Chain Complexes as Graded C[epsilon] modules (2)

by

< See Part 1

In this post, we look at the existance of long exact sequences for a given crude homology functor. I will give a sufficient (but possibly not necessary) condition for when such a functor will have always have them.

Long Exact Sequences

In the previous post, we talked about how to translate some basic homological algebra into the language of categories of modules. The most interesting construction was expressing the homology functor from \mathbb{C}[\epsilon]-mod to \mathbb{C}-mod as the image of a natural transformation between two functors. This construction could be duplicated any time you had a fully faithful functor between two abelian categories with a left and right adjoint; I am calling this new homology functor the ‘crude homology’ of the original functor. This was heartening news, because one of the main purposes of all of this was to try to understand what sort of contexts homology could be generalized to.

However, what good is this notion of the ‘crude homology’? Of course, one could just as easily ask what good the classical notion of homology is. A huge part of what makes homology so useful is the existance of long exact sequences of homology, and so an important test of the usefulness of crude homology is whether or not something like long exact sequences of homology exist.

I don’t have a complete answer to this question. All I have done so far is see exactly when the techniques that work in the chain complex case work in the general case; this is what I will show now.

The important thing in the chain complex case is that there is a natural transformation \delta:f^*\rightarrow f^! which comes from the action of \delta. Remember that f^*(M)=M/im(\delta) and f^!(M)=ker(\delta), so this map is well defined, and both its kernel and cokernel is H(M). Letting h denote the natural transformation from f^!\rightarrow f^* whose image is H, what this means is that h\circ \delta=0 and \delta\circ h=0 and that they are exact at both f^! and f^*.

Now, take an exact sequence,
0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0
and apply \delta: f^*\rightarrow f^! to this sequence to get

\begin{array}{ccccccccc} && f^*(A) & \rightarrow & f^*(B) & \rightarrow & f^*(C) & \rightarrow & 0 \\ && \downarrow && \downarrow && \downarrow && \\ 0 & \rightarrow & f^!(A) & \rightarrow & f^!(B) & \rightarrow & f^!(C) && \end{array}
This uses the fact that f^* is right exact and that f^! is left exact; it is a basic fact of adjoint functors in abelian categories that a functor that is left adjoint is right exact and vice-versa.

Ah, but the above diagram is the exact diagram one needs to apply the Snake Lemma (not a coincidence). It says that there exists a map from the kernel of the right-most vertical map, which is H(C), to the cokernel of the left-most vertical map, which is H(A). This map sits in a triangle with the induced maps from H(A)\rightarrow H(B) and from H(B)\rightarrow H(C), and further applications of the Snake Lemma show that this triangle is exact at each object. (Remember that because I’m ignoring gradings, long exact sequences become exact triangles.)

As I mentioned before, the information that made all of this work was the natural transformation \delta:f^*\rightarrow f^!, and the fact that its kernel and cokernel were both H. This will happen if and only if the kernel and cokernel of the natural transformation h:f^!\rightarrow f^* are naturally isomorphic.

Sadly, this seems to happen quite rarely. For instance, I think it is not true in the case I mentioned as a possible generalization of homology: when I have a scheme X and a choice of a infinitesmal line bundle Y over it, and I think about the push-forward functor f_*:Mod(X)\rightarrow Mod(Y). This is a fully faithful functor with a left and right adjoint, so a crude homology exists, but it would seem that there are no long exact sequences. Hopefully I’m wrong.

I will stop here, since I think bicomplexes will fit in better with the next chunk of things I want to talk about: superalgebras!

> Continue to Part 3

Advertisements

Tags:

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s


%d bloggers like this: