# Tag Info

34

If I understand the question correctly, the search is for a calculation of the asymptotic expansion of Gaussian integrals using concepts and techniques from category theory. Here is one such calculation: Feynman diagrams via graphical calculus (2001) There is a very close connection between the graphical formalism for ribbon categories and Feynman ...

32

The infinitude of the input data is deceptive. For example, an infinite finitely presented group may appear to be infinite input data but it's fully specified by a finite alphabet and a finite set of words in it. The same sort of thing happens here; everything is "finitely presented" in a suitable sense. First, let's recall the statement of Tannaka-Krein ...

19

I believe the following is a counterexample. Let $\mathcal{A}$ and $\mathcal{B}$ be closed symmetric monoidal abelian categories such that the unit object $1\in\mathcal{B}$ is projective and let $F:\mathcal{A}\to\mathcal{B}$ be a non-exact strong symmetric monoidal functor which has a right adjoint $G:\mathcal{B}\to\mathcal{A}$. For instance, if $A$ is a ...

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(Edited to correct mistakes signaled in comments below). I don't know much about the first steps on the theory, Krein and Tannaka. I can just say their works answer a question that seems very natural now, and that I think was natural even then. Since the beginning of the 20th century, representations of groups had been studied, used in many part of ...

17

In a nonabelian setting the correct notion of kernel is given by the kernel pair, and the correct notion of cokernel is given by the cokernel pair. For example, in any category, a morphism $f : a \to b$ is a monomorphism iff its kernel pair exists and is trivial, and dually $f$ is an epimorphism iff its cokernel pair exists and is trivial. By comparison, the ...

17

The paper: When projective does not imply flat, and other homological anomalies, Theory and Applications of Categories, Vol 5, pp. 202-250, 1999, available here by Gaunce Lewis shows that this behaviour is quite common in categories of Mackey functors for compact Lie groups. These categories arise very naturally in equivariant stable homotopy category. ...

15

The validity of string diagram equalities should be viewed as a form of coherence. What does an equality of two string diagrams tell us? Well, given such an equality, we can fix an arbitrary parenthesization and unitization of the input and the output. The associators and unitors are suppressed in a string diagram with the understanding that any two valid ...

15

First, having seen the edited version of Martin's question, let's quickly dispose of the construction of the free symmetric monoidal category generated by a category $C$. Objects are tuples $(x_1, \ldots, x_n)$ of objects of $C$. Morphisms are labeled permutations, where permutations are conveniently visualized as string diagrams, each string being labeled ...

15

If you look at books on (not just Abelian) abstract harmonic analysis, such as Hewitt-Ross, they tell that Tannaka-Krein duality was originally a non-Abelian version for compact topological groups of Pontryagin duality for locally compact Abelian topological groups $G$. The Pontryagin dual $\widehat{G}$ is also (unfortunately, not necessarily locally compact)...

15

Yes, the Walker-Wang model is related to the Crane-Yetter-Kauffman TQFT in the same way the Levin-Wen model is related to the Turaev-Viro TQFT. See, for example, the table on page 14 of the notes from the talk "Premodular TQFTs" found on this page In general, given an $n$-category with the right sort of duality, there is a standard procedure for ...

15

In a certain sense a monoidal version of a slice category is a category of comodules over a cocommutative comonoid object. If $C$ is a cocommutative comonoid object in a monoidal category, then the category of comodules over it $\mathbf{Comod}(C)$ has a monoidal structure with the monoidal product defined in a standard way as a tensor product $\otimes_C$ ...

15

Interesting question ! As far as I know, there are at least two secretly equivalent answers. You somehow already gave the first one: a modular tensor category is the same as a modular functor (though the precise statement is quite subtle, see the beautiful introduction to this paper : https://arxiv.org/abs/1509.06811). A modular functor is, roughly, a ...

15

To start with, I just want to make sure no one gets the impression that the categorical notion of trace was introduced by the paper you linked to; however "semi-famous" it might or might not be, it's only an exposition of material that's been well-known for some time. As to your question, your category $\bf Prop$ is also known as the Lindenbaum-Tarski ...

14

Marco Grandis has done some work on this, and you can extract answers for 1-3 from these papers Finite Sets and Symmetric Simplicial Sets - M Grandis - TAC (pdf) Higher Fundamental Functors for Simplicial Sets - M Grandis - CTGDC (pdf) See also An Alternative Presentation of the Symmetric-Simplicial Category - Eric R. Antokoletz - arxiv (link) Question 1 ...

13

The Drinfeld center $Z(C)$ of a braided category "contains" $C$ and $\bar C$ (the category with opposite braiding) and therefore also $C\boxtimes \bar C$ and you can show that the following is equivalent 1) $C$ is modular 2) $Z(C)$ is equivalent with $C\boxtimes \bar C$. In other words, for a braided category $C$ there is a natural notion of a center $Z(C)... 13 If a dualizing object exists, there is a bijection between isomorphism classes of dualizing objects and isomorphism classes of$\otimes$-invertible objects (i.e. the Picard group), given by tensoring your favorite dualizing object by a$\otimes$-invertible object. So the groupoid of$\ast$-autonomous structures, if nonempty, is equivalent to the groupoid of$...

12

Your example suggests the characterization of those $C$ which are already stabilized: they are exactly the monoidal categories whose monoidal structure is cocartesian. To see this, it suffices to observe that the monoidal structure on $\mathrm{CMon}(C)=\mathrm{Mon}(\mathrm{Mon}(C))$ is always cocartesian, and that cocartesianness makes every object into a ...

12

If $C$ is not assumed to be symmetric, then the answer to questions 1 and 2 is no. Let $p^* \dashv p_* \dashv p^!$ be a fully faithful adjoint triple with $p_* : A \to B$. (For instance, $A= \mathrm{Top}$ and $B=\mathrm{Set}$ with $p_*$ the forgetful functor, $p^*$ the discrete topology, and $p^!$ the indiscrete one.) Then $F = p^* p_*$ is an idempotent ...

12

First of all, a nitpick: the condition "$[D,C] = D^\vee\otimes C$" should be stated more precisely as "the canonical map $D^\vee\otimes C \to [D,C]$ is an isomorphism". Now as you mentioned in a comment, it's well-known that the $\forall C$ version of this condition is equivalent to dualizability of $D$, and indeed the $C=D$ case is already equivalent to ...

12

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 it also means the unit object is simple. Occasionally it also means symmetric. You just have to look at the definition used in each particular paper.

12

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 Ostrik, has appeared. They define a tensor category as follows: Let $k$ be an algebraically closed field, and $C$ a locally finite $k$-linear abelian rigid ...

11

The answer is yes; in fact, this is mentioned in a footnote to the definition of Tannakian category in Milne's corrected version of the Deligne-Milne paper, on page 32. This is proved in Deligne's Catégories tannakiennes, Theorem 1.12.

11

There are a couple of issues: The first thing is that isomorphism classes don't form a category. You may force them into being a category by picking a representative in each isomophrpism class, and then looking at the full subcategory that they span. But this construction depends on the choice of representative: If you change your mind, and pick another ...

11

In general, taking a "center" of a higher category is looking at the endomorphisms of the identity functor. For example, if you think of a monoid as a 1-category, then the endomorphisms of the identity functor are exactly the center of the monoid. Thinking of a tensor category as a 2-category with one object this also gives the Drinfeld center. If you ...

11

An answer only to question 5): Neantro Saavedra was a chilean mathematician who did his Ph.D. at IHES under the direction of Grothendieck - in fact he was the last of Grothendieck's student at IHES (he defended in 1972). I would guess that his motivation was essentially to get his degree, and that the ideas came from Grothendieck. After his thesis he wrote a ...

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This isn't quite an answer to your question either, but Appendix A of my 2003 book Higher Operads, Higher Categories contains something close. First it defines commutative monoids in the style you describe: as a set $A$ together with a function $A^S \to A$ for each finite set $S$, satisfying some axioms. Then it defines symmetric multicategories in the ...

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It depends on what you mean by "the 2-category of monoidal categories" and also what you mean by "complete". The 2-category of monoidal categories and strict monoidal functors is complete as a Cat-enriched category in the sense of enriched category theory, hence also complete as a bicategory. The 2-category of monoidal categories and strong monoidal ...

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One of the simplest examples of a non-spacial category is $\mathrm{End}(\mathrm{Vec}^{\oplus 2})$, the category of $2\times2$ matrices with vector space coefficients. Working over a field $\mathbb k$, this is a multifusion category with four simple objects:  \begin{pmatrix} \mathbb k & 0 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & \mathbb k \\ ...

11

I think that there is indeed no such finite group $G$, whether simple or otherwise. Note first that the representation $5_{1}$ can be assumed to be faithful ( for if $K$ is its kernel, then the group $G/K$ has the same property), so from now on, we assume it faithful. Note next that $Z(G) = 1$, since if $5_{1}$ lies over a linear character $\lambda$ of \$Z(G)...

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