# Tag Info

## Hot answers tagged braided-tensor-categories

### 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 ...
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### 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 ...
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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 ...
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### 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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### 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 ...
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### 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 ...
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### 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 ...
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### 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 $\mathcal{M}\boxtimes\mathcal{N}$ ...
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### 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. ...
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### 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 ...
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### 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 ...
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### 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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### 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 ...
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### 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 ...
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### 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 ...
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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 ...
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### 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.
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