A Cursed Space in the Wild

Published June 03, 2024

Recently, I had the misfortune of coming across the Sierpiński 2-point space in the wild. And while I’m sure that I must have seen it within minutes of learning the definition of a topological space, seeing it arise naturally in a linear algebra problem (of all places!) was horrifying.

Consider the problem of classifying linear maps between two finite dimensional complex vector spaces, \(A : V \to W\), up to change of basis. If you are very smart, you will eventually conclude that this is done completely by the set of numbers \((\dim V,\) \(\dim W,\) \(\dim \ker A,\) \(\dim \mathrm{im}\, A)\), subject to the rank–nullity theorem.

But if you are geometrically minded, you might think that linear maps are just matrices, so the space of all linear maps must be \(R_{m, n} = \mathrm{Mat}_{n \times m}(\mathbb{C}) \cong \mathbb{C}^{mn}\), where \(\dim V = m\) and \(\dim W = n\). And identifying linear maps that are equivalent under change of bases is done by quotienting this space by the group \(G_{m, n}\) \(= \mathrm{GL}(V) \times \mathrm{GL}(W) \cong\) \(\mathrm{GL}_m(\mathbb{C}) \times \mathrm{GL}_n(\mathbb{C})\). And then, you will conclude that the classification problem is solved completely by the space \(M_{m, n} = R_{m, n} / G_{m, n}\).

Fantastic! Let us try to compute this space in some simple cases.

First, note that for \(A \in R_{m, n}\) and \((S, T) \in G_{m, n}\), the action is given by

\[(S, T)\cdot A = T A S^{-1},\]

which is exactly a change of basis.

Now, let us do the simplest case: \(m = n = 1\). We have \(R_{1, 1} \cong \mathbb{C}\) and \(G_{1, 1} \cong \mathbb{C}^* \times \mathbb{C}^*\), and the action \(G_{1, 1} \curvearrowright R_{1, 1}\) is given by \((s, t)\cdot a = (t / s) a\) for every \(a \in \mathbb{C}\) and \(s, t \in \mathbb{C}^*\). It is an easy exercise to show that \(M_{1, 1} = \{[0], [1]\}\).

The good news is that \(M_{1, 1}\) has two points—corresponding to the two orbits of \(G_{1, 1}\) in \(R_{1, 1}\)—which matches with the linear algebra answer. The bad news is that the quotient topology makes \(M_{1, 1}\) the Sierpiński 2-point space.

Indeed, the point \(\{[0]\} \subset M_{1, 1}\) is closed because the orbit is a singleton, whose complement is open in \(\mathbb C.\) It is not open because the orbit is not open in \(\mathbb C\). \(\{[1]\} \subset M_{1, 1}\) is open because its orbit \(\mathbb C \setminus \{0\} \subset \mathbb C\) is open. It is not closed because the complement of its orbit is a singleton and therefore not open in \(\mathbb C\). Furthermore, as a result, the closure of \({\{[1]\}}\) is equal to the whole space, \(M_{1, 1}\).

The only things I knew about this space before seeing this example were that

  1. it is the only topology that can be given to a two element set other than the trivial and discrete topologies,
  2. it has very bad separation properties: \(M_{1, 1}\) is not Hausdorff, it is not even \(T_1\), and
  3. its only function is to be a counterexample that instructors use to torture undergrads in their first topology course.

And that’s why I was horrified when I saw it come up in, what is essentially, an elementary linear algebra problem.

I learned about this in Marcus Reineke’s great series of lectures on Quivers and Cohomological Hall Algebras at the Galileo Galilei Institute in May 2024.