propagating form $\omega$ of $(C_2(R), \tau)$ and for any $n \in \mathbb{N}$, thanks to Proposition 7.25. When $\omega(i) = \omega$ for all $i$, this is also the expression of

$$ w_C(\check{Z}_{n,3n-2}(\check{R}, K, (\omega(i))_{i \in \underline{3n-2}})). $$

We prove the following theorem at the end of Section 9.3. It implies Proposition 1.7.

**Theorem 7.32.** Let $K: S^1 \hookrightarrow \mathbb{R}^3$ be a knot embedding. For $i \in \underline{3}$, let $\omega_{S^2}(i)$ be a volume-one form of $S^2$, then $w_C(\check{Z}_{2,3}(\mathbb{R}^3, K, (p_{S^2}^*(\omega_{S^2}(i)))_{i \in \underline{3}}))$ is independent of the chosen $\omega_{S^2}(i)$. It is an isotopy invariant of $K$, which coincides with $w_C\check{Z}_2(S^3, K)$ and $w_C p^c \check{Z}_2(S^3, K)$.

**PROOF OF PROPOSITION 1.7 ASSUMING THEOREM 7.32:** The only degree 2 Jacobi diagrams on $S^1$ without looped edges, with at most three edges, and with no trivalent component are $\mathbb{X}$, $\mathbb{I}$, and $\mathbb{K}$. We have $w_C([\mathbb{X}]) = 1$, $w_C([\mathbb{I}]) = 0$, $w_C([\mathbb{K}]) = w_C([\mathbb{I}]) - w_C([\mathbb{X}]) = -1$. There are six elements of $\mathcal{D}_{2,3}^e(S^1)$ isomorphic to $\mathbb{X}$, one for each permutation $\sigma$ of $\underline{3}$. They may be drawn as

$$ \Gamma(\sigma) = \begin{tikzpicture}[baseline=(current bounding box.center), scale=0.5, every node/.style={scale=0.5}] 
  \node (A) at (0,0) {};
  \node (B) at (1,0) {};
  \node (C) at (0,1) {};
  \node (D) at (1,1) {};
  \draw[->] (A) -- (B) node right[above, sloped]{\small $\sigma(2)$};
  \draw[->] (B) -- (C) node left[below]{\small $\sigma(3)$};
  \draw[->] (C) -- (D) node right[above]{\small $\sigma(4)$};
  \draw[->] (D) -- (A) node left[below]{\small $\sigma(1)$};
\end{tikzpicture} $$

and we have $\zeta_{\Gamma(\sigma)} = \frac{1}{3! \times 4}$. For such a graph $\Gamma(\sigma)$, we also have

$$ I(S^3, K, \Gamma(\sigma), (p_{S^2}^*(\omega_{S^2}(i)))_{i \in \underline{3}}) \\ = \int_{(S^2)^3} \deg(\sigma_* (1_{S^2} \times G_{\mathbb{X}})) \wedge^3_{i=1} p_i^*(\omega_{S^2}(i)). $$
There are 16 elements of $\mathcal{D}_{2,3}^e(S^1)$ isomorphic to $\tilde{\mathcal{C}}$. They are obtained from the two diagrams

<figure><img src="image_71.png" /></figure>

by reversing the directions of some edges. For these diagrams $\zeta_{\Gamma} = \frac{1}{3! \times 8}$. Theorem 7.32 implies that for any embedding $K: S^1 \hookrightarrow \mathbb{R}^3$, we have

$$w_C \tilde{Z}_2(S^3, K) = \int_{(S^2)^3} \tilde{w}_2(K) \wedge_{i=1}^3 p_i^*(\omega_{S^2}(i))$$

for the locally constant degree map $\tilde{w}_2(K)$ of Proposition 1.7 defined on an open dense subset of $(S^2)^3$. Any point $(X_1, X_2, X_3)$ of this open dense subset of regular values of $(S^2)^3$, has an open connected neighborhood $\prod_{i=1}^3 W_i$ of regular values, and there are volume-one 2-forms $\omega_{S^2}(i)$ supported on $W_i$. For such forms, $w_C \tilde{Z}_2(S^3, K) = \tilde{w}_2(K)(X_1, X_2, X_3)$. So Theorem 7.32 implies that $\tilde{w}_2(K)$ is constant. $\square$

## 7.5 Straight links

A one-cycle $c$ of $S^2$ is algebraically trivial if, for any two points $x$ and $y$ outside its support, the algebraic intersection of an arc from $x$ to $y$ transverse to $c$ with $c$ is zero, or equivalently if the integral of any one-form of $S^2$ along $c$ is zero.

Let $(\tilde{R}, \tau)$ be an asymptotic rational homology $\mathbb{R}^3$. Let $K: S^1 \hookrightarrow \tilde{R}$ be a knot embedding. Recall the notation from Proposition 3.7 and Section 7.3. Set $C_K = C(K; \xleftarrow{\hspace{2em}} S^1)$. We have $C_K \subset C_2(R)$.

**Lemma 7.33.** If $p_{\tau}(\partial C_K)$ is algebraically trivial, then for any propagating chain $F$ of $(C_2(R), \tau)$ transverse to $C_K$ and for any propagating form $\omega_p$ of $(C_2(R), \tau)$, we have

$$\int_{C_K} \omega_p = \langle C_K, F \rangle_{C_2(R)} = I_{\theta}(K, \tau),$$

with respect to the definition of $I_{\theta}(K, \tau)$ in Lemma 7.15. In particular, we have $I_{\theta}(K, \tau) \in \mathbb{Q}$, and $I_{\theta}(K, \tau)$ is an integer when $R$ is an integral homology sphere.

**PROOF:** According to Lemma 7.13, changing the propagating form $\omega_p$ to $\omega'_p$ adds some $\int_{\partial C_K} p_{\tau}^*(\eta_S) = \int_{p_{\tau}(\partial C_K)} \eta_S$ for some one-form on $S^2$. Then by definition, $\int_{C_K} \omega_p$ is independent of the propagating form $\omega_p$ of $(C_2(R), \tau)$.
For a propagating chain $F$ of $(C_2(R), \tau)$ transverse to $C_K$, one can choose a propagating form dual to $F$ and supported near $F$ such that $\int_{C_K} \omega_p = \langle C_K, F \rangle_{C_2(R)}$. (See the end of Section B.2 and Lemma B.4 in particular, for details.) The rationality of $I_\theta(K, \tau)$ follows from the rationality of $F$. Since $F$ can be chosen to be an integral chain when $R$ is a $\mathbb{Z}$-sphere, we get $I_\theta(K, \tau) \in \mathbb{Z}$ in this case. $\square$

**Remark 7.34.** One could have proved that $\langle C_K, F \rangle_{C_2(R)}$ is independent of the chosen propagating chain $F$ of $(C_2(R), \tau)$ transverse to $C_K$ directly, as follows. Let $F$ and $F'$ be two such propagating chains with respective boundaries $p_\tau^{-1}(a)$ and $p_\tau^{-1}(a')$, where $a$ and $a'$ are in $S^2$. Let $[a, a']$ be a path from $a$ to $a'$ in $S^2$ transverse to $p_\tau(\partial C_K)$. There exists a rational chain $W$ of $C_2(R)$ transverse to $C_K$ and to $\partial C_2(R)$, whose boundary is $F' - F - p_\tau^{-1}([a, a'])$. Then $\partial C_K \cap W \subset C_K \cap \partial W$, and $\partial(C_K \cap W) = \pm C_K \cap \partial W$. So we have

$$ \langle C_K, \partial W \rangle_{C_2(R)} = \langle C_K, F' - F - p_\tau^{-1}([a, a']) \rangle_{C_2(R)} = 0. $$

So we get

$$ \begin{aligned} \langle C_K, F' \rangle_{C_2(R)} - \langle C_K, F \rangle_{C_2(R)} &= \langle C_K, p_\tau^{-1}([a, a']) \rangle_{C_2(R)} \\ &= \langle \partial C_K, p_\tau^{-1}([a, a']) \rangle_{\partial C_2(R)} \\ &= \langle p_\tau(\partial C_K), [a, a'] \rangle_{S^2} = 0. \end{aligned} $$

**Lemma 7.35.** Assume that $p_\tau(\partial C_K)$ is algebraically trivial. Let $Y \in S^2 \setminus p_\tau(\partial C_K)$. Let $Z: K \to S^2$ map $k \in K$ to the vector $Z(k)$ of $S^2$ orthogonal to $p_\tau(T_k K)$ in the half great circle of $S^2$ that contains $p_\tau(T_k K)$, $Y$, and $p_\tau(-T_k K)$. Define the parallel $K_{\parallel, \tau, Y}$ by pushing a point $k$ of $K$ in the direction $\tau(Z(k))$. Then we have

$$ I_{\theta}(K, \tau) = lk(K, K_{\parallel, \tau, Y}). $$

**PROOF:** Thanks to Lemma 3.12, $lk(K, K_{\parallel, \tau, Y})$ is the evaluation of any propagator of $C_2(R)$ on $K \times K_{\parallel, \tau, Y}$. Let $\omega_p$ be a propagating form of $C_2(R)$ (as in Definition 3.11), which may be expressed as $p_\tau^*(\omega_{-Y})$ on $UN(K)$, for a 2-form $\omega_{-Y}$ of $S^2$ supported in a geometric disk in $S^2 \setminus p_\tau(\partial C_K)$ centered at $(-Y)$. Observe that the intersection of such a disk with the half great circle of $S^2$ that contains $p_\tau(T_k K)$, $Y$, and $p_\tau(-T_k K)$ is empty for any $k \in K$.

Compute $lk(K, K_{\parallel, \tau, Y})$ as the limit of $I(K \xrightarrow{\phi} K^{\parallel, \tau, Y}, \omega_p)$, when $K_{\parallel, \tau, Y}$ tends to $K$. The configuration space $K \times K_{\parallel, \tau, Y}$ is a torus of $C_2(R)$. When $K_{\parallel, \tau, Y}$ tends to $K$, this torus tends to the union of the annulus $C(K; \phi \xrightarrow{\phi} K)$ and an annulus $K \times_{\tau, Y} J$ contained in $UR|_K$, which fibers over $K$ and whose
fiber over $k \in K$ contains all the limit directions from $k$ to a close point on $K_{\parallel, \tau, Y}$. This fiber is the half great circle of $UR|_k$ that $p_\tau$ maps to the half great circle of $S^2$ that contains $p_\tau(T_k K)$, $Y$, and $p_\tau(-T_k K)$. Thus $K \times K_{\parallel, \tau, Y}$ is homologous to the torus

$$T = C_K \cup (K \times_{\tau, Y} J).$$

The integral of $\omega_p$ on $K \times_{\tau, Y} J$ is the integral of $\omega_{-Y}$ along $p_\tau(K \times_{\tau, Y} J)$. It is zero since $p_\tau(K \times_{\tau, Y} J)$ does not meet the support of $\omega_{-Y}$. Therefore, we have $lk(K, K_{\parallel, \tau, Y}) = \int_{C_K} \omega_p = I_\theta(K, \tau)$, thanks to Lemma 7.33. $\square$

An isotopy class of parallels of a knot is called a *framing* (or a *parallelization*) of a knot.

**Corollary 7.36.** A knot embedding $K$ such that $p_\tau(\partial C_K)$ is algebraically trivial, with respect to a parallelization $\tau$, has a canonical framing induced by $\tau$, which is the framing induced by a parallel $K_{\parallel, \tau, Y}$ for an arbitrary $Y \in S^2 \setminus p_\tau(\partial C_K)$. For such a knot embedding $K$, for any propagating chain $F$ of $(C_2(R), \tau)$ transverse to $C_K$, and for any propagating form $\omega_p$ of $(C_2(R), \tau)$, we have

$$\int_{C_K} \omega_p = \langle C_K, F \rangle_{C_2(R)} = I_\theta(K, \tau) = lk(K, K_{\parallel, \tau, Y}).$$

**Proof:** Since the linking number $lk(K, K_{\parallel, \tau, Y})$ determines the framing, the corollary is a direct consequence of Lemmas 7.33 and 7.35. $\square$

**Definition 7.37.** A knot embedding $K: S^1 \hookrightarrow \tilde{R}$ is *straight* with respect to $\tau$ if the curve $p_\tau(U^+K)$ of $S^2$ is algebraically trivial (with the notation from Proposition 3.7 and Section 7.3). A link embedding is *straight* with respect to $\tau$ if all its components are.

Straight knot embeddings and almost-horizontal knot embeddings in $\mathbb{R}^3$ (defined before Lemma 1.6) are examples of knot embeddings $K$ such that $p_\tau(\partial C_K)$ is algebraically trivial. Therefore, Lemma 1.6 is a particular case of the above corollary. As a second corollary, we get the following lemma.

**Lemma 7.38.** For any knot embedding $K$ in $\tilde{R}$, there exists an asymptotically standard parallelization $\tilde{\tau}$ homotopic to $\tau$ such that the first vector $\tilde{\tau}(.; (1, 0, 0))$ of $\tilde{\tau}$ is tangent to $K$ and orients $K$. In this case, let $K_{\tilde{\tau}}$ be the parallel of $K$ obtained by pushing $K$ in the direction of the second vector $\tilde{\tau}(.; (0, 1, 0))$ of $\tilde{\tau}$. Then we have

$$I_\theta(K, \tilde{\tau}) = lk(K, K_{\tilde{\tau}}).$$
PROOF: In order to obtain $\tilde{\tau}$, it suffices to perform a homotopy of $\tau$ around the image of $K$ so that the first vector of $\tilde{\tau}$ becomes tangent to $K$ along $K$. Thus, $K$ is straight with respect to $(R, \tilde{\tau})$. Apply Corollary 7.36. $\square$

**Lemma 7.39.** Let $K_0$: $\{0\} \times S^1 \to \tilde{R}$ be a straight embedding with respect to $\tau$. Let $K$: $[0, 1] \times S^1 \to \tilde{R}$ be an embedding such that its restriction $K_1$: $\{1\} \times S^1 \to \tilde{R}$ is straight with respect to an asymptotically standard parallelization $\tau_1$ homotopic to $\tau$. Then $(I_\theta(K_1, \tau_1) - I_\theta(K_0, \tau))$ is an even integer. Furthermore, for any real number $x$ congruent to $I_\theta(K_0, \tau)$ modulo 2, there exists a straight embedding $K_1$ isotopic to $K_0$ such that $I_\theta(K_1, \tau) = x$.

PROOF: Let $H: t \mapsto \tau_t$ be a smooth homotopy from $\tau = \tau_0$ to $\tau_1$. Let $p_H: [0, 1] \times \partial C_2(R) \to S^2$ be the smooth map that restricts to $\{t\} \times \partial C_2(R)$ as $p_{\tau_t}$. There is a closed 2-form $\omega$ on $[0, 1] \times C_2(R)$ that restricts to $[0, 1] \times \partial C_2(R)$ as $p_H^*(\omega_{S^2})$. (Such a form may be obtained by modifying $p_{C_2(R)}^*p_\tau(\omega_{S^2})$ in a collar neighborhood of $U\tilde{R}$ using the homotopy $H$.) Then the integral of $\omega$ over

$$ \partial(\cup_{t \in [0,1]} C(K; \stackrel{\leftrightarrow}{\to} K_t)) = C(K; \stackrel{\leftrightarrow}{\to} K_1) - C(K; \stackrel{\leftrightarrow}{\to} K_0) - \cup_{t \in [0,1]} \partial C(K; \stackrel{\leftrightarrow}{\to} K_t) $$

vanishes. So we have

$$ I_{\theta}(K_1, \tau_1) - I_{\theta}(K_0, \tau) = \int_{\cup_{t \in [0,1]} \partial C(K; \stackrel{\leftrightarrow}{\to} K_t)} \omega. $$

This is the area in $S^2$ of the integral cycle $\cup_{t \in [0,1]} p_{\tau_t}(\partial C(K; \stackrel{\leftrightarrow}{\to} K_t))$. This cycle is the union of the two integral cycles

$$ \cup_{t \in [0,1]} p_{\tau_t}(U^+(K_t)) \text{ and } \cup_{t \in [0,1]} p_{\tau_t}(-U^-(K_t)) $$

which have the same integral area. So $(I_\theta(K_1, \tau_1) - I_\theta(K_0, \tau))$ is an even integer.

Adding two small almost-horizontal kinks in a standard ball as in the end of Subsection 1.2.4 which turn in opposite direction and contribute with the same crossing sign like ($\nearrow$ and $\swarrow$) or ($\searrow$ and $\swarrow$) preserves straightness and adds $\pm 2$ to $I_\theta$. $\square$

## 7.6 Second definition of $Z$

Let us state another version of Theorem 7.20 using straight links instead of homogeneous propagating forms. Recall $\zeta_\Gamma = \frac{(3n-|E(\Gamma)|)!}{(3n)!2^{|E(\Gamma)|}}$.
**Theorem 7.40.** Let $(\tilde{R}, \tau)$ be an asymptotic rational homology $\mathbb{R}^3$. Let $\mathcal{L} = \bigsqcup_{j=1}^k S_j^1$ be a disjoint union of $k$ circles. Let $L: \mathcal{L} \hookrightarrow \tilde{R}$ be a straight embedding with respect to $\tau$. For any $i \in \underline{3n}$, let $\omega(i)$ be a propagating form of $(C_2(R), \tau)$. Then the element $Z_n(\tilde{R}, L, (\omega(i))_{i \in \underline{3n}})$ of $\mathcal{A}_n(\mathcal{L})$ defined in Notation 7.16 is independent of the chosen propagating forms $\omega(i)$ of $(C_2(R), \tau)$. It is denoted by $Z_n^s(\tilde{R}, L, \tau)$. In particular, with the notation of Theorem 7.20, we have

$$Z_n^s(\tilde{R}, L, \tau) = Z_n(\tilde{R}, L, \tau).$$

We will give the proof of this theorem in Section 10.4. Theorem 16.9 shows how $Z_n(\tilde{R}, L, (\omega(i))_{i \in \underline{3n}})$ depends on the propagators without the straightness assumption.

Straight links $L$ with respect to $\tau$ are framed links $(L, L_{\parallel, \tau})$ according to Corollary 7.36 and Lemma 7.35. So we can keep the information from the link framing and define the invariant of straight links

$$\mathcal{Z}^f(\tilde{R}, L, L_{\parallel, \tau}) = \exp\left(-\frac{1}{4}p_1(\tau)\beta\right)Z(\tilde{R}, L, \tau),$$

which depends only on $(\tilde{R}, L)$ and on the $lk(K_j, K_{j\parallel, \tau})$ for the components $K_j$ of $L$, according to Theorem 7.20 and Corollary 7.36.

**Definition 7.41.** Recall the invariant $\mathcal{Z}$ of Theorem 7.20. Define the framed link invariant $\mathcal{Z}^f$ to satisfy

$$\mathcal{Z}^f(\tilde{R}, \bigsqcup_{j=1}^k K_j, \bigsqcup_{j=1}^k K_{j\parallel}) = \prod_{j=1}^k (\exp(lk(K_j, K_{j\parallel})\alpha)\#_j)\mathcal{Z}(R, \bigsqcup_{j=1}^k K_j),$$

for a link $\bigsqcup_{j=1}^k K_j$ equipped with a parallel $K_{j\parallel}$ for each component $K_j$.

Thanks to Theorem 7.20, Corollary 7.36, and Theorem 7.40, both definitions coincide for straight framed links $(L, L_{\parallel, \tau})$.

Again, we can reduce our averaging process when projecting to $\check{\mathcal{A}}(\mathcal{L})$ and get the following theorem, also proved in Section 10.4.

**Theorem 7.42.** Under the assumptions of Theorem 7.40, we have

$$\check{p}(Z_n^s(\tilde{R}, L, \tau)) = \sum_{\Gamma \in \mathcal{D}_{n, 3n-2}^e(\mathcal{L})} \zeta_{\Gamma} I(R, L, \Gamma, (\omega(i))_{i \in \underline{3n-2}}) \check{p}([\Gamma]).$$
# Chapter 8

## Compactifications of
configuration spaces

We use compactifications of our open configuration spaces $\tilde{C}(R, L; \Gamma)$ associated to links $L$ in a $Q$-sphere $R$ to study the behavior of our integrals and their dependence on the choice of propagating forms. More specifically, we prove the convergence of the integrals involved in the definitions of $Z$ by finding a smooth compactification (with boundary and ridges), to which the integrated forms extend smoothly. The variation of an integral under the addition of an exact form $d\eta$ is the integral of $\eta$ on the codimension-one faces of the boundary, which need to be identified precisely. Therefore, the proofs of Theorems 7.40 and 7.20 (and their variants with reduced averages) require a deep knowledge of configuration spaces. We give all the useful statements in Sections 8.2 to 8.4. We prove all of them in Sections 8.5 to 8.8.

Before giving all the required general statements, we present the main features of the involved compactifications in some examples. William Fulton and Robert MacPherson introduced similar compactifications in [FM94]. Scott Axelrod and Isadore Singer [AS94, Section 5], Maxim Kontsevich [Kon94], Raoul Bott and Clifford Taubes [BT94], and Dev Sinha [Sin04] also investigated these compactifications.

### 8.1 An informal introduction

Our first example of a compactification $C(R, L; \Gamma)$ of a configuration space $\tilde{C}(R, L; \Gamma)$ is the closed annulus $C(K; \leftarrow) = C(S^3, K; \leftarrow)$ of Subsection 1.2.4. Our (more interesting) second example is the compactification $C_2(R) = C(R, \emptyset; \to)$ studied in Section 3.2. Note that $C(K; \leftarrow)$ is the closure of $\tilde{C}(K; \leftarrow)$ in $C_2(R)$.