Homotopy of complex projective space

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 $$\pi_k(\mathbb{P}^n(\mathbb{C})) \cong \pi_k(S^{2n + 1})$$.