The n-Category Café

OK I have a question I didn’t see in the notes. Where does the loop space story (that’s supposed to rotate these clocks) come in, in this picture?

Do you really need the *-rep stuff? If you skip it, allowing non-*-reps, I’m sure your fibers will be different – but will they be different homotopically?

Allen wrote:

What are the ways to make a universal enveloping algebra $U(\mathfrak{g})$ a $\ast$-algebra?

The most reasonable way to make the universal enveloping algebra $U(\mathfrak{g})$ of the Lie algebra $\mathfrak{g}$ of a compact Lie group $G$ into a $\ast$-algebra is to make all the elements of $\mathfrak{g}$ skew-adjoint. Then a finite-dimensional unitary representation of $G$ gives a finite-dimensional complex $\ast$-representation of $U(\mathfrak{g})$ and vice versa.

Where does the loop space story (that’s supposed to rotate these clocks) come in, in this picture?

Good question — it’s been cleverly absorbed into the algebra! If you read Milnor’s book Morse Theory or my distillation here you’ll see he approximates the $k$-fold loop space of $\mathrm{O}(n)$ by a space $\Omega_k(n)$, which is a certain space geodesics in a certain space of geodesics in a certain space of geodesics… in $\mathrm{O}(n)$. And he cleverly sets up these spaces of $k$-fold iterated geodesics in such a way that you can describe them using $\ast$-representations of $\mathrm{Cliff}_k$ on the real Hilbert space $\mathbb{R}^n$.

How does this work? Milnor explains it with his usual clarity, but the core fact is that a unit-speed geodesic loop of length $2\pi$ in $\mathrm{O}(n)$ exactly corresponds to what Milnor and I call a complex structure on $\mathbb{R}^n$, namely a linear operator $J : \mathbb{R}^n \to \mathbb{R}^n$ with

$$J^2 = -1, \quad J^\dagger = J^{-1}$$

The first equation is what most people mean when they say ‘complex structure’, but Milnor and I want to restrict attention to orthogonal operators $J$, which the second equation accomplishes. These two equations are necessary and sufficient for

$$\exp(t J)$$

to be a 1-parameter subgroup of $\mathrm{O}(n)$ that first gets back to the identity when $t = 2\pi$.

So, all the hard topology in Bott periodicity comes from using Morse theory to show that the tractable space $\Omega_k(n)$ has the same fundamental group as the $k$-fold loop space of $\mathrm{O}(n)$ when $k$ is small enough compared to $n$.

I want to avoid all that stuff, since to me the spaces $\Omega_k(n)$ are more interesting! The way I do things, which is slightly different than how Milnor does it, these spaces become essential fibers of forgetful functors — and it’s easy to show they are disjoint unions of symmetric spaces, and it turns out we get all the compact symmetric spaces in Cartan’s ten infinite families this way. (You may recall that when I followed Milnor, I only got some.)

Do you really need the $\ast$-rep stuff?

I need it to get compact symmetric spaces!

If you skip it, allowing non-$\ast$-reps, I’m sure your fibers will be different – but will they be different homotopically?

If I skipped the $\ast$-stuff I’d get real affine varieties, not compact Riemannian symmetric spaces. I haven’t carefully checked, but I believe they would be homotopy equivalent. It’s clear in a bunch of cases. In some vague sense the $\ast$-stuff lets me pick out the ‘maximal compacts’ in what I’d get if I skipped the $\ast$-stuff. But I’m not sure what that means. For example, in one case I get

$$\mathrm{O}(m+n)/\mathrm{O}(m) \times \mathrm{O}(n)$$

if I use the $\ast$-stuff, but

$$GL(m+n,\mathbb{R})/GL(m,\mathbb{R}) \times GL(n,\mathbb{R})$$

if I don’t. These are isomorphic as real affine varieties, but I think only the former comes with a god-given Riemannian metric!

I don’t know about *-algebra structures on U(g) in general, but if instead we ask about *-Hopf algebra structures then they’re the same thing as real forms of g.

Thanks very much for catching that numbering mistake! That would have been very annoying to correct while giving my talk. I also found another and fixed that too.

“constant/varying signature”

that was not clear at all.

I mean, fixing $p+q$ and varying $p-q$, or fixing $p-q$ and varying $p+q$.

For example, I have in mind

$$\text{Cliff}_{p+1,q+1} = M_2\left( \text{Cliff}_{p,q}\right)$$

for any $p-q$ except $1\text{ mod }4$ (the split ones).

Malo wrote:

I am used to Clifford algebras being defined with arbitrary signature ($p$ square roots of 1 and $q$ square roots of -1).

Right, me too! But if you look at my slides, you’ll see I’m getting the compact symmetric spaces not directly from the Clifford algebras but from their categories of representations, and the forgetful functors between these. Thanks to the fact you mentioned:

$$\mathrm{Cliff}_{p+1,q+1} \cong \mathrm{M}_2\mathrm{Cliff}_{p,q})$$

we have an equivalence of categories

$$\mathsf{Rep}(\mathrm{Cliff}_{p+1,q+1}) \simeq \mathsf{Rep}(\mathrm{Cliff}_{p,q})$$

This means we can simplify the situation to the case where either $p = 0$ or $q = 0$. Then we have

$$\mathsf{Rep}(\mathrm{Cliff}_{p,0}) \simeq \mathsf{Rep}(\mathrm{Cliff}_{0,8-p})$$

so we might as well just consider Clifford algebras with only square roots of $-1$.

I think square roots of $-1$ are more important geometrically than $+1$, but to some extent that’s a matter of taste. I could have used square roots of $+1$. Then we would go around this ‘clock’ in the opposite direction, counterclockwise:

I’m not sure what you mean by that, since I don’t know how to use Clifford algebras to get universal enveloping algebras.

In fact $f^\ast$ is always monadic. On the Category Theory Community Server we saw how this can be shown using Beck’s monadicity theorem, and also that it’s a special case of a much more general theorem for arbitrary single-sorted Lawvere theories… but Mike Shulman wrote:

I often find it easier to show a functor is monadic by directly comparing its domain to the category of algebras for the induced monad, rather than by mucking around with split or reflexive coequalizers. In this case an $f^*f_!$-algebra is an $A$-module with a map $B\otimes _A M \to M$ satisfying some conditions, or equivalently an $A$-bilinear map $(B,M) \to M$ satisfying some conditions. If you forget about $A$-bilinearity and just regard this map as $\mathbb{Z}$-bilinear, the conditions are precisely those that make it define a $B$-module, and then the extra $A$-bilinearity is equivalent to saying that the given $A$-module structure is the one induced from this $B$-module structure via $f$.