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Background:

In Hatcher's Algebraic Topology Chapter 2, reduced homology with coefficient $R$ of a space $X$ is defined as the homology groups of a chain complex $$ ...\to C_{2}(X) \overset{\partial_{2}}\to C_{1}(X) \overset{\partial_{1}}\to C_{0}(X) \overset{\epsilon}\to R \to 0 $$ where $\epsilon(\Sigma_{i}n_{i}\sigma_{i}:=\Sigma_i n_{i})$.

If my understanding is correct, reduced homology is a better-behaved alternative for usual simplicial/singular homology.

Detail

I came up with a further modification of a singular homology for some $n$-dimensional orientable, compact, manifold $X$ defined as the homology of the chain complex

$$ R \overset{\eta}\to C_{n}(X) \to ...\to C_{2}(X) \overset{\partial_{2}}\to C_{1}(X) \overset{\partial_{1}}\to C_{0}(X) \overset{\epsilon}\to R \to 0\\ \text{where }\eta(r) := r\cdot [X] $$

which I refer to as reduced-reduced homology $H_{\bullet}'(X)$ of $X$.

Dually, one can also define the reduced-reduced cohomology ${H^{\bullet}}'(X)$ of $X$ by modifying the de Rham complex of $X$;

$$ R \overset{\Delta}\to\Omega^{0}(X) \to ...\overset{d_{n-1}}\to \Omega_{n-1}(X) \overset{d_{n}}\to \Omega_{n}(X) \overset{\int}\to R \to 0 $$ where $\Delta(r)$ is a constant scalar field on $X$ and $\int$ is an usual integration of a valume form on $X$.

It is easily checked that the above defined (co)chains are indeed (co)chain complexes.

Questoin:

My question is:

  1. Do the reduced (co)homologies well-behave? In particular, does Poincare duality $H_{n-p}'(X)\cong {H^{p}}'(X)$ holds?
  2. Are there any reference/article about this concept?

Thank you in advance.

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    $\begingroup$ Reduced homology isn’t ‘more’ well behaved in general. It might be more useful for certain classes of spaces (such as wedge sums and quotients). In other cases you’d probably want to stick to regular homology (such as when dealing with products). Also I’m assuming question 1 is about whether Poincaré duality holds for $H’(X)$ that you define right? $\endgroup$ Jun 26, 2020 at 13:33
  • $\begingroup$ Also, do you mean $\eta(r) = r[X]$? $X$ by itself is not a chain. $\endgroup$ Jun 26, 2020 at 13:39
  • $\begingroup$ Osama, thank you. I edited my post. $\endgroup$
    – Yuta
    Jun 26, 2020 at 13:43

1 Answer 1

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I think you lose functoriality altogether, but I’m not entirely sure (feel free to correct me if I’m wrong). Suppose you have a map $X \to Y$. Then you have chain maps and in particular a homomorphism $C_i(X) \to C_i(Y)$ and this also works for the augmented $\mathbb{Z}$ (the square with $C_0(X), C_0(Y)$ and the two $\mathbb{Z}$’s commute). However it isn’t apparent to me what happens with the square on the left. In the case where the dimensions of $X$ and $Y$ are the same, then we are trying to show commutativity of a square with two $R$’s, and $C_n(X)$ and $C_n(Y)$. For this to commute you might need some really strong conditions, such as that $f$ maps the fundamental class of $X$ to the fundamental class of $Y$. Now when $X$ and $Y$ are of different dimensions, the left square doesn’t even make sense because you have mismatched groups.

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  • $\begingroup$ Thanks, and I realized that the square induced by $\int$ also does not commute. $\endgroup$
    – Yuta
    Jul 1, 2020 at 23:52

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