August 11, 2019

Even-Dimensional Balls

Posted by John Baez Some of the oddballs on the $n$-Café are interested in odd-dimensional balls, but here’s a nice thing about even-dimensional balls: the volume of the $2n$-dimensional ball of radius $r$ is

$\frac{(\pi r^2)^n}{n!}$

Dillon Berger pointed out that summing up over all $n$ we get

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

It looks nice. But what does it mean?

First, why is it true?

We can show the volume of the $2n$-dimensional ball of radius $r$ is $(\pi r^2)^n/n!$ inductively. How? It’s enough to show that the volume of the unit $2(n+1)$-ball is $\pi /(n+1)$ times the volume of the unit $2n$-ball. Wikipedia gives a proof, and Greg Egan recently summarized the argument in a couple of tweets.

The idea is to map the unit $2(n+1)$-ball to the unit disk in the plane via a projection $\mathbb{R}^{2n} \to \mathbb{R}^2$. Imagine a $2(n+1)$-ball hovering over a disk! I can’t draw it, because the first interesting example is 4-dimensional. But imagine it.

The points over any point in the disk at distance $r$ from the origin form a $2n$-ball of radius $\sqrt{1 - r^2}$. We can get the volume of the $2(n+1)$-ball by integrating the volumes of these $2n$-balls over the disk.

That’s the idea. The rest is just calculus. Let the volume of the unit $2n$-ball be $V_{2n}$. We do the integral using polar coordinates:

$V_{2(n+1)} = \int_0^1 V_{2n} (1 - r^2)^n \; 2 \pi r d r$

Make the substitution $u = 1 - r^2$, so $d u = -2 r d r$. Some minus signs and 2’s cancel and we get

$V_{2(n+1)} = \pi V_{2n} \int_0^1 u^n \, d u = \frac{\pi}{n+1} V_{2n}$

as desired!

So that’s simple enough: since $V_0 = 1$ we get

$V_{2n} = \frac{\pi^n}{n!}$

And maybe that’s all that needs to be said. But suppose we follow Dillon Berger and sum the volumes of all even-dimensional balls of radius $r$. We get

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

This looks cute, but what does it mean?

It seems odd to sum the volumes of shapes that have different dimensions. But we do it in classical statistical mechanics when considering a system of particles with a variable numbers of particles: a so-called ‘grand canonical ensemble’, like an open container of gas, where molecules can flow in and out.

If we have a single classical harmonic oscillator, with momentum $p$ and position $q$, its energy is

$\frac{1}{2}(p^2 + q^2)$

where I’m choosing units that get rid of distracting constants.

If we impose the constraint that the energy is $\le E$ for some number $E$, the state of the oscillator is described by a point $(p,q)$ with

$p^2 + q^2 \le 2 E$

This is a point in the disk of radius $r = \sqrt{2E}$.

Now suppose we have an arbitrary collection of such oscillators—and we don’t know how many there are! Then the state of the system is described first by a natural number $n$, the number of oscillators, and then momenta $p_1, \dots, p_n$ and positions $q_1, \dots, q_n$ obeying

$p_1^2 + q_1^2 + \cdots + p_n^2 + q_n^2 \le 2 E$

In other words, it’s a point in the union of all even-dimensional balls of radius $r = \sqrt{2E}$.

Physicists would call this union the ‘phase space’ of our collection of oscillators. Its volume is the sum of the volumes of the balls:

$\sum_{n=0}^\infty V_{2n} r^{2n} = e^{\pi r^2} = e^{2 \pi E}$

Actually physicists would include the constants that I’m hiding, and this is actually somewhat interesting. For example, to make the area 2-form $d p \wedge d q$ dimensionless, they divide it by Planck’s ‘quantum of action’. This is what allows them to add up volumes of balls (or other symplectic manifolds) of different dimensions and get something that doesn’t break the rules of dimensional analysis. But this also brings Planck’s constant into a question of classical statistical mechanics—a surprise, but a surprise that’s well-known among experts: it shows up in computations such as the entropy of an ideal gas.

And indeed, entropy is exactly what I’m interested in now! Why would a physicist care about the volume of phase space for a collection of arbitrarily many oscillators with total energy $\le E$? Here’s why: if all we know about this collection of oscillators is that it’s energy is $\le E$, its entropy is just the logarithm of this phase space volume! So, it’s

$\ln (e^{2 \pi E}) = 2 \pi E$

Nice!

Physicists will not like how I tucked all the constants under the rug, getting a formula for entropy that’s missing things like Boltzmann’s constant and Planck’s constant, as well as the mass of the particles in our harmonic oscillators, and their spring constant. I leave it as a challenge to restore all these constants correctly. I suspect that the $2\pi$ will get eaten by the $2 \pi$ hiding in Planck’s constant $\hbar = h / 2\pi$. This I know: we will get something proportional to $E$, but with units of entropy.

This is what I’ve got so far. It’s not yet a new proof that the volumes of all even-dimensional balls of radius $r$ sum to

$e^{\pi r^2}$

but rather a physical interpretation of this fact. Maybe someone can keep the ball rolling….

Posted at August 11, 2019 3:13 AM UTC

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Re: Even-Dimensional Balls

This is a bit of a diversion that’s orthogonal to the physical interpretation, but the basic formula generalises in a nice way to the $p$-norm.

The volume of a ball in $\mathbb{R}^n$ under the $p$-norm is:

$V_{p,n}(r) = 2^n\frac{\Gamma(1+1/p)^n}{\Gamma(1+n/p)} r^n$

We then have:

$V_{p, k p}(r) = \frac{V_{p,p}(r)^k}{k!}$

and hence:

$\sum_{k=0}^{\infty} V_{p,k p}(r) = e^{V_{p,p}(r)}$

If you put $p=1$ and take out the factor of $2^n$, this includes the fact that the sum of the volumes of the $n$-dimensional simplices where $n$ orthogonal edges of length $L$ meet at one vertex is just $e^L$.

Posted by: Greg Egan on August 11, 2019 7:04 AM | Permalink | Reply to this

Re: Even-Dimensional Balls

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?
Posted by: John Baez on August 11, 2019 7:45 AM | Permalink | Reply to this

Re: Even-Dimensional Balls

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

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:

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

Posted by: Greg Egan on August 11, 2019 11:59 AM | Permalink | Reply to this

Re: Even-Dimensional Balls

The formula for the volume of the $n$-dimensional ball is like a great poem: tremendous thoughts expressed in just a few words. I can’t resist sharing some further thoughts:

The complex $n$-dimensional polydisc is defined to be the cartesian product of $n$ copies of the complex 1-dimensional polydisc:

$D_n = \{(z_1, \ldots, z_n) \in \mathbb{C}^n: \max_{i = 1, \ldots, n} {|z_i|} \leq 1\}.$

Its volume is $\pi^n$, and the symmetric group $S_n$ acts on $D_n$ by permuting coordinates. The volume of the orbifold $D_n/S_n$ should therefore be $\pi^n/n!$, the same as the volume of the complex ball

$B_n = \{(z_1, \ldots, z_n) \in \mathbb{C}^n: \sum_{i=1}^n {|z_i|}^2 \leq 1\}.$

Is there a structural reason for that, i.e., might there be a volume-preserving transformation $D_n/S_n \to B_n$?

I believe it was at the 1995 Category Theory Conference in Halifax that I watched Andreas Blass and Steve Schanuel construct such a map at the blackboard in real time. They wrote a short note about it which can be found here.

It goes like this: 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 $\phi: D_n \to B_n$ 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

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

Then $\phi$ induces a continuous well-defined map $D_n/S_n \to B_n$. Furthermore, when restricted to the set $P_n$ of $(z_1, \ldots, z_n)$ for which the $r_j$ are all distinct, $\phi$ induces a smooth symplectic isomorphism mapping $P_n/S_n$ onto the set $Q_n$ of $(w_1, \ldots, w_n) \in B_n$ for which $w_j \neq 0$ for $1 \leq j \leq n-1$.

In other words, writing $z_j = x_j + i y_j$ the symplectic 2-form

$\sum_{j=1}^n d x_j \wedge d y_j = \sum_{j=1}^n r_j d r_j \wedge d\theta_j$

is preserved by pulling back along $\phi \colon P_n/S_n \to Q_n$. Since symplectic maps are locally volume-preserving, and since $P_n$ and $Q_n$ are almost all of $D_n$ and $B_n$ respectively, this gives a new proof that the volume of $B_n$ is $\pi^n/n!$, independent from the more purely computational methods such as the one found on Wikipedia.

Posted by: Todd Trimble on August 11, 2019 12:55 PM | Permalink | Reply to this

Re: Even-Dimensional Balls

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.

Posted by: Greg Egan on August 11, 2019 11:30 PM | Permalink | Reply to this

Re: Even-Dimensional Balls

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.

Posted by: John Baez on August 12, 2019 3:53 AM | Permalink | Reply to this

Re: Even-Dimensional Balls

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 $\theta_i$.)

Posted by: John Baez on August 12, 2019 4:19 AM | Permalink | Reply to this

Re: Even-Dimensional Balls

Posted by: Greg Egan on August 12, 2019 7:41 AM | Permalink | Reply to this

Re: Even-Dimensional Balls

Interesting. I think the HyperDarts game is more like the map Blass and Schanuel came up with before André Joyal asked them to make it symplectic: that’s when they started messing with the angles.

Posted by: John Baez on August 12, 2019 8:18 AM | Permalink | Reply to this

Re: Even-Dimensional Balls

When I tweeted my “HyperDarts” game, Christopher Long tweeted an alternative version where the darts need to land with radii in descending order.

So his game would have the same probability of scoring after $n$ throws of the dart, and it’s easier to see where the $1/n!$ comes from, but it would need to use something like Todd’s map to specify a point in the $(2n)$-ball, whereas in my version of the game the Cartesian product of the $n$ successive dart locations will immediately be a point in the $(2n)$-ball.

Posted by: Greg Egan on August 12, 2019 8:22 AM | Permalink | Reply to this

Re: Even-Dimensional Balls

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.

Posted by: John Baez on August 12, 2019 5:20 AM | Permalink | Reply to this

Re: Even-Dimensional Balls

Already the construction $D_n/S_n$ isn’t quite a manifold (leaving aside symplectic), is it? I would have thought it’s more like an orbifold.

Might there be a good theory of symplectic orbifolds?

Posted by: Todd Trimble on August 12, 2019 12:55 PM | Permalink | Reply to this

Re: Even-Dimensional Balls

Already the construction $D_n/S_n$ isn’t quite a manifold (leaving aside symplectic), is it? I would have thought it’s more like an orbifold.

Good point. But the Chevalley–Shephard–Todd theorem may help. This implies that the ring of polynomials $S_n$-invariant polynomials on $\mathbb{C}^n$ isomorphic to a polynomial ring on $n$ generators (namely the elmentary symmetric functions). So, as a variety, $\mathbb{C}^n/S_n$ is smooth, and isomorphic to $\mathbb{C}^n$.

I don’t know if this counts as ‘cheating’, but it seems to give the open unit ball in $\mathbb{C}^n$ mod the action of $S_n$ the structure of a smooth manifold.

Posted by: John Baez on August 12, 2019 2:23 PM | Permalink | Reply to this

Re: Even-Dimensional Balls

Okay, that’s nice. So for example, the map

$\mathbb{C}^2 \to \mathbb{C}^2: (z, w) \mapsto (z+w, z w)$

realizes a homeomorphism $\mathbb{C}^2/2! \cong \mathbb{C}^2$, and so by the old trick we can transport along this homeomorphism the smooth structure on $\mathbb{C}^2$ to a diffeomorphic one on $\mathbb{C}^2/2!$. (Which reminds me of a funny “Math Fail”, where the test asks “Are the groups $\mathbb{Z}_2 \times \mathbb{Z}_3$ and $\mathbb{Z}_6$ isomorphic?”, and the student answers, “$\mathbb{Z}_2 \times \mathbb{Z}_3$ is, but $\mathbb{Z}_6$ isn’t.”)

Without calculating, my guess is that the displayed map is not symplectic though, and it might be tricky getting one that works. So at this point, I got nothin’.

Posted by: Todd Trimble on August 12, 2019 3:31 PM | Permalink | Reply to this

Re: Even-Dimensional Balls

Just to spell this out, you simply need a version of the fundamental theorem of algebra: every monic, degree $n$, complex polynomial is uniquely determined by its $n$ roots.

$(z_1, \dots, z_n)\mapsto \prod_i (z-z_i)$

gives the homeomorphism

$\mathbb{C}^n/S_n \to P_n(\mathbb{C})\cong \mathbb{C}^n,$

where $P_n(\mathbb{C})$ is the monic degree $n$ polynomials of degree $n$.

Posted by: Simon Willerton on August 12, 2019 10:32 PM | Permalink | Reply to this

Re: Even-Dimensional Balls

Simon wrote:

Just to spell this out, you simply need a version of the fundamental theorem of algebra.

Stop trying to demystify it! I was so proud to remember that result that the algebra of polynomials invariant under a finite reflection group form another polynomial algebra! But seriously: yes, it’s good to demystify this result. But now let me try to mystify it some more. We can say it like this: in the category of affine schemes over $\mathbb{C}$, the free commutative monoid on the line — that is, the affine line — is the coproduct

$point + line + plane + \cdots$

since we’d expect it to be

$1 + line + line^2/2! + line^3/3! + \cdots$

but this works out to be the same thing. Or, to use notation like the big boys, where $\mathbb{A}^n$ is affine $n$-space, the free commutative monoid on $\mathbb{A}^1$ is the coproduct

$\sum_{n=0}^\infty \mathbb{A}^n$

or if you want to freak out the big boys:

$\frac{1}{1 - \mathbb{A}^1}$

This is weird because it looks just like the free monoid on $\mathbb{A}^1$. I don’t think I’ve plumbed the depths of the fact that

$\mathbb{A}^n/n! \cong \mathbb{A}^n$

So far we’ve been using affine schemes over $\mathbb{C}$. I think it works over $\mathbb{Q}$ too, despite the failure of the Fundamental Theorem of Algebra, because that Chevalley–Shephard–Todd theorem still holds. But I think it fails over $\mathbb{Z}$.

Todd brings us down to earth:

Without calculating, my guess is that the displayed map is not symplectic though, and it might be tricky getting one that works. So at this point, I got nothin’.

Hmm. I’d really like to understand the symplectic map that Blass and Schanuel came up with. There should be a good reason for its existence. You laud it compared to the ‘more purely computational’ approach, but checking from their formula that this map is symplectic seems to require about as much work as the calculation I did in the blog entry. It could be the route to a purely conceptual proof that the volume of the unit $2n$-ball is $\pi^n/n!$. But the trick would be finding a purely conceptual proof that this map — or some variant — is symplectic.

Category theorists: never satisfied.

Posted by: John Baez on August 13, 2019 4:52 AM | Permalink | Reply to this

Re: Even-Dimensional Balls

You laud it compared to the ‘more purely computational’ approach

Actually, I think I was speaking factually and objectively where I said “this gives a new proof that the volume of $B_n$ is $\pi^n/n!$, independent from the more purely computational methods”, not lauding particularly in that sentence. In fact, I happen to love the computational proof that starts from the observation that

$\int_{-\infty}^\infty e^{-x^2}\;d x = \pi^{1/2}.$

But I also think the Blass-Schanuel proof is pretty damned cool, obviously. While agreeing with you that I want more understanding still.

Posted by: Todd Trimble on August 13, 2019 6:05 AM | Permalink | Reply to this

Re: Even-Dimensional Balls

I’m not sure exactly how this fits in with the Blass-Schanuel map, but it’s interesting to me that the map $f_r$ of the interior of the disk of radius $r$ minus the origin, to itself, such that $x$ and $f_r(x)$ are diagonally opposite vertices of a rectangle whose other two vertices are the origin and a point on the boundary: is area-preserving. I can compute the Jacobian determinant fairly easily, but there might be a way to make this even more obvious.

This property of $f_r$ means that if we pick a sequence of $n$ points in the unit disk such that their coordinates strung together all lie in the unit ball, we can also use $f_r$ to get a sequence of $n$ points in the unit disk with radii in descending order: In that animation, the black dots are the points corresponding to pairs of coordinates of a vector that lies within the unit ball in $\mathbb{R}^{2n}$, while the blue dots lie in the orbifold $D_n/S_n$, with radii in descending order.

So the fact that $f_r$ is measure-preserving seems like the key to there being a measure-preserving map between the ball and the orbifold.

Posted by: Greg Egan on August 13, 2019 6:55 AM | Permalink | Reply to this

Re: Even-Dimensional Balls

Neat, Greg! My first vague guess about how HyperDarts was related to a volume-preserving isomorphism $D^n \cong (D^2)^n/n!$ was wrong, and this idea corrects it.

I see several angles of attack:

1) What is the meaning of this map from the punctured disc to itself, that takes any vertex of a rectangle whose other two vertices are the origin and a point on the boundary, and sends it to the diagonally opposite vertex?

It ‘turns the disc inside out’, and when you do it twice you get back where you started. So it reminds me of conformal inversion. But unlike conformal inversion, it’s not conformal — that is, angle-preserving. It’s symplectic — that is, area-preserving!

2) What is the meaning of the Blass–Schanuel map that takes an $n$-tuple of points $z_j$ in the disk, first permutes them so that $r_1 \geq r_2 \geq \ldots \geq r_n$, and then maps them to

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

3) How is the map in 2) related to the map in 1)? You seem to be saying that they do similar jobs.

Posted by: John Baez on August 13, 2019 8:22 AM | Permalink | Reply to this

Re: Even-Dimensional Balls

Posted by: Greg Egan on August 13, 2019 8:51 AM | Permalink | Reply to this

Re: Even-Dimensional Balls

Thanks! Personally I’d set $r = 1$ and write $f$ instead of $f_r$ to make mathematicians enjoy this question more. I have the power to edit questions on MathOverflow, but power corrupts so I won’t do this unless you ask.

Posted by: John Baez on August 13, 2019 10:08 AM | Permalink | Reply to this

Re: Even-Dimensional Balls

I started out with just $f$ with $r=1$, but then I realised that to apply this properly to the map from the unit ball to the orbifold requires the generalisation to the case of arbitrary $r$. If you try to write everything you need in terms of $f_1$, you end up with a mess of ugly rescaling.

Posted by: Greg Egan on August 13, 2019 10:24 AM | Permalink | Reply to this

Re: Even-Dimensional Balls

Remind me why the volume of the unit 0-ball is 1 and not 0.

Posted by: Gavin Wraith on August 11, 2019 8:12 PM | Permalink | Reply to this

Re: Even-Dimensional Balls

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

Posted by: Todd Trimble on August 11, 2019 8:42 PM | Permalink | Reply to this

Re: Even-Dimensional Balls

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 $1$.

Posted by: Tom Leinster on August 11, 2019 9:43 PM | Permalink | Reply to this

Re: Even-Dimensional Balls

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!

Posted by: John Baez on August 12, 2019 4:09 AM | Permalink | Reply to this

Re: Even-Dimensional Balls

Thank you. I am convinced.

Posted by: Gavin Wraith on August 12, 2019 10:19 AM | Permalink | Reply to this

Mark

Summing over the volumes of even-dimensional spheres, and this interpretation is a very nice find! Now, how about taking the alternating sum?

$\sum_{n=0}^{\infty}(-1)^n\frac{(\pi r)^n}{n!} = e^{-\pi r^2}$

You get Gauss’ curve with unit area. In a way this is nice because I always wondered if there was a geometric intuition for the $\pi$ popping up in the $\sqrt{\pi}$ surface area under the Gauss curve.

The standard derivation by change of coordinates is not satisfying since it does not tell you what the ratio of circumference and radius have to do with $e^{-x^2}$. The interpretation of this expansion in terms of volumes of spheres does!

Posted by: Mark van der Loo on August 12, 2019 3:14 PM | Permalink | Reply to this

Re: Even-Dimensional Balls

The exponential function as a shadow of the free abelian monoid functor is quite an old idea. I certainly remember Bill Lawvere talking about it. This thread reminds me of the fact that the Booleanization of the unit interval, seen as a topological distributive lattice, is the infinite-dimensional sphere.

Posted by: Gavin Wraith on August 12, 2019 4:30 PM | Permalink | Reply to this

Re: Even-Dimensional Balls

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?

Posted by: Todd Trimble on August 12, 2019 6:50 PM | Permalink | Reply to this

Re: Even-Dimensional Balls

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

Posted by: John Baez on August 13, 2019 7:30 AM | Permalink | Reply to this

Re: Even-Dimensional Balls

“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

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

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

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.

Posted by: Todd Trimble on August 13, 2019 1:43 PM | Permalink | Reply to this

Re: Even-Dimensional Balls

It may well have appeared in print somewhere, but I am not sure where. My memory is rather cloudy about it. Also my skills laying out MathML are wanting, so be patient if the following is somewhat turgid to read.

Consider the topos SS of semisimplicial sets; that is, presheaves on the category of finite nonempty ordered sets and monotone maps. We have a geometric-realization functor from it to the category of topological spaces and continuous maps that preserves finite limits, and hence algebraic structure. The unit interval I is the geometric realization of the representable object D given by the ordered set ={0<1}. Hence the Booleanization of I, B(I), will be the geometric realization of the Booleanization of D, B(D). For an ordered set [n], D([n]) is the set of monotone functions from [n] to , whereas B(D)([n]) is the set of all functions from [n] to . It remains to show that this semisimplicial set has the infinite sphere as its geometric realization, with the antipodal map given by switching the two elements of .

The semisimplicial n-sphere can be described as a colimit of a small pushout diagram involving two copies of [n+1] (the north and south hemispheres) and an (n-1)-sphere, the equator and its embeddings into the two hemispheres. I am hoping that someone more adroit than me will see a tidy way of identifying the colimit of the spheres with B(D). I think it is a question of book-keeping, but maybe there are some slicker approaches.

Posted by: Gavin Wraith on August 12, 2019 10:44 PM | Permalink | Reply to this

Re: Even-Dimensional Balls

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.

Posted by: Todd Trimble on August 13, 2019 12:41 AM | Permalink | Reply to this

Re: Even-Dimensional Balls

Thanks for the encouragement. A monotone map from [n] to  is determined by the inverse image of {1}, a top-segment of [n]. How can we express an arbitrary subset of [n] in terms of top-segments? Evidently we need negation (to get bottom-segments) and then we can use intersection to get mid-segments. We get union from intersection using De Morgan’s laws, and then an arbitrary subset is a union of mid-segments. This is just a roundabout way of describing the theory of Boolean algebras as an extension of the theory of semi-lattices. This extension is essentially the adjunction of an involution, to give negation. The tower of spheres is a coordinate-free way of describing this extension. 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.

I suspect there is a much slicker way of dealing with all this by looking at finite non-empty ordered sets-with-involution, and the apparatus of semisimplicial sets within the topos of sets-with-involution.

Posted by: Gavin Wraith on August 13, 2019 11:02 AM | Permalink | Reply to this

Re: Even-Dimensional Balls

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.

Posted by: Todd Trimble on August 13, 2019 3:36 PM | Permalink | Reply to this

Re: Even-Dimensional Balls

A propos my last remark, I seem to remember that Springer Lecture Notes 283, Complexe Cotangent et Deformations, by Illusie, had lots of nice stuff about semisimplicial objects.

Of course we should be thinking of [n] as a category, and Hom(_,) as a dualizing functor which is the source of dimension-shifting. The canonical map [n] -> Hom(Hom([n],),) embeds [n] as a mid-segment of [n+2]. Very suggestive.

Posted by: Gavin Wraith on August 13, 2019 3:26 PM | Permalink | Reply to this

Re: Even-Dimensional Balls

A propos Myles Tierney’s remark: If $\Omega$ is the subobject classifier in the topos of semisimplicial sets, then $\Omega([n])$ is the set of subfunctors of the representable functor given by $[n]$, or sieves of the slice category over $[n]$. If $S$ is a subset of $\{0, ...,n\}$ we have the sieve of those arrows into $[n]$ which send elements only into $S$. Thus $\Omega$ is the booleanization of $$.

Posted by: Gavin Wraith on August 13, 2019 8:29 PM | Permalink | Reply to this

Re: Even-Dimensional Balls

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

Posted by: Todd Trimble on August 13, 2019 8:42 PM | Permalink | Reply to this

Re: Even-Dimensional Balls

Sorry, I was mistaken. $\Omega[n]$ is not defined by all subsets S of $\{0,..,n\}$, but only the down-segments. So $\Omega$ is not a Boolean algebra, and Myles was either wrong or misrecalled

Posted by: Gavin Wraith on August 14, 2019 12:22 PM | Permalink | Reply to this

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