Where Do Linearly Compact Vector Spaces Come From?
Posted by Tom Leinster
Where do what come from?
Linearly compact vector spaces are what fill in this blank:
sets are to vector spaces as compact Hausdorff spaces are to ???
I’ll explain the exact sense in which that’s true, as part of the continuing story of codensity monads. Of course, I’ll also give you the actual definition.
Here’s the idea. Compactness of a topological space is something like finiteness of a set; indeed, the archetypal example of a compact space is a finite set with the discrete topology. Can we find a decent notion of “compactness” for topological vector spaces, such that the archetypal example of a “compact” vector space is a finite-dimensional vector space with the discrete topology?
Compactness itself won’t do: for example, $\mathbb{R}^2$ is finite-dimensional but not compact (with either the euclidean or the discrete topology). But linear compactness will fulfil all our desires.
Before I go further, I want to thank Todd Trimble. I’d never heard of linearly compact vector spaces until Todd told me about them in a MathOverflow answer. Thanks, Todd! More on that answer later.
Here’s a quick summary of my last two posts:
- Every functor $G: \mathbf{B} \to \mathbf{A}$ has a so-called codensity monad, which is a monad on $\mathbf{A}$. (Well: not quite every functor, but subject only to the existence of certain limits.)
- The codensity monad of the inclusion $\mathbf{FinSet} \hookrightarrow \mathbf{Set}$ is the ultrafilter monad $U$. Here $\mathbf{FinSet}$ is the category of finite sets, and the ultrafilter monad sends a set $X$ to the set $U(X)$ of ultrafilters on $X$.
This time, we’re going to think about algebras for codensity monads.
What, in particular, are the algebras for the ultrafilter monad? This question is answered by a well-known theorem of Manes:
Theorem The algebras for the ultrafilter monad are the compact Hausdorff spaces.
Last time I said “ultrafilters are inevitable”, because they arise via a general categorical construction from the inclusion $\mathbf{FinSet} \hookrightarrow \mathbf{Set}$. In the same sense, compact Hausdorff spaces are inevitable.
Let me take a moment to explain roughly why Manes’s theorem is true. An ultrafilter on a topological space $X$ can be viewed as something like sequence in $X$. In particular, one can say what it means for an ultrafilter on $X$ to “converge” to a point of $X$. (I won’t say exactly what this means — it’s not difficult, but I don’t want to digress too much.) Here are some appealing facts about ultrafilter convergence:
- $X$ is compact iff every ultrafilter on $X$ converges to at least one point.
- $X$ is Hausdorff iff every ultrafilter on $X$ converges to at most one point.
- $X$ is, therefore, compact Hausdorff iff every ultrafilter on $X$ converges to exactly one point.
So if $X$ is a compact Hausdorff space, there is a map of sets $U(X) \to X$ assigning to each ultrafilter its unique limit. Moreover, if you know which ultrafilters converge to which points, you can recover the whole topology. So a compact Hausdorff space can be viewed as a set $X$ together with a map $U(X) \to X$, satisfying some axioms. Those axioms turn out to be exactly the axioms for an algebra for a monad: hence Manes’s theorem.
So, general categorical machinery turns the inclusion $\mathbf{FinSet} \hookrightarrow \mathbf{Set}$ into the category of compact Hausdorff spaces. The game now is to feed other, similar, functors into the same machine. For example: what happens if we feed in the inclusion functor
$\mathbf{FDVect} \hookrightarrow \mathbf{Vect}?$
Here $\mathbf{Vect}$ is the category of vector spaces over some field $k$, and $\mathbf{FDVect}$ is the category of finite-dimensional vector spaces. The questions we have to answer are:
- What is the codensity monad of $\mathbf{FDVect} \hookrightarrow \mathbf{Vect}$? (This is a monad on $\mathbf{Vect}$, analogous to the ultrafilter monad on $\mathbf{Set}$.)
- What are the algebras for this codensity monad? (Since it’s a monad on $\mathbf{Vect}$, these are vector spaces equipped with some extra structure. They are the linear analogue of compact Hausdorff spaces.)
Without further ado, the answer to Question 1 is:
Theorem The codensity monad of $\mathbf{FDVect} \hookrightarrow \mathbf{Vect}$ is the double dualization monad.
As you’d guess, the double dualization monad sends a vector space $X$ to its double dual $X^{**}$. So, I’m inviting you to think of elements of a double dual space as the linear analogues of ultrafilters.
If you want the full proof, see Theorem 7.5 of my recent paper. I’ll just sketch it here.
It’s similar to the proof that the codensity monad of $\mathbf{FinSet} \hookrightarrow \mathbf{Set}$ is the ultrafilter monad, which I sketched last time. The key idea there was integration against an ultrafilter. To recap: given a set $X$, an ultrafilter $\Omega$ on $X$, and a finite set $B$, we have a canonical map
$\int_X - \, d\Omega \colon \mathbf{Set}(X, B) \to B.$
Analogously, in the linear situation, we can integrate against an element of the double dual. That is, given a vector space $X$, an element $\Omega$ of $X^{**}$, and a finite-dimensional vector space $B$, we have a canonical map
$\int_X - \, d\Omega \colon \mathbf{Vect}(X, B) \to B.$
I won’t explain what this integration is. You can find it in Proposition 7.1 of my paper. Or maybe you can find a slicker construction yourself — there’s probably only one sensible way to assign to each element of $X^{**}$ and finite-dimensional vector space $B$ a map $\mathbf{Vect}(X, B) \to B$.
The upshot is that for each $X$ we have a canonical map
$X^{**} \to \int_{B \in \mathbf{FDVect}} [\mathbf{Vect}(X, B), B]$
given by
$\Omega \mapsto \int_X - \, d\Omega.$
The second-to-last integral sign denotes an end. This end is actually $T(X)$, where $T$ is the codensity monad of $\mathbf{FDVect} \hookrightarrow \mathbf{Vect}$. So, we’ve just defined a natural transformation $( )^{**} \to T$.
Just as in the case of sets and ultrafilters, elements of the double dual space $X^{**}$ can be thought of as something like measures on $X$. The map $X^{**} \to T(X)$ then assigns to each measure on $X$ an integration operator. As before, this correspondence between measures and integrals turns out to be one-to-one; and that, in essence, proves the theorem.
So, the double dualization monad is the linear analogue of the ultrafilter monad. This answers Question 1. But Question 2 asks: what are its algebras?
My train is about to reach its destination, so I’ll wrap this up quickly. Here’s the answer to Question 2:
Theorem The algebras for the codensity monad of $\mathbf{FDVect} \hookrightarrow \mathbf{Vect}$ are the linearly compact vector spaces.
I learned this from Todd Trimble: see this MathOverflow answer or Theorem 7.8 of my paper. (There are also similar results in the paper by Kennison and Gildenhuys that I mentioned last time.)
I won’t explain the proof, but I will tell you what a linearly compact vector space is.
A linearly compact vector space is a topological vector space with certain properties. Our ground field $k$ has no topology, so I have to be clear about the meaning of “topological vector space”: it’s either a $k$-vector space internal to $\mathbf{Top}$, or it’s a topological vector space with respect to the discrete topology on $k$. They amount to the same thing.
A linearly compact vector space over a field $k$ is a topological vector space over $k$ such that:
- the topology is linear: the open affine subspaces form a basis for the topology
- any family of closed affine subspaces with the finite intersection property has nonempty intersection
- the topology is Hausdorff.
The first condition might strike you as really bizarre. For example, it’s not true of $\mathbb{R}^n$ with the Euclidean topology, which is probably the first topological vector space that pops into your head. In fact, a finite-dimensional vector space can be made into a linearly compact vector space in one and only one way: by giving it the discrete topology. This is like the fact that a finite set can be made into a compact Hausdorff space in one and only one way — again, by giving it the discrete topology.
The second condition is the analogue of compactness. “Finite intersection property” means that the intersection of any finite number of members of the family is nonempty. It’s just like the fact that a topological space is compact iff every family of closed subsets with the finite intersection property has nonempty intersection.
As for the third condition, well: as so often happens, “Hausdorff” is taken for granted to such an extent that the terminology ignores it. They’re not called “linearly compact Hausdorff vector spaces”, though perhaps they should be.
In summary, we have the following table of analogues:
sets vector spaces finite sets finite-dimensional vector spaces ultrafilters elements of the double dual compact Hausdorff spaces linearly compact vector spaces
Can this be extended to theories other than vector spaces? I don’t know. But I hope someone will find out!
In the final installment: are ultraproducts part of the codensity story?
Re: Where Do Linearly Compact Vector Spaces Come From?
This stuff is wicked cool!