A counterexample to finitistic-dimension finiteness

The explanation

Start with a small algebra: numbers a + bx, where x² = 0. It is a two-dimensional space with a multiplication. Take the one-dimensional module k, where x acts as zero. Try to rebuild k from free pieces, which are copies of the algebra itself. Map the algebra onto k by sending a + bx to a. The part that is lost, all the bx, is one-dimensional. It is again a copy of k. So the next step faces the same problem, and the process never stops.

Such a sequence of steps is called a projective resolution. A free piece is the simplest kind of projective module. The projective dimension of a module is the number of steps its resolution needs, when that number is finite. For k above it is infinite. In this small algebra, every module with finite projective dimension turns out to be free already, so the bound is 0.

The little finitistic dimension of an algebra collects only the modules whose resolutions stop. It is the largest projective dimension among finite-dimensional modules where that dimension is finite. Modules like k, with endless resolutions, are left out. The little finitistic-dimension conjecture asserts that this number is finite for every finite-dimensional algebraSource 1. Hyman Bass publicized these questions in 1960Source 2.

Much was known in favour. The conjecture held for monomial algebras and for algebras whose radical cubed is zeroSource 3. Igusa and Todorov proved it for algebras of representation dimension at most three. Huisgen-Zimmermann had shown that the little and big versions of the dimension can differSource 4. But each of her examples had a finite bound.

This paper gives a negative answerSource 5. It builds one finite-dimensional algebra over the complex numbersSource 6. For every whole number m, it has a module whose projective dimension is at least 2m − 2 and still finiteSource 6. So m = 1, 2, 3, 4 give lower bounds 0, 2, 4, 6, and the bounds keep growing. The algebra is chosen once, before m variesSource 7.

The idea is a countdown. The paper first builds an infinite algebra, the group algebra of a specially chosen groupSource 8. On it a fixed operation acts on modules. For each m there is a module that survives exactly m − 1 rounds of this operation and vanishes on round mSource 9. The paper then copies this countdown into a finite-dimensional algebra. There each round adds steps to a resolution. Vanishing means the resolution stops, and survival keeps it goingSource 10.

The one obstacle is making the copy work the same way for every module at onceSource 11. The paper builds a single fixed piece of machinery, a complex of bimodules, that imitates the operation uniformly. Relations that first hold only approximately are corrected by an extra layer of data. Then a bound on how long resolutions get over a helper algebra makes the result split cleanly.

There are side results. For this algebra, injective modules fail to generate a certain large category of modules, by a theorem of RickardSource 12. A related algebra has these dimensions infinite on the left and zero on the rightSource 13. A companion paper gives a separate example in characteristic 2, using another methodSource 14.

How sure is this? A computer proof checker called Lean has checked a formal statement of the main result. It gives a finite-dimensional complex algebra with modules of finite projective dimension at least 2m − 2, for every m. Lean also checks the left–right lopsided example. No specialist review of the paper has been published.

What it doesn't show. This does not say anything about modules whose resolutions never stop, which were already allowed.