3
$\begingroup$

Reduced homology

In my understanding, the reduced homology is better-behaved than the usual singular homology because the $0$th reduced homology

  1. counts the non-trivial "closed" $0$-chain and
  2. reflects the idea of "orientation is the volume form". (see below for detail)

Poincare duality for singular (co)homologies

On the other hand, the Poincare duality for the usual singular (co)homologies: $$H_i(X) \cong H^{n-i}(X)$$ holds for an oriented closed manifold $X$.

Question

So I thought there should be a correspondence of the Poincare duality for the reduced homology. That is, let $\tilde{H}'_\bullet, \tilde{H}^\bullet$ be reduced homology and the "reduced cohomology" thing (I don't know what's this). Then, $$\tilde{H}_i(X) \cong \tilde{H}^{n-i}(X)$$ holds for an oriented closed manifold $X$. Is there anything like this?

I Googled and found a "reduced cohomology" but found nothing about the duality between them similar to the Poincare duality.

Two reasons why I think the reduced homology is better-behaved than the usual one;

  1. we can regard "closed" $0$-chain as the one whose weight is $0$. So it matches well to the idea "n-th homology count the number of nontrivial(non null-homologous) closed n-chain" at $n=0$.

  2. By definition of n-simplex ($n\geq 0$), we think them as n-triangle with orientation but without orientation when $n=0$. This is odd. The orientation is (an equivalence class of) the volume form, so the orientation of $0$-triangle should be the scalar field on the $0$-triangle. This naturally leads to the definition of reduced homology.

Any reference would be appreciated.

Thanks in advance.

$\endgroup$
0

1 Answer 1

1
$\begingroup$

No. We have $\tilde H_k(X) = H_k(X)$ for $k > 0$ and $\tilde H_0(X) \oplus \mathbb Z \approx H_0(X)$, similarly for cohomology. Thus

$$\tilde H_i(X) \approx \tilde H^{n-i}(X)$$ for all $i$ with $0 < i < n$. For $i = 0, n$ it is wrong.

$\endgroup$
4
  • $\begingroup$ Yes, the reduced homology is different from the usual one, so the duality obviously does not hold. Are there any "reduced cohomology" thing and duality between them? $\endgroup$
    – Yuta
    Jul 4, 2020 at 10:43
  • $\begingroup$ I do not know what you could mean by a reduced cohomology thing. But if want a version of Poincare duality for $n$-dimensional manifolds based on reduced (co)homology, you must be aware that the "reduced theories" have a special behavior in a certain dimension $n$. Thus you will not find "universal" theories for abritrary spaces (or at least for arbitrary manifolds) with the desired behavior. You could try to develop theories only for $n$-dimensional manifolds, but in my opinion that doesn't make much sense. $\endgroup$
    – Paul Frost
    Jul 4, 2020 at 22:36
  • $\begingroup$ Thanks. Do you mean the reduced homology is in some sense ill-behaved? I edited my question could you look at it? $\endgroup$
    – Yuta
    Jul 5, 2020 at 1:27
  • 2
    $\begingroup$ Reduced theories and unreduced theories both have their advantages and disadvantages. It depends on the context which behaves "better". For duality in manifolds you better work with unreduced theories, also the Mayer-Vietoris sequence in dimension $0$ should be based on unreduced homology. The suspension isomorphism needs reduced homology. The idea of reduced groups is to eliminate the trivial part $H_*(point)$ which is contained in all homology groups. In fact, one has $H_*(X) \approx \tilde H_*(X) \oplus H_*(point)$ for all $X$. $\endgroup$
    – Paul Frost
    Jul 5, 2020 at 13:08

You must log in to answer this question.

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