Homotopy of complex projective space

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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}).