# Induced isomorphisms on cohomology of double differential complexes

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.

• Did you try a spectral sequence argument? (First quadrant makes things work here)
– Pedro
Aug 10, 2016 at 1:12
• @PedroTamaroff I don't know what that is. Aug 10, 2016 at 1:14
• Well, you have a double complex. This gives rise to a spectral sequence and you have an isomorphism on the E^1 page, so you get one on the next pages. Because you're on the first quadrant, your ss converges to those homology groups. You get an isomorphism on the limit page, and hence you then get the graded map asociated to f is a isomorphsim, so f must be one.
– Pedro
Aug 10, 2016 at 1:20
• @PedroTemaroff I don't know what a spectral sequence is. Aug 10, 2016 at 3:14
• @RyanUnger Do you know where I can find a proof without using spectral squences when the rows are exact? Thanks! Jul 29, 2021 at 11:07

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 $L^{\bullet,\bullet}$.

• It says in the OP that $p,q\ge0$. Aug 10, 2016 at 0:25
• @0celo7 : Ah! I missed that. That will help a lot. Aug 10, 2016 at 0:27

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.

• Would you like to double check my argument in the comments above?
– Pedro
Aug 10, 2016 at 1:37
• @PedroTamaroff: I might be missing something, but your argument seems incomplete to me. We have an isomorphism of $E^2$ terms, but how do we know that it is compatible with differentials (and therefore imposes a map of $E^3$ terms), and even if we know this, how do we know that map is an isomorphism, etc? Aug 10, 2016 at 1:40
• If a map of spectral sequences induces an isomorphism on one page, it induces an isomorphism on subsequent pages. In this case the spectral sequence associated to the double complexes has first terms exactly the $p,q$ terms appearing in the double complexes, and the $E^1$ page is where the OP is describing $f$ induces an isomorphism.
– Pedro
Aug 10, 2016 at 1:52
• @PedroTamaroff : Yes, you are right, of course, so your argument is or is not complete depending on whether one already knows that a map of SSs inducing an isomorphism on one page induces isomorphisms on all. (I myself either did not know this or had forgotten it.) (And by the same token, my argument is or is not complete depending on whether one already knows a bit about spectral sequences and mapping cones.) For the record, you posted your comment while I was typing my answer and I never saw it till you pointed it out; otherwise I'd have mentioned it. Aug 10, 2016 at 1:57
• @0celo7 : I'm sure that with a little effort, one could eliminate the spectral sequences from this argument and replace them with the hard computations that spectral sequences are designed to avoid. But then you'd have to do a lot of work following those computations, and with the same amount of effort, you could be learning about spectral sequences. I'm sorry not to have a simpler answer for you. Aug 10, 2016 at 3:03

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.

• Why is horizontal homology zero? Aug 11, 2016 at 13:10
• Oh---I should have said this: this is a depiction of the mapping cone of $f$. Aug 11, 2016 at 14:06
• Ok, but why is the horizontal homology zero? Aug 11, 2016 at 14:28
• Because of the long exact homology sequence for the mapping cone. Aug 11, 2016 at 14:40
• @0celo7: Explicitly, index the mapping cone in the horizontal direction (so that $C_{ij}=L_{ij}\oplus K_{i,j+1}$. Then each row of the mapping cone is the mapping cone of the horizontal row and hence has zero homology. (Your last comment appeared while I was typing this. Let me gently suggest that it's going to be a lot easier to learn a few of the basic notions and then apply them to your problem than it would be to try solving your problem without using any of the basic notions.) Aug 11, 2016 at 15:05