# Morphisms of Spectral Sequences and alternating products

Let $E_{a,b}^{r}, F_{a,b}^{r}$ be two (co)homologica first quadrant spectral sequences of vector spaces over a field $K$, and $f : E \to F$ be a morphism of spectral sequences.

Assume that morphisms $f_{a,b}^{1} : E_{a,b}^{1} \to F_{a,b}^{1}$ of the first page are all isomoprhisms.

If the numbers of non-zero terms appearing in the first pages of $E$ and $F$ are finite, then one can show that the following equation holds:

$\prod_{n}(\det (f_{n}))^{(-1)^{n}} = \prod_{a,b}(\det (f_{a,b}^{1}))^{(-1)^{a+b}}$.

Here, $f_{n}$ denotes the morphism $f_{n} : E_{n} \to F_{n}$, where $E_{n}$ and $F_{n}$ are the $n$th terms to which $E$ and $F$ converge(i.e. $E_{n}$ is a vector space equipped with filtration whose $i$th congruent is isomorphic to $E^{\infty}_{i,n-i}$).

(In order to define $\det$, one has to fix bases, or assume $F=E$ and $f$ is an endomorphism of $E$.)

My question is :

QUESTION

Assume that $E_{n}$ and $F_{n}$ are zero for any $n$ large enough.

Then, the left hand side of the above equation is well-defined.

If the numbers of non-zero terms appearing in the first pages of $E$ and $F$ are not necessarily finite, is there any way to calculate the left hand side by using $f_{a,b}^{1}$s?

If so, then how can it be done?

# Maybe

this question can be essentially reduced to the following:

Let $C_{\ast}$ be a chain complex of $K$-vector spaces of the form $\cdots \to A_{2} \to A_{1} \to A_{0} \to 0$.

And let $f_{\ast} : C_{\ast} \to C_{\ast}$ be an endomorphism of the chain complex $C_{\ast}$.

Assume that the homology $H_{n}(C_{\ast})$ is zero for any $n>>0$.

Then, enoting by $H_{n}(f_{\ast})$ the homomorphism $H_{n}(C_{\ast}) \to H_{n}(C_{\ast})$ induced by $f$, the alternating product $\prod_{n} \det (H_{n}(f_{\ast}))^{(-1)^{n}}$ is well-defined if each $\det (H_{n}(f_{\ast}))$ is non-zero.

Under this situation, can the alternating product $\prod_{n} \det (H_{n}(f_{\ast}))^{(-1)^{n}}$ be expressed as "$\prod_{n}\det (f_{n})^{(-1)^{n}}$"?

The latter product is a priori an infinite product, so one will need to make some modifications.

Hiro -- what exactly are $E_n, f_n$ and $F_n$? If $E_n=\bigoplus_{a+b=n}E^1_{a,b}$ and similarly for $F_n$ then, under your conditions, where can infinitely many terms appear from? –  algori Dec 29 '12 at 17:05
Thanks for your question. $E_{n}$ denotes the $n$th term to which the spectral sequence $E$ converges i.e. $E_{a,b}^{1} \Longrightarrow E_{n}$. –  Hiro Dec 29 '12 at 18:32
To define $\det$ you need to fix bases in the various vector spaces.. The answer to your question is, yes, you can compute the determinant (or torsion) ofvthe morphism $f$ but you need to fix bases, and you need to deal with a rathervtricky sign problem. –  Liviu Nicolaescu Dec 29 '12 at 20:11
Tanks for your comment. Yes, I should heve fixed the bases. Or, if necessary, let $F=E$ and $f$ be a endomorphism of $E$. Then det is well-defined. What do you exactly mean by "sign problem" ? Is there any reference to learn the method? –  Hiro Dec 29 '12 at 21:18