Induced isomorphisms on cohomology of double differential complexes

Question

Let $K^{\bullet,\bullet}=\bigoplus_{p,q\ge0}K^{p,q}$ be a double differential complex, i.e. we have differential operators $$\cdots\stackrel{d}{\to} K^{p,q-1}\stackrel{d}{\to} K^{p,q}\stackrel{d}{\to} K^{p,q+1}\stackrel{d}{\to}\cdots $$ and $$\cdots\stackrel{\delta}{\to} K^{p-1,q}\stackrel{\delta}{\to} K^{p,q}\stackrel{\delta}{\to} K^{p+1,q}\stackrel{\delta}{\to}\cdots.$$

We may define a single differential complex $$K^\bullet=\bigoplus_n\left(\bigoplus_{p+q=n}K^{p,q}\right)$$ with differential operator $D=\delta+(-1)^pd$. Let $L^{\bullet,\bullet}$ be a second double complex, also with differential operators $d$ and $\delta$. Let $f:K^{\bullet,\bullet}\to L^{\bullet,\bullet}$ be a complex homomorphism in the following sense: $f$ is a vector space homomorphism, $f(K^{p,q})\subset L^{p,q}$, and $f$ commutes with $d$ and $\delta$.

Since $f$ commutes with $d$, it descends to a map $f^*:H_d(K^{p,\bullet})\to H_d(L^{p,\bullet})$ of cohomologies, where $$K^{p,\bullet}=\bigoplus_q K^{p,q},$$ with differential $d$ and similarly for $L^{p,\bullet}$. Suppose that $f$ induces an isomorphism $H_d(K^{p,\bullet})\cong H_d(L^{p,\bullet})$ for all $p$ in this manner. Does it follow that this $f^*$ also gives an isomorphism between $H_D(K^\bullet)$ and $H_D(L^\bullet)$?

I know this to be true when the $\delta$-cohomologies are zero, i.e. the $q$ rows are all exact. In Bott & Tu, this general case is claimed, but not proven.

Answer 1

I believe the following is a counterexample:

Let $K^{p,q}$ be ${\mathbb Z}$ if $p+q=0$ or $p+q=1$, but zero otherwise.

Let the maps $K^{p,-p}\rightarrow K^{p+1,-p}$ and $K^{p,-p}\rightarrow K^{p,-p+1}$ be the identity . (All other maps are automatically zero.)

Let $L^{\bullet,\bullet}$ be the zero complex, and map $f:L^{\bullet,\bullet}\rightarrow K^{\bullet,\bullet}$ in the only possible way.

Then the horizontal homology of $K^{p,q}$ is everywhere zero, so $f$ induces an isomorphism on horizontal (and vertical!) homology.

But the element $(...,1,1,1,1,1,1,....)$ in $\oplus_pK^{p,-p}$ represents a nontrivial homology class in $H_D(K^{\bullet,\bullet})$, which of course cannot come from the homology of the zero complex $L^{\bullet,\bullet}$.

Answer 2

My earlier answer gives a counterexample that ignores your condition that $K$ and $L$ live in the first quadrant. Given that condition, what you want is true:

Let $C^{\bullet,\bullet}$ be the mapping cone of $f$. Then the long exact cohomology sequence for $f^{p,\bullet}$ shows that the horizontal cohomology of $C^{\bullet,\bullet}$ is zero. Therefore the first spectral sequence for the cohomology of $C^{\bullet,\bullet}$ collapses at $E^2$. But because $C^{\bullet,\bullet}$ lives in the first quadrant, the spectral sequence must converge, necessarily to zero, which gives your desired result.

Answer 3

With mild apologies for the proliferation of answers, here is what the spectral sequence argument is trying to tell you, without the spectral sequences:

A pretty typical element in total homology of the mapping cone of $f$ is represented by something like $a-b+c$ in the diagram below (which maps to $x-(x+y)+y=0$):

(Of course, it might also be represented by a much longer chain. Accounting for all the possibilities gets tedious, and the spectral sequence is designed to do the accounting for you. But this will give you the idea of what it's doing.)

We can always assume that $a$ is at the left margin of the quadrant.

Now: Horizontal homology is zero, so $c$ lifts horizontally to some $p$. In turn, $p$ maps vertically to some $b'$, which maps horizontally to $-y$, so $b+b'$ maps horizontally to zero. Now use again the fact that horizontal homology is zero to lift $b+b'$ horizontally to $q$. In turn, $q$ maps vertically to some $a'$, which maps horizontally to $-x$, as does $-a$. Because we've now hit the left end of the quadrant, the map that takes $a$ to $x$ is injective, so $a'=-a$.

Thus $-q+p$ maps (counting both horizontal and vertical components) to $-(a'+b+b')+(b'+c)=a-b+c$, which is to say that it kills $a-b+c$ in the homology of the total complex, as desired.

Similar considerations will kill any homology class, using the vanishing of horizontal homology over and over.

And now you've made the first step both toward understanding spectral sequences and toward understanding why we'd prefer to work with them.