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 $k\geq 2$, $\pi _{k}(S^{1})$ 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 $2n+1<k,k\neq 4n+1$: We get that $\pi _{k}(\mathbb {P} ^{n}(\mathbb {C} ))\cong \pi _{k}(S^{2n+1})$.