## January 19, 2015

### The Univalent Perspective on Classifying Spaces

#### Posted by Mike Shulman I feel like I should apologize for not being more active at the Cafe recently. I’ve been busy, of course, and also most of my recent blog posts have been going to the HoTT blog, since I felt most of them were of interest only to the HoTT crowd (by which I mean, “people interested enough in HoTT to follow the HoTT blog” — which may of course include many Cafe readers as well). But today’s post, while also inspired by HoTT, is less technical and (I hope) of interest even to “classical” higher category theorists.

In general, a classifying space for bundles of $X$’s is a space $B$ such that maps $Y\to B$ are equivalent to bundles of $X$’s over $Y$. In classical algebraic topology, such spaces are generally constructed as the geometric realization of the nerve of a category of $X$’s, and as such they may be hard to visualize geometrically. However, it’s generally useful to think of $B$ as a space whose points are $X$’s, so that the classifying map $Y\to B$ of a bundle of $X$’s assigns to each $y\in Y$ the corresponding fiber (which is an $X$). For instance, the classifying space $B O$ of vector bundles can be thought of as a space whose points are vector spaces, where the classifying map of vector bundle assigns to each point the fiber over that point (which is a vector space).

In classical algebraic topology, this point of view can’t be taken quite literally, although we can make some use of it by identifying a classifying space with its representable functor. For instance, if we want to define a map $f:B O\to B O$, we’d like to say “a point $v\in B O$ is a vector space, so let’s do blah to it and get another vector space $f(v)\in B O$. We can’t do that, but we can do the next best thing: if blah is something that can be done fiberwise to a vector bundle in a natural way, then since $Hom(Y,B O)$ is naturally equivalent to the collection of vector bundles over $Y$, our blah defines a natural transformation $Hom(-,B O) \to Hom(-,B O)$, and hence a map $f:B O \to B O$ by the Yoneda lemma.

However, in higher category theory and homotopy type theory, we can really take this perspective literally. That is, if by “space” we choose to mean “$\infty$-groupoid” rather than “topological space up to homotopy”, then we can really define the classifying space to be the $\infty$-groupoid of $X$’s, whose points (objects) are $X$’s, whose morphisms are equivalences between $X$’s, and so on. Now, in defining a map such as our $f$, we can actually just give a map from $X$’s to $X$’s, as long as we check that it’s functorial on equivalences — and if we’re working in HoTT, we don’t even have to do the second part, since everything we can write down in HoTT is automatically functorial/natural.

This gives a different perspective on some classifying-space constructions that can be more illuminating than a classical one. Below the fold I’ll discuss some examples that have come to my attention recently.

Posted at 6:25 PM UTC | Permalink | Followups (15)

## January 9, 2015

### The AMS Must Justify Its Support of the NSA

#### Posted by Tom Leinster That’s the title of a letter I’ve just had published in the Notices of the AMS (Feb 2015, out yesterday). Text follows. There’s also a related letter from Daniel Stroock of MIT.

Plus, there’s an article by the NSA’s director of research, Michael Wertheimer. I have a few points to make about that — read on.

Posted at 7:11 AM UTC | Permalink | Followups (20)

## January 5, 2015

### Mathematics and Magic: the de Bruijn Card Trick

#### Posted by Simon Willerton A mathematician hands out a pack of cards to a group of five people. They repeatedly cut the deck and then take a card each. The mathematician tries to use telepathy to divine the cards that the people are holding but unfortunately due to solar disturbances, the mind waves are a bit scrambled and the mathematician has to ask a few questions to help unscramble the images being received. “Who had porridge for breakfast?” “Who is holding a red card?” “Is anyone a Pisces?” “Who has a dog called Stanley?” The answers to these questions are sufficient to allow the mathematician to name the card that each person is holding. The audience applaud wildly.     The first thing to note is that the authors are both respected mathematicians, so it is perhaps not surprising to learn that the mathematics involved is actually non-trivial. In my undergraduate course on Knots and Surfaces I do a few knot and rope tricks to enliven the lectures and to demonstrate some of the ideas in the course, but these are generally sleight-of-hand tricks unlike the tricks in this book which all have some interesting mathematics underlying them.

In this post I want to wear my mathematician’s hat and explain how the above card trick is based on the existence of de Bruijn sequences. Of course, if I were wearing my magician’s hat, I wouldn’t be allowed to reveal how the trick works!

Posted at 7:36 PM UTC | Permalink | Followups (8)

## January 1, 2015

### Braided 2-Groups from Lattices

#### Posted by John Baez I’d like to tell you about a cute connection between lattices and braided monoidal categories.

We’ve been looking at a lot of lattices lately, like the $\mathrm{E}_8$ lattice and the Leech lattice. A lattice is an abelian group under addition, so we can try to categorify it and construct a ‘2-group’ with points in the lattice as objects, but also some morphisms. Today I’ll show you that for a lattice in a vector space with an inner product, there’s a nice 1-parameter family of ways to do this, each of which gives a ‘braided 2-group’. Here the commutative law for addition in our lattice:

$a + b = b + a$

is replaced by an isomorphism:

$a + b \cong b + a$

And this has some fun spinoffs. For example: for any compact simple Lie group $G$, the category of representations $Rep(T)$ of any maximal torus $T \subseteq G$, with its usual tensor product, has a 1-parameter family of braidings that are invariant under the action of the Weyl group.

What is this good for? I don’t know! I hope you can help me out. The best clues I have seem to be lurking here:

Posted at 1:17 AM UTC | Permalink | Followups (14)