13 votes

What is a tensor category?

There is no single accepted definition of “tensor category” that matches all uses. Almost always it means abelian (or a similar cocomplete condition) and k-linear. Usually it also means rigid. Often ...
  • 26.7k
13 votes
Accepted

What is a tensor category?

There seem to be many different definitions in the literature, based on individual papers. But, I think that might change, now that the textbook Tensor Categories, by Etingof, Gelaki, Nikshych, and ...
  • 22.6k
12 votes
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When is the endofunctor category of a monoidal category braided? When is it ribbon? Fusion? Modular?

The category $Func(C,C)$ is very rarely braided. It's a bit like asking "when is the endomorphism algebra of a vector space commutative?" For example, if $C=Vect\oplus Vect$, then $Func(C,C)$ is ...
11 votes
Accepted

Twists, balances, and ribbons in pivotal braided tensor categories

Question 2: Given a pivotal braided category $\mathcal{C}$, there are 2 ways to endow $\mathcal{C}$ with twists under which $\mathcal{C}$ is a rigid balanced category. Conversely, given a rigid ...
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11 votes
Accepted

Is the category $\operatorname{sVect}$ an "algebraic closure" of $\operatorname{Vect}$?

$\newcommand\sVec{\mathrm{sVec}}\newcommand\Vec{\mathrm{Vec}}$Yes. Over an algebraically closed field of characteristic $0$, $\sVec$ is the algebraic closure of $\Vec$. By "algebraic closure" of $K$ I ...
10 votes
Accepted

The tensor product of two monoidal categories

The book Tensor Categories discusses, with many variations, the details of Robert McRae's answer. Just like for vector spaces, there are a number of related but inequivalent "tensor products" of ...
9 votes
Accepted

Why does a tetracategory with one object, one 1-morphism and one 2-morphism give a symmetric monoidal category

It will come from a compatibility between different ways of composing interchangers. (I'm going to use = to mean iso/homotopy in a HoTT-like way throughout, for ease of notation. I will also confuse ...
  • 26.7k
9 votes

The tensor product of two monoidal categories

If your categories are locally finite abelian, I think you are looking for the Deligne tensor product of $\mathcal{M}$ and $\mathcal{N}$. The Deligne tensor product $\mathcal{M}\boxtimes\mathcal{N}$ ...
8 votes

Why does a tetracategory with one object, one 1-morphism and one 2-morphism give a symmetric monoidal category

The most easily referenced definition of a tetracategory - due to Todd Trimble - is in this paper by a former student of mine: Alex Hoffnung, Spans in 2-categories: a monoidal tricategory. ...
  • 19.8k
8 votes

The dual of a dual in a rigid tensor category

TL;DR: $X\cong (X^*)^*$ is not necessarily true, and this seems to be a folklore result, mentioned e.g. in these notes by Müger on p.9 (found by Eduardo Pareja Tobes). However, finding explicit ...
  • 4,271
8 votes
Accepted

The dual of a dual in a rigid tensor category

As Tobias said in his answer, a good place to look for examples is in endofunctor categories with composition as the monoidal product, where duals are adjoints. But another way to get a rigid ...
  • 60.3k
8 votes
Accepted

Under what conditions is a symmetric tensor category equivalent to $\operatorname{\mathsf{Rep}}G$ for some group $G$?

If you are in characteristic zero and the dimension of every object is a positive integer then it admits a fiber functor to $Vec$ and is equivalent to $Rep(G)$ for some (pro-algebraic) group $G$. ...
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7 votes

When modular tensor categories are equivalent?

A tensor category includes the information of a tensor product, which is something that takes objects and returns objects. This means that a tensor functor can't just "preserve tensor product" it ...
  • 26.7k
7 votes
Accepted

Do we have a braided tensor category for vertex algebra modules by using conformal blocks on an arbitary compact Riemann Surface?

In general, you won't get a vertex tensor category, because you don't get well-defined unit behavior when you use conformal blocks on higher genus surfaces. Huang-Lepowsky assume the vertex operator ...
  • 43.3k
7 votes
Accepted

Does the functor $\mathcal{C} \to \mathcal{Z}(\mathcal{C})$ have adjoints?

Short answer: Yes, it can possibly have an adjoint. Longer answer: Assume that $\mathcal{C}$ is rigid, and that the coend $L = \int^{X \in \mathcal{C}} X^* \otimes X$ exists. It is a coalgebra. Your ...
  • 338
7 votes
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Motivating quantum groups from knot invariants

Let $\mathcal{C}$ be the category of finite-dimensional representations of a semisimple Lie algebra $\mathfrak{g}$ and $\mathcal{C}[\![\hbar]\!]$ the ribbon category you mention (which depends on the ...
6 votes
Accepted

On the existence of a square root for a unitary modular tensor category

A characterization of Drinfeld centers of fusion categories is given in this paper as braided fusion categories containing a so-called Lagrangian algebra.
  • 7,517
6 votes

When is the endofunctor category of a monoidal category braided? When is it ribbon? Fusion? Modular?

This is in some sense an unnatural question to ask. Endofunctors only form a monoidal category in general, and if you want a braiding, that's not just an extra property: it's extra structure. Where ...
6 votes
Accepted

Braided Hopf algebras and Quantum Field Theories

Some particular braided Hopf algebras known as Nichols algebras are useful in conformal field theories. Here you have some references: Semikhatov, A. M.; Tipunin, I. Yu. Logarithmic $\widehat{s\ell}(...
6 votes

What is a true invariant of $G$-crossed braided fusion categories?

Modular tensor categories give (via Reshetikhin-Turaev) a 321 oriented TFT. This gives a huge source of invariants, in particular any closed oriented 3-manifold gives a numerical invariant of MTCs. ...
  • 26.7k
6 votes
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Nonbraided rigid monoidal category where left and right duals coincide

The simplest example is G-graded vector spaces where G is a non-abelian group.
  • 26.7k
6 votes
Accepted

Ordered logic is the internal language of which class of categories?

Yes, ordered logic is the internal language of non-symmetric monoidal categories. As with linear and nonlinear logic, if the ordered logic contains function-types then they correspond to internal-...
  • 60.3k
6 votes
Accepted

Drinfeld center of $\mathrm{Mod}_R$

Let $(X,\Phi)$ be an object of the Drinfeld center. We'd like to prove that $(X,\Phi)$ is isomorphic to $(X,$ standard symmetry isomorphism $)$, which would prove the equivalence as you say we already ...
  • 9,655
6 votes
Accepted

Constructing the inverse of a braiding in a braided pivotal category

As Tim says this should perhaps be an answer: yes this is true, and doesn't require pivotal. As long as the category is right rigid, the braiding is automatically invertible. See e.g. Prop 1.3 in Day, ...
  • 7,517
6 votes

Deligne Tensor Product of Categories, Explicit Equivalence of $A\otimes_\mathbb{C} B\text{-Mod} \cong A\text{-Mod}\boxtimes B\text{-Mod}$

The explicit definition you give is not the Deligne tensor product, it’s what’s often called the “naive” tensor product. The naive tensor product typically won’t be abelian. The Deligne tensor ...
  • 26.7k
5 votes
Accepted

Cyclic structure on a balanced (or ribbon) monoidal category

In https://arxiv.org/abs/2010.10229 we characterize cyclic framed little 2-disks algebras in any symmetric monoidal bicategory. In the case that this symmetric monoidal bicategory is given by finite ...
  • 1,342
5 votes
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Balanced monoidal and homotopy symmetric

No, a balanced monoidal structure is something different. One way to think about these things is in terms of the relevant operads. Braided monoidal means an algebra over the $E_2$ operad, while ...
5 votes

Is there a non-degenerate quadratic form on every finite abelian group?

Yes. It's necessary and sufficient to show that every finite abelian group admits a nondegenerate quadratic form valued in a finite cyclic group. The following slightly stronger statement is true: ...
5 votes
Accepted

Connection between braided tensor categories and local systems on moduli of stable marked genus zero curves

Welcome to MO. One place where this idea is somewhat explained is Bezrukavnikov-Finkelberg- Schechtman, Factorizable sheaves and quantum groups (https://arxiv.org/abs/q-alg/9712001) but maybe this is ...
  • 7,517
5 votes
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What is the proof of the compatibility of a braiding with the unitors?

Using the notation of Joyal and Street §2, here’s a proof of $\newcommand{\r}{\rho}\newcommand{\l}{\lambda}\newcommand{\x}{\otimes}\newcommand{\comp}{\!\!\cdot\!}\r_A = \l_A \comp c_{A,I}$. Since $\l$...

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