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 $k=4n+1$: We get that ${\pi }_{4n+1}({\mathbb{P}}^n(\mathbb{C}))\cong \mathbb{Z}$.
• Case $2n+1<k,k\ne 4n+1$: We get that ${\pi }_k({\mathbb{P}}^n(\mathbb{C}))\cong {\pi }_k(S^{2n+1})$.