14
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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 ...
- 25.5k
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 ...
- 27.6k
13
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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 ...
- 52.2k
12
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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 ...
- 42.2k
11
votes
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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 ...
- 5,192
10
votes
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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 ...
- 52.2k
9
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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 ...
- 64.1k
9
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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 $\mathcal{M}\boxtimes\mathcal{N}$ ...
- 311
8
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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 ...
- 5,596
8
votes
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On the existence of a square root for a 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.
- 7,952
8
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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 ...
- 27.6k
8
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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 $G$. ...
- 1,992
7
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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}(...
- 3,038
7
votes
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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 ...
- 44.9k
7
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.
- 27.6k
7
votes
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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 ...
- 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 ...
- 3,287
6
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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 ...
- 114k
6
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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. ...
- 27.6k
6
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: ...
- 114k
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-...
- 64.1k
6
votes
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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 ...
- 12.5k
6
votes
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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, ...
- 7,952
6
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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 ...
- 27.6k
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 ...
- 114k
5
votes
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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
...
- 1,372
5
votes
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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 ...
- 7,952
5
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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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5
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What is the proof of the compatibility of a braiding with the unitors?
A completely spelled out diagrammatic proof is given in Proposition 1.3.21 of Volume II of Bimonoidal Categories, $E_n$-Monoidal Categories, and Algebraic K-Theory by Niles Johnson and Donald Yau.
5
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Easy example of a non-symmetric braiding of $\operatorname{Rep}(G)$?
Since you mention classification results for $R$-matrices:
For finite abelian groups, there is a bijection between the set of universal $R$-matrices of the group hopf algebra $\mathbb C[G]$, the set ...
- 7,618
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