Homotopy of complex projective space

This article describes the value (and the process used to compute it) of some homotopy invariant(s) for a topological space or family of topological spaces. The invariant is homotopy group and the topological space/family is complex projective space
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Statement

This article describes the homotopy groups, including the set of path components $\pi_0$, the fundamental group $\pi_1$, and the higher homotopy groups $\pi_k$ of $\mathbb{P}^n(\mathbb{C})$.

Case $n = 0$

For $n = 0$, $\mathbb{P}^n(\mathbb{C})$ is the one-point set. Hence, all its homotopy groups are the trivial group. The set of path components $\pi_0$ is a one-point set and can be considered the trivial group.

Case $n = 1$

For $n = 1$, $\mathbb{P}^1(\mathbb{C}) \cong S^2$ (a homeomorphism), i.e., it is the 2-sphere. Its homotopy groups are hence the same as those of the 2-sphere. Specifically, they are as follows:

• $\pi_0(\mathbb{P}^1(\mathbb{C}))$ is a one-point set.
• $\pi_1(\mathbb{P}^1(\mathbb{C}))$ is the trivial group.
• $\pi_2(\mathbb{P}^1(\mathbb{C})) \cong \mathbb{Z}$, i.e., it is isomorphic to the group of integers, with the identity map being the generator.
• $\pi_3(\mathbb{P}^1(\mathbb{C})) \cong \mathbb{Z}$, i.e., it is isomorphic to the group of integers, with the map being the Hopf fibration.
• $\pi_4(\mathbb{P}^1(\mathbb{C})) \cong \mathbb{Z}/2\mathbb{Z}$.

Higher homotopy groups are the same as those of the 2-sphere.

Case of higher $n$

For this case, we use the fiber bundle of sphere over projective space $S^{2n + 1} \to \mathbb{P}^n(\mathbb{C})$ with fiber $S^1$. We get the following long exact sequence of homotopy of a Serre fibration:

$\dots \to \pi_k(S^1) \to \pi_k(S^{2n + 1}) \to \pi_k(\mathbb{P}^n(\mathbb{C})) \to \pi_{k-1}(S^1) \to \dots$

For $k \ge 2$, $\pi_k(S^1)$ is trivial. Thus we get the following:

• Case $k = 0$: $\pi_0(\mathbb{P}^n(\mathbb{C}))$ is a one-point space.
• Case $k = 1$: We get $\pi_1(\mathbb{P}^n(\mathbb{C}))$ is trivial.
• Case $k = 2$: We get $\pi_2(\mathbb{P}^n(\mathbb{C})) \cong \mathbb{Z}$..
• Case $2 < k < 2n + 1$: We get that $\pi_k(\mathbb{P}^n(\mathbb{C}))$ is the trivial group.
• Case $k = 2n + 1$: We get that $\pi_{2n+1}(\mathbb{P}^n(\mathbb{C})) \cong \mathbb{Z}$.
• Case $2n + 1 < k, k \ne 4n + 1$: We get that $\pi_k(\mathbb{P}^n(\mathbb{C})) \cong \pi_k(S^{2n + 1})$.