The n-Category Café

Before I can describe this categorical viewpoint on potential functions, I have to explain what potential functions are in the magnitude context, and why they’re very useful. That’s what I’ll do today.

This part of the story is about metric spaces. For now I’ll assume they satisfy all the classical axioms, including symmetry of the metric, meaning that $d(a, a') = d(a', a)$ for all points $a$ and $a'$. When metric spaces are viewed as enriched categories, symmetry isn’t automatic — but we’ll come to that next time.

A weighting on a finite metric space $A$ is a function $w: A \to \mathbb{R}$ such that for all $a \in A$,

$$\sum_{a' \in A} e^{-d(a, a')} w(a') = 1.$$

Everyone who sees this formula for the first time asks where the exponential comes from. Ultimately it’s because of the enriched category viewpoint (which again we’ll come to next time), but the short story is that exponential is essentially the only reasonable function that converts addition into multiplication.

For simplicity, I’ll assume here that every finite metric space $A$ has a unique weighting, which I’ll call $w_A$. Since the definition of weighting involves the same number of equations as unknowns, this is generically true (and it’s always true for subspaces of $\mathbb{R}^n$), even though there are exceptions.

The magnitude of $A$ is

$$|A| = \sum_{a \in A} w_A(a) \in \mathbb{R}.$$

That’s for finite metric spaces. To extend the definition to compact metric spaces, there are various ideas you might try. You could define the magnitude of a compact space as the supremum of all the magnitudes of its finite subspaces. Or, you could take an ever-denser sequence of finite subsets of your compact space, then define its magnitude to be the limit of the magnitudes of the approximating subsets. Or, you could try replacing the sums in the formulas above by integrals, somehow.

Mark Meckes showed that all these approaches are equivalent. They all give the same definition of the magnitude of a compact space. (At least, this is true subject to a condition called “positive definiteness” which I won’t discuss and which always holds for subspaces of $\mathbb{R}^n$.)

How do we actually calculate magnitude, say for compact subspaces $A$ of $\mathbb{R}^n$?

In principle, you already know how do it. You run through all the finite subsets of $A$, calculate the magnitude of each using the definition above, and take the sup. The trouble is, this procedure is incredibly hard to work with. It’s completely impractical.

A slightly more practical approach is to look for a “weight measure”, that is, a signed measure $w$ on $A$ such that for all $a \in A$,

$$\int_A e^{-d(a, a')}\, d w(a') = 1.$$

As Mark showed, if $w$ is a weight measure then the magnitude of $A$ is given by $|A| = w(A)$. This is the analogue of the formula $|A| = \sum_a w(a)$ for finite spaces.

Example   Take an interval $[c, d]$ in $\mathbb{R}$. It’s an elementary exercise to check that

$$\tfrac{1}{2}(\delta_c + \lambda_{[c, d]} + \delta_d)$$

is a weight measure on $[c, d]$, where $\delta_c$ and $\delta_d$ denote the Dirac deltas at $c$ and $d$, and $\lambda_{[c, d]}$ is Lebesgue measure on $[c, d]$. It follows that $[c, d]$ is the total mass of this measure, which is

$$1 + \tfrac{1}{2}(d - c).$$

In other words, it’s $1$ plus half the length of the interval.

The trouble with weight measures is that very few spaces have them. Even Euclidean balls don’t, beyond dimension one. It turns out that we need something more general than a measure, something more like a distribution.

In another paper, Mark worked out exactly what kind of measure-like or distribution-like thing a weighting should be. The answer is very nice, but I won’t explain it here, because I want to highlight the formal aspects above all.

So I’ll simply write the weighting on a metric space $A$ as $w_A$, without worrying too much about what kind of thing $w_A$ actually is. All that matters is that it behaves something like a measure: we can pair it with “nice enough” functions $\phi: A \to \mathbb{R}$ to get a real number, which I’ll write as

$$\int_A \phi(a) \, d w_A(a)$$

or simply

$$\int_A \phi \, d w_A.$$

And I’ll assume that all the spaces $A$ that we discuss do have a unique weighting $w_A$.

That’s the background. But back to the question: how do we actually calculate magnitude?

First main idea   Don’t look at metric spaces $A$ in isolation. Instead, consider spaces embedded in some big space $B$ that we understand well.

Think of the big space $B$ as fixed. The typical choice is $\mathbb{R}^n$. One sense in which we “understand” $\mathbb{R}^n$ well is that we have a weight measure on it: it’s just Lebesgue measure divided by a constant factor $c_n$. This follows easily from the fact that $\mathbb{R}^n$ is homogeneous. (It doesn’t matter for anything I’m going to say, but the constant $c_n$ is $\int_{\mathbb{R}^n} e^{-\|x\|}\, d x$, for which there’s a standard formula involving $\pi$s and factorials.)

The potential function of a subspace $A \subseteq B$ is the function $h_A: B \to \mathbb{R}$ defined by

$$h_A(b) = \int_A e^{-d(a, b)} \, d w_A(a).$$

By definition of weighting, $h_A$ has constant value $1$ on $A$. But outside $A$, it could be anything, and it turns out that its behaviour outside $A$ is very informative.

Although $w_A$ is a measure-like thing on the subspace $A$ of $B$, we typically extend it by $0$ to all of $B$, and then the definition of the potential function becomes

$$h_A(b) = \int_B e^{-d(x, b)} \, d w_A(x)$$

Examples   In all these examples, I’ll take the big space $B$ to be $\mathbb{R}$.

• Let $A = \{0\}$. Then the weighting $w_A$ on $A$ is the Dirac delta at $0$, and the potential function $h_A: \mathbb{R} \to \mathbb{R}$ is given by

  $$h(b) = e^{-|b|}$$

  ($b \in \mathbb{R}$), whose graph looks like this:

  

• Let $A = [-2, 2]$. As we’ve already seen,

  $$w_A = \tfrac{1}{2}(\delta_{-2} + \lambda_{[-2, 2]} + \delta_2),$$

  and a little elementary work then reveals that the potential function $h_A: \mathbb{R} \to \mathbb{R}$ is

  $$h(b) = \begin{cases} e^{-(-2 - b)} &\text{if}   b \leq -2, \\ 1 &\text{if}   -2 \leq b \leq 2, \\ e^{-(b - 2)} &\text{if}   b \geq 2. \end{cases}$$

  So the potential function looks like this:

  

• In the two examples we just did, the potential function of $A$ is just the negative exponential of the distance between $A$ and the argument $b$. That’s not exactly coincidence: as we’ll see in the next part, the function just described corresponds to a strict colimit, whereas the potential function corresponds to a lax colimit. So it’s not surprising that they coincide in simple cases.

  But this only happens in the very simplest cases. It doesn’t happen for Euclidean balls above dimension $1$, or even for two-point spaces. For example, taking the subset $A = \{-1, 1\}$ of $B = \mathbb{R}$, we have

  $$h_A(x) = \frac{e^{-|x + 1|} + e^{-|x - 1|}}{1 + e^{-2}},$$

  which looks like this:

  

  whereas $b \mapsto e^{-d(A, b)}$ looks like this:

  

  Similar, but different!

And now we come to the:

Second main idea   Work with potential functions instead of weightings.

To explain what this means, I need to tell you three good things about potential functions.

• The potential function determines the magnitude:

  $$|A| = \int_B h_A \, d w_B.$$

  At the formal level, proving this is a one-line calculation: substitute the definition of $h_A$ into the right-hand side and follow your nose.

  For example, we saw that $\mathbb{R}^n$ has weighting $\lambda/c_n$, where $\lambda$ is Lebesgue measure and $c_n$ is a known constant. So

  $$|A| = \frac{1}{c_n} \int_{\mathbb{R}^n} h_A(x) \, d x$$

  for compact $A \subseteq \mathbb{R}^n$. Here $d x$ refers to ordinary Lebesgue integration.

  (For $\mathbb{R}^n$, and in fact a bit more generally, this result appears as Theorem 4.16 here.)

• You can recover the weighting from the potential function. So, you don’t lose any information by working with one rather than the other.

  How do you recover it? Maybe it’s easiest to explain in the case when the spaces are finite. If we write $Z_B$ for the $B \times B$ matrix $(e^{-d(b', b)})_{b', b \in B}$ then the definition of the potential function can be expressed very succinctly:

  $$h_A = Z_B w_A.$$

  Here $h_A$ and $w_A$ are viewed as column vectors with entries indexed by the points of $B$. (For $w_A$, those entries are $0$ for points not in $A$). Assuming $Z_B$ is invertible, this means we recover $w_A$ from $h_A$ as $Z_B^{-1} h_A$. And something similar is true in the non-finite setting.

  However, what really makes the technique of potential functions sing is that when $B = \mathbb{R}^n$, there’s a much more explicit way to recover the weighting from the potential function:

  $$w_A = (I - \Delta)^{(n + 1)/2} h_A$$

  (up to a constant factor that I’ll ignore). Here $I$ is the identity and $\Delta$ is the Laplace operator, $\sum_{i = 1}^n \frac{\partial^2}{\partial x_i^2}$. This is Proposition 5.9 of Mark’s paper.

  How much more of this bullet point you’ll want to read depends on how interested you are in the analysis. The fundamental point is simply that $(I - \Delta)^{(n + 1)/2}$ is some kind of differential operator. But for those who want a bit more:

  • To make sense of everything, you need to interpret it all in a distributional sense. In particular, this allows one to make sense of the power $(n + 1)/2$, which is not an integer if $n$ is even.

  • Maybe you wondered why someone might have proved a result on the magnitude of odd-dimensional Euclidean balls only, as I mentioned at the start of the post. What could cause the odd- and even-dimensional cases to become separated? It’s because whether or not $(n + 1)/2$ is an integer depends on whether $n$ is odd or even. When $n$ is odd, it’s an integer, which makes $(I - \Delta)^{(n + 1)/2}$ a differential rather than pseudodifferential operator. Heiko Gimperlein, Magnus Goffeng and Nikoletta Louca later worked out lots about the even-dimensional case, but I won’t talk about that here.

  • Finally, where does the operator $(I - \Delta)^{(n + 1)/2}$ come from? Sadly, I don’t have an intuitive explanation. Ultimately it comes down to the fact that the Fourier transform of $e^{-\|\cdot\|}$ is $\xi \mapsto (1 + \|\xi\|)^{-(n + 1)/2}$ (up to a constant). But that itself is a calculation that’s really quite tricky (for me), and it’s hard to see anything beyond “it is what it is”.

• The third good thing about potential functions is that they satisfy a differential equation, in the situation where our big space $B$ is $\mathbb{R}^n$. Specifically:

  $$\begin{cases} h_A \equiv 1 &\text{on}   A \\ (I - \Delta)^{-(n + 1)/2} h_A \equiv 0&\text{on}   \mathbb{R}^n \setminus A. \end{cases}$$

  Indeed, the definition of weighting implies that $h_A \equiv 1$ on $A$, and the “second good thing” together with the fact that $w_A$ is supported on $A$ give the second clause.

  Not only do we have a differential equation for $h_A$, we also have boundary conditions. There are boundary conditions at the boundary of $A$, because of something I’ve been entirely vague about: the functions we’re dealing with are meant to be suitably smooth. There are also boundary conditions at $\infty$, because our functions are also meant to decay suitably fast.

  Maybe Mark will read this and correct me if I’m wrong, but I believe there are exactly the right number of boundary conditions to guarantee that there’s (typically? always?) a unique solution. In any case, the following example — also taken from Mark’s paper — illustrates the situation.

Example   Let’s calculate the magnitude of a real interval $A = [c, d]$ using the potential function method.

Its potential function $h_A$ is a function $\mathbb{R} \to \mathbb{R}$ such that $h_A \equiv 1$ on $[c, d]$ and $h_A - h''_A = 0$ on the rest of the real line. (That expression comes from taking $(I - \Delta)^{-(n + 1)/2}$ in the case $n = 1$.)

The functions $f$ satisfying $f'' = f$ are those of the form $k e^{\pm x}$ for some constant $k$, and in our case we’re choose the constant and the $\pm$ sign differently on the two connected components of $\mathbb{R} \setminus A$. So there are lots of solutions. But $h_A$ is required to be continuous and to converge to $0$ at $\pm \infty$, and that pins it down uniquely: the one and only solution is

$$h_A(x) = \begin{cases} e^{-(c - x)} &\text{if}   x \leq c, \\ 1 &\text{if}   c \leq x \leq c, \\ e^{-(x - d)} &\text{if}   d \leq x. \end{cases}$$

I showed you the graph of this potential function above, in the case where $A = [-2, 2]$.

So the magnitude of $A = [c, d]$ is

$$\frac{1}{c_1} \int_\mathbb{R} h_A(x)\, d x,$$

where $c_1$ is the constant $\int_\mathbb{R} e^{-|x|} \, d x = 2$. This gives the answer:

$$|A| = 1 + \tfrac{1}{2}(d - c) = 1 + \tfrac{1}{2}length(A).$$

 

The crucial point is that in this example, we didn’t have to come up with a weighting on $A$. The procedure was quite mechanical. And that’s the attraction of the method of potential functions.

Next time, I’ll put all this into a categorical context using the notion of the magnitude of a functor, which I introduced here.