Reference request for equality of torsion of H1 and H2

Question

I have heard that for a surface $X$ (algebraic? smooth? compact?) the torsion part of $H_1(X,\mathbb{Z})$ is the same as that of $H_2(X,\mathbb{Z})$. Please could you give me a correct statement? I would like also a general theorem and some reference if possible. Thank you very much.

Answer 1

Compact surfaces are completely classified.

$H_2(X,\mathbb{Z})$ is either zero or $\mathbb{Z}$ depending on whether your surface is orientable.

There are non-orientable surfaces that don't satisfy your condition, for example $\mathbb{R}P^2$:

$H_1(\mathbb{R}P^2,\mathbb{Z}) = \mathbb{Z}_2$

If you restrict yourself to compact orientable surfaces then there's no torsion in any of the homology groups: $H_2(X,\mathbb{Z}) \cong H^0(X,\mathbb{Z}) \cong \mathrm{Ext}^1(H_{-1}(X,\mathbb{Z})) \oplus \mathrm{Hom}_\mathbb{Z}(H_0(X,\mathbb{Z}), \mathbb{Z})$

First by the Poincare duality, then by the universal coefficient theorem. Now you don't really need to know what $\mathrm{Ext}^1$ is, only that it sends $0$ to $0$, and $H_0(X,\mathbb{Z}) \cong \bigoplus_{\pi_0(X)} \mathbb{Z}$, where $\pi_0(X)$ is the number of path components of $X$.

Therefore $\mathrm{Hom}_\mathbb{Z} (H_0(X,\mathbb{Z}), \mathbb{Z}) \cong \prod_{\pi_0{X}} \mathbb{Z}$, which is certainly torsion-free.

$H_1(X,\mathbb{Z})$ is also torsion free. You get it by abelianising the fundamental group or cellular homology.

Answer 2

As krey says, this is false for compact real surfaces, and $\mathbb{RP}^2$ gives a counterexample. A general statement along these lines is that if $X$ is any reasonable space (this includes all compact manifolds, not just surfaces), then there's an identification between the torsion part of $H_i(X, \mathbb{Z})$ and of $H^{i+1}(X, \mathbb{Z})$ (so for the second group I need to take cohomology, not homology). Setting $i = 1$ gives an identification between the torsion part of $H_1(X, \mathbb{Z})$ and the torsion part of $H^2(X, \mathbb{Z})$.

This comes from the universal coefficient theorem, which gives a short exact sequence

$$0 \to \text{Ext}^1(H_i(X, \mathbb{Z}), \mathbb{Z}) \to H^{i+1}(X, \mathbb{Z}) \to \text{Hom}(H_{i+1}(X, \mathbb{Z}), \mathbb{Z}) \to 0$$

which splits (noncanonically), so

$$H^{i+1}(X, \mathbb{Z}) \cong \text{Ext}^1(H_i(X, \mathbb{Z}), \mathbb{Z}) \oplus \text{Hom}(H_{i+1}(X, \mathbb{Z}), \mathbb{Z}).$$

Now, if $H_i(X, \mathbb{Z})$ is finitely generated (this is all I need for "reasonable"), then $\text{Ext}^1(H_i(X, \mathbb{Z}), \mathbb{Z})$ is (noncanonically) isomorphic to the torsion subgroup of $H_i(X, \mathbb{Z})$, while $\text{Hom}(H_{i+1}(X, \mathbb{Z}), \mathbb{Z})$ is necessarily torsion-free. So the conclusion follows.

But there's hope! If in addition $X$ is a compact oriented manifold of dimension $n$, then Poincaré duality furnishes an isomorphism

$$H^{i+1}(X, \mathbb{Z}) \cong H_{n-i-1}(X, \mathbb{Z}).$$

It follows that in this case we get a (noncanonical) isomorphism between the torsion subgroups of $H_i(X, \mathbb{Z})$ and $H_{n-i-1}(X, \mathbb{Z})$. (A more canonical way to describe this identification is via the torsion linking form.)

If $n = 4, i = 1$, this gives a noncanonical isomorphism between the torsion subgroups of $H_1(X, \mathbb{Z})$ and $H_2(X, \mathbb{Z})$. Hence your original statement holds if $X$ is a compact oriented $4$-manifold, and in particular if $X$ is a compact complex surface.