2
$\begingroup$

I'm studying a geometrical problem where an (even, negative-definite) lattice $L$ arises. Roughly, as an intersection pairing for curves on a surface. In fact, the problem naturally leads me to consider the shifted lattices $L + a$. In principle, you can certainly shift by any fractional vector $a \in L \otimes_{\mathbb{Z}} \mathbb{Q}$. But it turns out that I only ever have to shift by elements in the discriminant group, i.e. $a \in L^{*}/L$. I didn't require this, it came naturally from the geometry.

Moreover, the geometry tells me to consider a shifted lattice for each orbit of $L^{*}/L$ under the action of the isometry group of $L$. So the problem somehow spans all possible shifted lattices of this type, up to isometry.

Basically, I am looking for some context. Is there a structural or theoretical reason why one "should" shift by elements of the discriminant group as opposed to some general fractional vector? Like might this be hinting at something, or are these special kinds of shifted lattices that show up in other areas of math? One idea is that the discriminant group somehow measures non-unimodularity. Unimodular lattices show up in a closely related problem, so maybe this family of shifted lattices is naturally what you should consider in the non-unimodular case.

$\endgroup$
2
  • $\begingroup$ Just to clarify: are you looking at context from the perspective of lattices, or also from the perspective of topology? $\endgroup$ Commented Jun 8, 2021 at 7:28
  • $\begingroup$ I would be very interested in both! $\endgroup$
    – Benighted
    Commented Jun 8, 2021 at 11:04

1 Answer 1

1
$\begingroup$

From a topological perspective, to me what you're looking at is the inclusion of the middle-dimensional homology group into cohomology, in the case of an oriented manifold with boundary. The quotient is then injected into the homology of the boundary, by the long exact sequence of the pair. To be a bit more precise, let $W$ be a $4n$-dimensional manifold, with boundary $M$. Then from the long exact sequence of the pair you have: $$ H_{2n}(W) \to H_{2n}(W,M) \to H_{2n-1}(M) $$ If we throw away the torsion in the first two groups, $H_{2n}$ is a lattice $L$ (with respect to the usual intersection product), and $H_{2n}(W,M) = H^{2n}(W)$ (by Poincaré–Lefschetz duality) is its dual lattice $L^*$. The quotient $L^*/L$ then injects into $H_{2n-1}(M)$. (Things are cleaner if you assume that $W$ is constructed only using handles of index at most $2n$, and even cleaner if you suppose that there are no $2n-1$-handles.) Looking at the inclusion of homology in cohomology automatically restricts the shift you can do.

What you have is an extension/boundary problem: if you have a class in $A \in H_{2n-1}(M)$, your coset is the set of relative homology classes in $H_{2n}(W,M)$ whose boundary is $A$. (Dually, this corresponds to extending $2n$-cochains.)

From the algebraic perspective (which I know a lot less about), I think you're looking at a shadow of your lattice. I've seen shadows studied in the unimodular case: Elkies has two papers on "lattices with long shadows" (Math. Res. Lett. 1995), but more work has been done in the past 25 years. Related results in the non-unimodular case due to Owens and Strle (Amer. J. Math. 2012).

Since you mention that your lattice is even and that you're working with curves on surfaces: this reminds me of extending spin$^c$ structures from the boundary of a 4-manifold to the interior, and your dual lattice $L^*$ is $H^2$ of the 4-manifold (with $L$ being $H_2$, again). Translates of $L$ in $L^*$ correspond (up to doubling) to first Chern classes of spin$^c$ structures that restrict to a given spin$^c$ structure on the boundary. The paper by Owens and Strle on non-unimodular lattices has a bit of background on this, as far as I remember (it's the motivation that lead them to studying the problem in the first place).

$\endgroup$
2
  • $\begingroup$ Thanks a lot for the reply. My surfaces are projective, so it is not totally obvious to me where a manifold with boundary might enter the picture. But some of these papers on shadows of a lattice look intriguing and relevant somehow! $\endgroup$
    – Benighted
    Commented Jun 9, 2021 at 1:28
  • $\begingroup$ If your surfaces are singular, removing a neighbourhood of their singularities will turn them into 4-manifolds with boundary, which could explain the failure of unimodularity. If you're looking at Pic instead, then my comment is less relevant. But then again there's a lot of work on, for instance, K3 lattices, as initiated by Nikulin. $\endgroup$ Commented Jun 9, 2021 at 7:44

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .