Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

One of the common definitions of homology using the singular chains, i.e. maps from the simplex into your space. The free abelian group on these can be made into a chain complex and one can take the homology of this. The result is usually called singular homology.

However, one can use a smaller chain complex instead, by taking the quotient with the degenerate chains (those that are the image of a degeneracy map $\sigma$). This will give the same result in homology.

In some articles, I've seen authors replace the singular chains with the normalized singular chains, often claiming that this is "for technical reasons" (a example is Costello's article on the Gromov-Witten potential associated to a TCFT). What are the important technical differences between these two functors? In which situations is there a preferred one?

share|improve this question
add comment

3 Answers

up vote 6 down vote accepted

Before going to linearity, note a technical advantage of normalized geometric realization is that it preserves products. That probably has linear consequences.

The simplest technical advantage of normalized chains is the Dold-Kan theorem that the normalized chains give an equivalence of categories between simplicial abelian groups and chain complexes in positive degrees. But the tensor structures on the two categories (detailed in Tom's answer) do not match under this equivalence. The normalized functor is both lax monoidal and oplax monoidal. I think the fat chain functor is still oplax monoidal. The technical advantage is that only the normalized functor is lax monoidal. My guess is that this is what Costello wants, in particular the consequence Tom mentions that a simplicial ring becomes a dga. Greg's parenthetical is another multiplicative property that I would expect to demonstrate the failure of the lax monoidal property, but I don't see that it does.

I have never seen an example where abnormal chains are technically advantageous; degenerate simplices tend only to get in the way of precise considerations. But if they don't get in the way, thinking about them may be a distraction, as Tom says.

share|improve this answer
    
You say that the normalized chain functor $N$ is lax monoidal and the unnormalized one $U$ is not. I disagree. The functors $U$ and $N$ come with natural transformations $U\to N$ and $N\to U$ (in fact, $N$ is a direct summand in $U$). So the monoidal structure $NA\otimes NB\to N(A\otimes B)$ of $N$ gives us a monoidal structure for $U$: $UA\otimes UB\to NA\otimes NB\to N(A\otimes B)\to U(A\otimes B)$. All arrows are quasi-isomorphisms. Am I wrong? –  Semen Podkorytov Aug 14 '13 at 12:22
add comment
  • I don't know Costello's reasons, but for example I find it convenient that the normalized chains on the one-point space are concentrated in degree zero (so the normalized chains functor takes the unit objet to unit object). Generally, when two construction have all the same good properties one tends to prefer the smaller one...
share|improve this answer
add comment

The question was about chains on a space, presumably singular chains. If what you're up to is, for example, learning or teaching the basic facts about singular homology, like homotopy-invariance and excision, then the switch to normalized chains seems like an unnecessary complication. In that context, you could say that non-normalized chains have the advantage.

To expand on Ben's comments:

For A and B simplicial abelian groups and NA and NB the associated normalized chain complexes, there is a quasi-isomorphism $NA\otimes NB\rightarrow N(A\otimes B)$, where the tensor product of simplicial groups is defined levelwise $(A\otimes B)_n=A_n\otimes B_n$ and the tensor product of chain complexes is the usual $(C\otimes D)_n=\oplus_p (C_p\otimes D_{n-p})$. An associative simplicial ring leads in this way to an associative differential graded ring.

Another technical advantage of the normalized complex is its role in proving that homotopy groups of a simplicial abelian group are homology groups (Dold-Thom Theorem).

share|improve this answer
    
My first version of the above answer had some nonsense, now edited out. Thanks, Ben, for the private heads-up. –  Tom Goodwillie Jun 11 '10 at 3:12
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.