# 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 comes equipped with various operations on its power set . For instance, there are the maps (interior), (closure), and (complement). These interact with each other in nontrivial ways; for instance .

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

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

Proof. The relations , , and are clear, and conjugating the first by shows that is already implied by these. Note also that and are monotone, and for all . A straightforward induction shows that if and are words in and , then . We conclude that

so (this is the well-known fact that is a regular open set). Conjugating by also gives the relation (saying that is a regular closed set).

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

To show that all 14 differ, one has to construct, for any in the list above, a set in some topological space such that . In fact we will construct a single set in some topological space where all 14 sets differ.

Call the sets for the noncomplementary sets obtained from , and their complements the complementary sets. By the arguments above, the noncomplementary sets always satisfy the following inclusions:
It suffices to show that the noncomplementary sets are pairwise distinct: this forces (otherwise ), so each noncomplementary set contains and therefore cannot agree with a complementary set.

For our counterexample, consider the 5 element poset given by

and let be the disjoint union of (the Alexandroff topology on , see the previous post) with a two-point indiscrete space . Recall that the open sets in are the upwards closed ones and the closed sets the downward closed ones. Let . Then the diagram of inclusions becomes We see that all 7 noncomplementary sets defined by are pairwise distinct.

Remark. Steen and Seebach [SS, Example 32(9)] give a different example where all 14 differ, namely a suitable subset . 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 needs at least 6 points. We claim that 6 is not possible either:

Lemma. Let be a finite topological space, and any subset. Then and .

But in a 6-element counterexample , the diagram of inclusions shows that any point occurs as the difference for some . Since each of and is either open or closed, we see that the naive constructible topology on is discrete, so is by the remark from the previous post. So our lemma shows that cannot be a counterexample.

Proof of Lemma. We will show that ; the reverse implication was already shown, and the result for follows by replacing with its complement.

In the previous post, we saw that is the Alexandroff topology on some finite poset . If is any nonempty subset, it contains a maximal element , and since is a poset this means for all . (In a preorder, you would only get .)

The closure of a subset is the lower set , and the interior is the upper set . So we need to show that if and , then .

By definition of , we get , i.e. there exists with . Choose a maximal with this property; then we claim that is a maximal element in . Indeed, if , then so , meaning that there exists with . Then and , which by definition of means . Thus is maximal in , hence since and . From we conclude that , finishing the proof.

The lemma fails for non- spaces, as we saw in the example above. More succinctly, if is indiscrete and , then and . The problem is that is maximal, but and .

References.

[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

satisfying , , and (characteristic only) can be realised as the modulo 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

Proof. If , we have , with equality for all if and only if the Hodge–de Rham spectral sequence degenerates and is torsion-free for all . Because contains an ample class, we must have equality on , hence everywhere because of how spectral sequences and universal coefficients work.

Thus, in any characteristic, we conclude that , so and the same for . Thus, is a fibration, so a fibre and a relatively ample divisor are linearly independent in the Néron–Severi group, contradicting the assumption .

Remark. In characteristic zero, the Hodge diamonds

cannot occur for any , by essentially the same argument. Indeed, the only thing left to prove is that the image cannot be a surface. If it were, then 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

is realised by , where is a curve of genus . 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).

References.

[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 and 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 and are not isomorphic, even as rings. With this example in mind, it’s tempting to try the following:

Example. Let be a field, and let . Let

be an infinite-dimensional polynomial ring over , and let

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

isomorphic to by shifting all the indices by 1. To see that and are not isomorphic as rings, note that is closed under addition, and the same is not true in .

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.