15
$\begingroup$

I originally posted this question on MSE (link), but was suggested to post here instead.

While learning about spectral sequences a friend of mine found a proof of the snake lemma using spectral sequences. We noticed that the proof works equally well for larger bicomplexes. Particularly if you have an exact (anti)-commutative diagram

enter image description here you get an exact sequence.

enter image description here

We also have a little write-up of the proof here. We looked around, but couldn't find any reference to this anywhere, and no one else we talked to had really thought about it before. While toying with this we realized that the hypothesis is quite strong. That is, it is pretty difficult to find any interesting exact bicomplexes of the right size. I'm starting to suspect that there might not really be any interesting examples of this, and that that is the reason we haven't found anything about it anywhere.

So we're wondering, has anyone seen this before? Can anyone think of any non-trivial examples or applications of this?

Improve this question
$\endgroup$
1
  • 2
    $\begingroup$ (I would call it the long snake lemma, rather. For some reason "anaconda" sounds... weird.) $\endgroup$
    – Pedro
    Commented May 4, 2020 at 13:53

1 Answer 1

Reset to default
9
$\begingroup$

Spectral sequences are obtained by applying the snake lemma ad nauseam, essentially. You are doing that (finitely many times) in your spectral sequence argument.

Let me illustrate. Consider the case you have a diagram that has three exact rows and four exact columns, and start with the top right corner space $C_{1,4}$, and pick $x$ there.

If $d_v(x)=0$, pull this back to some $x'\in C_{1,3}$ and apply $d_v(x')\in C_{2,3}$. Then $d_h$ of this is zero, so there is some $x''$ in $C_{2,2}$ such that $d_h(x'') = d_v(x')$. Apply $d_v$ again to fall into $C_{3,2}$. Then take the class of this, and this is your map.

So, in general, you start with $x_{1,n} \in C_{1,n}$ in the kernel of $d_v$, pull it back to $x_{1,n-1}$, apply $d_v$, pull it back, apply $d_v$, keep going, until you reach $C_{n+1,1}$. You can do this because your rows are exact, and this (I think) is all that you need, you don't need your columns to be exact (as in the snake lemma).

You can do this in general for any $n$, with no spectral sequences, its just a longer version of the snake lemma. The spectral sequence formalism is packaging neatly, but this is what is going on.

Improve this answer
$\endgroup$
3
  • 1
    $\begingroup$ Thank you for your response. I do understand what is going on in the theorem. My question is whether it is at all significant or useful. Are you saying that it is not because it is just a trivial consequence of the snake lemma? $\endgroup$
    – Jacob FG
    Commented May 4, 2020 at 15:43
  • 5
    $\begingroup$ My interpretation of Pedro's answer is something like """ part of the point of using spectral sequences is to make applications of the snake lemma invisible, and when working with spectral sequences it is already common to "apply finitely many snake lemmas" without comment. """ $\endgroup$
    – Jalex Stark
    Commented May 4, 2020 at 16:49
  • $\begingroup$ @JalexStark Agreed. That's a good way to put it. :) $\endgroup$
    – Pedro
    Commented May 5, 2020 at 11:30

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .