The n-Category Café

To start with, recall from complex analysis the following facts:

• For any power series $\sum_{n=0}^\infty a_n z^n$ with complex coefficients, if $R = (\limsup_{n\to\infty} |a_n|^{1/n})^{-1}$ then the series converges to a complex-analytic function in the open disc of radius $R$ about the origin.
• Conversely, if $f$ is complex-analytic in the open disc of radius $R\gt 0$ about the origin, then it is equal to a unique power series there.

The following generalization of this is also standard, though it doesn’t always make it into first complex analysis courses:

• For any Laurent series $\sum_{n=-\infty}^\infty a_n z^n$ with complex coefficients, if $R = (\limsup_{n\to\infty} |a_n|^{1/n})^{-1}$ and $r = \limsup_{n\to \infty} |a_{-n}|^{1/n}$ then the series converges to a complex-analytic function in the open annulus of inner radius $r$ and outer radius $R$ centered at the origin.
• Conversely, if $f$ is complex-analytic in a nonempty open annulus centered at the origin, then it is equal to a unique Laurent series there.

Thus, there is a sort of bijection between Laurent series $\sum_{n=-\infty}^\infty a_n z^n$, by which formally I mean just functions $\mathbb{Z} \to \mathbb{C}$, and analytic functions defined on annuli centered at $0$. I say a “sort of” bijection because it may happen that $r=R$, in which case the corresponding open annulus is empty and thus the Laurent series can’t be recovered from any analytic function.

(Note that these Laurent series are infinite in both directions. In particular, this means that two such Laurent series can’t necessarily be multiplied to obtain a new one. For this reason the field of formal Laurent series requires $a_n\neq 0$ for only finitely many $n\lt 0$; analytically this means restricting to functions $f$ analytic in a punctured disc with a pole at the origin. Arbitrary bi-infinite Laurent series also allow essential singularities at the origin, as well as the case $r\gt 0$ of an annulus rather than a punctured disc.)

I recently learned that this operation taking a Laurent series (i.e. a function $a:\mathbb{Z}\to\mathbb{C}$) to its corresponding analytic function $f:\mathbb{C}\to \mathbb{C}$ is known as the Z-transform — except that for some reason people stick in a minus sign, writing $\mathcal{Z}(a_n) = \sum_{n=-\infty}^\infty a_n z^{-n}$. (Wikipedia says that the more mathematically natural convention $\sum_{n=-\infty}^\infty a_n z^n$ is used in geophysics.)

Note that the coefficients of the Laurent series corresponding to an analytic function can be obtained from Cauchy’s integral formula, also known (modulo a minus sign) as the “inverse Z-transform”: if $f(z) = \sum_{n=-\infty}^\infty a_n z^n$ then

$$a_n = \frac{1}{2\pi i} \oint f(z) z^{-n-1} \, dz$$

where the integral is around a counterclockwise circle contained in the annulus of definition. Now suppose we require the unit circle to be contained in the annulus, i.e. that $r\lt 1\lt R$, which on the Laurent series side amounts to imposing bounds on how fast $a_n$ can grow as $n\to\pm\infty$. If we additionally parametrize the unit circle by $t\in [0,1]$ as $z = e^{2\pi i t}$, with $g(t) = f(e^{2\pi i t})$, then the Laurent series expression on the unit circle becomes

$$g(t) = \sum_{n=-\infty}^\infty a_n e^{2 \pi i t n}$$

while Cauchy’s integral formula becomes

$$a_n = \frac{1}{2\pi i} \int_{t=0}^1 g(t) e^{- 2\pi i t n} \, dt.$$

(I’m sure I’m going to mess up the constants somewhere. When I do, please correct me in the comments and I’ll fix it in the post.)

Thus we have sort of recovered the discrete-time Fourier transform, relating functions $\mathbb{Z}\to \mathbb{C}$ with functions $S^1 \to \mathbb{C}$. Of course the general Fourier transform involves functions $S^1 \to \mathbb{C}$ that may not extend analytically to any annulus, and there is a lot more subtlety there. But I like seeing it as arising naturally from Laurent series expansions of analytic functions.

Now, what if we want to consider “power series” with non-integral exponents? This may seem like a weird thing to do, but there are various contexts in which it matters. The one that drew me into it was thinking about magnitude of metric spaces, which naturally involves inverting a matrix whose entries are of the form $q^{d(x,y)}$ for some $q$. The standard choice of $q$ is $e^{-1}$, but then people also scale the metric space by a variable $t$, which amounts to also looking at $q = e^{-t}$. So it seems natural to just treat $q$ as a formal variable, working over some kind of formal polynomial-like ring, and only plug in numbers for it afterwards.

If all the distances $d(x,y)$ in the metric space are integers (e.g. if it is a graph with the shortest-path metric), then $q^{d(x,y)}$ actually lives in a polynomial ring; but otherwise, it has to live somewhere more general. There is a ring $\mathbb{C}[q^{[0,\infty)}]$ of “generalized polynomials”, i.e. finite sums $\sum_{\beta\in S} a_\beta q^\beta$ where $S$ is a finite set of nonnegative real numbers, and it has a fraction field $\mathbb{C}(q^{[0,\infty)})$ of “generalized rational functions”. And for purposes of comparing magnitude to magnitude homology, we want to “do long division” to these generalized rational functions and obtain some kind of power series $\sum_{\beta\in S} a_\beta q^\beta$ where $S\subseteq \mathbb{R}$ can be infinite.

If we want these things to form a field, we have to impose some restrictions on $S$. If we require $S$ to be well-ordered, we get the Hahn series field $\mathbb{C}((q^{\mathbb{R}}))$. This is bigger than necessary for the purpose of magnitude, since the series corresponding to generalized rational functions always have order type $\omega$. However, arbitrary Hahn series of order type $\omega$ aren’t closed under multiplication; e.g. the product of $\sum_{n\in \mathbb{N}} q^n$ by $\sum_{n\in \mathbb{N}} q^{1-1/n}$ has order type $\omega^2$. But instead of expanding to the entire field of Hahn series, we can also restrict to those Hahn series of order type $\omega$ whose exponents approach $\infty$, or equivalently have only finitely many exponents in any interval $(-\infty,N)$ (thereby excluding $\sum_{n\in \mathbb{N}} q^{1-1/n}$). I learned recently that this is a sort of Novikov field, but I don’t know a good notation for it.

Anyway, today we’re not talking about magnitude, and we don’t need our formal series to form a field, but we do want to “evaluate” them at complex numbers and ask whether they converge. It seems to me that a natural restriction for this purpose, analogous to bi-infinite Laurent series $\sum_{n=-\infty}^\infty a_n z^n$, is to require that there are only finitely many exponents in any finite interval $[-N,N]$; but there are certainly other choices one could make. Specifically, for any series $\sum_{\beta\in S} a_\beta q^\beta$ with this property we can evaluate the terms at any given $q$, write down partial sums (either treating both directions $\pm\infty$ separately, or with a “conditional convergence” version that goes in both directions at once), and ask whether they converge.

However… what sort of thing can $q$ be? By analogy with Laurent series, we’d like it to be a complex number; but for a non-integer $\beta$, the function $q^\beta$ of a complex $q$ is multi-valued. Specifically, we can define it as $q^\beta = e^{\beta \log q}$, so every value of $\log q$ gives us a different value of $q^\beta$. Since the infinitely many values of $\log q$ differ by integer multiples of $2\pi i$, the infinitely many values of $q^\beta$ are related by factors of $e^{2\pi i\beta}$ — which is $1$ if $\beta$ is an integer, so that in that case $q^\beta$ is uniquely defined, but not in general otherwise. (If $\beta$ is rational, then $e^{2\pi i\beta}$ is a root of unity, so that $q^\beta$ has finitely many distinct values; but for irrational $\beta$ it has infinitely many.)

The simplest category-theoretically principled way to make sense of a “multi-valued function” $X\to Y$ is to consider it to be an ordinary function on a larger space, i.e. a span $X \leftarrow H \to Y$ such that $H\to X$ is surjective. In the case of analytic functions like $\log q$ and $q^\beta$, we have $X=Y=\mathbb{C}$, while $H$ is a Riemann surface. Since $q^\beta$ is defined in terms of $\log q$, for our purposes it suffices to fix $H$ as the Riemann surface of $\log$, which is a sort of helix fibred over $\mathbb{C}$ with fiber $\mathbb{Z}$ over every nonzero point:

I find it pleasing to simply use the variable $q$ to denote a point of this surface $H$. Thus $q$ is not, properly speaking, a complex number itself; but we can take its logarithm (as long as it’s nonzero), and raise it to real exponents, getting a complex number as an answer. If we write $\log q = x+ y i$, then $e^x$ is the absolute value $|q|$, while $y$ is the argument/angle of $q$: note that since $q\in H$ rather than $\mathbb{C}$, its argument is uniquely defined (rather than up to $2\pi$).

Now we can interpret a sum like $\sum_{\beta\in S} a_\beta q^\beta$ as making sense (and possibly converging) for any $q\in H$. I expect that there is a suitable definition of $r$ and $R$ in terms of $a_\beta$ such that this sum converges on the “helical strip” $r\lt |q|\lt R$ to a function analytic on $H$ (maybe someone in the comments will confirm or refute that). However, recovering the series from the function is a thornier problem, since if we “go around a circle” once in $H$ we don’t get back to where we started, so we can’t integrate around a loop as in the Cauchy integral formula.

Before tackling that problem, let’s consider a further generalization. A Laurent series allows us to add up terms $a_n z^n$ for all integers $n$; can we generalize our Novikov series to add up terms $a_\beta q^\beta$ for all real numbers $\beta$ rather than just a countable family of them? (For instance, we might be interested in magnitude of infinite metric spaces.) For this to make sense (and in particular, for it to have a chance of being finite), we ought to take account of the topology of $\mathbb{R}$ and do an integral rather than a sum; thus we’re thinking about “formal integrals” of the form

$$\int_{-\infty}^\infty \alpha(\beta)\, q^\beta \,d\beta.$$

Our coefficient function $a:\mathbb{Z}\to \mathbb{C}$ has now, of course, been replaced by a function $\alpha:\mathbb{R}\to \mathbb{C}$. But actually, it’s even better to allow $\alpha$ to be a distribution. For one thing, this is the only way the continous version can include the discrete version: given $a:\mathbb{Z}\to \mathbb{C}$, we can take $\alpha = \sum_{n\in \mathbb{Z}} a_n \delta_n$ to be a modulated Dirac comb distribution to make the integral reduce to the sum.

Thus, we can take a complex-valued distribution on $\mathbb{R}$, representing the “coefficients” of a “continuous Novikov series”, and integrate it against $q^\beta$ to obtain an analytic function on some region in $H$ (wherever the integral converges), probably a helical strip $r\lt |q|\lt R$.

This is the Laplace transform!

Unfortunately, people don’t usually write the Laplace transform in terms of $H$ and $q$. Instead, they fix a particular coordinatization of $H$ and work entirely with those coordinates, which (in my view) obscures this nice geometric/analytic picture. To obtain this coordinatization, note that $\log : H \setminus \{0\} \to \mathbb{C}$ is in fact an isomorphism. The inverse of a logarithm is, of course, an exponential map; this means that for $t\in \mathbb{C}$ we can regard $e^t$ as being, not a complex number, but a point of $H$ (the argument of this point “remembers” the whole imaginary part of $t$, as opposed to the complex $e^t\in \mathbb{C}$ which only remembers $Im(t)$ modulo $2\pi$).

If we now stick in another random minus sign, we obtain the coordinatization $(t\mapsto e^{-t}) : \mathbb{C} \to H \setminus \{0\}$ that people generally use. Writing our integral in terms of $q = e^{-t}$, we obtain

$$F(t) = \int_{-\infty}^\infty \alpha(\beta)\, e^{-\beta t} \,d\beta.$$

which is the standard formula for the two-sided Laplace transform. (The “one-sided” version just restricts $\alpha$ to have support in $[0,\infty)$, giving a sort of continuous analogue of power series rather than Laurent series.)

Note that under this coordinatization, $q\to 0$ corresponds to $Re(t)\to\infty$. Thus, if the $q$-integral converges in a helical strip $r\lt |q|\lt R$ on $H$, the corresponding $t$-integral converges in a vertical strip $-\log(R) \lt Re(t) \lt -\log(r)$ on $\mathbb{C}$. (I say “if” because I expect that it always does, but I haven’t been able to find references on the Internet about the convergence properties of the Laplace transform when $\alpha$ is only a distribution, and since this is just a lark for me I haven’t actually tried to look it up in textbooks.)

Remember that we got the discrete-time Fourier transform from Laurent series by looking at the unit circle? The corresponding thing to look at in this case is the “unit helix” $|q|=1$ in $H$, which corresponds to the imaginary axis $Re(t)=0$. And indeed, if we assume these are contained in the region of convergence, we obtain upon restriction to the line $Re(t)=0$ the standard continuous-time Fourier transform of $\alpha$.

The fiddliest part of the picture is the inverse Laplace transform, and I’m not going to give any details here (partly because I don’t fully understand them myself). Instead of integrating around a circle contained in the annulus of convergence, we integrate along a helix in $H$ contained in the helical strip of $q$-convergence, or equivalently a vertical line contained in the vertical strip of $t$-convergence. In the discrete case, we used the fact that the integrals of $z^n$ around a loop are equal to zero except when $z=-1$, in order to build a sort of Kronecker-delta and extract the coefficients $a_n$ of a Laurent series one by one. In the continuous case, I believe that the integral of $q^\beta$ along the helix actually doesn’t exist for any $\beta$, but it fails to exist in different ways: for $\beta\neq -1$ it oscillates, while for $\beta=-1$ it diverges. Thus, if we are careful about how we pass to the limit in computing the integrals (the formulas I’ve seen appear to be a version of Cauchy principal value), we can use them to build a sort of Dirac delta and extract the “coefficients” function $\alpha(\beta)$.

This method allows us to recover $\alpha$ from its Laplace transform, so that the transform is injective. But characterizing the image of the Laplace transform seems to be much harder than for the Z-transform. In the discrete case, as I recalled above, every analytic function in an annulus is equal to a convergent Laurent series in that annulus. But it seems that the same is not true for analytic functions on helical strips in $H$ — or if it is, I haven’t found anyone pointing it out. (If it were true, I expect it would require the generality of $\alpha$ being a distribution, or at least a measure.) Maybe someone will drop by and give an answer in the comments!