The n-Category Café

Wow, that’s much more surprising than the formula for $p = 2$. I don’t know what to make of it.

Here’s a puzzle about the physics of the $p = 2$ case. In my post we’re dealing with a system of arbitrarily many classical harmonic oscillators, with their total energy constrained to be $\le E$. I worked out the entropy of this system and showed it was exactly $2 \pi E$ (in the units I’m using, which were chosen to suppress lots of constants).

But we can ask other questions. For example:

• What’s the expected number of oscillators?

This sounds funny, but since the number of oscillators is variable this question makes sense.

Hint: the probability that there are $n$ oscillators is the volume of the $2n$-ball of radius $r = \sqrt{2 E}$, divided by the sum over all $n$ of such volumes.

Or:

• What is the most probable number of oscillators?

Given your hint, we have:

$$P(n) = \frac{(2\pi E)^n}{n! e^{2\pi E}}$$

So I think the expected number of oscillators is:

$$\sum_{n=0}^\infty n P(n) = \sum_{n=0}^\infty n \frac{(2\pi E)^n}{n! e^{2\pi E}} = 2\pi E$$

And I think the most probable number of oscillators is:

$$\lfloor 2\pi E\rfloor$$

based on the ratio:

$$\frac{P(n)}{P(n+1)} = \frac{n+1}{2 \pi E}$$

That’s a nice construction! I was hoping there was something along these lines, with $n$ complex variables and some reduction of the polydisc, but I couldn’t see what it was myself.

Wow! Last night I decided to post this question:

Is there a natural volume-preserving map from the $n$th power of the unit disk mod $S_n$ to the unit $2n$-ball?

I couldn’t think of one. But now my question has been answered before I could even ask it!

I remember being surprised, back when I was trying to learn algebraic geometry from Griffiths and Harris, that the polydisk is a better generalization of the unit disk for the purposes of multivariable complex analysis than the unit ball in $\mathbb{C}^n$. (Now I can’t even remember what basic theorems work for the polydisk but not the ball.)

It’s nice to see that in some funny sense we are just taking the unit ball and modding out by permutations of the coordinate axes. I haven’t actually understood Todd’s post yet, but I’m looking forward to it.

By the way, Greg: when you created the game of HyperDarts, were you deliberately using the construction Todd described:

Given $(z_1, \ldots, z_n) \in \mathbb{C}^n$, write coordinates $z_j$ in polar coordinate form $z_j = r_j e^{i \theta_j}$, and define an $S_n$-invariant map $\phi \colon D_n \to B_n$ by first permuting the $z_j$ so that $r_1 \geq r_2 \geq \ldots \geq r_n$ and then mapping $(z_1, \ldots, z_n)$ to

$$(\sqrt{r_1^2 - r_2^2}\; e^{i\theta_1}, \sqrt{r_2^2 - r_3^2} \; e^{i(\theta_1 + \theta_2)}, \ldots, \sqrt{r_{n-1}^2-r_n^2} \; e^{i(\theta_1 + \theta_2 + \ldots + \theta_{n-1})}, r_n\; e^{i(\theta_1 + \theta_2 + \ldots + \theta_n)})$$

Or was it just a coincidence that you ‘gamified’ this map?

(The relation between your game and this map is still a bit confusing to me, but I think they use the same idea, though your game doesn’t use anything about the angles $\theta_i$.)

So Blass and Schanuel seem to have showed something like this: the free commutative monoid on the unit disk is the disjoint union of all even-dimensional unit balls… in the symmetric monoidal category of measure spaces and measure-preserving maps, with the usual product of measure spaces giving the monoidal structure.

But that’s a wimpy way to phrase their result. I would like to use the category of symplectic manifolds, but there seems to be some niggly problem occurring on a set of measure zero. So, maybe we should use symplectic manifolds and almost-everywhere-defined symplectomorphisms!

If we take the $n$th power of the disk and mod out by $S_n$, we’re getting the space of $n$-element multisets of the disk. This is isomorphic to the space of degree-$n$ polynomials with all roots in the disk, mod multiplication by nonzero constants. Is there some nice way to see why this should be (almost) a $2n$-ball?

There should be something nice going on here, maybe connected to the $A_{n+1}$ Coxeter group.

Wouldn’t the $0$-dimensional volume just count a number, i.e., the cardinality of $\{0\} \subseteq \mathbb{R}^0$ is $1$?

Another way to justify it is as follows: Lebesgue measure on $\mathbb{R}^n$ is normalized so that $[0, 1]^n$ has volume $1$. Taking $n = 0$, this says that $\{0\} \subseteq \mathbb{R}^0$ has volume $1$.

To higher-dimensional beings a point may seem like almost nothing, but a point ain’t nothing if you’re 0-dimensional yourself: it’s quite substantial!

Yes! I’ve heard Lawvere mention this Booleanization as well. I imagine it’s related to the fact that the total space of the classifying bundle $E(\mathbb{Z}/2) \to B(\mathbb{Z}/2)$, using for example the Milgram construction, is an infinite-dimensional sphere $S^\infty$. Here the action of $\mathbb{Z}/2$ on $S^\infty$ is realized by the antipodal map, now viewed as the logical negation operator on this Booleanization.

Do you have any idea where $S^\infty$ as Booleanization might appear in print?

What does “booleanization” mean in this discussion? — as in “booleanization of the unit interval”, or booleanization of various other objects.

“Booleanization” here means the left adjoint to the forgetful functor

$$Bool_C \to DistLat_C$$

where $C$ can be one of various categories; the ones of interest in this discussion here are $C = Set^{\Delta^{op}}$ and $C =$ some suitable version of $Top$, namely a “convenient category” of spaces that is cartesian closed, to ensure on general theoretical grounds that geometric realization

$$R: Set^{\Delta^{op}} \to C$$

preserves finite limits (it’s products that are the real issue). For example, the category of compactly generated weak Hausdorff spaces is a popular choice for $C$.

There’s a bit going on under the hood here; the following may be more than you want or need to hear, but here goes anyway. To be very explicit about Booleanization, one can construct it as an example of computing relative adjoints. If we have a morphism of Lawvere theories $T \to T'$ like $Th(DistLat) \to Th(Bool)$, and if for the sake of convenience $C$ is cocomplete and cartesian closed, then first of all the forgetful functors

$$Alg_T(C) \to C, \qquad Alg_{T'}(C) \to C$$

are monadic, with left adjoints that I can write down as

$$F_T(c) = \int^{n \in Fin} T(n, 1) \cdot c^n$$

where $T(n, 1)$ is a hom-set for the Lawvere theory $T$, and $n$ by abuse of language is the image of a finite set $n$ under the canonical product-preserving functor $Fin^{op} \to T$. Second of all, in this situation the forgetful functor

$$Alg_{T'}(C) \to Alg_T(C)$$

will itself be monadic, where the left adjoint takes a $T$-algebra $(c, \xi: M_T c \to c)$ [here $M_T$ is the monad attached to $T$] to a kind of “tensor product”, i.e., a $T'$-algebra formed as a reflexive coequalizer of a diagram of the form

$$M_{T'} M_T c \rightrightarrows M_{T'} c$$

which I can leave to the imagination, with the hint that $M_T$ acts on the left on $c$ and on the right on $M_{T'}$. Putting all this together, we have an explicit Booleanization functor which is this left adjoint $DistLat_C \to Bool_C$.

Again, provided that geometric realization preserves finite products and colimits, it will preserve all the above constructions, i.e., it will “preserve Booleanization”. Actually, it might be possible to simplify all this: that realization commutes with Booleanization means we have a square of left adjoints, and one might see more directly that the corresponding square of right adjoints (involving relative forgetful functors and “singularization” which is right adjoint to realization) commutes for some simple reason, but I don’t feel like pursuing that direction at the moment.

I’ll add too that even though we want something like $CGWH$ in place of $Top$ on general theoretical grounds, it will often be the case that the relevant coends are the same whether computed in $CGWH$ or in $Top$, and that turns out to be the case in the specific example we are discussing. I think I could probably nail down the relevant technical statement in some appendix of the book by Hatcher.

But it’s also fun to try to imagine intuitively how the Booleanization of $I$ as distributive lattice is built up in the topological case. You take a copy of $I$ and then formally negate all its elements, and then glue $I$ and $\neg I$ head to tail to form a circle. But then you have to add in more conjunctions and disjunctions of elements of this circle, and perhaps these fill out an upper and lower hemisphere of an $S^2$, and keep on going. If at some stage you’ve filled in a higher-dimensional ball or hemisphere, then the formal negations fill out the antipodal hemisphere, and formally adjoined conjunctions of those fill out something higher-dimensional still. This is admittedly terribly sketchy and probably Gavin has a clearer picture of this, if you follow all his instructions carefully.

A very nice argument; thanks Gavin! The thing I had not considered was passing through the simplicial Booleanization of the simplicial distributive lattice you denote as $D$ (which is the nerve of the poset $0 \leq 1$), and then identifying in simple terms what this $Bool(D)$ actually is.

To write in LaTeX, you need to select a Markdown filter (I usually select ‘itex to MathML with parbreaks’), and then for the most part it’s straight LaTeX.

In principle this seems slick enough to me, although I haven’t yet digested some of your hints like “We can get a sphere in two ways: either as the boundary of a ball or by glueing two balls together along their boundaries. The equivalence of these two reflect the identities which hold for the extension of algebraic theories.”

Another thing all this reminds me of (with apologies to John – weren’t we discussing volumes of even-dimensional balls or something?) is something I thought I remembered Myles Tierney saying in the classroom once, that the geometric realization of the subobject classifier in the topos of simplicial sets is also an infinite-dimensional sphere. Which seems believable to me, but I haven’t tried to work it out in detail. One very easy consistency check is that whatever it is, it must be contractible, since it carries the structure of a meet-lattice with top.

I’m confused. Simplicial sets is not a Boolean topos. How can $\Omega$ be a Booleanization of anything?