The 14 closure operations

I’ve been getting complaints that my lemmas have not been so lovely (or little) lately, so let’s do something a bit more down to earth. This is a story I learned from the book Counterexamples in topology by Steen and Seebach [SS].

A topological space X comes equipped with various operations on its power set \mathcal P(X). For instance, there are the maps S \mapsto S^\circ (interior), S \mapsto \bar S (closure), and S \mapsto S^{\text{c}} (complement). These interact with each other in nontrivial ways; for instance (\bar S)^{\text{c}} = (S^{\text{c}})^\circ.

Consider the monoid M generated by the symbols a (interior), b (closure), and c (complement), where two words in a, b, and c are identified if they induce the same action on subsets of an arbitrary topological space X.

Lemma. The monoid M has 14 elements, and is the monoid given by generators and relations

    \[M = \left\langle a,b,c\ \left|\ \begin{array}{cc} a^2=a, & (ab)^2=ab, \\ c^2=1, & cac=b.\end{array}\right.\right\rangle.\]

Proof. The relations a^2 = a, c^2=1, and cac=b are clear, and conjugating the first by c shows that b^2=b is already implied by these. Note also that a and b are monotone, and a(T) \subseteq T \subseteq b(T) for all T \subseteq X. A straightforward induction shows that if v and w are words in a and b, then vaw(S) \subseteq vw(S) \subseteq vbw(S). We conclude that

    \[abab(S) \subseteq abb(S) = ab(S) = aab(S) \subseteq abab(S),\]

so (ab)^2=ab (this is the well-known fact that ab(S) is a regular open set). Conjugating by c also gives the relation (ba)^2 = ba (saying that ba(S) is a regular closed set).

Thus, in any reduced word in M, no two consecutive letters agree because of the relations a^2=a, b^2=b, and c^2=1. Moreover, we may assume all occurrences of c are at the start of the word, using ac=cb and bc=ca. In particular, there is at most one c in the word, and removing that if necessary gives a word containing only a and b. But reduced words in a and b have length at most 3, as the letters have to alternate and no string abab or baba can occur. We conclude that M is covered by the 14 elements

    \[\begin{array}{ccccccc}1, & a, & b, & ab, & ba, & aba, & bab,\\c, & ca, & cb, & cab, & cba, & caba, & cbab. \end{array}\]

To show that all 14 differ, one has to construct, for any i\neq j in the list above, a set S_{i,j} \subseteq X_{i,j} in some topological space X_{i,j} such that i(S_{i,j}) \neq j(S_{i,j}). In fact we will construct a single set S \subseteq X in some topological space X where all 14 sets i(S) differ.

Call the sets i(S) for i \in \langle a,b\rangle the noncomplementary sets obtained from S, and their complements the complementary sets. By the arguments above, the noncomplementary sets always satisfy the following inclusions:
Inclusions between different setsIt suffices to show that the noncomplementary sets are pairwise distinct: this forces a(S) \neq \varnothing (otherwise a(S) = ba(S)), so each noncomplementary set contains a(S) and therefore cannot agree with a complementary set.

For our counterexample, consider the 5 element poset P given by

    \[\begin{array}{ccccc}p\!\!\! & & & & \!\!\!q \\ \uparrow\!\!\! & & & & \!\!\!\uparrow \\ r\!\!\! & & & & \!\!\!s \\ & \!\!\!\nwarrow\!\!\! & & \!\!\!\nearrow\!\!\! & \\ & & \!\!\!t,\!\!\!\!\! & & \end{array}\]

and let X be the disjoint union of P^{\operatorname{Alex}} (the Alexandroff topology on P, see the previous post) with a two-point indiscrete space \{u,v\}. Recall that the open sets in P^{\operatorname{Alex}} are the upwards closed ones and the closed sets the downward closed ones. Let S = \{p,s,u\}. Then the diagram of inclusions becomes Lattice of inclusions for the exampleWe see that all 7 noncomplementary sets defined by S are pairwise distinct. \qedsymbol

Remark. Steen and Seebach [SS, Example 32(9)] give a different example where all 14 differ, namely a suitable subset S \subseteq \mathbf R. That is probably a more familiar type of example than the one I gave above.

On the other hand, my example is minimal: the diagram of inclusions above shows that for all inclusions to be strict, the space X needs at least 6 points. We claim that 6 is not possible either:

Lemma. Let X be a finite T_0 topological space, and S \subseteq X any subset. Then aba(S) = ab(S) and bab(S) = ba(S).

But in a 6-element counterexample X, the diagram of inclusions shows that any point occurs as the difference i(S) \setminus j(S) for some i,j \in \langle a,b\rangle\setminus 1. Since each of i(S) and j(S) is either open or closed, we see that the naive constructible topology on X is discrete, so X is T_0 by the remark from the previous post. So our lemma shows that X cannot be a counterexample.

Proof of Lemma. We will show that ab(S) \subseteq aba(S); the reverse implication was already shown, and the result for bab(S) follows by replacing S with its complement.

In the previous post, we saw that X is the Alexandroff topology on some finite poset (X,\leq). If T \subseteq X is any nonempty subset, it contains a maximal element t, and since X is a poset this means u \geq t \Rightarrow u=t for all u \in T. (In a preorder, you would only get u \geq t \Rightarrow u \leq t.)

The closure of a subset T \subseteq X is the lower set \bigcup_{t \in T} X_{\leq t}, and the interior is the upper set \{t \in T\ |\ X_{\geq t} \subseteq T\}. So we need to show that if x \in ab(S) and y \geq x, then y \in ba(S).

By definition of ab(S), we get y \in b(S), i.e. there exists z \in S with z \geq y. Choose a maximal z with this property; then we claim that z is a maximal element in X. Indeed, if u \geq z, then u \geq x so u \in b(S), meaning that there exists v \in S with v \geq u. Then v \geq z \geq y and v \in S, which by definition of z means v=z. Thus z is maximal in X, hence z \in a(S) since z \in S and X_{\geq z} = \{z\}. From z \geq y we conclude that y \in ba(S), finishing the proof. \qedsymbol

The lemma fails for non-T_0 spaces, as we saw in the example above. More succinctly, if X = \{x,y\} is indiscrete and S = \{x\}, then aba(S) = \varnothing and ab(S) = X. The problem is that x is maximal, but X_{\geq x} \supsetneq \{x\} and x \not\in a(S).


[SS] L.A. Steen, J.A. Seebach, Counterexamples in Topology. Reprint of the second (1978) edition. Dover Publications, Inc., Mineola, NY, 1995.

Hodge diamonds that cannot be realised

In Paulsen–Schreieder [PS19] and vDdB–Paulsen [DBP20], the authors/we show that any block of numbers

    \[\left(\begin{array}{ccccc} & & h^{n,n} & & \\ & \iddots & & \ddots & \\ h^{n,0} & & & & h^{0,n} \\ & \ddots & & \iddots & \\ & & h^{0,0} & & \end{array}\right) \in \big(\mathbf Z/m\big)^{(n+1)^2}\]

satisfying h^{0,0} = 1, h^{p,q} = h^{n-p,n-q}, and h^{p,q} = h^{q,p} (characteristic 0 only) can be realised as the modulo m reduction of a Hodge diamond of a smooth projective variety.

While preparing for a talk on [DBP20], I came up with the following easy example of a Hodge diamond that cannot be realised integrally, while not obviously violating any of the conditions (symmetry, nonnegativity, hard Lefschetz, …).

Lemma. There is no smooth projective variety (in any characteristic) whose Hodge diamond is

    \[\begin{array}{ccccc} & & 1 & & \\ & 1 & & 1 & \\ 0 & & 1 & & 0 \\ & 1 & & 1 & \\ & & 1.\!\! & & \end{array}\]

Proof. If \operatorname{char} k > 0, we have \sum_{p+q = i}h^{p,q} \geq h_{\operatorname{dR}}^i \geq h_{\operatorname{cris}}^i, with equality for all i if and only if the Hodge–de Rham spectral sequence degenerates and H_{\operatorname{cris}}^i is torsion-free for all i. Because H^2_{\operatorname{cris}} contains an ample class, we must have equality on h^2, hence everywhere because of how spectral sequences and universal coefficients work.

Thus, in any characteristic, we conclude that h^1_{\operatorname{Weil}}(X) = 2, so \dim \mathbf{Pic}_X^0 = 1 and the same for \dim \mathbf{Alb}_X. Thus, X \to \mathbf{Alb}_X is a fibration, so a fibre and a relatively ample divisor are linearly independent in the Néron–Severi group, contradicting the assumption h^{1,1}(X) = 1. \qedsymbol

Remark. In characteristic zero, the Hodge diamonds

    \[\begin{array}{ccccc} & & 1 & & \\ & a & & a & \\ 0 & & 1 & & 0 \\ & a & & a & \\ & & 1 & & \end{array}\]

cannot occur for any a \geq 1, by essentially the same argument. Indeed, the only thing left to prove is that the image X \to \mathbf{Alb}_X cannot be a surface. If it were, then X would have a global 2-form; see e.g. [Beau96, Lemma V.18].

This argument does not work in positive characteristic due to the possibility of an inseparable Albanese map. It seems to follow from Bombieri–Mumford’s classification of surfaces in positive characteristic that the above Hodge diamond does not occur in positive characteristic either, but the analysis is a little intricate.

Remark. On the other hand, the nearly identical Hodge diamond

    \[\begin{array}{ccccc} & & 1 & & \\ & a & & a & \\ 0 & & 2 & & 0 \\ & a & & a & \\ & & 1 & & \end{array}\]

is realised by C \times \mathbf P^1, where C is a curve of genus a. This is some evidence that the full inverse Hodge problem is very difficult, and I do not expect a full classification of which Hodge diamonds are possible (even for surfaces this might be out of reach).


[Beau96] A. Beauville, Complex algebraic surfaces. London Mathematical Society Student Texts 34 (1996).

[DBP20] R. van Dobben de Bruyn and M. Paulsen, The construction problem for Hodge numbers modulo an integer in positive characteristic. Forum Math. Sigma (to appear).

[PS19] M. Paulsen and S. Schreieder, The construction problem for Hodge numbers modulo an integer. Algebra Number Theory 13.10, p. 2427–2434 (2019).

Rings that are localisations of each other

This is a post about an answer I gave on MathOverflow in 2016. Most people who have ever clicked on my profile will probably have seen it.

Question. If A and B are rings that are localisations of each other, are they necessarily isomorphic?

In other words, does the category of rings whose morphisms are localisations form a partial order?

In my previous post, I explained why k[x] and k[x,x^{-1}] are not isomorphic, even as rings. With this example in mind, it’s tempting to try the following:

Example. Let k be a field, and let K = k(x_1, x_2, \ldots). Let

    \[A = K[x_0,x_{-1},\ldots]\]

be an infinite-dimensional polynomial ring over K, and let

    \[B = A\left[\frac{1}{x_0}\right].\]

Then B is a localisation of A, and we can localise B further to obtain the ring


isomorphic to A by shifting all the indices by 1. To see that A and B are not isomorphic as rings, note that A^\times \cup \{0\} is closed under addition, and the same is not true in B. \qed

Is there a moral to this story? Not sure. Maybe the lesson is to do mathematics your own stupid way, because the weird arguments you come up with yourself may help you solve other problems in the future. The process is more important than the outcome.