I have a very basic problem when trying to understand "You could have invented spectral sequences" by Timothy Chow (but I will index cohomologically since I'm more interested in cohomology). This is basic linear algebra and probably I miss something quite obvious.

We take a filtered complex a complex $\dots \to C^n \stackrel{d^n}{\to} C^{n+1} \to \dots $ of finite dimensional vector spaces, with a one-step filtration $F^1C \subset C$.

We define $E_0^{0,n} = C^n/F^1C^n$ and $E_0^{1,n -1} = F^1C^n$. We define $$E_1^{0,n} = \frac{\ker (E_0^{0,n} \to E_0^{0,n+1})}{\text{im}(E_0^{0,n-1} \to E_0^{0,n})}$$ where the maps are induced by $d$ and similarly $$E_1^{1,n} = \frac{\ker (E_0^{1,n} \to E_0^{1,n+1})}{\text{im}(E_0^{1,n-1} \to E_0^{1,n})}$$

Now I would like to check that $d$ induces map $E_1^{0,n} \to E_1^{1,n}$. I am not convinced by this. Indeed this map should be a map $$ \frac{\ker (C^n/F^1C^n \to C^{n+1}/F^1C^{n+1})}{\text{im}(C^{n-1}/F^1C^{n-1} \to C^n/F^1C^n)} \to \frac{\ker (F^1C^{n+1} \to F^1C^{n+2})}{\text{im}(F^1C^n \to F^1C^{n+1})} $$

I don't see how to get an element of $F^1C^{n+1}$ from the kernel of $\overline d : C^n/F^1C^n \to C^{n+1}/F^1C^{n+1}$. If $\overline d$ lifts to a map $d : C^n/F^1C^n \to C^{n+1}$ then elements of $\overline d$ would be canonically send to some elements in $F^1C^{n+1}$. But I believe this is not the case, since there is no reasons that $F^1C^n$ goes to zero by $d$. So taking $dz$ just gives me an element of $C^3$. It would work if e.g we fix an arbitrary complement $K^n \oplus F^1C^n = C^n$ for each $n$ but we didn't do this, so I'm probably missing something really obvious or wrote something wrong.


Let $[x] \in \ker(C^n / F^1 C^n \to C^{n+1} / F^1 C^{n+1})$.

This means, by definition of the kernel, that you have a $x$ in $C^n$ such that $[dx] = 0$ in $C^{n+1}/F^1 C^{n+1}$. By definition of the quotient, this means that $dx$ is actually in $F^1 C^{n+1}$. So there's your element of $F^1 C^{n+1}$: it's just $dx$.

Next you need to check that $dx$ is actually in $\ker(d : F^1 C^{n+1} \to F^1 C^{n+2})$. But of course this follows from $d^2 = 0$. So from $x$ you do get an element of $\ker(F^1 C^{n+1} \to F^1 C^{n+2})$.

Moreover, suppose that $[x] = 0$, i.e. $x \in F^1 C^n$. You want to check that $dx = 0$. But in the target you mod out by the image $\operatorname{im}(F^1 C^n \to F^1 C^{n+1})$, so you're good.

At this point, we have a well defined linear map $$\ker(C^n / F^1 C^n \to C^{n+1} / F^1 C^{n+1}) \to \frac{\ker(F^1 C^{n+1} \to F^1 C^{n+2})}{\operatorname{im}(F^1 C^n \to F^1 C^{n+1})}.$$ The final step is to check that if $x$ is in the image of $C^{n-1}/F^1 C^{n-1} \to C^n / F^1 C^n$, then $[dx] = 0$. Again this follows from $d^2 = 0$ and we're done building the linear map.

  • $\begingroup$ Ok, I forgot that moding out by $im(F^1C^n \to F^1C^{n+1})$ was making the map well-defined. Thank you ! $\endgroup$ – student Apr 3 '18 at 13:23

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.