# Deformation of reducible representations of link group in SL (2,C)

Michael Heusener, Clément Legrand

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Michael Heusener, Clément Legrand. Deformation of reducible representations of link group in SL (2,C). 2025. hal-05124560

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https://hal.science/hal-05124560v1

Preprint submitted on 22 Jun 2025

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Distributed under a Creative Commons Attribution 4.0 International License
# DEFORMATION OF REDUCIBLE REPRESENTATIONS OF LINK GROUP IN SL₂(C)

MICHAEL HEUSENER AND CLÉMENT LEGRAND

ABSTRACT. Let L be a link in the tree-dimensional sphere. We study deformations of reducible representation of the link group associated with zeros of the multi-variable Alexander polynomial. Under some hypothesis, we also describe the local structure of the representation variety at such reducible representations.

## INTRODUCTION

We consider an oriented link $L = L_1 \cup \dots \cup L_\mu$ with $\mu$ components and $X_L := S^3 \setminus \text{int} V(L)$ the link exterior of L in $S^3$. We suppose L is non-splittable such that $X_L$ is a compact connected orientable irreducible 3-manifold with boundary. The aim of this article is to study the local structure of the SL₂(C)-representation scheme of the fundamental group $\Gamma_L := \pi_1(X_L)$ at abelian, non-central representations.

The homology group $H_1(X_L) := H_1(X_L; \mathbb{Z}) \cong \mathbb{Z}^\mu$ is a free abelian group of rank $\mu$. The link L determines a basis $(t_1, \dots, t_\mu)$ of $H_1(X_L) \cong \langle t_1 \rangle \times \dots \times \langle t_\mu \rangle$ where the generator $t_i$ is represented by a meridian of the component $L_i$ of L. For each component $L_i$ we obtain a homomorphism $\varphi_i: \Gamma_L \to (\mathbb{Z}, +)$ given by the linking number $\varphi_i(\gamma) = \text{lk}(L_i, \gamma)$, and the Hurewicz homomorphism $\varphi: \Gamma_L \to \langle t_1 \rangle \times \dots \times \langle t_\mu \rangle$ is given by $\varphi(\gamma) = t_1^{\varphi_1(\gamma)} \dots t_\mu^{\varphi_\mu(\gamma)}$.

By our choice of a basis we can identify $\widehat{H_1(X_L)} := \text{Hom}(H_1(X_L), \mathbb{C}^*)$ with $(\mathbb{C}^*)^\mu$ by mapping $\alpha$ to $(\alpha(t_1), \dots, \alpha(t_\mu)) \in (\mathbb{C}^*)^\mu$. Each abelian homomorphism $\alpha \in \widehat{H_1(X_L)}$ induces an homomorphism $\alpha = \alpha \circ \varphi: \Gamma_L \to \mathbb{C}^*$, and we can define an abelian representation $\rho_\alpha: \pi_1(X_L) \to \text{SL}_2(\mathbb{C})$ by

$$ \rho_{\alpha}(\gamma) = \begin{pmatrix} \alpha^{1/2}(\gamma) & 0 \\ 0 & (\alpha^{1/2}(\gamma))^{-1} \end{pmatrix}, \quad \forall \gamma \in \Gamma_L. $$

Here $\alpha^{1/2}: \Gamma_L \to \mathbb{C}^*$ is a *root* of $\alpha$ i.e. any homomorphism such that for all $\gamma \in \Gamma_L$ we have $\alpha(\gamma) = (\alpha^{1/2}(\gamma))^2$. Notice that $\alpha^{1/2}$ is unique up to an homomorphism $\varepsilon: \Gamma_L \to \{\pm 1\}$. The choice of a fixed root is irrelevant, and it does not influence the results. To simplify the notation the choice will be not reflected in the notation.

The representation $\rho_\alpha$ is reducible and abelian. The first question is whether we can deform $\rho_\alpha$ into a reducible non-abelian representation. The result is well known for knots

<footnote>Date: June 20, 2025.</footnote>
as shown by G. Burde [Bur67] and G. de Rham [dR67]. We can deform $\rho_\alpha$ into a reducible
non-abelian representation if and only if for a meridian $m$ in the knot group $\alpha(m)$ is a root
of the Alexander polynomial of the knot. More recently, L. Bénard and A. Conway showed
a similar result for links in [BC20]. The Alexander polynomial of the knot is replaced by the
multi-variable Alexander polynomial $\Delta_L \in \mathbb{Z}[t_1^{\pm 1}, \dots, t_\mu^{\pm 1}]$ of the link $L$. By the identification
of $\widehat{H_1(X_L)}$ with $(\mathbb{C}^*)^\mu$ we obtain the following

**Theorem 1** ([BC20, Theorem 2.4], Lemma 5). Let $\alpha \in \widehat{H_1(X_L)}$ be a nontrivial homomorphism. Then there exists a reducible non-abelian representation $\varrho: \Gamma_L \to \text{SL}_2(\mathbb{C})$ which has the same character as $\rho_\alpha$ if and only if $\alpha$ is a zero of the multi-variable Alexander polynomial $\Delta_L$.

The second question we study in this article is the geometry of the representation scheme $\mathcal{R}(\Gamma_L) = \text{Hom}(\Gamma_L, \text{SL}_2(\mathbb{C}))$ near a reducible non-abelian representation. In [HPSP01] it is shown that for knots a reducible non-abelian representation is a smooth point of the representation scheme if $\alpha(m)$ is a simple root of the Alexander polynomial.

In what follows we let $V(\Delta_L) \subset (\mathbb{C}^*)^\mu$ denote the vanishing locus of the multi-variable Alexander polynomial $\Delta_L$.

**Theorem 2** (Theorem 11). Let $L$ be a link with nontrivial multi-variable Alexander polynomial. Let $\alpha \in \widehat{H_1(X_L)}$ be a nontrivial homomorphism. Assume $\alpha$ is a smooth point of $V(\Delta_L)$. Let $\varrho$ be a reducible non-abelian representation such that $\varrho$ and $\rho_\alpha$ have the same character. Then $\varrho$ is a smooth point of the representation scheme $\mathcal{R}(\Gamma_L)$ and is contained in a unique algebraic component of dimension $\mu + 3$.

Finally we study the local geometry near the abelian representation $\rho_\alpha$.

**Theorem 3.** Let $L$ be a link with nontrivial multi-variable Alexander polynomial. Let $\alpha \in \widehat{H_1(X_L)}$ be a nontrivial homomorphism. Assume $\alpha$ is a smooth point of $V(\Delta_L)$. Then, the representation $\rho_\alpha$ is contained in exactly two algebraic components one of dimension $\mu + 2$ containing only abelian representations and one of dimension $\mu + 3$ containing irreducible ones. Moreover, $\rho_\alpha$ is a smooth point of both components, and the two components intersect transversally at $\rho_\alpha$.

These results will be used in a forthcoming article to study the local structure of the
$\text{SL}_2(\mathbb{C})$-character scheme of link groups.

# 1. NOTATION AND FACTS

In this section we introduce the notation and we recall some known results which we will
use throughout this article. First of all modules will always be considered as left modules.
We put $\llbracket 1, n \rrbracket := \{1, \dots, n\}$ for the set of integer between 1 and $n$ and $\delta_{ij}$ for the Kronecker
delta.
1.1. **Group cohomology.** The main reference for this part is the book of Brown [Bro76]. Let $\Gamma$ be a group and $M$ be a $\mathbb{C}[\Gamma]$-module. We consider the normalized cochain complex $(C^*(\Gamma; M), \delta^*)$ given by

$$ C^q(\Gamma; M) = \{f: \Gamma^q \to M \mid \text{such that } f(\gamma_1, \dots, \gamma_q) = 0 \text{ whenever some } \gamma_i = 1\} $$

The map $\delta^{q+1}: C^q(\Gamma; M) \to C^{q+1}(\Gamma; M)$ given by

$$ \delta^{q+1} f(\gamma_1, \dots, \gamma_{q+1}) = \gamma_1 \cdot f(\gamma_2, \dots, \gamma_{q+1}) + \sum_{i=1}^{q} (-1)^i f(\gamma_1, \dots, \gamma_i \gamma_{i+1}, \dots, \gamma_{q+1}) + (-1)^{q+1} f(\gamma_1, \dots, \gamma_q). $$

We let $B^q(\Gamma; M) = \text{Im} \delta^q$ and $Z^q(\Gamma; M) = \text{ker} \delta^{q+1}$ denote the coboundary and cocycle space respectively. It is well known that the cohomology of this complex coincides with $H^*(\Gamma; M)$, the cohomology of the group $\Gamma$ with coefficient in $M$ [Bro76, III.1]. Notice also that for an Eilenberg-MacLane complex $X$ we have an isomorphism

$$ H^*(X; M) \cong H^*(\pi_1 X; M). $$

Here $H^*(X; M)$ denotes the cohomology of $X$ with twisted coefficients.

We identify $C^0(\Gamma; M)$ with $M$. Then the 0-th cohomology group is the group of invariant elements under the action $H^0(\Gamma; M) = M^\Gamma = \{x \in M \mid \gamma \cdot x = x, \forall \gamma \in \Gamma\}$. The first cocycle group is given by $Z^1(\Gamma; M) = \{d: \Gamma \to M \mid d(\gamma_1\gamma_2) = d(\gamma_1) + \gamma_1 \cdot d(\gamma_2), \forall \gamma_1, \gamma_2 \in \Gamma\}$ and the first coboundary group by $B^1(\Gamma; M) = \{d \in Z^1(\Gamma; M) \mid \exists b \in M, d(\gamma) = (\gamma - 1) \cdot b \forall \gamma \in \Gamma\}$. We call a cocycle $d \in Z^1(\Gamma; M)$ principal if $d$ is a coboundary and non-principal otherwise.

A short exact sequence:

$$ 0 \to M_1 \xrightarrow{i} M_2 \xrightarrow{p} M_3 \to 0 $$

of $\Gamma$-modules gives rise to a short exact sequence of cochain complexes:

$$ 0 \to C^*(\Gamma; M_1) \xrightarrow{i^*} C^*(\Gamma; M_2) \xrightarrow{p^*} C^*(\Gamma; M_3) \to 0. $$

We will make use of the corresponding long exact cohomology sequence (see [Bro76, III. Proposition 6.1]):

$$ (1) \qquad 0 \to H^0(\Gamma; M_1) \to H^0(\Gamma; M_2) \to H^0(\Gamma; M_3) \xrightarrow{\beta^0} H^1(\Gamma; M_1) \to \dots $$

Recall that the connection homomorphism $\beta^n: H^n(\Gamma; M_3) \to H^{n+1}(\Gamma; M_1)$ is determined by the snake lemma: if $z \in Z^n(\Gamma; M_3)$ is a cocycle and if $\tilde{z} \in (p^*)^{-1}(z) \subset C^n(\Gamma; M_2)$ is any lift of $z$ then $\delta_2(\tilde{z}) \in \text{Im}(i^*)$ where $\delta_2$ denotes the coboundary operator of $C^*(\Gamma; M_2)$. It follows that any cochain $z' \in C^{n+1}(\Gamma; M_1)$ such that $i^*(z') = \delta_2(\tilde{z})$ is a cocycle and that its cohomology class does only depend on the cohomology class represented by $z$. The cocycle $z'$ represents the image of the cohomology class represented by $z$ under $\beta^n$.
Let $M_1, M_2, M_3$ be $\mathbb{C}[\Gamma]$-modules. The cup product between two cochains $u \in C^p(\Gamma; M_1)$ and $v \in C^q(\Gamma; M_2)$ is the cochain $u \smile v \in C^{p+q}(\Gamma; M_1 \otimes M_2)$ defined by

$$u \smile v(\gamma_1, \dots, \gamma_{p+q}) := u(\gamma_1, \dots, \gamma_p) \otimes \gamma_1 \cdots \gamma_p \circ v(\gamma_{p+1}, \dots, \gamma_{p+q})$$

where $M_1 \otimes M_2$ is a $\Gamma$-module with the diagonal action. We can combine the cup product of two cocycles with a linear map $M_1 \otimes M_2 \to M_3$ compatible by the action of $\Gamma$. In particular, each bilinear $\Gamma$-invariant map $\beta: M_1 \times M_2 \to M_3$ defines a linear map $M_1 \otimes M_2 \to M_3$ and hence a cup product

$$\beta: C^p(\Gamma; M_1) \otimes C^q(\Gamma; M_2) \to C^{p+q}(\Gamma; M_3).$$

Let $z_i \in Z^1(\Gamma; M_i)$ with $i \in [1, 2]$ be cocycles. If we define $f: \Gamma \to M_3$ by

$$f(\gamma) = \beta(z_1(\gamma) \otimes z_2(\gamma)),$$

we obtain by a direct calculation that

$$\delta^1 f(\gamma_1, \gamma_2) + \beta(z_1(\gamma_1) \otimes \gamma_1 \cdot z_2(\gamma_2)) + \beta(\gamma_1 \cdot z_1(\gamma_2) \otimes z_2(\gamma_1)) = 0 \text{ for all } \gamma_1, \gamma_2 \in \Gamma.$$

If we denote by $\tau: M_2 \otimes M_1 \to M_1 \otimes M_2$ the twist operator i.e. $\tau(x_2 \otimes x_1) = x_1 \otimes x_2$, then $\beta \circ \tau: M_2 \otimes M_1 \to M_3$ is also bilinear and compatible with the action of $\Gamma$. This gives

$$(2) \qquad \delta^1 f + z_1 \stackrel{\beta}{\smile} z_2 + z_2 \stackrel{\beta \circ \tau}{\smile} z_1 = 0.$$

In low dimension the cocycles are related to the Fox-calculus. Let $F_n$ be the free group of rank $n$ and $i \in [1, n]$. The Fox derivative $\partial_{S_i}: \mathbb{Z}[F_n] \to \mathbb{Z}[F_n]$ satisfies

$$\partial_{S_i}(\eta_1 \eta_2) = \partial_{S_i}(\eta_1) \epsilon(\eta_2) + \eta_1 \partial_{S_i}(\eta_2) \quad \text{and} \quad \partial_{S_i}(S_j) = \delta_{ij}, \quad \forall \eta_1, \eta_2 \in \mathbb{Z}[F_n], \forall j \in [1, n]$$

where $\epsilon: \mathbb{Z}[F_n] \to \mathbb{Z}$ denotes the augmentation map.

We can use Fox calculus to compute the first cocycle space in the following way. Suppose that $(s_1, \dots, s_n)$ is a system of generators $\Gamma$ and let $\pi: F(S_1, \dots, S_n) \to \Gamma$ be the canonical projection. The free group $F_n := F(S_1, \dots, S_n)$ acts on $M$ via $\pi$, and every derivation $z_\psi: F_n \to M$ is uniquely determined by a map $\psi: \{S_1, \dots, S_n\} \to M$,

$$z_{\psi}(w) = \sum_{i=1}^{n} \partial_{S_i} w \cdot \psi(S_i) \quad \text{for } w \in F_n.$$

The derivation $z_\psi$ factors through $\Gamma$ if and only if for all $R \in \ker \pi$ we have $z_\psi(R) = 0$.

In the case that $\Gamma \cong \langle S_1, \dots, S_n \mid R_1, \dots, R_m \rangle$ is finitely presented, we can identify $Z^1(\Gamma; M)$ with the following submodule of $Z^1(F_n; M) \cong M^n$

$$(3) \qquad Z^1(\Gamma; M) \cong \{ (x_1, \dots, x_n) \in M^n \mid \forall j \in \{1, \dots, m\}, \sum_{i=1}^{n} \partial_{S_i} R_j \cdot x_i = 0 \}.$$
In what follows we extend a normalized 2-cochain linearly to the first component in the following sense. If $c: \Gamma \times \Gamma \to M$ is a 2-cochain, $\gamma_0 \in \Gamma_L$ and $\eta = \sum_{\gamma \in \Gamma_L} n_\gamma \gamma \in \mathbb{C}[\Gamma]$ we put

$$c(\eta, \gamma_0) := \sum_{\gamma \in \Gamma} n_{\gamma} c(\gamma, \gamma_0).$$

Following Proposition 6.3 in [HP05] we obtain that a normalized 2-cocycle $c: \Gamma \times \Gamma \to M$ is a coboundary if and only if there exists $(x_1, \dots, x_n) \in M^n$ such that

$$(4) \qquad \sum_{i=1}^{n} \pi(\partial_{S_i} R_j) \cdot x_i + c(\pi(\partial_{S_i} R_j), \pi(S_i)) = 0 \quad \text{holds for all } j \in [1, m].$$

1.2. **Alexander polynomial.** We consider an oriented link $L = L_1 \cup \dots \cup L_\mu$ in the 3-sphere $S^3$ with $\mu \ge 2$ components, and we let $X_L := S^3 \setminus V(L)$ denote the link exterior. We suppose $L$ is non-splittable such that $X_L$ is a compact connected orientable irreducible 3-manifold with boundary. It follows from the sphere theorem that $X_L$ is an Eilenberg-MacLane space. The fundamental group of the link exterior $X_L$ is denoted by $\Gamma_L := \pi_1 X_L$. We choose a Wirtinger presentation of $\Gamma_L$, and because $L$ is non split the deficiency of $\Gamma_L$ is one

$$(5) \qquad \Gamma_L \cong \langle S_1, \dots, S_n \mid R_1, \dots, R_{n-1} \rangle.$$

We write $F_n = \langle S_1, \dots, S_n \mid - \rangle$, for the free group of rank $n$ and $\pi: F_n \to \Gamma_L$ for the canonical projection. Each generator $S_i$ of the presentation (5) represents a meridian of a component of $L$. This gives a map

$$\kappa: [1, n] \to [1, \mu] \text{ given by } \kappa(i) = k \text{ if } S_i \text{ represents a meridian of } L_k.$$

In what follows we will put $\kappa_i := \kappa(i)$ for all $i \in [1, n]$.

The first homology group $H_1(X_L) := H_1(X_L, \mathbb{Z})$ is a free abelian group of rank $\mu$. A basis of $H_1(X_L)$ is given by $(t_1, \dots, t_\mu)$ where $t_k$ is represented by a meridian of the component $L_k$ of $L$. This gives us an isomorphism of $H_1(X_L)$ with the multiplicative free abelian group $\langle t_1 \rangle \times \dots \times \langle t_\mu \rangle$. Hence the group ring $\mathbb{Z}[H_1(X_L)]$ is isomorphic to the ring of Laurent polynomials in $\mu$ variables $\mathbb{Z}[t_1^{\pm 1}, \dots, t_\mu^{\pm 1}]$. In what follows we will use these identifications.

With this notation the Hurewicz epimorphism $\varphi: \Gamma_L \to H_1(X_L)$ is given by

$$\varphi(\gamma) = t_1^{\varphi_1(\gamma)} \cdots t_\mu^{\varphi_\mu(\gamma)}.$$

where the homomorphisms $\varphi_k: \Gamma_L \to (\mathbb{Z}, +)$ denotes the linking number $\varphi_k(\gamma) = \text{lk}(\gamma, L_k)$. To simplify notations, will make use of the *multi-index notation*, for $\mathbf{t} = (t_1, \dots, t_\mu)$, $\varphi = (\varphi_1, \dots, \varphi_\mu)$, and $\gamma \in \Gamma_L$ we put

$$\bar{\gamma}(\mathbf{t}) := \mathbf{t}^{\varphi(\gamma)} := \varphi(\gamma).$$

If no confusion can arise, we always use the same symbol for a group homomorphism and its linear extension to the corresponding group rings.
The main references for the multi-variable Alexander polynomial are the books [Kaw96, BZH14]. We consider the universal abelian cover over $X_L$ denoted by $p: \hat{X}_L \to X$. We have $p_*(\pi_1 \hat{X}_L) = [\Gamma_L, \Gamma_L]$. The deck transformation group of $\hat{X}_L$ acts on $\hat{X}_L$ and is isomorphic to $H_1(X_L)$. A linearization of this action gives a structure of $\mathbb{Z}[t^{\pm 1}]$-module on $H_1(\hat{X}_L)$ and on $H_1(\hat{X}_L, p^{-1}(x))$, where $x \in X_L$ is a point.

By considering the augmentation map $\epsilon: \mathbb{Z}[t^{\pm 1}] \to \mathbb{Z}$, we have the following $\mathbb{Z}[t^{\pm 1}]$-exact sequence

$$ (6) \qquad 0 \to H_1(\hat{X}_L) \to H_1(\hat{X}_L, p^{-1}(x)) \to \ker \epsilon \to 0. $$

If $L$ is a knot, the $\mathbb{Z}[t^{\pm 1}]$-module $\ker \epsilon$ is free, so the sequence (6) splits. Then a presentation matrix for $H_1(\hat{X}_L)$ is obtained by removing one column of the Alexander matrix. For links this is not the case. (See Section 9 in [BZH14]).

The Alexander matrix $J(t)$ is a presentation matrix for the $\mathbb{Z}[t^{\pm 1}]$-module $H_1(\hat{X}_L, p^{-1}(x))$. It is obtained in the following way. Let $J(t) = J(t_1, \dots, t_\mu) \in \mathcal{M}_{(n-1)n}(\mathbb{Z}[t^{\pm 1}])$ be the matrix defined by

$$ J(t)_{ji} = \overline{\pi(\partial_{S_i} R_j)(t)}, \quad i \in [1, n], \quad j \in [1, n-1]. $$

The matrix $J(t)$ is called the Alexander matrix of the Wirtinger presentation (5). For $i \in [1, n]$ we put $J_i(t) \in \mathcal{M}_{(n-1)}(\mathbb{Z}[t^{\pm 1}])$ for the square matrix obtained by removing the $i$-th column of the Alexander matrix.

Notice that the fundamental formula of Fox-calculus implies that for each relation $R$ in the normal subgroup generated by the defining relations we obtain

$$ (7) \qquad \sum_{i=1}^{n} \overline{\pi(\partial_{S_i} R)(t)}(t_{\kappa_i} - 1) = 0. $$

For a link there is no general procedure for obtaining a presentation matrix for $H_1(\hat{X}_L)$ in terms of the Alexander matrix. However, we can still derive invariants of the oriented link $L$ from the Alexander matrix. We are interested in the multi-variable Alexander polynomial $\Delta_L(t)$ which is defined by

$$ \Delta_L(t) = \text{pgcd} (\text{minors of order } n-1 \text{ of } J(t)). $$

There is a simple relation between the minors of order $n-1$ of the Alexander matrix and the Alexander polynomial [Kaw96, Lemma 7.3.2], for any $i \in [1, n]$ we have

$$ (8) \qquad \det J_i(t) = \pm(t_{\kappa_i} - 1)\Delta_L(t). $$

The Alexander polynomial is symmetric up to a unit. There is a unit $\pm t_1^{l_1} \cdots t_\mu^{l_\mu} \in \mathbb{Z}[t^{\pm 1}]$ such that

$$ (9) \qquad \Delta_L(t^{-1}) = \pm t_1^{l_1} \cdots t_\mu^{l_\mu} \Delta_L(t). $$
Every homomorphism $\alpha: H_1(X_L) \to \mathbb{C}^*$ defines a multi-index $\alpha = (\alpha_1, \dots, \alpha_\mu)$ where
$\alpha_k = \alpha(t_k)$ for $k \in [1, \mu]$. This allows us to identify $\widehat{H_1(X_L)} := \text{Hom}(H_1(X_L), \mathbb{C}^*)$ with $(\mathbb{C}^*)^\mu$.

If $\Delta_L(\boldsymbol{\alpha}) = 0$ we say that $\boldsymbol{\alpha}$ is a zero of the Alexander polynomial. Following [McM02] we
let $V(\Delta_L)$ denote the vanishing locus of the Alexander polynomial in $(\mathbb{C}^*)^\mu$,

$$
(10) \qquad V(\Delta_L) := \{\boldsymbol{\alpha} \in (\mathbb{C}^*)^\mu \mid \Delta_L(\boldsymbol{\alpha}) = 0\}.
$$

Because of the symmetry, we obtain that **α** is a zero of the Alexander polynomial if and only
if the homomorphism **α**⁻¹ is a zero of the Alexander polynomial.

For $k \in [1, \mu]$, we write $\partial_{t_k} := \frac{\partial}{\partial t_k}$ for the partial derivative with respect to the variable $t_k$.
Let grad $\Delta_L(\mathbf{t}) = (\partial_{t_1} \Delta_L(\mathbf{t}), \dots, \partial_{t_\mu} \Delta_L(\mathbf{t})) \in \mathbb{Z}[t^{\pm 1}]^\mu$ denote the gradient of the Alexander polynomial.

For a given $\boldsymbol{\alpha} \in \widehat{H_1(X_L)}$, the complex numbers $\mathbb{C}$ turn into a $\mathbb{C}[\Gamma_L]$-module via $\boldsymbol{\alpha} := \boldsymbol{\alpha} \circ \varphi$
i.e. for $\gamma \in \Gamma_L$ and $z \in \mathbb{C}$ we define

$$
\gamma \cdot z = \alpha(\gamma) z = \boldsymbol{\alpha}(\varphi(\gamma)) z = \bar{\gamma}(\boldsymbol{\alpha}) z.
$$

The resulting $\Gamma_L$ module will be simply denoted by $\mathbb{C}_\alpha$, and we let $\mathbb{C}$ denote the trivial
$\Gamma_L$-module. In what follows we will make use the following proposition.

**Proposition 1.** Let $\boldsymbol{\alpha} \in \widehat{H_1(X_L)}$ be non-trivial homomorphism. If $\boldsymbol{\alpha}$ is a smooth point of $V(\boldsymbol{\alpha})$ then $\dim H^1(\Gamma_L; \mathbb{C}_\alpha) = 1$.

*Proof.* If follows from Equation (3) that we can identify $Z^1(\Gamma_L, \mathbb{C}_\alpha)$ with the right-kernel of $J(\boldsymbol{\alpha})$. Notice that $B^1(\Gamma_L, \mathbb{C}_\alpha)$ is one-dimensional if and only if $\boldsymbol{\alpha}$ is nontrivial. Therefore, we obtain for a non-trivial homomorphism $\boldsymbol{\alpha}: H_1(X_L) \to \mathbb{C}^*$ such that $\dim H^1(\Gamma_L, \mathbb{C}_\alpha) > 0$ if and only if $\Delta_L(\boldsymbol{\alpha}) = 0$ (see also [McM02, Thm. 3.1]).

If we assume that $\dim H^1(\Gamma_L; \mathbb{C}_{\pm}) > 1$ then $\text{rg } J(\boldsymbol{\alpha}) < n - 2$. This implies that the $n-2$ minors of $J(\boldsymbol{\alpha})$ are all zero. On the other hand, the coefficients of the adjugate matrix $J_i^*(\boldsymbol{\alpha})$ are up to sign $n-2$ minors of $J(\boldsymbol{\alpha})$. Hence $J_i^*(\boldsymbol{\alpha})$ is the zero matrix. By Jacobi's formula we obtain for all $t_k$ that

$$
\partial_{t_k} (\det \circ J_i)(\boldsymbol{\alpha}) = \text{tr} \left( J_i^*(\boldsymbol{\alpha}) \partial_{t_k} J_i(\boldsymbol{\alpha}) \right) = 0.
$$

Now observe that by Equation (8) we have

$$
\partial_{t_k} (\det \circ J_i)(\mathbf{t}) = \begin{cases} \Delta_L(\mathbf{t}) + (t_k - 1)\partial_{t_k} \Delta_L(\mathbf{t}) & \text{if } k = \kappa_i, \\ (t_{\kappa_i} - 1)\partial_{t_k} \Delta_L(\mathbf{t}) & \text{if } k \neq \kappa_i. \end{cases}
$$

Since $\Delta_L(\boldsymbol{\alpha}) = 0$, and if we fix $i \in [1, n]$ such that $\alpha_{\kappa_i} \neq 1$, then it follows that $\partial_{t_k} \Delta_L(\boldsymbol{\alpha}) = 0$.
Hence $\text{grad} \Delta_L(\boldsymbol{\alpha}) = 0$. This contradicts the assumption, and $\dim H^1(\Gamma_L; \mathbb{C}_{\pm}) = 1$. $\square$
Remark 2. It follows from the symmetry of the Alexander polynomial (see Equation (9)) that for a given $\alpha \in (\mathbb{C}^*)^\mu$ we have

$$ \Delta_L(\alpha) = 0 \text{ and } \mathrm{grad} \Delta_L(\alpha) \neq 0 \iff \Delta_L(\alpha^{-1}) = 0 \text{ and } \mathrm{grad} \Delta_L(\alpha^{-1}) \neq 0. $$

1.3. **Representations space.** The main references for this part are [AH24] and in [HP23].

For a finitely generated group we let $\mathcal{R}(\Gamma)$ denote the $SL_2(\mathbb{C})$-representation scheme of $\Gamma$.

As observed by A. Weil in [Wei64] for $\rho \in \mathcal{R}(\Gamma)$ the scheme theoretic Zariski tangent space $T_\rho \mathcal{R}(\Gamma)$ is isomorphic to $Z^1(\Gamma; \mathfrak{sl}_2)$. Informally speaking, given a smooth path $\rho_t$ with $\rho_0 = \rho$ one gets a cocycle $D: \Gamma \to \mathfrak{sl}_2^\rho$ by

$$ (11) \qquad D(\gamma) := \frac{d\rho_t(\gamma)}{dt}\bigg|_{t=0} \rho(\gamma)^{-1}, \quad \forall \gamma \in \Gamma_L. $$

This leads to the following inequalities

$$ (12) \qquad \dim_{\rho} \mathcal{R}(\Gamma) \le \dim T_{\rho} \mathcal{R}(\Gamma) = \dim Z^1(\Gamma; \mathfrak{sl}_2^{\rho}). $$

The surjection $\Gamma_L \to H_1(X_L) \cong \mathbb{Z}^\mu$ gives a closed immersion $\mathcal{R}(\mathbb{Z}^\mu) \hookrightarrow \mathcal{R}(\Gamma_L)$, and the image will be denoted by $\mathcal{R}_{ab}(\Gamma_L)$. Notice that $\mathcal{R}_{ab}(\Gamma_L)$ is irreducible and has dimension $\mu + 2$. In what follows we will make use of the following result (see [BC20, Corollary 2.6]).

**Lemma 3.** Let $L \subset S^3$ be a link, and $\alpha \in \widehat{H_1(X_L)}$ be a nontrivial representation. If $\Delta_L(\alpha) \neq 0$, then $\rho_\alpha$ is a smooth point of the representation variety $\mathcal{R}(\Gamma_L)$. In particular, if $\Delta_L \neq 0$, then the abelian representations $\mathcal{R}_{ab}(\Gamma_L)$ form a $\mu + 2$-dimensional component of $\mathcal{R}(\Gamma_L)$.

*Proof.* First notice that $\rho_\alpha \in \mathcal{R}_{ab}(\Gamma_L)$, and $\dim \mathcal{R}_{ab}(\Gamma_L) = \mu + 2$. Hence the local dimension $\dim_{\rho_\alpha} \mathcal{R}(\Gamma_L) \ge \mu + 2$. Next we calculate the dimension of the tangent space $T_{\rho_\alpha} \mathcal{R}(\Gamma_L)$. The Lie algebra $\mathfrak{sl}_2^{\rho_\alpha}$ splits as a direct sum

$$ \mathfrak{sl}_2^{\rho_\alpha} \cong \mathbb{C}_\alpha \oplus \mathbb{C} \oplus \mathbb{C}_{\alpha^{-1}}. $$

Therefore, $\Delta_L(\alpha) \neq 0$ implies that $H^1(\Gamma_L; \mathbb{C}_{\alpha^{\pm 1}}) = 0$, and since $\alpha$ is nontrivial we obtain that $Z^1(\Gamma_L; \mathfrak{sl}_2^{\rho_\alpha})$ has dimension $\mu + 2$. Therefore, Equation (12) implies that

$$ \mu + 2 \le \dim_{\rho_\alpha} \mathcal{R}(\Gamma_L) \le \dim T_{\rho_\alpha}^{\mathrm{Zar}} \mathcal{R}(\Gamma_L) = \dim Z^1(\Gamma_L; \mathfrak{sl}_2^{\rho_\alpha}) = \mu + 2, $$

and $\rho_\alpha$ is a smooth point of $\mathcal{R}(\Gamma_L)$. Next, we have $\rho_\alpha \in \mathcal{R}_{ab}(\Gamma_L)$, and $\dim \mathcal{R}_{ab}(\Gamma_L) = \mu + 2$ implies that $\rho_\alpha$ is a smooth point of $\mathcal{R}_{ab}(\Gamma_L)$. $\square$

Following [Mag88] we call a cocycle $D \in Z^1(\Gamma; \mathfrak{sl}_2^\rho)$ a *curve tangent* if there exists an analytic path of representation $\rho_t$ such that $\rho_0 = \rho$ and Equation (11) is satisfied. A curve tangent is contained in the tangent cone $TC_\rho \mathcal{R}(\Gamma_L)$, and in particular, if each cocycle $D \in Z^1(\Gamma; \mathfrak{sl}_2^\rho)$ is a curve tangent then the tangent cone $TC_\rho \mathcal{R}(\Gamma_L)$ and the tangent space $T_\rho \mathcal{R}(\Gamma_L)$
coincide (see [Leg25, Chap. 3]). In this case $\rho$ is a smooth point of the representation scheme $\mathcal{R}(\Gamma)$.

Using results from [HPSP01, HP05, HM16] on can show that for the group of a link all the cocycles are curve tangents under the condition that the first cohomology group $H^1(\Gamma_L; \mathfrak{s}\mathfrak{l}_2^\rho)$ is $\mu$-dimensional.

**Proposition 4** ([Heu16, Proposition 3.2]). Let $\rho \in \mathcal{R}(\Gamma_L)$ be a representation. We have $\dim H^1(\Gamma_L; \mathfrak{s}\mathfrak{l}_2^\rho) \ge \mu$. If $\dim H^1(\Gamma_L; \mathfrak{s}\mathfrak{l}_2^\rho) = \mu$, then all the cocycles $D \in Z^1(\Gamma_L; \mathfrak{s}\mathfrak{l}_2^\rho)$ are curve tangents and $\rho$ is a smooth point of the representations scheme $\mathcal{R}(\Gamma_L)$. Moreover $\rho$ is contained in a unique algebraic component of dimension $3 + \mu - \dim H^0(\Gamma_L; \mathfrak{s}\mathfrak{l}_2^\rho)$.

## 2. REDUCIBLE NON-ABELIAN REPRESENTATIONS

Recall that for $\alpha \in \widehat{H_1(X_L)}$ the complex numbers $\mathbb{C}$ turn into a $\mathbb{C}[\Gamma_L]$-module via $\alpha = \alpha \circ \varphi$. In what follows this $\mathbb{C}[\Gamma_L]$-module will be denoted by $\mathbb{C}_\alpha$.

Each $\alpha \in \widehat{H_1(X_L)}$ admits a square root i.e. an homomorphism $\beta \in \widehat{H_1(X_L)}$ such that $\beta^2 = \alpha$. A square root is unique up to an homomorphism $\varepsilon: H_1(X_L) \to \{\pm 1\}$. In order to simplify notations we will write $\alpha^{1/2}$ for a fixed square root of $\alpha$.

By fixing a square root $\alpha^{1/2}$ of $\alpha$ we obtain a diagonal representation $\rho_\alpha: \Gamma_L \to \text{SL}_2(\mathbb{C})$ given by

$$ \rho_{\alpha}(\gamma) = \begin{pmatrix} \bar{\gamma}(\alpha^{1/2}) & 0 \\ 0 & \bar{\gamma}(\alpha^{-1/2}) \end{pmatrix}. $$

The study of non-abelian reducible representations $\varrho: \Gamma_L \to \text{SL}_2(\mathbb{C})$ which have the same character as $\rho_\alpha$ goes back to G. Burde and de G. Rham [Bur67, dR67]. A version for links was recently given by L. Bénard and A. Conway [BC20]. For completeness, we state the following result.

**Lemma 5.** Let $\alpha \in \widehat{H_1(X_L)}$ be a nontrivial homomorphism. Then there exists a reducible non-abelian representation $\varrho: \Gamma_L \to \text{SL}_2(\mathbb{C})$ such that $\chi_\varrho = \chi_{\rho_\alpha}$ if and only if $\alpha$ is a zero of the Alexander polynomial.

*Proof.* For each map $d: \Gamma_L \to \mathbb{C}$ we can define $\rho_\alpha^d: \Gamma_L \to \text{SL}_2(\mathbb{C})$ given by

$$ \rho_{\alpha}^{d}(\gamma) = \begin{pmatrix} 1 & d(\gamma) \\ 0 & 1 \end{pmatrix} \rho_{\alpha}(\gamma). $$

Notice that $\rho_\alpha^d$ is a representation if and only if $d \in Z^1(\Gamma_L; \mathbb{C}_\alpha)$ is a derivation, and that $\rho_\alpha^d$ is abelian if and only if $d \in B^1(\Gamma_L; \mathbb{C}_\alpha)$ is principal derivation.

Hence, if $\alpha$ is a zero of the Alexander polynomial then it follows from Proposition 1 that a non-abelian representation $\rho_\alpha^d$ exists.

On the other hand, if $\varrho: \Gamma_L \to \text{SL}_2(\mathbb{C})$ is a reducible non-abelian representation such that $\chi_\varrho = \chi_{\rho_\alpha}$ then $\varrho$ is conjugate to a representation of the form $\rho_{\alpha+}^d$ or $\rho_{\alpha-1}^d$ (see Section 2
in [Boy02]), where $d_±: \Gamma_L \to \mathbb{C}_{\alpha \pm 1}$ are non-principal derivations. Hence it follows from Proposition 1, and the symmetry of the Alexander polynomial that $\alpha$ is a zero of the Alexander polynomial. □

Remark 6. Notice that the diagonal representation $\rho_\alpha$ is contained in the closure of the orbit of the upper triangular representation $\rho_\alpha^d$. More precisely we have for all $\gamma \in \Gamma_L$ that

$$ \lim_{t \to 0} \begin{pmatrix} t & 0 \\ 0 & t^{-1} \end{pmatrix} \rho_{\alpha}^{d}(\gamma) \begin{pmatrix} t^{-1} & 0 \\ 0 & t \end{pmatrix} = \rho_{\alpha}(\gamma). $$

### 3. COHOMOLOGICAL COMPUTATIONS

Throughout this section we consider a link $L$ with nontrivial multi-variable Alexander polynomial $\Delta_L$, and a non-trivial homomorphism $\alpha \in \widehat{H_1(X_L)}$. Moreover, we will suppose that $\alpha$ is a smooth point of the vanishing locus $V(\Delta_L)$. In other words, we suppose that $\alpha$ is a zero of the Alexander polynomial and $\mathrm{grad} \, \Delta(\alpha) \neq 0$.

By Proposition 1 and Remark 2 we have

$$ \dim H^1(\Gamma_L; \mathbb{C}_\alpha) = \dim H^1(\Gamma_L; \mathbb{C}_{\alpha-1}) = 1. $$

Let $d \in Z^1(\Gamma_L; \mathbb{C}_\alpha)$ be a non-principal cocycle and $\rho_+ = \rho_\alpha^d$. The aim of this section is to prove that under the conditions above we have $\dim H^1(\Gamma_L; \mathfrak{sl}_2^{\rho_+}) = \mu$. By Proposition 4, it follows that $\rho_+$ is a scheme smooth representation in $\mathcal{R}(\Gamma_L)$.

#### 3.1. Exact sequences. We use exact sequences in cohomology to make some computations.

The link group $\Gamma_L$ acts on $\mathfrak{sl}_2(\mathbb{C})$ via $\mathrm{Ad}\,\rho_+$. In what follow this $\Gamma_L$-module will be denoted by $\mathfrak{sl}_2^{\rho_+}$. The subspace of upper triangular matrices $\mathfrak{b}$ is invariant under this action, and the corresponding $\Gamma_L$-module will be denoted by $\mathfrak{b}^{\rho_+} \subset \mathfrak{sl}_2^{\rho_+}$. Notice that the quotient $\mathfrak{sl}_2^{\rho_+}/\mathfrak{b}^{\rho_+}$ is isomorphic to $\mathbb{C}_- := \mathbb{C}_{\alpha-1}$. Hence we obtain an exact sequence of $\Gamma_L$-modules

$$ (13) \qquad 0 \to \mathfrak{b}^{\rho_+} \to \mathfrak{sl}_2^{\rho_+} \to \mathbb{C}_- \to 0. $$

The matrices of $\mathfrak{b}$ with zero diagonal are also invariant under the $\mathrm{Ad}\,\rho_\alpha^d$ action. This submodule is isomorphic to $\mathbb{C}_+$, and we obtain a second exact sequence of $\Gamma_L$-modules

$$ (14) \qquad 0 \to \mathbb{C}_+ \to \mathfrak{b}^{\rho_+} \to \mathbb{C} \to 0 $$

where $\mathfrak{b}^{\rho_+}/\mathbb{C}_+$ is isomorphic to the trivial $\Gamma_L$-module $\mathbb{C}$.

From the sequence (13) and from the long exact sequence (1) in cohomology we obtain the following exact sequence

$$ (15) \qquad 0 \to H^1(\Gamma_L; \mathfrak{b}^{\rho_+}) \to H^1(\Gamma_L; \mathfrak{sl}_2^{\rho_+}) \to H^1(\Gamma_L; \mathbb{C}_-) \xrightarrow{\nabla_1} H^2(\Gamma_L; \mathfrak{b}^{\rho_+}) \to \dots $$

Notice that $H^0(\Gamma_L; \mathfrak{b}^{\rho_+})$, $H^0(\Gamma_L; \mathbb{C}_-)$ and $H^0(\Gamma_L; \mathfrak{sl}_2^{\rho_+})$ are trivial. By the snake lemma, a direct calculation gives that the connecting homomorphism $\nabla_1: H^1(\Gamma_L; \mathbb{C}_-) \to H^2(\Gamma_L; \mathfrak{b}^{\rho_+})$
is given by

$$ (16) \qquad \nabla_1([d_-])(\gamma_1, \gamma_2) = \begin{bmatrix} \alpha(\gamma_1)^{-1} d(\gamma_1) d_-(\gamma_2) & \alpha(\gamma_1)^{-1} d(\gamma_1)^2 d_-(\gamma_2) \\ 0 & -\alpha(\gamma_1)^{-1} d(\gamma_1) d_-(\gamma_2) \end{bmatrix}. $$

In the same way the short exact sequence (14) gives the following exact sequence in cohomology

$$ (17) \qquad 0 \to H^1(\Gamma_L; \mathfrak{b}^{\rho+}) \to H^1(\Gamma_L; \mathbb{C}) \xrightarrow{\nabla_2} H^2(\Gamma_L; \mathbb{C}_+) \to H^2(\Gamma_L; \mathfrak{b}^{\rho+}) \to H^2(\Gamma_L; \mathbb{C}) \to \dots $$

since $H^0(\Gamma_L; \mathbb{C}) \to H^1(\Gamma_L; \mathbb{C}_+)$ is an isomorphism. The connection homomorphism $\nabla_2$ is given in the following way. If $h: \Gamma_L \to \mathbb{C}$ is a homomorphism then $\nabla_2(h)$ is represented by the 2-cocycle $\nabla(h): \Gamma_L \times \Gamma_L \to \mathbb{C}_+$ given by

$$ \nabla(h)(\gamma_1, \gamma_2) = -2d(\gamma_1)h(\gamma_2). $$

**Proposition 7.** If the connection homomorphism $\nabla_2$ in sequence (17) is nontrivial then the first connection homomorphism $\nabla_1$ is trivial and $\dim H^1(\Gamma_L; \mathfrak{sl}_2^{\rho+}) = \mu$.

*Proof.* First notice that the link complement $X_L$ is aspherical, and hence the cohomology groups of $\Gamma_L$ and $X_L$ are isomorphic. Next recall that $H^0(\Gamma_L; \mathbb{C}_\pm)$ is trivial, and $H^1(\Gamma_L; \mathbb{C}_\pm)$ is one-dimensional. This, together with an Euler characteristic argument, implies that $H^2(\Gamma_L; \mathbb{C}_\pm) \cong \mathbb{C}$. Since $\dim H^1(\Gamma_L; \mathbb{C}) = \mu$, we obtain that $\nabla_2$ is nontrivial if and only if $\dim H^1(\Gamma_L; \mathfrak{b}^{\rho+}) = \mu - 1$.

From the exact sequence (15) we have

$$ \dim H^1(\Gamma_L; \mathfrak{sl}_2^{\rho+}) = \dim H^1(\Gamma_L; \mathfrak{b}^{\rho+}) + \dim \ker \nabla_1 = \mu - 1 + \dim \ker \nabla_1. $$

Now $\dim H^1(\Gamma_L; \mathfrak{sl}_2^{\rho+}) \ge \mu$ implies that $\dim \ker \nabla_1 \ge 1$, and hence $\nabla_1$ is the zero map. It follows that $\dim H^1(\Gamma_L; \mathfrak{sl}_2^{\rho+}) = \mu$. $\square$

## 3.2. Connection homomorphism.
The aim of this section is to find an equivalent condition for the nontrivially of the connection homomorphism $\nabla_1$ in the exact sequence (15). To this end, we fix a non-principal cocycle $d \in Z^1(\Gamma_L; \mathbb{C}_+)$ and we consider the map $\nabla: H^1(\Gamma_L; \mathbb{C}) \to Z^2(\Gamma_L; \mathbb{C}_+)$ given by

$$ \nabla(h): \Gamma_L \times \Gamma_L \to \mathbb{C}_+, (\gamma_1, \gamma_2) \mapsto -2d(\gamma_1)h(\gamma_2). $$

We want to understand under which conditions $\nabla(h)$ is a coboundary. The first step is to write $\nabla$ in terms of the cup product. The linear map $\beta_+$: $\mathbb{C} \otimes \mathbb{C}_+ \to \mathbb{C}_+$, $z_0 \otimes z_+ \mapsto z_0 z_+$ is compatible with the $\Gamma_L$-action and we have that for all $\gamma_1, \gamma_2 \in \Gamma_L$ that

$$ h \stackrel{\beta_+}{\rightleftharpoons} d(\gamma_1, \gamma_2) = h(\gamma_1)(\gamma_1 \cdot d(\gamma_2)) \implies \nabla(h) = -2d \stackrel{\beta_+ \circ \tau}{\rightleftharpoons} h. $$
From Equation (2) we have that the cup product is cohomologous to the opposite of his twisted version

$$0 = [h \overset{\beta_+}{\rightleftharpoons} d] + [d \overset{\beta_+ \circ \tau}{\rightleftharpoons} h] \in H^2(\Gamma_L; \mathbb{C}_+).$$

In what follows we will use the following setup. Each vector $h = (h_1, \dots, h_\mu) \in \mathbb{C}^\mu$ and each $\alpha = (\alpha_1, \dots, \alpha_\mu) \in (\mathbb{C}^*)^\mu$ determine homomorphism $h: \Gamma_L \to (\mathbb{C}, +)$ and $\alpha: \Gamma_L \to \mathbb{C}^*$ respectively. More precisely, for $\gamma \in \Gamma_L$ and $\varphi(\gamma) = t_1^{k_1} \cdots t_\mu^{k_\mu}$ we obtain

$$h(\gamma) = h_1 k_1 + \cdots + h_\mu k_\mu \quad \text{and} \quad \alpha(\gamma) = \bar{\gamma}(\alpha) = \alpha_1^{k_1} \cdots \alpha_\mu^{k_\mu}.$$

Next view $\alpha \in \widehat{H_1(X_L)}$ as an element of $(\mathbb{C}^*)^\mu$ and $h \in T_1(\mathbb{C}^*)^\mu$ as a tangent vector at the trivial representation $1 \in \widehat{H_1(X_L)}$. Then the component wise product $\alpha h \in T_\alpha(\mathbb{C}^*)^\mu$ is a tangent vector at $\alpha$ in $(\mathbb{C}^*)^\mu$. Moreover, the directional derivative of a multivariable differentiable function $f$ along a given vector $v$ at a given point $x$ will be denoted by $D_v f(x)$. With this notation we obtain

**Proposition 8.** Let $h \in \mathbb{C}^\mu$, $\alpha \in (\mathbb{C}^*)^\mu$ and $d \in Z^1(\Gamma_L, \mathbb{C}_\alpha)$ be given. We put $d = (d(S_1), \dots, d(S_n))^t \in \mathbb{C}^\mu$.

Then the normalized 2-cocycle $h \overset{\beta_+}{\rightleftharpoons} d$ is a coboundary if and only if the non-homogeneous linear system

$$J(\alpha)x = D_{\alpha h} J(\alpha)d$$

has a solution $x \in \mathbb{C}^n$.

**Proof.** By Equation (4), the 2-cocycle $h \overset{\beta_+}{\rightleftharpoons} d$ is a coboundary if and only if there exists $x = (x_1, \dots, x_n) \in \mathbb{C}_+^n$ such that for all $j \in [1, n-1]$

$$\sum_{i=1}^{n} \overline{\pi(\partial_{S_i} R_j)}(\alpha) x_i + h \overset{\beta_+}{\rightleftharpoons} d(\pi(\partial_{S_i} R_j), \pi(S_i)) = 0.$$

Now for $\gamma, \gamma_0$ in $\Gamma_L$ we obtain $h \overset{\beta_+}{\rightleftharpoons} d(\gamma, \gamma_0) = h(\gamma) \alpha(\gamma)d(\gamma_0)$ and for $\varphi(\gamma) = \bar{\gamma}(\mathbf{t}) = t_1^{k_1} \cdots t_\mu^{k_\mu}$ we have grad $\bar{\gamma}(\mathbf{t}) = (k_1 t_1^{k_1-1}, \dots, k_\mu t_\mu^{k_\mu-1})$. Therefore,

$$h(\gamma) \alpha(\gamma) = (h_1 k_1 + \cdots + h_\mu k_\mu) \alpha_1^{k_1} \cdots \alpha_\mu^{k_\mu} \\ = (\alpha_1 h_1, \dots, \alpha_\mu h_\mu) \cdot \text{grad} \bar{\gamma}(\alpha) = D_{\alpha h} \bar{\gamma}(\alpha).$$

Hence $h \overset{\beta_+}{\rightleftharpoons} d(\gamma, \gamma_0) = D_{\alpha h} \bar{\gamma}(\alpha) d(\gamma_0)$, and by linear extension we obtain for every $\eta \in \mathbb{C}[\Gamma_L]$ that

$$h \overset{\beta_+}{\rightleftharpoons} d(\eta, \gamma_0) = D_{\alpha h} \bar{\eta}(\alpha) d(\gamma_0).$$

Therefore,

$$\sum_{i=1}^{n} h \overset{\beta_+}{\rightleftharpoons} d(\pi(\partial_{S_i} R_j), \pi(S_i)) = \sum_{i=1}^{n} D_{\alpha h}(\overline{\pi(\partial_{S_i} R_j)})(\alpha)d(\pi(S_i)),$$
and $h \stackrel{\beta_+}{\rightleftharpoons} d$ is a coboundary if and only if there exists $\mathbf{x} \in \mathbb{C}^n$ such that

$$J(\boldsymbol{\alpha})\mathbf{x} + D_{\alpha h}J(\boldsymbol{\alpha})\mathbf{d} = 0,$$

where $\mathbf{d} = (d(S_1), \dots, d(S_n))^t \in \mathbb{C}^n$.

<figure><img src="image_2.png" /></figure>

Next we will show that if the non-homogeneous linear system $J(\boldsymbol{\alpha})\mathbf{x} = D_{\alpha h}J(\boldsymbol{\alpha})\mathbf{d}$ has a solution then the directional derivative $D_{\alpha h}\Delta_L(\boldsymbol{\alpha})$ vanishes. This allows us to prove Theorem 2. First we start with a lemma.

**Lemma 9.** Let $A: \mathbb{C}^\mu \to \mathcal{M}_{n-1}(\mathbb{C})$, $\mathbf{t} \mapsto A(\mathbf{t})$ be a smooth map. Suppose that there is $\mathbf{t}_0 \in \mathbb{C}^\mu$ such that $\det A(\mathbf{t}_0) = 0$ and let $\mathbf{v} \in \ker A(\mathbf{t}_0)$ be a nontrivial vector in the kernel. Then for all $\mathbf{w} \in \mathbb{C}^\mu$ the following assertion holds.

If the linear system $A(\mathbf{t}_0)\mathbf{x} = D_w A(\mathbf{t}_0)\mathbf{v}$ has a solution $\mathbf{x} \in \mathbb{C}^{n-1}$ then

$$D_w \det A(\mathbf{t}_0) = 0.$$

*Proof.* We let $A^*: \mathbb{C}^\mu \to \mathcal{M}_{n-1}(\mathbb{C})$ denote the map $A^* = \mathrm{adj} \circ A$ where $\mathrm{adj}: \mathcal{M}_{n-1}(\mathbb{C}) \to \mathcal{M}_{n-1}(\mathbb{C})$ is the map which sends a matrix to its adjugate.

We have $A^*(\mathbf{t}) A(\mathbf{t}) = \det A(\mathbf{t}) I_{n-1}$. By taking the directional derivative $D_w$ on both sides, and by multiplying with $\mathbf{v}$ we obtain the following relation

$$(18) \qquad D_w A^*(\mathbf{t}_0) A(\mathbf{t}_0) \mathbf{v} + A^*(\mathbf{t}_0) D_w A(\mathbf{t}_0) \mathbf{v} = D_w \det A(\mathbf{t}_0) \mathbf{v}.$$

If there exists $\mathbf{x} \in \mathbb{C}^{n-1}$ such that $A(\mathbf{t}_0)\mathbf{x} = D_w A(\mathbf{t}_0)\mathbf{v}$, then the left hand side of Equation (18) vanishes since $\mathbf{v} \in \ker A(\mathbf{t}_0)$ and $A^*(\mathbf{t}_0) A(\mathbf{t}_0)$ is the zero matrix. Finally, we obtain $D_w \det A(\mathbf{t}_0) = 0$ since $\mathbf{v}$ is nontrivial.

**Corollary 10.** Let $\mathbf{h} \in \mathbb{C}^\mu$, $\boldsymbol{\alpha} \in (\mathbb{C}^*)^\mu$ and $d \in Z^1(\Gamma_L, \mathbb{C}_\alpha)$. We let $\mathbf{d} = (d(S_1), \dots, d(S_n))^t \in \mathbb{C}^n$ denote the column vector given by $d \in Z^1(\Gamma_L; \mathbb{C}_\alpha)$. We assume that $d$ is a non-principal cocycle and that $\boldsymbol{\alpha} \neq \mathbf{1}$.

If the linear system

$$J(\boldsymbol{\alpha})\mathbf{x} = D_{\alpha h}J(\boldsymbol{\alpha})\mathbf{d}$$

has a solution, then the directional derivative $D_{\alpha h}\Delta_L(\boldsymbol{\alpha}) = 0$ vanishes.

*Proof.* The fundamental formula of Fox-calculus gives that

$$J(\mathbf{t})(t_{\kappa_1} - 1, \dots, t_{\kappa_n} - 1)^t = 0$$

From this we obtain directly that

$$J(\boldsymbol{\alpha})\mathbf{b} = 0 \quad \text{and} \quad J(\boldsymbol{\alpha})\mathbf{y} + D_{\alpha h}J(\boldsymbol{\alpha})\mathbf{b} = 0$$

where

$$\mathbf{b} = (\alpha_{\kappa_1} - 1, \dots, \alpha_{\kappa_n} - 1)^t \quad \text{and} \quad \mathbf{y} = (\alpha_{\kappa_1} h_{\kappa_1}, \dots, \alpha_{\kappa_n} h_{\kappa_n})^t.$$
Suppose that there exists $x \in \mathbb{C}^n$ such that $J(\alpha)x = D_{\alpha h}J(\alpha)d$. After changing the order of the generators, we can assume that $\alpha_{\kappa_n} \neq 1$. This allows us to add a principal derivation to $d$ to obtain a non-principal derivation $d'$ with $d'(\pi(S_n)) = 0$ and a solution $x' \in \mathbb{C}^n$ of the system $J(\alpha)x' = D_{\alpha h}J(\alpha)d'$. Now we can add a multiple of $b$ to $x'$ to obtain a solution of the system $J(\alpha)x' = D_{\alpha h}J(\alpha)d'$ such that the last component of $x'$ vanishes. Finally, we obtain a solution $\tilde{x} \in \mathbb{C}^{n-1}$ of the system

$$J_n(\alpha)\tilde{x} = D_{\alpha h}J_n(\alpha)\tilde{d}$$

where $\tilde{x}$ and $\tilde{d}$ are obtained from $x'$ and $d'$ by removing the last component respectively. Notice that $\tilde{d}$ is nontrivial and contained in the kernel of $J_n(\alpha)$. To conclude we can apply Lemma 9, and obtain from Equation (8) that

$$0 = D_{\alpha h} \det J_n(\alpha) = \alpha_{\kappa_n} h_{\kappa_n} \Delta_L(\alpha) + (\alpha_{\kappa_n} - 1) D_{\alpha h} \Delta_L(\alpha).$$

Finally, $\Delta_L(\alpha) = 0$ and $\alpha_{\kappa_n} \neq 1$ imply that $D_{\alpha h} \Delta_L(\alpha) = 0$ □

**Theorem 11.** Let $L$ be a link with nontrivial multi-variable Alexander polynomial, and $\alpha \in \widehat{H_1(X_L)}$ a nontrivial homomorphism. Assume that $\alpha \in V(\Delta_L)$ is a smooth point. Let $\varrho$ be a reducible non-abelian representation such that $\chi_{\varrho} = \chi_{\rho_{\alpha}}$. Then $\varrho$ is a smooth point of the representation scheme and is contained in a unique algebraic component of dimension $\mu + 3$.

*Proof.* We choose $k \in [1, \mu]$ such that $\partial_{t_k} \Delta_L(\alpha) \neq 0$. From the proof of Lemma 5 it follows that $\varrho$ is similar to a representation of the form $\rho_{\alpha}^{d_+}$ or to $\rho_{\alpha^{-1}}^{d_-}$ with $d_{\pm} \in Z^1(\Gamma_L; \mathbb{C}_{\pm})$ a non-principal cocycle.

Suppose that $\varrho = \rho_{\alpha}^{d_+}$. It follows from Corollary 10 that there is no $x \in \mathbb{C}^n$ such that

$$J(\alpha)x = \partial_{t_k} J(\alpha)d_+.$$

From Proposition 8, we obtain that the cocycle $\varphi_k \stackrel{\beta_+}{\rightleftharpoons} d_+$ is not a coboundary. Hence the connection homomorphism $H^1(\Gamma_L; \mathbb{C}) \xrightarrow{\nabla_2} H^2(\Gamma_L; \mathbb{C}_+)$ in the sequence (17) is non-trivial, and Proposition 7 implies $\dim H^1(X; \mathfrak{sl}_2^{\rho_{\alpha}^d}) = \mu$. Finally, Proposition 4 gives the result.

The case $\varrho = \rho_{\alpha^{-1}}^{d_-}$ is similar. □

**Lemma 12.** The $(\mu+3)$-dimensional algebraic component from Theorem 11 contains irreducible representations. In particular, the representation $\varrho$ is limit of irreducible representations.

*Proof.* First observe that if $A_t$ is an analytic path of matrices in $SL_2(\mathbb{C})$ such that $A_0 = \text{diag}(\alpha, \alpha^{-1})$ is diagonal with $\alpha \neq \alpha^{-1}$. Then there exists an analytic path of matrices $P_t$ in $SL_2(\mathbb{C})$ and an analytic path in $\alpha_t$ in $\mathbb{C}$ such that $P_t^{-1}A_tP_t = \text{diag}(\alpha_t, \alpha_t^{-1})$ such that $\alpha_0 = \alpha$, $P_0 = I_2$ and $t \in \mathbb{C}$ is in a neighborhood of 0 (see [Leg25, Chap. 4]).

Let $\rho_t$ be an analytic path of reducible representation such that $\rho_0 = \varrho$. Up to conjugation we can assume that $\varrho$ is an upper triangular representation. Moreover, since $\alpha$ is non-trivial
we can assume, up to conjugation by an analytic path of matrices $P_t$, that there exists a generator $\pi(S_i)$ of $\Gamma_L$ such that $\rho_t(\pi(S_i)) = \text{diag}(\alpha_t^{\kappa_i}, (\alpha_t^{\kappa_i})^{-1})$ is diagonal. Now, all the matrices $\rho_t(S_j)$ share a common stable line which is one of the two stable lines of $\rho_t(\pi(S_i))$. This implies that all matrices $\rho_t(\pi(S_j))$, $j \in [1, n]$, are simultaneous upper triangular or lower triangular. On the other hand, $\varrho$ is upper triangular, and hence $\rho_t$ is upper triangular.

We have $\dim H^1(\Gamma_L; sl_2^{\rho+}) = \mu$, and by Proposition 4 all the cocycle $D \in Z^1(\Gamma_L; sl_2^{\rho+})$ are integrable. Suppose that there are only reducible representations in the component $\mathcal{R}_+$. Then we consider $\rho_t$, an analytic path of representations such that $\rho_0 = \rho_+$ and taking $t$ close enough to 0 we can suppose that $\rho_t$ is in $\mathcal{R}_+$. Now we use the preceding argument to write

$$ \rho_t = P_t \begin{pmatrix} \alpha_t^{1/2} & \alpha_t^{-1/2} d_t \\ 0 & \alpha_t^{-1/2} \end{pmatrix} P_t^{-1}. $$

We can use the chain rule to compute the associated cocycle in $Z^1(\Gamma_L; sl_2^{\rho+})$

$$ D(\gamma) := \left. \frac{d\rho_t(\gamma)}{dt} \right|_{t=0} \rho_+(\gamma^{-1}) = (\gamma - 1) \cdot \left. \frac{dP_t}{dt} \right|_{t=0} + \left. \frac{d}{dt} \begin{pmatrix} \alpha_t^{1/2}(\gamma) & \alpha_t^{-1/2} d_t(\gamma) \\ 0 & \alpha_t^{-1/2}(\gamma) \end{pmatrix} \right|_{t=0}. $$

We obtain the following homology class

$$ [D] = \left[ \left. \frac{d}{dt} \begin{pmatrix} \alpha_t^{1/2} & \alpha_t^{-1/2} d_t \\ 0 & \alpha_t^{-1/2} \end{pmatrix} \right|_{t=0} \right] \in H^1(\Gamma_L; sl_2^{\rho+}). $$

And we deduce an isomorphism

$$ H^1(\Gamma_L; \mathfrak{b}^{\rho+}) \cong H^1(\Gamma_L; sl_2^{\rho+}) $$

which is in contradiction with the exact sequence (15). □

**Corollary 13.** The representations $\rho_{\alpha}^{d+}$ and $\rho_{\alpha^{-1}}^{d-}$ are contained in the same algebraic component of $\mathcal{R}(\Gamma_L)$. This algebraic component will be denoted by $\mathcal{R}_0$.

*Proof.* Let $\mathcal{R}_+$, respectively $\mathcal{R}_-$ denote the components of dimension $\mu + 3$ containing $\rho_{\alpha}^{d+}$ and $\rho_{\alpha^{-1}}^{d-}$ respectively. According to Lemma 12 we can choose a sequence of irreducible representations $\rho_p$ in $\mathcal{R}_+$ which converges to $\rho_{\alpha}^{d+}$. We use a result from S. Boyer in [Boy02, Proposition 2.9] stating that there is a subsequence $(\rho_{p_q})$ and another sequence $(\rho'_{p_q})$ such that for all $q \in \mathbb{N}$, $\rho'_{p_q}$ is similar to $\rho_{p_q}$ and $\rho'_{p_q}$ converges to $\rho_{\alpha^{-1}}^{d-}$. The algebraic components of $\mathcal{R}(\Gamma_L)$ are stable under conjugation [CS83, Proposition 1.1.1]. Because $\rho'_{p_q}$ is similar to $\rho_{p_q}$ we have that $\rho'_{p_q} \in \mathcal{R}_+$ and because $\mathcal{R}_+$ is closed we have $\rho_{\alpha^{-1}}^{d-} \in \mathcal{R}_+$. Thus $\mathcal{R}_- = \mathcal{R}_+$. □

Theorem 11 and Lemma 12 prove Theorem 2.
## 4. TANGENT CONE FOR THE ABELIAN REPRESENTATION

Throughout this section we suppose that $L$ is a link with nontrivial Alexander polynomial, and $\alpha$ is a smooth point of $V(\Delta_L)$. The aim of this section is to determine the tangent cone $TC_{\rho_\alpha}\mathcal{R}(\Gamma_L)$ at $\rho_\alpha$.

### 4.1. The quadratic cone.
The quadratic cone was first introduced by W. Goldman in [Gol84] and subsequently used by W. Goldman and J. Millson [GM88]. It is a second order approximation of the representation variety near $\rho_\alpha$.

We suppose that $\alpha$ is a nontrivial root of the Alexander polynomial and $\mathrm{grad}\,\Delta_L(\alpha) \neq 0$. We choose a pair $(i, k) \in [1, n] \times [1, \mu]$ such that $\alpha_{\kappa_i} \neq 1$ and $\partial_{t_k}\Delta_L(\alpha) \neq 0$. Let $d_\pm \in Z^1(\Gamma_L; \mathbb{C}_\pm)$ be two non-principal cocycles such that $d_\pm(S_i) = 0$ and $b_\pm: \Gamma_L \to \mathbb{C}, \gamma \mapsto (\alpha^{\pm 1}(\gamma) - 1)$ the non trivial coboundary. We denote by $\mathfrak{sl}_2^\alpha$ the $\Gamma_L$-module $\mathfrak{sl}_2^{\rho_\alpha}$.

The Lie bracket $[\cdot, \cdot]: \mathfrak{sl}_2^\alpha \otimes \mathfrak{sl}_2^\alpha \to \mathfrak{sl}_2^\alpha$ is compatible with the action of $\Gamma_L$. This induces a cup product $\sqcup := [\cdot, \cdot]$. The quadratic cone $QC_{\rho_\alpha}\mathcal{R}(\Gamma_L)$ at $\rho_\alpha$ is a subscheme of the tangent space $T_{\rho_\alpha}\mathcal{R}(\Gamma_L)$ given by the ideal generated by the equations

$$0 = [u \sqcup u] \in H^2(\Gamma_L; \mathfrak{sl}_2^\alpha) \quad \text{with} \quad u \in Z^1(\Gamma_L; \mathfrak{sl}_2^\alpha).$$

We have that $\mathfrak{sl}_2^\alpha = \mathbb{C}_+ \oplus \mathbb{C} \oplus \mathbb{C}_-$ is a direct sum of $\Gamma_L$-modules. This induces an isomorphism of $\mathbb{C}$-vector spaces

$$H^k(\Gamma_L; \mathbb{C}_+) \oplus H^k(\Gamma_L; \mathbb{C}) \oplus H^k(\Gamma_L; \mathbb{C}_-) \xrightarrow{\cong} H^k(\Gamma_L; \mathfrak{sl}_2^\alpha)$$

Let $p_\pm: Z^*(\Gamma_L; \mathfrak{sl}_2^\alpha) \to Z^*(\Gamma_L; \mathbb{C}_\pm)$ and $p: Z^*(\Gamma_L; \mathfrak{sl}_2^\alpha) \to Z^*(\Gamma_L; \mathbb{C})$ denote the projection onto each factor. We put $u_\pm := p_\pm(u)$ and $u_0 := p(u)$. With these notations we have the formulas

$$ \begin{align} p_+(u \sqcup u)(\gamma_1, \gamma_2) &= 2u_0(\gamma_1)(\gamma_1 \cdot u_+(\gamma_2)) - 2u_+(\gamma_1)u_0(\gamma_2) \tag{19} \\ p(u \sqcup v)(\gamma_1, \gamma_2) &= u_+(\gamma_1)(\gamma_1 \cdot u_-(\gamma_2)) - u_-(\gamma_1)(\gamma_1 \cdot u_+(\gamma_2)) \nonumber \\ p_-(u \sqcup v)(\gamma_1, \gamma_2) &= 2u_-(\gamma_1)u_0(\gamma_2) - 2u_0(\gamma_1)(\gamma_1 \cdot u_-(\gamma_2)). \nonumber \end{align} $$

We consider the linear maps given by

$$\beta_\pm: \mathbb{C} \otimes \mathbb{C}_\pm \to \mathbb{C}_\pm, \quad z_0 \otimes z_\pm \mapsto z_0z_\pm \quad \text{and} \quad \beta: \mathbb{C}_+ \otimes \mathbb{C}_- \to \mathbb{C}, \quad z_+ \otimes z_- \mapsto z_+z_-.$$

These maps are compatible with the action of $\Gamma_L$ and they induce cup products. We obtain using Equations (2) and (19)

$$[p_+(u \sqcup u)] = 4[u_0 \stackrel{\beta_+}{\rightleftharpoons} u_+], \quad [p(u \sqcup u)] = 2[u_+ \stackrel{\beta}{\rightleftharpoons} u_-] \quad \text{and} \quad [p_-(u \sqcup u)] = -4[u_0 \stackrel{\beta_-}{\rightleftharpoons} u_-].$$

Hence

$$[u \sqcup u] = (4[u_0 \stackrel{\beta_+}{\rightleftharpoons} u_+], 2[u_+ \stackrel{\beta}{\rightleftharpoons} u_-], -4[u_0 \stackrel{\beta_-}{\rightleftharpoons} u_-]).$$
We know that the space of cocycles $Z^1(\Gamma_L; \mathbb{C}_\pm)$ is two-dimensional and a basis is given by $(d_\pm, b_\pm)$. The space of cocycles $Z^1(\Gamma_L; \mathbb{C})$ is $\mu$-dimensional and a basis is given by the family $(\varphi_l)_{1\le l\le\mu}$. Writing $u \in Z^1(\Gamma_L; \mathfrak{sl}_2^\alpha)$ in this basis there is a unique $(x_+, y_+, x_1, \dots, x_\mu, x_-, y_-) \in \mathbb{C}^{\mu+4}$ such that

$$u_{\pm} = x_{\pm}d_{\pm} + y_{\pm}b_{\pm} \quad \text{and} \quad u_0 = \sum_{l=1}^{\mu} x_l\varphi_l.$$

By the previous computation and the usual properties of cup products we obtain

$$(20) \quad [u \sqcup u] = \left( 4x_+ \sum_{l=1}^{\mu} x_l[\varphi_l \stackrel{\beta_+}{\rightleftharpoons} d_+], 2x_+x_-[d_+ \stackrel{\beta}{\rightleftharpoons} d_-], -4x_- \sum_{l=1}^{\mu} x_l[\varphi_l \stackrel{\beta_-}{\rightleftharpoons} d_-] \right).$$

**Proposition 14.** With the notation above, the two-cocycle $d_+ \stackrel{\beta}{\rightleftharpoons} d_-$ is a coboundary and

$$[u \sqcup u] = 4 \sum_{l=1}^{\mu} x_l \frac{\alpha_l \partial_{t_l} \Delta_L(\boldsymbol{\alpha})}{\alpha_k \partial_{t_k} \Delta_L(\boldsymbol{\alpha})} \left( x_+ [\varphi_k \stackrel{\beta_+}{\rightleftharpoons} d_+], 0, -x_- [\varphi_k \stackrel{\beta_-}{\rightleftharpoons} d_-] \right).$$

*Proof.* First we show that the 2-cocycle $d_+ \stackrel{\beta}{\rightleftharpoons} d_- \in Z^2(\Gamma_L; \mathbb{C})$ is a coboundary. Recall the connection homomorphism $\nabla_1: H^1(X_L; \mathbb{C}_-) \to H^2(X_L; \mathfrak{b}^{\rho+})$ in the exact sequence (15) is given by

$$\nabla_1([d_-]) = \begin{bmatrix} d_+ \stackrel{\beta}{\rightleftharpoons} d_- & d_+^2 \stackrel{\beta}{\rightleftharpoons} d_- \\ 0 & -d_+ \stackrel{\beta}{\rightleftharpoons} d_- \end{bmatrix}.$$

We know by Proposition 7 that $\nabla_1([d_-])$ is a coboundary and by the exact sequence (15) the arrow $H^2(\Gamma_L; \mathfrak{b}^{\rho+}) \to H^2(\Gamma_L; \mathbb{C})$ is an isomorphism. Thus $d_+ \stackrel{\beta}{\rightleftharpoons} d_- \in B^2(\Gamma_L; \mathbb{C})$.

We have $\dim H^2(\Gamma_L; \mathbb{C}_+) = 1$ because the Euler characteristic $\chi(X_L) = 0$. Next, we choose $k \in [1, n]$ such that $\partial_{t_k} \Delta_L(\alpha) \ne 0$. By Proposition 10 we have that $[\varphi_k \stackrel{\beta_+}{\rightleftharpoons} d_+]$ is a basis of $H^2(\Gamma_L; \mathbb{C}_+)$. Hence, for $l \in [1, \mu]$ there is a unique $a_l \in \mathbb{C}$ such that

$$[\varphi_l \stackrel{\beta_+}{\rightleftharpoons} d_+] = a_l [\varphi_k \stackrel{\beta_+}{\rightleftharpoons} d_+].$$

Therefore, the cocycle $(\varphi_l - a_l\varphi_k) \stackrel{\beta_+}{\rightleftharpoons} d_+$ is a coboundary, and by Proposition 8 we obtain for $\mathbf{h} = a_l\mathbf{h}_k - \mathbf{h}_l$ that

$$(21) \qquad 0 = D_{\alpha h} \Delta(\alpha) = a_l \alpha_k \partial_{t_k} \Delta(\alpha) - \alpha_l \partial_{t_l} \Delta(\alpha) \implies a_l = \frac{\alpha_l \partial_{t_l} \Delta(\alpha)}{\alpha_k \partial_{t_k} \Delta(\alpha)}.$$

By the symmetry of the Alexander polynomial (9), there exists $c \in \mathbb{C}^*$ such that for all $l \in [1, \mu]$ the equation

$$(22) \qquad \alpha_l^{-2} \partial_{t_l} \Delta_L(\alpha^{-1}) = c \partial_{t_l} \Delta_L(\alpha)$$

holds. Hence $\partial_{t_k} \Delta_L(\alpha) \ne 0$ implies $\partial_{t_k} \Delta_L(\alpha^{-1}) \ne 0$ and $[h_k \stackrel{\beta_-}{\rightleftharpoons} d_-]$ is a basis of $H^2(\Gamma_L; \mathbb{C}_-)$. In the same way as above, there exists for each $l \in [1, \mu]$ a unique $b_l \in \mathbb{C}$ such that

$$[\varphi_l \stackrel{\beta_-}{\rightleftharpoons} d_-] = b_l [\varphi_k \stackrel{\beta_-}{\rightleftharpoons} d_-].$$
The cocycle $(\varphi_l - b_l\varphi_k) \stackrel{\beta_-}{\rightleftharpoons} d_-$ is a coboundary, and as above we obtain

$$b_l = \frac{\alpha_l^{-1} \partial_{t_l} \Delta(\boldsymbol{\alpha}^{-1})}{\alpha_k^{-1} \partial_{t_k} \Delta(\boldsymbol{\alpha}^{-1})}.$$

Finally, Equation (22) implies that for all $\epsilon \in [1, \mu]$ the equation

$$\alpha_l^{-1} \partial_{t_l} \Delta_L(\boldsymbol{\alpha}^{-1}) = c \alpha_l \partial_{t_l} \Delta_L(\boldsymbol{\alpha})$$

holds. This shows that $a_l = b_l$.

□

**Corollary 15.** The quadratic cone $QC_{\rho\alpha}\mathcal{R}(\Gamma_L) \subset T_{\rho\alpha}\mathcal{R}(\Gamma_L)$ is the union of two transversal linear subspaces $A_0$ and $A_{ab}$. More precisely, we have $\dim A_0 = \mu + 3$, $\dim A_{ab} = \mu + 2$ and $\dim A_0 \cap A_{ab} = \mu + 1$. Hence, $QC_{\rho\alpha}\mathcal{R}(\Gamma_L)$ is a reduced subscheme of the tangent space $T_{\rho\alpha}\mathcal{R}(\Gamma_L)$.

*Proof.* The tangent space $T_{\rho\alpha}\mathcal{R}$ is isomorphic to $Z^1(\Gamma_L; \mathfrak{s}\mathfrak{l}_2^\alpha) \cong \mathbb{C}^{\mu+4}$, and it contains the quadratic cone as a subscheme. The coordinate ring of $Z^1(\Gamma_L; \mathfrak{s}\mathfrak{l}_2^\alpha)$ is isomorphic to the polynomial ring $\mathbb{C}[x_+, y_+, x_1, \dots, x_\mu, x_-, y_-]$. By Proposition 14 the ideal of the quadratic cone is given by

$$I = (x_+ \sum_{l=1}^{\mu} a_l x_l, x_- \sum_{l=1}^{\mu} a_l x_l)$$

where $a_l$ is as in Equation (21).

It is not hard to see that $I = I_1 \cap I_2$ is the intersection of the two prime ideals

$$I_1 = (\sum_{l=1}^{\mu} a_l x_l) \quad \text{and} \quad I_2 = (x_+, x_-).$$

Now, $I_1$ is the ideal of a hyperplane $A_0$ since $a_k \neq 0$, and $I_2$ is the ideal of an $(\mu + 2)$-dimensional affine subspace $A_{ab}$, and $I_0 + I_1$ is the ideal of the intersection $A_0 \cap A_{ab}$ which is an $\mu + 1$ dimensional subspace.

Finally, the quadratic cone is reduced if and only if the ideal $I$ is radical. This is the case since prime ideals are radical, and the intersection of radical ideals is radical. □

**Theorem 16.** The tangent cone $TC_{\rho\alpha}\mathcal{R}(\Gamma_L)$ and quadratic cone $QC_{\rho\alpha}\mathcal{R}(\Gamma_L)$ coincide as subschemes of the tangent space $T_{\rho\alpha}\mathcal{R}(\Gamma_L)$. Hence the tangent cone $TC_{\rho\alpha}\mathcal{R}(\Gamma_L)$ is reduced, and consequently the diagonal representation $\rho_\alpha$ is a reduced point of the representation scheme.

*Proof.* We will show that the tangent cone equals the quadratic cone which is reduced by Corollary 15.

The tangent cone $TC_{\rho\alpha}\mathcal{R}_0$ has only components of dimension $\mu + 3$, and is contained in the tangent cone $TC_{\rho\alpha}\mathcal{R}(\Gamma_L)$. On the other hand, $TC_{\rho\alpha}\mathcal{R}(\Gamma_L)$ is contained in $QC_{\rho\alpha}\mathcal{R}(\Gamma_L)$, and therefore $TC_{\rho\alpha}\mathcal{R}(\Gamma_L)$ is contained in $A_0$ since this is the unique $(\mu + 3)$-dimensional component of maximal dimension of $QC_{\rho\alpha}\mathcal{R}(\Gamma_L)$. Hence $TC_{\rho\alpha}\mathcal{R}_0 = A_0$ is an affine subspace
of $T_{\rho_{\alpha}}\mathcal{R}(\Gamma_L)$. Moreover, it is easy to see that $TC_{\rho_{\alpha}}\mathcal{R}_{ab}(\Gamma_L) = A_0$. For more details see [Leg25, Chap. 4]. □

## 5. LOCAL STRUCTURE NEAR $\rho_{\alpha}$

In this section we present a proof of Theorem 3. Let $L$ be a link with nontrivial multi-variable Alexander polynomial, and $\alpha \in \widehat{H_1(X_L)}$ a nontrivial homomorphism which is a smooth point of $V(\Delta_L)$. We put $\alpha = \alpha \circ \varphi$, and we choose a couple $(i, k) \in [1, n] \times [1, \mu]$ such that $\alpha(S_i) = \alpha_{\kappa_i} \neq 1$ and $\partial_{t_k}\Delta(\alpha) \neq 0$. Let $d_{\pm} \in Z^1(\Gamma_L; \mathbb{C}_{\pm})$ be two non-principal cocycles such that $d_{\pm}(S_i) = 0$, we consider the two representations $\rho_{\pm} = \rho_{\alpha}^{d_{\pm}}$ and $\rho_{\pm} = \rho_{\alpha}^{\alpha}$.

**Theorem 17.** Let $\alpha \in \widehat{H_1(X_L)}$ be a nontrivial homomorphism. Assume $\alpha$ is a smooth point of $V(\Delta_L)$. Then, the representation $\rho_{\alpha}$ is contained in exactly two algebraic components one of dimension $\mu + 2$ containing only abelian representations and one of dimension $\mu + 3$ containing irreducible ones. Moreover, $\rho_{\alpha}$ is a smooth point of both components and the two components intersect transversely at $\rho_{\alpha}$.

*Proof.* It was shown in Lemma 3 that $\mathcal{R}_{ab}(\Gamma_L)$ is an algebraic component of $\mathcal{R}(\Gamma_L)$ which contains $\rho_{\alpha}$. Moreover, $\rho_{\alpha}$ is a non-central representation and it follows from [HP23, Section 7] that $\rho_{\alpha}$ is a smooth point of $\mathcal{R}_{ab}(\Gamma_L)$.

Let $\mathcal{R}_0$ be the component of dimension $\mu + 3$ from Corollary 13. By Remark 6 we have that $\rho_{\alpha}$ is contained in $\mathcal{R}_0$. Next we show that $\rho_{\alpha}$ is a smooth point of $\mathcal{R}_0$. The tangent cone at $\rho_{\alpha}$ of the component $\mathcal{R}_0$ is included in the quadratic cone

$$TC_{\rho_{\alpha}}\mathcal{R}_0 \subset TC_{\rho_{\alpha}}\mathcal{R}(\Gamma_L) \subset QC_{\rho_{\alpha}}\mathcal{R}(\Gamma_L).$$

Moreover, by Corollary 15 the quadratic cone is the union of $A_{ab}$ and $A_0$, two affine spaces of dimension $\mu + 2$ and $\mu + 3$, respectively. We have

$$\mu + 3 = \dim \mathcal{R}_0 = \dim_{\rho_{\alpha}} \mathcal{R}_0 = \dim TC_{\rho_{\alpha}}\mathcal{R}_0.$$

By [Mum99, III.§3. Corollary] all the components of $TC_{\rho_{\alpha}}\mathcal{R}_0$ have the same dimension $\mu + 3$. Thus we have $TC_{\rho_{\alpha}}\mathcal{R}_0 \subset A_0$ and because of the dimension we conclude $TC_{\rho_{\alpha}}\mathcal{R}_0 = A_0$. On the other hand, the tangent space $T_{\rho_{\alpha}}\mathcal{R}_0$ is the smallest affine space containing $TC_{\rho_{\alpha}}\mathcal{R}_0$ (see [Mum99, III.§4]) thus the tangent cone $TC_{\rho_{\alpha}}\mathcal{R}_0$ is an affine scheme and hence $\rho_{\alpha}$ is a smooth point of $\mathcal{R}_0$.

We suppose that $W$ is a component of $\mathcal{R}(\Gamma_L)$ containing the representation $\rho_{\alpha}$. If $W$ contains only abelian representations then $W \subset \mathcal{R}_{ab}$, and hence $W = \mathcal{R}_{ab}$.

If $W$ contains a representation $\rho$ such that $H^1(\Gamma_L; sl_2) = 0$ then it follows that $\dim W \ge \mu + 3$ (see [AH24, Lemma 2.3]). If $W$ contains an irreducible representation $\rho$ then there is a sequence of irreducible representations in $W$ converging to $\rho_{\alpha}$. Let $\rho$ irreducible and close enough to $\rho_{\alpha}$. We consider the map $t: \mathcal{R}(\Gamma_L) \to \mathcal{X}(\Gamma_L)$. By [CS83, Proposition
1.5.3] we obtain $dim t(W) = dim W - 3$. Now we use the theorem of the fiber [Sha77, I.I.6.3.Theorem 7] which gives $t^{-1}(\chi_{\rho_{\alpha}}) \ge dim W - dim t(W) = 3$. If $t^{-1}(\chi_{\rho_{\alpha}})$ contains only abelian representations then by a result from Boyer [Boy02, Proposition 2.4] we would have $dim t^{-1}(\chi_{\alpha}) = 2$. Thus there is a reducible non-abelian representation $\rho \in W$ such that $\chi_{\rho_{\alpha}} = \chi_{\rho}$. But $\rho$ is either similar to $\rho_{+}$ or to $\rho_{-}$, which are both contained in $R_0$. Thus $W = R_0$.

If $W$ contains an non-abelian reducible representation there exists a sequence of non-abelian representations $(\rho_p)$ in $W$ converging to $\rho_{\alpha}$. If this sequence contains an irreducible representation then the previous argument applies. Suppose there are only reducible non-abelian representation in the sequence $(\rho_p)$. Then we have

$$ \rho_p = C_p \begin{pmatrix} \beta_p & \beta_p^{-1} d_p \\ 0 & \beta_p^{-1} \end{pmatrix} C_p^{-1} $$

where $C_p \in SL_2(\mathbb{C})$, $\beta_p: \Gamma_L \to \mathbb{C}^*$ is a homomorphism and $d_p \in Z^1(\Gamma_L; \mathbb{C}_{\beta_p^2})$ is a non-principal cocycle. Up to extracting a sub-sequence we can suppose that $\beta_p$ converges to $\alpha^{1/2}$ or to $\alpha^{-1/2}$. Suppose we are in the first case. Then there exists a rank $p_0$ and two indices $i \in [1, n]$ and $k \in [1, \mu]$ such that $\Delta_L(\beta_{p_0}^2) = 0$, $\beta_{p_0}(\pi(S_i)) \neq 1$ and $\partial_{t_k} \Delta_L(\beta_{p_0}) \neq 0$. Then, according to Theorem 2 the point $\rho_{p_0}$ is a smooth point of $\mathcal{R}(\Gamma_L)$ and $\rho_{p_0}$ is in $W$. Then, by Lemma 12 there exist irreducible representations in $W$ and we are back to the previous case. $\square$

## 6. EXAMPLES

In this section we try to illustrate our theorem on several examples.

### 6.1. Whitehead link.
We consider the Whitehead link $L = L_1 \cup L_2$ with the following diagram

<figure><img src="image_3.png" /></figure>

This diagram allow to compute a Wirtinger presentation of the fundamental group

$$ \Gamma_L = \langle S_1, S_2 | S_1 S_2 S_1^{-1} S_2 S_1 S_2^{-1} S_1 S_2 = S_2 S_1 S_2^{-1} S_1 S_2 S_1^{-1} S_2 S_1 \rangle. $$

We set $\varphi \circ \pi(S_1) = t_1$ and $\varphi \circ \pi(S_2) = t_2$. The Jacobian matrix is given by

$$ J(t_1, t_2) = ((t_1 - 1)(t_2 - 1)^2 \quad (t_1 - 1)^2(t_2 - 1)). $$
From this we obtain the Alexander polynomial of the Whitehead link

$$ \Delta_L(t_1, t_2) = (t_1 - 1)(t_2 - 1). $$

The partial derivatives are given by

$$ \partial_{t_1} \Delta_L(t_1, t_2) = (t_2 - 1) \quad \partial_{t_2} \Delta_L(t_1, t_2) = (t_1 - 1). $$

Suppose $\alpha_1 = 1$ and $\alpha_2 \in \mathbb{C}^* \setminus \{1\}$ then

$$ \Delta_L(1, \alpha_2) = 0 \quad \text{and} \quad \partial_{t_1} \Delta_L(1, \alpha_2) \neq 0. $$

Then we can use Theorem 2, and obtain that every non-abelian reducible representation $\rho$ such that $\chi_{\rho}(S_1) = \pm 2$ and $\chi_{\rho}(S_2) = \alpha_2^{1/2} - \alpha_2^{-1/2}$ is a smooth point of $\mathcal{R}(\Gamma_L)$. Moreover a detailed study from Boden and Curtis [BC16, Lemma 2.7] shows that there is only two components in the representation variety. This illustrates Theorem 3.

### 6.2. Torus link.
For $n \ge 1$ let $L_n = \sigma_1^{2n}$ be the torus link with two components.

A presentation of the link group is given by

$$ \pi_1 X_{L_n} = \langle S_1, S_2 \mid (S_1 S_2)^n = (S_2 S_1)^n \rangle. $$

The Alexander polynomial is given by

$$ \Delta_{L_n}(t_1, t_2) = 1 + t_1 t_2 + \dots + (t_1 t_2)^{n-1} = \prod_{k=1}^{n-1} (t_1 t_2 - e^{i \frac{2k\pi}{n}}). $$

The set of $(\alpha_1, \alpha_2) \in \mathbb{C}^{*2}$ such that $\Delta_L(\alpha_1, \alpha_2) = 0$ is the disjoint union of $(n-1)$ hyperbolas given by the equations

$$ \alpha_1 \alpha_2 = e^{i \frac{2\pi k}{n}} \quad 1 \le k \le n-1. $$

The partial derivatives are given by

$$ \partial_{t_1} \Delta_L(t_1, t_2) = t_2 \sum_{l=1}^{n-1} \prod_{\substack{1 \le j \le n-1 \\ j \ne l}} (t_1 t_2 - e^{i \frac{2j\pi}{n}}) \quad \partial_{t_2} \Delta_L(t_1, t_2) = t_1 \sum_{l=1}^{n-1} \prod_{\substack{1 \le j \le n-1 \\ j \ne l}} (t_1 t_2 - e^{i \frac{2j\pi}{n}}). $$

Let $(\alpha_1, \alpha_2) \in \mathbb{C}^{*2}$ and $k \in \{1, \dots, n-1\}$ such that $\alpha_1 \alpha_2 = e^{i \frac{2k\pi}{n}}$ we obtain

$$ \partial_{t_1} \Delta_L(\alpha_1, \alpha_2) = \alpha_2 \prod_{\substack{1 \le j \le n-1 \\ j \ne k}} (e^{i \frac{2k\pi}{n}} - e^{i \frac{2j\pi}{n}}) \ne 0 $$

$$ \partial_t \Delta_L(\alpha_1, \alpha_2) = \alpha_1 \prod_{\substack{1 \le j \le n-1 \\ j \ne k}} (e^{i \frac{2k\pi}{n}} - e^{i \frac{2j\pi}{n}}) \ne 0. $$

Suppose $\rho$ is a reducible non-abelian representation such that $\chi_{\rho} = \chi_{\rho_{\alpha}}$ with $\alpha = (\alpha_1, \alpha_2)$ then $\rho$ is a smooth point of the representation variety $\mathcal{R}(\pi_1 X_{L_n})$.
Suppose $\rho$ is an irreducible representation then we have $\mathbb{C}[\rho(\Gamma_L)] = M_2(\mathbb{C})$ by the Burnside Theorem. We have $\rho(S_1S_2)^n \in Z(\mathbb{C}[\rho(\Gamma_L)])$ thus $\rho(S_1S_2)^n = \pm I_2$. From this equation we have up to conjugation that $\rho(S_1S_2) = \begin{pmatrix} e^{i\frac{k\pi}{n}} & 0 \\ 0 & e^{-i\frac{k\pi}{n}} \end{pmatrix}$ for a $k \in \{0, \dots, 2n-1\}$. The case $\rho(S_1S_2) = \pm I_2$ is not possible otherwise $\rho$ is reducible. Thus we have

$$ \mathrm{tr}(\rho(S_1S_2)) = 2\Re(e^{i\frac{k\pi}{n}}), \quad k \in \{1, n-1\}. $$

Let $z = e^{i\frac{k\pi}{n}}$ for a $k \in \{1, n-1\} \cup \{n-1, 2n-1\}$. Suppose $\rho(S_1) \in \mathrm{SL}_2(\mathbb{C})$. We can choose $\rho(S_2) = \rho(S_1)^{-1} \begin{pmatrix} z & 0 \\ 0 & \bar{z} \end{pmatrix}$ the relation is satisfied and we can choose arbitrary the trace of each matrix.

Hence $\mathcal{R}(\Gamma_{L_n})$ has $n$ components, the component of abelian representations, and $n-1$ components each containing irreducible representations.

**6.3. Borromean rings.** We are interested with the Borromean rings a link with three component $L = L_1 \cup L_2 \cup L_3$. A presentation of the link group is given by

$$ \Gamma_L = \langle S_1, S_2, S_3 \mid [S_1, [S_2, S_3^{-1}]], [S_2, [S_3, S_1^{-1}]] \rangle. $$

We can compute the Jacobian matrix

$$ J(t_1, t_2, t_3) = \begin{pmatrix} 0 & (t_1-1)(t_3-1) & (t_1-1)(t_2-1) \\ (t_3-1)(t_2-1) & 0 & (t_1-1)(t_2-1) \end{pmatrix}. $$

Then the Alexander polynomial is given by

$$ \Delta_L(t_1, t_2, t_3) = (t_1-1)(t_2-1)(t_3-1). $$

And the partial derivatives

$$ \partial_{t_1} \Delta_L(t_1, t_2, t_3) = (t_2-1)(t_3-1) \quad \partial_{t_2} \Delta_L(t_1, t_2, t_3) = (t_1-1)(t_3-1) \quad \partial_{t_3} \Delta_L(t_1, t_2, t_3) = (t_1-1)(t_2-1). $$

Let $\alpha: \Gamma_L \to \mathbb{C}^*$ be a homomorphism such that $\alpha_1 = 1$ and $\alpha_2, \alpha_3 \in \mathbb{C}^* \setminus \{1\}$ then

$$ \Delta_L(1, \alpha_2, \alpha_3) = 0 \text{ and } \partial_{t_1} \Delta_L(1, \alpha_2, \alpha_3) \neq 0. $$

Every non-abelian reducible representation $\rho$ such that $\chi_\rho(S_1) = \pm 2$, $\chi_\rho(S_2) = \alpha_2 - \alpha_2^{-1}$ and $\chi_\rho(S_3) = \alpha_3 - \alpha_3^{-1}$ is a smooth point of $\mathcal{R}(\Gamma_L)$.

Apart the component of abelian representations, there is the component of representations which maps $S_1$ to a central element (see [HP23, Section 9]). This identifies the two components passing through the point $\rho_\alpha$.

Suppose now we have $\alpha_1 = \alpha_2 = 1$ and $\alpha_3 \in \mathbb{C}^* \setminus \{1\}$. In this case our results do not apply since $\Delta(1, 1, \alpha_3) = 0$ and grad $\Delta(1, 1, \alpha_3)$ is trivial.
A detailed investigation (see [Leg25]), by using results from [HP23, Section 9], shows that in this case the representation $\rho_{\alpha}$ is contained in four components of $\mathcal{R}(\Gamma_L)$.

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UNIVERSITÉ CLERMONT AUVERGNE, CNRS, LABORATOIRE DE MATHÉMATIQUES BLAISE PASCAL, F-63000 CLERMONT-FERRAND, FRANCE.

Email address: michael.heusener@uca.fr

UNIVERSITÉ CLERMONT AUVERGNE, CNRS, LABORATOIRE DE MATHÉMATIQUES BLAISE PASCAL, F-63000 CLERMONT-FERRAND, FRANCE.

Email address: michael.heusener@uca.fr