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I'm reading a text on homology, and it shows that the differential complex of the projective plane ($\mathbb P^2$) is given by $\mathbb Z \overset{\times 2}{\longrightarrow}\mathbb Z \overset{0}{\longrightarrow}\mathbb Z \longrightarrow 0$. It then goes on and says that this implies that:

$H_0(\mathbb P^2, \mathbb Z) \cong \mathbb Z,\quad H_1(\mathbb P^2, \mathbb Z) \cong \mathbb Z/2\mathbb Z,\quad H_2(\mathbb P^2, \mathbb Z) \cong 0$

However, I'm afraid I don't see the connection. How exactly is the homology calculated given the the differential complex?

Frank Vel
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1 Answers1

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It follows directly from the definition of the homology of a chain complex. Given a chain complex

$$\dots \xrightarrow{\partial} C_{i+1}(X) \xrightarrow{\partial} C_i(X) \xrightarrow{\partial} C_{i-1}(X) \xrightarrow{\partial} \dots$$

its $n^{\text{th}}$ homology is, by definition,

$$H_n(X, \mathbb{Z}) := \dfrac{\ker \partial : C_n(X) \to C_{n-1}(X)}{\operatorname{im} \partial : C_{n+1}(X) \to C_n(X)}.$$

Here the chain complex is $0 \to \mathbb{Z} \xrightarrow{\times 2} \mathbb{Z} \xrightarrow{0} \mathbb{Z} \to 0$. That is, $C_0(\mathbb{P}^2) = C_1(\mathbb{P}^2) = C_2(\mathbb{P}^2) = \mathbb{Z}$ and $C_i(\mathbb{P}^2) = 0$ for $i > 2$. Moreover, the boundary map $\partial : C_2(\mathbb{P}^2) \to C_1(\mathbb{P}^2)$ amounts to multiplication by $2$, whereas the boundary map $\partial : C_1(\mathbb{P}^2) \to C_0(\mathbb{P}^2)$ is the zero map.

So, for example,

$$H_1(\mathbb{P}^2, \mathbb{Z}) = \dfrac{\ker \partial : C_1(X) \to C_0(X)}{\operatorname{im} \partial : C_2(X) \to C_1(X)} = \dfrac{\ker (\mathbb{Z} \xrightarrow{0} \mathbb{Z})}{\operatorname{im} (\mathbb{Z} \xrightarrow{\times 2} \mathbb{Z})} = \mathbb{Z}/2\mathbb{Z}.$$

Michael Albanese
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