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

it is the only topology that can be given to a two element set other than the trivial and discrete topologies,
it has very bad separation properties: $M_{1, 1}$ is not Hausdorff, it is not even $T_1$ , and
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.