**Proposition A.1.** The trace satisfies the following identities:

$$1. \operatorname{tr}_a(\lambda f + \mu g) = \lambda \operatorname{tr}_a(f) + \mu \operatorname{tr}_a(g) \text{ for } f, g: \mathbf{a} \to \mathbf{a} \text{ and } \lambda, \mu \in \mathbb{C}$$

$$2. \operatorname{tr}_{\mathbf{a} \otimes \mathbf{b}}(f \otimes g) = \operatorname{tr}_{\mathbf{a}}(f) \operatorname{tr}_{\mathbf{b}}(g) \text{ for } f: \mathbf{a} \to \mathbf{a} \text{ and } g: \mathbf{b} \to \mathbf{b}$$

$$3. \operatorname{tr}_{\mathbf{a}}(g \circ f) = \operatorname{tr}_{\mathbf{b}}(f \circ g) \text{ for } f: \mathbf{a} \to \mathbf{b} \text{ and } g: \mathbf{b} \to \mathbf{a}$$

Each of these properties is straightforward to prove, and they generalize the usual linearity and cyclicity properties of the matrix trace. Further, we define the *dimension* of **a** to be $\dim(\mathbf{a}) = \operatorname{tr}_{\mathbf{a}}(\mathrm{id}_{\mathbf{a}})$. This generalizes the usual notion of the dimension of a representation.

**Proposition A.2.** We have the following identities:

$$\dim(\mathbf{a} \otimes \mathbf{b}) = \dim(\mathbf{a})\dim(\mathbf{b}), \quad \dim(\mathbf{a} \oplus \mathbf{b}) = \dim(\mathbf{a}) + \dim(\mathbf{b}), \quad \dim(\bar{\mathbf{a}}) = \dim(\mathbf{a}). \tag{A.19}$$

### A.2.2 Implications of semisimplicity

Semisimplicity places many non-trivial constraints on a tensor category. We will now explore a few non-trivial consequences of semisimplicity.

**Proposition A.3.** For any simple object **a** in a symmetric tensor category *C*, $\dim(\mathbf{a}) \neq 0$.

*Proof.* Because **a** is simple, $\operatorname{Hom}(\mathbf{a} \to \mathbf{a})$ is one-dimensional. From this it follows that both $\operatorname{Hom}(\mathbf{1} \to \mathbf{a} \otimes \bar{\mathbf{a}})$ and $\operatorname{Hom}(\mathbf{a} \otimes \bar{\mathbf{a}} \to \mathbf{1})$ are also one-dimensional. Using semisimplicity we then find that

$$\mathbf{a} \otimes \bar{\mathbf{a}} = \mathbf{1} \oplus \dots \tag{A.20}$$

where the additional simple objects are not isomorphic to **1**. Since both $\operatorname{Hom}(\mathbf{1} \to \mathbf{a} \otimes \bar{\mathbf{a}})$ and $\operatorname{Hom}(\mathbf{a} \otimes \bar{\mathbf{a}} \to \mathbf{1})$ are one-dimensional, we can deduce that

$$\delta_{\mathbf{a},\bar{\mathbf{a}}} = \lambda \iota_1, \quad \delta^{\mathbf{a},\bar{\mathbf{a}}} = \mu \pi_1 \tag{A.21}$$

for non-zero $\lambda, \mu \in \mathbb{C}$. We can then use this to compute:

$$\dim(\mathbf{a}) = \operatorname{tr}_{\mathbf{a}}(\mathrm{id}_{\mathbf{a}}) = \delta^{\mathbf{a},\bar{\mathbf{a}}} \circ \delta_{\mathbf{a},\bar{\mathbf{a}}} = \lambda \mu(\pi_1 \circ \iota_1) = \lambda \mu \tag{A.22}$$

which is non-zero.

**Proposition A.4.** If $\mathbf{b} \approx \mathbf{a}_1 \oplus \dots \oplus \mathbf{a}_n$ then $\bar{\mathbf{b}} \approx \bar{\mathbf{a}}_1 \oplus \dots \oplus \bar{\mathbf{a}}_n$.

*Proof.* It suffices to prove this for the simpler case $\mathbf{b} \approx \mathbf{a}_1 \oplus \mathbf{a}_2$. In this case we have a pair of morphisms

$$\pi_k : \mathbf{b} \to \mathbf{a}_k, \quad \iota_k : \mathbf{a}_k \to \mathbf{b} \tag{A.23}$$

for $k=1,2$, satisfying the conditions defining the direct sum. Let us now define

$$\bar{\pi}_k = \bar{\pi}_k \begin{tikzpicture}[baseline=(current bounding box.center)] \node (top) at (0,0.5) {$\bar{\mathbf{a}}_k$}; \node (bottom) at (0,-0.5) {$\bar{\mathbf{b}}$}; \draw[->] (top) -- (bottom); \node[above] at (top) {$\iota_k$}; \node[below] at (bottom) {$\pi_k$}; \end{tikzpicture}, \quad \bar{\iota}_k = \bar{\iota}_k \begin{tikzpicture}[baseline=(current bounding box.center)] \node (top) at (0,0.5) {$\mathbf{b}$}; \node (bottom) at (0,-0.5) {$\mathbf{a}_k$}; \draw[->] (top) -- (bottom); \node[above] at (top) {$\bar{\pi}_k$}; \node[below] at (bottom) {$\bar{\iota}_k$}; \end{tikzpicture}. \tag{A.24}$$

It is easy to verify that these maps also satisfy the direct sum conditions, and so $\bar{\mathbf{b}} \approx \bar{\mathbf{a}}_1 \oplus \bar{\mathbf{a}}_2$.
Given any two morphisms $f: \mathbf{a} \to \mathbf{b}$ and $g: \mathbf{b} \to \mathbf{a}$, we can define the bilinear pairing

$$ \langle g, f \rangle = \mathrm{tr}_{\mathbf{a}}(g \circ f) = \mathrm{tr}_{\mathbf{b}}(f \circ g). \qquad (A.25) $$

**Proposition A.5.** The pairing $\langle \cdot, \cdot \rangle$ on $\mathrm{Hom}(\mathbf{a} \to \mathbf{b}) \times \mathrm{Hom}(\mathbf{b} \to \mathbf{a})$ is non-degenerate.

*Proof.* For any simple object **c** which appears in the decomposition of an object **a** we can choose a basis of projection and embedding morphisms $\iota_{\mathbf{a},\mathbf{c}}^i$ and $\pi_{\mathbf{a},\mathbf{c}}^i$ satisfying

$$ \pi_{\mathbf{a},\mathbf{c}}^i \circ \iota_{\mathbf{a},\mathbf{c}}^j = \delta^{ij} \mathrm{id}_{\mathbf{c}}. \qquad (A.26) $$

If **c** also appears in **b** we can likewise choose a basis of morphisms $\iota_{\mathbf{b},\mathbf{c}}^i$ and $\pi_{\mathbf{b},\mathbf{c}}^i$. We then construct a basis of $\mathrm{Hom}(\mathbf{a} \to \mathbf{b})$ comprising of morphisms $P_{\mathbf{c}}^{ij} = \iota_{\mathbf{b},\mathbf{c}}^i \circ \pi_{\mathbf{a},\mathbf{c}}^j$ and likewise a basis of $\mathrm{Hom}(\mathbf{b} \to \mathbf{a})$ of morphisms $Q_{\mathbf{c}}^{ij} = \iota_{\mathbf{a},\mathbf{c}}^i \circ \pi_{\mathbf{b},\mathbf{c}}^j$. Now we compute the pairing

$$ \langle Q_{\mathbf{c}}^{ji}, P_{\mathbf{c}}^{i'j'} \rangle = \mathrm{tr}_{\mathbf{a}}(Q_{\mathbf{c}}^{ji} \circ P_{\mathbf{c}}^{i'j'}) = \delta^{ii'} \mathrm{tr}_{\mathbf{a}}(\iota_{\mathbf{a},\mathbf{c}}^j \circ \pi_{\mathbf{a},\mathbf{c}}^{j'}) = \delta^{ii'} \mathrm{tr}_{\mathbf{c}}(\pi_{\mathbf{a},\mathbf{c}}^{j'} \circ \iota_{\mathbf{a},\mathbf{c}}^{j}) = \delta^{ii'} \delta^{jj'} \mathrm{dim}(\mathbf{c}). \qquad (A.27) $$

Since $\mathrm{dim}(\mathbf{c}) \neq 0$ is always non-zero by proposition A.3, the pairing $\langle \cdot, \cdot \rangle$ is non-degenerate.

### A.2.3 A note on definitions

As in the paper we only consider symmetric tensor categories, we have defined rigidity, trace and dimensions to be as simple as possible in this context. In a more general tensor category, if (A.14) is satisfied by $\mathbf{a}$ and $\bar{\mathbf{a}}$ then we say that $\bar{\mathbf{a}}$ is a left dual to $\mathbf{a}$, and that $\mathbf{a}$ is a right dual to $\bar{\mathbf{a}}$. Unlike the symmetric case, in a general tensor category the left dual of $\mathbf{a}$ (which we can denote $\mathbf{a}^-$) and the right dual of $\mathbf{a}$ (which we can denote $-\mathbf{a}$) do not have to be isomorphic.

A *pivotal* tensor category is a tensor category with a series of isomorphisms $-\mathbf{a} \to \mathbf{a}^-$ satisfying various naturalness conditions. This pivotal structure can be used to define left and right traces; if these traces are always equal we say that the tensor category is *spherical*. Even on symmetric tensor categories there may be multiple inequivalent pivotal structures, and hence inequivalent traces and dimensions. In some applications (such as to 3d TQFTs) distinguishing between different structures may be important.

In a symmetric tensor category, the braiding gives rise to a *canonical* pivotal structure which is automatically spherical. The potential existence of other pivotal structures is not important for our purposes.

## A.3 Further facts about Deligne categories

### A.3.1 Positive tensor categories

A *positive* symmetric tensor category is one for which all objects have non-negative dimension. For any group $G$ the category $\mathrm{Rep}G$ is clearly positive. On the other hand, as we have seen already, Deligne categories are in general not positive. To see a general reason, consider some object $\mathbf{a}$ in a symmetric tensor category $\mathcal{C}$ with dimension $\mathrm{dim}(\mathbf{a}) = \alpha$. We can then compute that the dimension of the antisymmetrization $\Lambda^k(\mathbf{a})$ is

$$ \mathrm{dim}(\Lambda^k(\mathbf{a})) = \frac{\alpha(\alpha - 1) \dots (\alpha - k + 1)}{k!}. \qquad (A.28) $$
If $\alpha$ is not an integer than this quantity will become negative for some sufficiently large values of $k$. So we conclude that all dimensions must be integers in a positive category, which excludes Deligne categories apart from the integer parameter case when they coincide with $\mathrm{Rep} G$.

### A.3.2 Universal properties of $\widetilde{\mathrm{Rep}} U(n)$, $\widetilde{\mathrm{Rep}} O(n)$, $\widetilde{\mathrm{Rep}} Sp(n)$ and $\widetilde{\mathrm{Rep}} S_n$

Irreducible representations of a group can be classified as real, pseudoreal, or complex. This classification generalizes to simple objects in a symmetric tensor category. Since the dual of a simple object is itself simple, either $\mathrm{Hom}(\mathbf{a} \to \bar{\mathbf{a}})$ is zero-dimensional, or there exists an isomorphism $i_{\mathbf{a}} : \mathbf{a} \to \bar{\mathbf{a}}$, unique up to a scale factor. In the latter case, we can define a morphism

$$B^{\mathbf{a},\mathbf{a}} \equiv \delta^{\mathbf{a},\bar{\mathbf{a}}} \circ (\mathrm{id}_{\mathbf{a}} \otimes i_{\mathbf{a}}) \in \mathrm{Hom}(\mathbf{a} \otimes \mathbf{a} \to \mathbf{1}). \quad (A.29)$$

Going the other way, given a $B^{\mathbf{a},\mathbf{a}}$ we can construct $i_{\mathbf{a}} = (B^{\mathbf{a},\mathbf{a}} \otimes \mathrm{id}_{\mathbf{a}}) \circ (\mathrm{id}_{\mathbf{a}} \otimes \delta_{\mathbf{a},\bar{\mathbf{a}}})$, so the two Hom-spaces are isomorphic, in particular $\mathrm{Hom}(\mathbf{a} \otimes \mathbf{a} \to \mathbf{1})$ is one-dimensional. This implies that the constructed morphism is either symmetric or antisymmetric:

$$B^{\mathbf{a},\mathbf{a}} \circ \beta_{\mathbf{a},\mathbf{a}} = \pm B^{\mathbf{a},\mathbf{a}}. \quad (A.30)$$

We hence have the following tripartite classification of simple objects:

1. If $\mathbf{a}$ and $\bar{\mathbf{a}}$ are not isomorphic, that is, if $\mathrm{Hom}(\mathbf{a} \to \bar{\mathbf{a}})$ is trivial, then we say that $\mathbf{a}$ is complex.

2. If $\mathbf{a}$ and $\bar{\mathbf{a}}$ are isomorphic and $B^{\mathbf{a},\mathbf{a}}$ is symmetric then we say that $\mathbf{a}$ is real.

3. If $\mathbf{a}$ and $\bar{\mathbf{a}}$ are isomorphic and $B^{\mathbf{a},\mathbf{a}}$ is pseudoreal, then we say that $\mathbf{a}$ is pseudoreal.

The categories $\widetilde{\mathrm{Rep}} U(n)$, $\widetilde{\mathrm{Rep}} O(n)$ and $\widetilde{\mathrm{Rep}} Sp(n)$ are special in the theory of symmetric tensor categories:

**Theorem A.6 (Deligne [77–79]).** Let $\mathcal{C}$ be a symmetric tensor category and $\mathbf{a} \in \mathcal{C}$ any object with $\mathrm{dim}(\mathbf{a}) = n$:

1. There is a unique (up to isomorphism) functor $F : \widetilde{\mathrm{Rep}} U(n) \to \mathcal{C}$ with $F([+]) = \mathbf{a}$.

2. If there is a symmetric isomorphism $\mathbf{a} \to \bar{\mathbf{a}}$, then there is a unique (up to isomorphism) functor $F : \widetilde{\mathrm{Rep}} O(n) \to \mathcal{C}$ with $F([1]) = \mathbf{a}$.

3. If there is a skew-symmetric isomorphism $\mathbf{a} \to \bar{\mathbf{a}}$, then there is a unique (up to isomorphism) functor $F : \widetilde{\mathrm{Rep}} Sp(n) \to \mathcal{C}$ with $F([1]) = \mathbf{a}$.

If $n \notin \mathbb{Z}$ these functors are all faithful.³⁷

Essentially, this says that out of all the tensor categories with an object $\mathbf{a}$ of dimension $\mathrm{dim}(\mathbf{a}) = n \notin \mathbb{Z}$, the category $\widetilde{\mathrm{Rep}} U(n)$ is the most symmetric, in the sense that it contains the minimal space of morphisms in $\mathrm{Hom}(\mathbf{a}^{\otimes k} \to \mathbf{a}^{\otimes k})$. In a similar vein, $\widetilde{\mathrm{Rep}} O(n)$ is the most symmetric tensor category with a real representation $\mathrm{dim}(\mathbf{a}) = n$ and $\widetilde{\mathrm{Rep}} Sp(n)$ is the most symmetric tensor category with a pseudoreal representation.
The category $\tilde{\mathrm{Rep}} S_n$ also satisfies a universality property; it is the universal category with a commutative Frobenius algebra of dimension $n$ [78]. To explain the precise meaning of this statement is beyond the scope of the appendix, so instead we shall simply note that Frobenius algebras arise naturally as the algebraic structure underlying 2d TQFTs (see [80] for a pedagogic introduction to this topic). While traditionally defined over vector spaces, they can be generalized to any category.<fnref n="45" /> The universality property of $\tilde{\mathrm{Rep}} S_n$ states that for any category $\mathcal{C}$, functors from $\tilde{\mathrm{Rep}} S_n \to \mathcal{C}$ are in one-to-one correspondence with commutative Frobenius algebras in $\mathcal{C}$.

## A.4 Deligne's theorem on classification of tensor categories

We have seen that the representations of any group $G$ are naturally described by a symmetric tensor category $\mathrm{Rep} G$. By taking this abstract approach to representation theory, we then exhibited several families of categories which smoothly interpolate between the representation categories for various groups such as $O(N)$, $U(N)$ and $S_N$ (section 9). By inspection, for non-integer $n$ these Deligne categories contain objects of non-integer and negative dimensions, and so cannot be equivalent to $\mathrm{Rep} G$ for some group $G$.

One wonder if there exist some tensor categorical symmetries which have all objects of integer and positive dimensions, and yet which are inequivalent to any group symmetry. Deligne's theorem on tensor categories says that this is impossible. This theorem describes the conditions under which a symmetric tensor category is equivalent to $\mathrm{Rep} G$. More generally, it classifies all symmetric tensor categories satisfying a certain finiteness condition, but for positive categories the statement is particularly simple:

**Theorem A.7 (Deligne [78, 79]).** Any positive symmetric tensor category $\mathcal{C}$ is equivalent to $\mathrm{Rep} G$ for some group $G$. If $\mathcal{C}$ is a fusion category,<fnref n="46" /> then $\mathcal{C}$ is equivalent to $\mathrm{Rep} G$ for some unique<fnref n="47" /> finite group $G$.

Because any category $\mathrm{Rep} G$ is automatically positive, we see that positivity is both necessary and sufficient for a symmetric tensor category to be of the form $\mathrm{Rep} G$.

We will now describe Deligne's theorem more generally. Given an $\mathbf{a}$ in a tensor category $\mathcal{C}$, we define length($\mathbf{a}$) to be the number of simple objects appearing in $\mathbf{a}$. We say that $\mathcal{C}$ has subexponential growth if for every $\mathbf{a} \in \mathcal{C}$ there exists a $C_{\mathbf{a}}$ such that for all $k > 0$, length($\mathbf{a}^{\otimes k}$) $\le (C_{\mathbf{a}})^k$.

All positive tensor categories satisfy the subexponential growth condition. Indeed, as discussed in section A.3.1, in any positive tensor category the dimensions of all objects must be integers and so:

$$ \mathrm{length}(\mathbf{a}^{\otimes k}) \le \mathrm{dim}(\mathbf{a})^k. \qquad (\mathrm{A.31}) $$

On the other hand, Deligne categories such as $\tilde{\mathrm{Rep}} O(n)$ do not satisfy the subexponential growth condition. For non-integer $n$ the number of linearly independent morphisms

<footnote n="45">⁴⁵See footnote 17 for further discussion.</footnote>

<footnote n="46">⁴⁶A fusion category is a tensor category with a finite number of simple objects.</footnote>

<footnote n="47">⁴⁷There exist isocategorical groups $G_1$ and $G_2$ for which $\mathrm{Rep} G_1$ and $\mathrm{Rep} G_2$ are equivalent as fusion categories, but which have different braidings and so are distinct as symmetric fusion categories [81–83].</footnote>
$\mathbf{n}^{\otimes k} \to \mathbf{n}^{\otimes k}$ grows factorially in $k$, and this in turn implies that the number of simple objects in $\mathbf{n}^{\otimes k}$ also grows factorially.

Let us begin with fusion categories. These always have subexponential growth, as proved in section 9.9 of [13]. Given a finite group $G$, let $z \in G$ be in the centre of $G$ and satisfy $z^2 = 1$. On any irreducible representation of $G$, $z$ acts either as the identity or minus the identity. For any simple object $\mathbf{a} \in \mathrm{Rep} G$ we define

$$ \mathrm{sgn}(\mathbf{a}) = \begin{cases} +1 & \text{if } z \text{ acts as the identity on } \mathbf{a} \\ -1 & \text{if } z \text{ acts as the minus the identity on } \mathbf{a}. \end{cases} \qquad (A.32) $$

We then define $\mathrm{Rep}(G, z)$ to be the category $\mathrm{Rep} G$ but with the modified braiding:

$$ \beta_{\mathbf{a},\mathbf{b}}(\mathbf{a} \otimes \mathbf{b}) = (-1)^{\mathrm{sgn}(\mathbf{a})\mathrm{sgn}(\mathbf{b})} \mathbf{b} \otimes \mathbf{a}. \qquad (A.33) $$

We can think of these as representations where if $\mathrm{sgn}(\mathbf{a}) = -1$ we use Grassmann rather than real variables. Indeed the pair $(G, z)$ can be thought of as a finite analogue of a supergroup [84]. We then have the following result:

**Theorem A.8 (Deligne [78, 79]).** Any symmetric fusion category $\mathcal{C}$ is equivalent to $\mathrm{Rep}(G, z)$ for a unique finite group $G$ and $z \in G$ as above.

Deligne's theorem for arbitrary subexponential symmetric tensor categories is more technical to state. Broadly speaking it shows that such categories are equivalent to representations of supergroups.

## A.5 Unitarity

Usually in quantum field theory we have some notion of unitarity. In order to define this for theories with a more general categorical symmetry we need to introduce some new structure.

As in section 7.8, conjugation $*$ is a anti-linear braided monoidal functor $*: \mathcal{C} \to \mathcal{C}$ which is involutive, $** = \mathrm{id}_\mathcal{C}$, and for which $\mathbf{a}^*$ is dual to $\mathbf{a}$.

Because $\mathbf{a}^*$ is dual to $\mathbf{a}$, there are cap and cocap maps $\delta^{\mathbf{a},\mathbf{a}^*}$ and $\delta_{\mathbf{a}^*,\mathbf{a}}$, and also $\delta^{\mathbf{a}^*,\mathbf{a}}$ and $\delta_{\mathbf{a},\mathbf{a}^*}$, satisfying the zig-zag relations. As in the previous subsections, we can always choose these maps to satisfy

$$ \delta_{\mathbf{a}^*,\mathbf{a}} = \delta_{\mathbf{a},\mathbf{a}^*} \circ \beta_{\mathbf{a}^*,\mathbf{a}}, \quad \delta^{\mathbf{a},\mathbf{a}^*} = \beta_{\mathbf{a}^*,\mathbf{a}} \circ \delta^{\mathbf{a}^*,\mathbf{a}}. \qquad (A.34) $$

**Proposition A.9.** For any simple object $\mathbf{a}$, there are cap and cocap maps $\delta^{\mathbf{a},\mathbf{a}^*}$ and $\delta_{\mathbf{a}^*,\mathbf{a}}$ such that

$$ (\delta^{\mathbf{a},\mathbf{a}^*})^* = \delta^{\mathbf{a}^*,\mathbf{a}}, \quad (\delta_{\mathbf{a}^*,\mathbf{a}})^* = \delta_{\mathbf{a},\mathbf{a}^*}. \qquad (A.35) $$

*Proof.* For simple object $\mathbf{a}$, $\mathrm{Hom}(\mathbf{a} \otimes \mathbf{a}^* \to \mathbf{1})$ and $\mathrm{Hom}(\mathbf{a}^* \otimes \mathbf{a} \to \mathbf{1})$ are one-dimensional and so

$$ (\delta^{\mathbf{a},\mathbf{a}^*})^* = \alpha \delta^{\mathbf{a}^*,\mathbf{a}} \qquad (A.36) $$

for some number $\alpha$. We thus find that

$$ (\delta^{\mathbf{a},\mathbf{a}^*})^{**} = \alpha (\delta^{\mathbf{a},\mathbf{a}^*} \circ \beta_{\mathbf{a},\mathbf{a}^*})^* = |\alpha|^2 \delta^{\mathbf{a},\mathbf{a}^*} \circ \beta_{\mathbf{a},\mathbf{a}^*} \circ \beta_{\mathbf{a}^*,\mathbf{a}} = |\alpha|^2 \delta^{\mathbf{a},\mathbf{a}^*}. \qquad (A.37) $$
Our definition of an adjoint does not make explicit use of the additive structure on $\mathcal{C}$, and indeed we could generalize the notion of an adjoint to any symmetric monoidal category. Semisimplicity will however prove very constraining:

**Proposition B.1.** Given an adjoint $(\mathfrak{g}, \tau)$ and any two objects $\mathbf{a}_1, \mathbf{a}_2 \in \mathcal{C}$,

$$ \tau_{\mathbf{a}_1 \oplus \mathbf{a}_2} = \iota_1 \circ \tau_{\mathbf{a}_1} \circ (\mathrm{id}_{\mathfrak{g}} \otimes \pi_1) + \iota_2 \circ \tau_{\mathbf{a}_2} \circ (\mathrm{id}_{\mathfrak{g}} \otimes \pi_2). \quad (B.4) $$

*Proof.* Let $\pi_i$ and $\iota_i$ be the projection and embedding morphisms from $\mathbf{a}_i$ into $\mathbf{a}_1 \oplus \mathbf{a}_2$. Applying naturality to $\iota_i$, we find that

$$ \tau_{\mathbf{a}_1 \oplus \mathbf{a}_2} \circ (\mathrm{id}_{\mathfrak{g}} \otimes \iota_i) = \iota_i \circ \tau_{\mathbf{a}_i} \implies \tau_{\mathbf{a}_1 \oplus \mathbf{a}_2} \circ (\mathrm{id}_{\mathfrak{g}} \otimes (\iota_i \circ \pi_i)) = \iota_i \circ \tau_{\mathbf{a}_i} \circ (\mathrm{id}_{\mathfrak{g}} \otimes \pi_{a_i}). \quad (B.5) $$

We thus find that

$$ \iota_1 \circ \tau_{\mathbf{a}_1} \circ (\mathrm{id}_{\mathfrak{g}} \otimes \pi_1) + \iota_2 \circ \tau_{\mathbf{a}_2} \circ (\mathrm{id}_{\mathfrak{g}} \otimes \pi_2) = \tau_{\mathbf{a}_1 \oplus \mathbf{a}_2} \circ (\mathrm{id}_{\mathfrak{g}} \otimes (\iota_1 \circ \pi_1 + \iota_2 \circ \pi_2)) = \tau_{\mathbf{a}_1 \oplus \mathbf{a}_2}, \quad (B.6) $$

which is what we set out to prove.

**Proposition B.2.** Any adjoint $(\mathfrak{g}, \tau)$ can be decomposed as $\mathfrak{g} \approx \mathbf{a}_1 \oplus \cdots \oplus \mathbf{a}_n$,

$$ \tau_{\mathbf{b}} = \sum_{i=1}^{n} (\sigma_i)_{\mathbf{b}} \circ (\pi_i \otimes \mathrm{id}_{\mathbf{b}}), \quad (B.7) $$

where each $\mathbf{a}_i$ is a simple object and each $(\mathbf{a}_i, \sigma_i)$ is an adjoint for all $i = 1, \dots, n$.

*Proof.* Using semisimplicity we can always decompose $\mathfrak{g} \approx \mathbf{a}_1 \oplus \cdots \oplus \mathbf{a}_n$, with project and embedding morphisms $\pi_i$ and $\iota_i$. For each $i$ we can then define

$$ (\sigma_i)_{\mathbf{b}} \equiv (\tau_{\mathbf{b}}) \circ (\iota_i \otimes \mathrm{id}_{\mathbf{b}}), \quad \text{so that} \quad \tau_{\mathbf{b}} = \sum_{i=1}^{n} (\sigma_i)_{\mathbf{b}} \circ (\pi_i \otimes \mathrm{id}_{\mathbf{b}}). \quad (B.8) $$

It is straightforward to check that $(\mathbf{a}_i, \sigma_i)$ satisfies naturality, (B.2) and (B.3). To prove non-degeneracy, we note that if $f : \mathbf{a}_i \to \mathbf{a}_i$ satisfies

$$ (\sigma_i)_{\mathbf{b}} \circ (f \otimes \mathrm{id}_{\mathbf{a}}) = 0 \quad (B.9) $$

for all $\mathbf{b} \in \mathcal{C}$, then

$$ \tau_{\mathbf{b}} \circ ((\iota_i \circ f \circ \pi_i) \otimes \mathrm{id}_{\mathbf{a}}) = 0. \quad (B.10) $$

The non-degeneracy of $\tau$ then implies that

$$ \iota_i \circ f \circ \pi_i = 0 \implies f = \pi_i \circ 0 \circ \iota_i = 0, \quad (B.11) $$

and so $(\mathbf{a}_i, \sigma_i)$ is an adjoint.
Taken together, these two propositions allows us to restrict our attention to simple objects. To specify an adjoint $(\mathfrak{g}, \tau)$ we need only specify $\tau_a$ for each simple object $a$, and to find all possible adjoints in a category we can restrict our search to simple adjoints.

We can introduce a partial ordering on the collection of adjoints in a category $\mathcal{C}$, writing $(\mathfrak{g}, \tau) \succeq (\mathfrak{h}, \sigma)$ if there exists a morphism $f : \mathfrak{h} \to \mathfrak{g}$ such that

$$ \sigma_a = \tau_a \circ (f \otimes \mathrm{id}_a) \text{ for all } a \in \mathcal{C}. \qquad (B.12) $$

We shall say that two adjoints are isomorphic, $(\mathfrak{g}, \tau) \approx (\mathfrak{h}, \sigma)$, if both $(\mathfrak{g}, \tau) \succeq (\mathfrak{h}, \sigma)$ and $(\mathfrak{h}, \sigma) \succeq (\mathfrak{g}, \tau)$.

**Proposition B.3.** Given isomorphic adjoints $(\mathfrak{g}, \tau) \approx (\mathfrak{h}, \sigma)$, there exists a unique isomorphism $f : \mathfrak{g} \to \mathfrak{h}$ such that

$$ \tau_a = \sigma_a \circ (f \otimes \mathrm{id}_a), \qquad \sigma_a = \tau_a \circ (f^{-1} \otimes \mathrm{id}_a). \qquad (B.13) $$

*Proof.* Let $f : \mathfrak{g} \to \mathfrak{h}$ and $g : \mathfrak{h} \to \mathfrak{g}$ be morphisms satisfying

$$ \tau_a = \sigma_a \circ (f \otimes \mathrm{id}_a), \qquad \sigma_a = \tau_a \circ (g \otimes \mathrm{id}_a), \qquad (B.14) $$

for all $a \in \mathcal{C}$. We can then compute

$$ \tau_a \circ ((\mathrm{id}_g - g \circ f) \otimes \mathrm{id}_a) = \tau_a - \sigma_a \circ (f \otimes \mathrm{id}_a) = 0. \qquad (B.15) $$

and so the non-degeneracy condition on $(\mathfrak{g}, \tau)$ implies that $g \circ f = \mathrm{id}_g$. An identical argument implies that $f \circ g = \mathrm{id}_h$ and thus that $g = f^{-1}$. As inverses in a category are unique, it follows that $f$ and $g$ are the unique morphisms satisfying (B.14).

Let us now consider the task of finding all adjoints, up to isomorphism, in a category which is *finitely generated*. Recall that, as defined in section 7.7, an object $\mathfrak{g}$ generates a category $\mathcal{C}$ if any simple object in $\mathcal{C}$ appears in $\mathfrak{g}^{\otimes k}$ for sufficiently large $k$. If such an object exists we say that $\mathcal{C}$ is *finitely generated*. The Deligne categories we have considered in this paper are all *finitely generated*, and so are the representation categories of finite group and of semisimple Lie algebras, so this condition is not very restrictive.

**Proposition B.4.** If a continuous category $\mathcal{C}$ is generated by an object $\mathfrak{g}$ and $(\mathfrak{h}, \tau)$ is an adjoint in $\mathcal{C}$, then the morphism

$$ \eta \equiv (\tau_{\mathfrak{g}} \otimes \mathrm{id}_{\bar{\mathfrak{g}}}) \circ (\mathrm{id}_{\mathfrak{h}} \otimes \delta_{\mathfrak{g}, \bar{\mathfrak{g}}}) \qquad (B.16) $$

embeds $\mathfrak{h}$ into $\mathfrak{g} \otimes \bar{\mathfrak{g}}$.

*Proof.* We will prove this by contradiction, assuming that $\eta$ is not an embedding morphism, i.e. has a non-trivial kernel. In this case there must exist a simple object $j$ in $\mathfrak{h}$ (with $\pi_j$ and $\iota_j$ projection and embedding morphisms) such that

$$ \eta \circ \iota_j = 0 \implies \tau_{\mathfrak{g}} \circ (\iota_j \otimes \mathrm{id}_{\mathfrak{g}}) = 0. \qquad (B.17) $$
We can then use (B.2) to show that for any $k$,

$$ \tau_{g^{\otimes k}} \circ (l_j \otimes \mathrm{id}_{g^{\otimes k}}) = 0. \qquad (B.18) $$

Now consider any other simple object $\mathbf{a} \in \mathcal{C}$. Because $\mathbf{g}$ generates $\mathcal{C}$ there exists embedding and projection operators $\iota_{\mathbf{a}} : \mathbf{a} \to \mathbf{g}^{\otimes k}$ and $\pi_{\mathbf{a}} : \mathbf{g}^{\otimes k} \to \mathbf{a}$ for some sufficiently large value of $k$. Applying (B.1) to $\pi_{\mathbf{a}}$ and then (B.18) we find that

$$ \tau_{\mathbf{a}} \circ (\mathrm{id}_{\mathbf{h}} \otimes \pi_{\mathbf{a}}) \circ (l_j \otimes \mathrm{id}_{\mathbf{g}^{\otimes k}}) = \pi_{\mathbf{a}} \circ \tau_{\mathbf{g}^{\otimes k}} \circ (l_j \otimes \mathrm{id}_{\mathbf{g}^{\otimes k}}) \circ (l_j \otimes \mathrm{id}_{\mathbf{g}^{\otimes k}}) = 0. \qquad (B.19) $$

If we now contract the l.h.s. sides of this equation with $\pi_j \otimes \iota_{\mathbf{a}}$ we find that

$$ \tau_{\mathbf{a}} \circ (\mathrm{id}_{\mathbf{h}} \otimes \pi_{\mathbf{a}}) \circ (l_j \otimes \mathrm{id}_{\mathbf{g}^{\otimes k}}) \circ (\pi_j \otimes \iota_{\mathbf{a}}) = \tau_{\mathbf{a}} \circ (l_j \circ \pi_j) \otimes (\pi_{\mathbf{a}} \circ \iota_{\mathbf{a}}) = \tau_{\mathbf{a}} \circ ((l_j \circ \pi_j) \otimes \mathrm{id}_{\mathbf{a}}), \qquad (B.20) $$

and so conclude that for every $\mathbf{a} \in \mathcal{C}$,

$$ \tau_{\mathbf{a}} \circ ((l_j \circ \pi_j) \otimes \mathrm{id}_{\mathbf{a}}) = 0. \qquad (B.21) $$

Because $\iota_j \circ \pi_j$ is a non-zero morphism from $\mathbf{h} \to \mathbf{h}$, we find that non-degeneracy condition for $(\mathbf{h}, \tau)$ has been violated, and so we have a contradiction.

By this proposition, to find all adjoints we may simply go through the list of all simple objects in $\mathbf{g} \otimes \bar{\mathbf{g}}$. As a simple application, $\widetilde{\mathrm{Rep}}\mathcal{O}(n)$ is generated by the object $\mathbf{n}$. Since $\mathbf{n}^{\otimes 2} \approx \mathbf{S} \oplus \mathbf{A} \oplus \mathbf{1}$, to find all of the adjoints in $\widetilde{\mathrm{Rep}}\mathcal{O}(n)$ we can simply check these three possibilities, and find that $\mathbf{A}$ is the only adjoint (for $\mathbf{S}$ and $\mathbf{1}$ the category does not contain morphisms $\tau_{\mathbf{a}}$ with the needed properties).

In a symmetric tensor category $\mathcal{C}$ we will define the maximal adjoint object $(\mathbf{m}, T)$, if it exists, to be an adjoint such that for any other adjoint $(\mathbf{g}, \tau)$, $(\mathbf{m}, T) \succeq (\mathbf{g}, \tau)$. Due to proposition B.3, if a maximal adjoint exists, it is unique up to unique isomorphism.

**Proposition B.5.** In a finitely generated tensor category $\mathcal{C}$ there exists a unique maximal adjoint.

*Proof.* We will begin with the following easy observation. Given any two adjoints $(\mathbf{h}, \tau_1)$ and $(\mathbf{h}, \tau_2)$ where $\mathbf{h}$ is simple, then any linear combination $(\mathbf{h}, \lambda_1 \tau_1 + \lambda_2 \tau_2)$ is also an adjoint so long as $\lambda_1 \tau_1 + \lambda_2 \tau_2 \neq 0$.

Let $\mathbf{g}$ generate $\mathcal{C}$ and let $\mathbf{a}_i$ be a simple object in $\mathbf{g} \otimes \bar{\mathbf{g}} \approx \mathbf{a}_1 \oplus \cdots \oplus \mathbf{a}_n$. We can then define $U_i$ to be the set of morphisms $f : \mathbf{a}_i \to \mathbf{g} \otimes \bar{\mathbf{g}}$ such that

$$ f = (\tau_{\mathbf{g}} \otimes \mathrm{id}_{\bar{\mathbf{g}}}) \circ (\mathrm{id}_{\mathbf{a}_i} \otimes \delta_{\mathbf{g}, \bar{\mathbf{g}}}) \qquad (B.22) $$

for some adjoint $(\mathbf{a}_i, \tau)$. Note that if such an adjoint exists it is unique; for if there were two such adjoints $(\mathbf{a}_i, \tau_1)$ and $(\mathbf{a}_i, \tau_2)$ then the $(\mathbf{a}_i, \tau_1 - \tau_2)$ would be an adjoint but would violate proposition B.4.

Since any linear combinations of adjoints is an adjoint, the set $U_i$ is a vector space. Choose a basis $f_1, \dots, f_{k_i}$ of $U_i$, corresponding to adjoints $(\mathbf{a}_i, \tau_1), \dots, (\mathbf{a}_i, \tau_{k_i})$. We can then define an adjoint $(\mathbf{a}_i^{\oplus k_i}, \sigma_i)$ with

$$ (\sigma_i)_{\mathbf{b}} = \sum_{j=1}^{k_i} \iota_j \circ (\tau_j)_{\mathbf{b}} \circ (\pi_j \otimes \mathrm{id}_{\mathbf{b}}), \qquad (B.23) $$
where $\iota_i$ and $\pi_i$ are the embedding and projection morphisms into $\mathbf{a}_i^{\oplus k_i}$. By construction $(\mathbf{a}_i^{\oplus k_i}, \sigma_i) \succeq (\mathbf{a}_i, \tau)$ for any adjoint $(\mathbf{a}_i, \tau)$.

Repeating this construction for each $\mathbf{a}_i \in \mathbf{g} \otimes \bar{\mathbf{g}}$, we can then define $\mathbf{m} = \bigoplus_{i=1}^n \mathbf{a}_i^{\oplus k_i}$ and

$$T_{\mathbf{b}} = \sum_{i=1}^{n} \iota_{\mathbf{a}_i} \circ (\sigma_i)_{\mathbf{b}} \circ (\pi_{\mathbf{a}_i} \otimes \mathrm{id}_{\mathbf{b}}). \qquad (\mathrm{B.24})$$

It is straightforward to verify that $(\mathbf{m}, T)$ is an adjoint, and, using propositions B.4 and B.2, that for any adjoint $(\mathbf{h}, \tau)$ in $\mathcal{C}$, $(\mathbf{m}, T) \succeq (\mathbf{h}, \tau)$.

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