4
$\begingroup$

For an additive function $\lambda$ and an exact sequence of modules

$0 \rightarrow M_1 \rightarrow M_2 \rightarrow M_3 \rightarrow 0$,

we have $\lambda(M_2) = \lambda(M_1) + \lambda(M_3)$ by definition. If the modules are graded and every morphism preserves degree, then the Poincaré series are also additive, simply by using the additivity of $\lambda$ on every summand of the power series. What if the first morphism merely carries homogeneous elements of degree $d$ to, say, elements of degree $d+k$ for instance? (and similarly for the second morphism) What would the formula for the Poincaré series be in this case?

I figured that I'd simply have to use additivity on a single sequence

$0 \rightarrow M_{1_n} \rightarrow M_{2_{n+deg(f)}} \rightarrow M_{3_{n+deg(f)+deg(g)}} \rightarrow 0$,

but I'm clueless on how to relate the Poincaré series' from here. (I deleted my old unanswered question since it was bloated and hid the main issue I was trying to point out)

$\endgroup$
4
$\begingroup$

If the first morphism has degree $k$ and the second has degree $n$, then you can instead work with the genuine exact sequence of graded modules $$0 \to \Sigma^{k+n} M_1 \to \Sigma^n M_2 \to M_3 \to 0$$

where $\Sigma^k M$ denotes $M$, but where elements of degree $d$ in $M$ are interpreted as elements of degree $d+k$. (By "genuine" I mean the morphisms preserve degree.) This has the effect of multiplying the Poincaré series by the appropriate powers of $t$ and the rest is straightforward. More precisely, letting $\chi$ denote the Poincaré series, we get $$t^n \chi(M_2) = t^{k+n} \chi(M_1) + \chi(M_3).$$

$\endgroup$
  • $\begingroup$ Neat, thanks a bunch! (upvote incoming when I have enough rep. [which I have as I've just seen.]) $\endgroup$ – jonny Mar 21 '12 at 19:41
  • $\begingroup$ No problem. For the sake of completeness, the more general fact is that you can define tensor products of graded modules (say where each piece is finite-dimensional over the base field $F$) and Poincaré series are multiplicative with respect to tensor products. To shift a graded module by $k$ just take the tensor product with the module which is $0$ in all degrees except $k$, where it is equal to $F$. $\endgroup$ – Qiaochu Yuan Mar 21 '12 at 19:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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