2
$\begingroup$

Is there some other way to characterize the functions $f:\mathbb Z\times \mathbb Z\to \mathbb Z$ which are expressible as $$f(x,y)=g(x)+g(y)-g(x+y)$$ for some $g:\mathbb Z\to\mathbb Z$?

Easy facts: (1) $f$ must satisfy $f(x,y)=f(y,x)$ and $f(x,0)=g(0)$ for all $x$. (2) Not all functions $f$ are expressible, since for $x,y\in\{1,\ldots,n\}$ the number of choices of $f(x,y)$ the left grows quadratically in $n$ and the number of possible values of $g$ on which they depend, grows just linearly.

$\endgroup$
5
  • 1
    $\begingroup$ $f(x,0)=g(0)$, not $0$. $\endgroup$ Apr 5, 2011 at 10:34
  • 1
    $\begingroup$ en.wikipedia.org/wiki/Quadratic_form $\endgroup$ Apr 5, 2011 at 10:41
  • 1
    $\begingroup$ Here is a necessary condition: $$f(x,y+z)+f(y,z)=f(y,x+z)+f(x,z).$$ Could it be sufficient ? $\endgroup$ Apr 5, 2011 at 11:24
  • $\begingroup$ @ chris: yes, that's right. I will edit the post. $\endgroup$
    – Mircea
    Apr 5, 2011 at 12:06
  • $\begingroup$ @ steve and denis: Thanks Steve, that's an interesting link! In fact my original motivation was apparently different: One has a family of operators $I_x$ with $I_x\circ I_y=F(x,y) I_{x+y}, F>0$, I wanted to know when can I normalize the whole family cancelling away the $F$-factor. The answer of Denis means that associativity of the composition is the requirement! I find that very nice. $\endgroup$
    – Mircea
    Apr 5, 2011 at 12:41

2 Answers 2

4
$\begingroup$

The necessary and sufficient condition is that $$f(x,y+z)+f(y,z)=f(y,x+z)+f(x,z),\qquad\forall x,y,z\in\mathbb Z.\qquad (1)$$ On the one hand, this is obviously necessary. On the other hand, the function $g$, if it exists, is given by $$g(k)=kg(1)-\sum_{j=1}^{k-1}f(j,1).$$ Then $g$ is suitable if and only if $$f(k,\ell)=\sum_{j=1}^{k-1}f(j,1)-\sum_{j=\ell}^{k+\ell-1}f(j,1),\qquad k,\ell\in\mathbb Z.$$ There remains to check that these relations are consequences of (1).

Note that (1) contains the fact that $x\mapsto f(x,0)$ is constant.

$\endgroup$
1
  • $\begingroup$ Thank you for the fast answer! I think that this settles the question. $\endgroup$
    – Mircea
    Apr 5, 2011 at 12:25
5
$\begingroup$

This is secretly a question about group cohomology. Suppose we want to compute $\operatorname{Ext}^*_{\mathbb{Z}G}(\mathbb{Z}, \mathbb{Z})$ when $G=\langle g \rangle$ is infinite cyclic and the action on the integers is trivial. We can easily write down a free resolution

$$ 0 \to \mathbb{Z}G(g-1) \to \mathbb{Z}G \to \mathbb{Z} \to 0 $$

and work it out from there: in particular it's clear that $\operatorname{Ext}^n$ vanishes for $n>1$. But we could also try using the bar resolution. On writing this down and applying $\hom_{\mathbb{Z}G} ( -, \mathbb{Z})$ you find that the ext groups are given by the cohomology of the following cocomplex:

$$ \hom_{\mathbb{Z}G} (\mathbb{Z}G, \mathbb{Z}) \to \hom_{\mathbb{Z}G} (\bigoplus_{g \in G} \mathbb{Z}G[g], \mathbb{Z}) \to \hom_{\mathbb{Z}G} (\bigoplus_{g,h \in G} \mathbb{Z}G[g|h],\mathbb{Z}) \to \ldots $$

Of course, $\hom_{\mathbb{Z}G} (\bigoplus_{g \in G} \mathbb{Z}G[g], \mathbb{Z})$ can be identified with the set of maps of sets $G \to \mathbb{Z}$ and so on. On doing this we get that $\operatorname{Ext}^2$ is the kernel of the map

$$\partial_2 : \operatorname{Map}(G\times G, \mathbb{Z}) \to \operatorname{Map}(G\times G\times G, \mathbb{Z}) $$

given by $\partial_2 (f)(g,h,k) = f(h,k) - f(gh,k) + f(g,hk) -f(g,h)$, modulo the image of the map $\partial_1: \operatorname{Map}(G , \mathbb{Z}) \to \operatorname{Map}(G\times G, \mathbb{Z})$ defined by $\partial_1(f)(g,h) = f(h) - f(gh) + f(h)$.

We already know that this cohomology group is zero. But this says exactly that the set of functions $f:G \times G \to \mathbb{Z}$ expressible as $g(x)-g(xy)+g(y)$ is equal to those functions in the kernel of $\partial_2$.

This is equivalent to Denis Serre's answer above - it may look as if there's a discrepancy, but this is fixable by noting that any such f is symmetric.

$\endgroup$
1
  • $\begingroup$ Thanks mt! This is certainly a very suggestive interpretation, and it also has a nice side effect: it indicates clearly how one could generalize the problem! $\endgroup$
    – Mircea
    Apr 6, 2011 at 20:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.