Re: integral homology

[John Cremona <john.cremona@gmail.com>]

Well, I no longer need an answer to my question since it turned out that in my specific application, the matrix A10 was so simply structured that I could simply replace all of steps 1-3 by a new preliminary step which just deleted certain rows from A21 without having to do the HNF or matrix inversion steps.

It would still be nice to know if there is a better way to do the general case,  For example, in the HNF computation of H and U from A10, I presume that a series of column operations are applied to A10, with U recording those operations; and then it would be easy to compute U^{-1} at the same time step by step, I think.  In some of the examples I was having difficulty with where the dimensions n0, n1, n2 are 36,14219, and 24599,  I was running out of memory even after setting PARISIZEMAX to 50000000000, i.e. 50G, and this occurred while in _ZM_inv_worker() so while inverting U.

John

On Fri, 15 Nov 2024 at 15:03, John Cremona <john.cremona@gmail.com> wrote:

I am computing some integral homology in a C++ program using libpari, and while what I have works fine it is possible that there is a better (simpler or faster) way to do this, possibly even a built-in function which I have not found which does what I want more simply.

I have Z-linear maps  maps Z^n2 -> Z^n1 -> Z^n0 whose composite is 0 with integer matrices A10 (size n0 x n1) and A21 (size n1 x n2), so acting on column vectors on the right, and A10*A21=0.  All I want is the abelian group structure of ker(A10)/im(A21).  A rough sketch of what I do is this:

1. Find the HNF of A10, say H = A10*U with U in GL(n1,Z) representing a change of basis for the module in the middle.

2. Compute M = U^{-1}*A21, the new matrix of the second map.

3.  Drop the last r rows of M, where r is the rank of H (also of A10).

4. Return the SNF of M.

Is there a better way?

John