65

First, there is indeed nothing mathematically very deep in this observation, and I agree that the word "breakthrough" might be exaggerated. But on the other hand lots of very deep ideas look trivial once spelled out explicitly. Moreover being younger than Voevosky I have never been really exposed to the idea that categories were sets of higher dimension (but ...

39

The other answers are quite good and nothing is wrong with them, but lest a wrong impression be given (e.g. by the implicit suggestion that the idea of "categories as sets in the next dimension" in early work on higher categories was "wrong"), I want to add that while Voevodsky is of course correct from a certain point of view, there is another valid point ...

answered Aug 31 '18 at 19:13

Mike Shulman

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27

I realize I'm dropping in long after all the excitement is over. However, I'd like to point out an implicit assumption which is (seemingly) inherent in the original question as well as some of the answers: namely, that "being strict" is a property of $\infty$-groupoids, rather than a structure put on an $\infty$-groupoid.
It is reasonable to suppose we ...

17

The main obstruction to this kind of duality is not so much that not every $C^*$-algebra is a convolution algebra (though, at least if we don't use twisted convolution algebra, there are known obstruction as mentioned in the comment), but rather that the construction that attach a convolution $C^*$-algebra to a groupoid is not 'injective' at all.
This is ...

16

Upon substituting $x=\frac{1}{q}$ we obtain
$$\sum_{n\geq 0}\frac{1}{|\mathrm{GL}_n(\mathbb F_q)|}=\sum_{n\geq 0}\frac{x^{n^2}}{(1-x)(1-x^2)\cdots(1-x^n)}$$
and this evaluates to the product $\prod_{i\geq 1}\frac{1}{(1-x^{5i-4})(1-x^{5i-1})}$ by the first Rogers-Ramanujan identity.
You can also interpret the second Rogers-Ramanujan identity as giving a ...

answered Oct 7 '20 at 1:14

Gjergji Zaimi

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14

The influence of Brandt's groupoid definition on the definition of category by Eilenberg and Mac Lane has been discussed on the category discussion list.
Bill Cockcroft told me in 1964-70 that there was an influence; he had visited Chicago for a year some time earlier. The use of groupoids in algebra was common knowledge in the 1940s, see the 1943 book on ...

14

Basically, you're asking which (weak) homotopy types can be modelled by simplicial sets which carry the structure of a simplicial abelian group.
Every simplicial abelian group splits as a product of Eilenberg MacLane spaces. $$X \simeq \prod_{n \in \mathbb N} K(\pi_i X, n).$$ This is a consequence of the Dold-Kan correspondence, and the fact that every ...

12

The answer is sort of well known. A simplicial set $X$ is the nerve of a groupoid if and only if any $n$-horn has a unique filler for all $n\geq 2$. The horn $\Lambda^k[n]$ is the simplicial subset of $\Delta[n]$ obtained by removing the non-degenerate $n$-simplex of $\Delta[n]$ and it's $k^{\text{th}}$ face. The previous statement means that any map $\...

12

Even if this answer can be seen as an expansion of the last two lines of Simon's answer, it does not really come in the same spirit.
From the point of view of enriched category theory, posets are categories enriched over the boolean algebra $2=\{\bot < \top\}$. Very often this enriched category theory is called $0$-category theory.
$$\text{Pos} = 0-\...

11

There will be no way to do what you'd like inside homotopy type theory, where the only equality visible is propositional equality. You could move to meta-level and provide a meta-theoretic definition of a strict structure which involves judgmental equalities, but that brings a new kind of trouble with it.
Let me put it another way. Suppose you have a ...

answered Oct 8 '13 at 20:52

Andrej Bauer

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11

As already pointed out, Buss and Sims have found an example of a $C^*$-algebra which is not isomorphic to its opposite, and
hence it is not a groupoid $C^*$-algebra. However twisted groupoid $C^*$-algebras are not necessarily self-opposite so, as the authors
point out, nothing so far prevents their example from being realized as a twisted convolution ...

10

Perhaps these slides will be helpful. I'll try to explain what happens in your special case.
Let $M$ be a monoid and let $\mathcal{B} M$ be the topos of right $M$-sets. The points of $\mathcal{B} M$ are the left $M$-sets $P$ that satisfy the following conditions:
$P$ is inhabited.
Given elements $p_0$ and $p_1$ of $P$, there exist an element $p$ of $P$ and ...

10

Everything that's been written so far about the classification of finite groupoids reducing to the classification of finite groups is true but, I think, misleading. In order to actually produce a list of finite groups from a finite groupoid $X$ you need to choose a basepoint in each connected component of $X$. For example, to produce $M_{12}$ from $M_{13}$ ...

answered Mar 27 '15 at 1:46

Qiaochu Yuan

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10

The discussion at the linked nlab page points to a few different accounts of topologies on the fundamental groupoid, but I'm not feeling energetic enough to track them all down.
So let me just consider the simplest thing. Let $X$ be a topological space and consider the following topologically-enriched groupoid, which we'll call $\Pi_1(X)$:
The objects of $\...

9

As you will see this is a very rich area, so my answer can do no more than give a sketch. The basic idea is that a good category of coverings, Cov(X), say, has certain good categorical properties and then one can show that those properties give the existence of a groupoid such that functors from that groupoid, G, to sets form a category equivalent to Cov(X)....

8

Strict $\infty$-groupoids are equivalent to crossed complexes, see paper available here, and the latter are often more convenient to handle, because of their analogies to chain complexes. The classifying space $BC$ of a crossed complex $C$ fibres over a $K(G,1)$ with fibre the classifying space of a chain complex, so up to homotopy a product of Eilenberg ...

8

First a notational issue: you shouldn't write $G$ for the one-object groupoid corresponding to $G$. A much better name for this groupoid is $BG$, or $\text{pt} / G$.
A natural way to write down a groupoid whose groupoid cardinality is $1$ is to write down a groupoid / homotopy quotient $X/G$ where $|X| = |G|$. In turn, a natural class of such quotients are ...

answered Jun 1 '16 at 5:20

Qiaochu Yuan

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8

Yes, but it's more complicated. First let me describe the special case where $\pi_1$ is trivial. If $G$ is such a 3-group, then $X = BG$ is a pointed connected simply connected space. It fits into a fibration
$$B^3 \pi_3 \to X \to B^2 \pi_2$$
which, as it turns out, can be delooped once into a fiber sequence
$$B^3 \pi_3 \to X \to B^2 \pi_2 \to B^4 \pi_3$$
...

answered Jun 10 '17 at 16:17

Qiaochu Yuan

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8

I cannot say what exactly Voevodsky meant but here is a wild guess.
Disclaimer in what follows I use heavily type theoretic notation, so you have trouble understanding feel free to ask in the comments.
In constructive, specifically type theoretic, settings one usually define sets as pairs of the form $(X,=)$ where $(=)$ is an equivalence relation, which in ...

8

This is true.
Since the $2$-category of groupoids is equivalent to the $2$-category of $1$-truncated spaces, the statement follows immediately from 5.5.1.8 and 5.5.6.21 in Lurie's Higher Topos Theory.

answered Jul 25 '19 at 18:24

Piotr Pstrągowski

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8

$\newcommand{\Z}{\mathbf{Z}} \newcommand{\Mod}{\mathrm{Mod}}$Note that if $\Mod^{\geq 0}_\Z$ denotes the category of connective $\mathrm{H}\Z$-module spectra, then by the Dold-Kan correspondence and the Schwede-Shipley theorem, there are equivalences of categories
$$\Mod^{\geq 0}_\Z \simeq \mathrm{Ch}_{\geq 0}(\Z) \simeq \mathrm{Fun}(\Delta^{op},\mathrm{Ab}) ...

8

I'm surprised this has been up for days with nobody telling Atticus the obvious. Let $I$ be the unit interval category with two objects, $0$ and $1$, and one non-identity arrow $0 \to 1$. A natural transformation $F\to G$ between functors $\mathcal C \to \mathcal D$ is the same thing as a functor $\mathcal C \times I \to D$ that restricts to $F$ on $\...

8

Two relevant papers are:
Dupont's Interchange of filtered 2-colimits and finite 2-limits.
Canevali's 2-filtered bicolimits and finite weighted bilimits commute in Cat.
The former proves that finite conical pseudolimits and filtered pseudolimits commute in $\mathbf{Cat}$; whereas the latter proves the analagous result for finite weighted bilimits and ...

7

Actions of a Lie groupoid are defined on p. 34 of K.C.H. Mackenzie, "General Theory of Lie Groupoids and Lie Algebroids" LMS Lecture Notes Series no 213, 2005.
Note that in general for a groupoid action of $G$ on sets it is often convenient to follow C. Ehresmann and to insist on having a function $f: E \to Ob(G)$ so that $g \in G(x,y)$ maps the fibre of ...

7

The holonomy groupoid of a foliation is another example of a useful groupoid
it is described here:
http://www.ams.org/journals/bull/2005-42-01/S0273-0979-04-01036-5/S0273-0979-04-01036-5.pdf
For a singular foliation see http://users.uoa.gr/~iandroul/AS-holgpd-final.pdf

7

I don't think it's a good idea to mix the slogans "topological spaces are locales" and "topological spaces are $\infty$-groupoids." I think the former slogan encapsulates what we ended up defining as topological spaces while the latter slogan encapsulates what we should've defined as topological spaces, at least if we're algebraic topologists (recall that $\...

answered Sep 18 '13 at 23:00

Qiaochu Yuan

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7

Sorry for answering so late, I just was pointed to this question. Before answering your questions in the case of a Lie group (i.e., one object) let me point out that the difference between $\mathbb{R}$-coefficients and $\mathbb{S}^1$-coefficients it not negligible but is somehow the whole heart of the story.
The cohomology of the continuous cochains is ...

7

I am a novice too, but I think that an equivalent perspective (which is the first I met) is far more enlightening than the Kan condition. Maybe it is the right moment to check if I'm right: things can't be so simple! :)
Start with your favourite small category $\bf C$. Denote its set of objects as ${\bf C}_0$.
Now consider the set of arrows $X\to Y$; ...

7

The fundamental groupoid does not preserve homotopy limits. The simplest examples can be obtained from the path-space sequence $\Omega X\to \operatorname{pt}\to X$ which shows that $\Omega X$ is the homotopy limit of $\operatorname{pt}\to X\leftarrow \operatorname{pt}$. If $\pi_1$ preserves homotopy limits, then the fundamental groupoid of $\Omega X$ should ...

answered Jul 12 '14 at 19:22

Matthias Wendt

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