How does Hermite normal form (over $Z$) vary in families? I.e. if I have an $n\times m$ matrix $M$ whose entries are integral polynomials in some integral variable $x$, how does the Hermite normal form of the integral matrix $M(p)$ (obtained by setting $x$ equal to $p$) vary as a function of $p$? What about the special case that the entries are (at most) linear in $x$?

The question is a bit open ended so answers could be of several kinds, eg: (i) how certain integer programming problems associated to $M(p)$ depend on $p$; (ii) by explaining how the answer can be expressed in a way that generalizes Ehrhart theory; (iii) by specializing to an important case that is well-understood; (iv) in some other interesting way.

I would also really appreciate a pointer to any relevant literature.

  • $\begingroup$ My guess is that for x sufficiently large the Hermite normal form itself consists of integral polynomials. One should be able to prove this by computing the analogue of the Hermite normal form over Z[x] and then showing that it satisfies the conditions of the the usual Hermite normal form for large x. $\endgroup$ – Qiaochu Yuan Nov 21 '09 at 3:46
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    $\begingroup$ Unfortunately, I don't think it is that simple. For example, a family of 2x2 upper triangular matrices with 2s on the diagonal and x in the upper right has "periodic" Hermite normal form over Z (the upper right entry alternates between 0 and 1). This is why I speculated about Ehrhart theory (eg. maybe the entries are eventually quasipolynomials) but this is just a guess too. I do think your guess is probably true for some "generic" families, in a sense that needs a bit more clarification . . . $\endgroup$ – Danny Calegari Nov 21 '09 at 7:22

Hi "DC". I think that I have worked out that the Hermite normal form is a "trichotomous quasipolynomial" in the variable $p$. If $f:\mathbb{Z} \to \mathbb{Z}$ is a function, then my definition is that $f$ is a trichotomous quasipolynomial if it is a quasipolynomial for $x \gg 0$, possibly a different quasipolynomial for $x \ll 0$, and in between finitely many unrestricted values.

I think that if $R$ is a canonically Euclidean ring — Euclidean with canonically chosen quotients and remainders — then there is a Hermite normal form for matrices over $R$. In particular, I think that $R$ does not have to be a Euclidean domain.

As a first try, let $A$ be the ring of all functions $f:\mathbb{Z} \to \mathbb{Z}$, using pointwise quotients and remainders. Hermite normal form over this ring is a model of computing Hermite normal form for any $\mathbb{Z}$ family of integer matrices. $A$ is sort-of a Euclidean ring, except it isn't Noetherian.

Let $B$ be the subring of $A$ consisting of trichotomous quasipolynomials. Then I believe that $B$ is Noetherian and it is a Euclidean ring. If that is correct, then you obtain a Hermite normal form that is also a trichotomous quasipolynomial.

It's not correct, at least not in any obvious way. It is easy to check that $B$ is not only a subring, but is also closed under quotients with pointwise remainders. In order to compute how $b(x)$ divides into $a(x)$, you can reduce to the case in which $a(x)$ and $b(x)$ are both polynomials. Then for starters there is a quotient and a remainder in $\mathbb{Q}[x]$: $$a(x) = q(x)b(x) + r(x).$$ Since $r(x)$ has lower degree than $b(x)$, its values are smaller than those of $b(x)$ when $x \gg 0$, so that part is okay; but it may be negative and $q(x)$ may not be integral. We can fix all that by rounding $q(x)$, which creates quasipolynomial behavior; and by changing it by 1 to make $r(x)$ positive. Also since $r(x)$ might have odd degree, there may be a different quasipolynomial solution when $x \ll 0$.

The part that is either not true or far from obvious is why $B$ is Euclidean. My argument for that fell apart. However, I can still show that the Euclidean algorithm for finitely many elements $a_1,\ldots,a_n$ of $B$ terminates in a finite number of steps, and consequently that the Hermite normal form stabilizes in a finite number of steps with trichotomous quasipolynomial entries. The proof is a two-stage induction. The outer stage is the sum of the degrees of $a_1,\ldots,a_n$. If $\deg a_j < \deg a_k$ for some $j$ and $k$, then dividing $a_j$ into $a_k$ reduces the degree of $a_k$. On the other hand, suppose that the degrees are all equal. Then we can pass to a congruence class for the input $x$ in $\mathbb{Z}$ and apply a linear change of variables so that the leading coefficients are all integers. Then (in the inner induction) the Euclidean algorithm on these polynomials, for $x \gg 0$, amounts to the Euclidean algorithm on their coefficients. It is important, in this inner inductive part, to only change the variable $x$ once; don't worry if the lower-order coefficients are not integers. Eventually a $0$ is produced and the degrees once again decrease.

Alas, this is a very informal writeup, but this time I think that it works.

  • $\begingroup$ OK, I will take away your "tick" and then put it back again. Thanks. $\endgroup$ – Danny Calegari Dec 1 '09 at 16:31
  • $\begingroup$ You don't need a Euclidean ring (nor Noetherian!) to get Hermite normal form, just a Hermite ring (Kaplansky, "Elementary Divisors and Modules", Trans. AMS, 1949). I believe it is almost trivial to show that your ring $B$ is Hermite: you just need that for any $(a,b)$, there is an invertible 2x2 matrix $Q$ over $B$ such that $(a,b)Q = (d,0)$ for some $d$. The key thing to show is that this is true for quasipolynomials; I think the trichotomousness (trichotomosity?) almost follows for free if $a,b$ are. $\endgroup$ – Joshua Grochow Nov 19 '18 at 15:00

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