Return to Answer

This is proved in the book Representations of compact Lie groups by Bröcker and tom Dieck and reviewed here. It is Proposition 7.5 in Chapter V. The proof is for compact, connected Lie groups, but any connected Lie group has the homotopy type of its maximal compact subgroup. (Everything here is for finite-dimensional Lie groups, of course.)

Edit: Maybe I should add, given that two of the other answers mention similar proofs, that the one in this book does not use Morse theory. It uses only basic covering space techniques once it is shown that $\pi_2(G)$ is isomorphic to $\pi_2(G_r)$, where $G_r$ are the regular elements, itself not a difficult lemma.

Edit: The following is wrong! It is only true rationally, which is why I didn't remember having seen a proof of the general case :)

Also if you believe that simply-connected compact Lie groups have the homotopy type of a product of odd spheres, then this follows. It is fairly easy to see that this is the case rationally, but do not remember whether the general statement is hard to prove.

This is proved in the book Representations of compact Lie groups by Bröcker and tom Dieck and reviewed in the Bulletin of the AMS. It is Proposition 7.5 in Chapter V. The proof is for compact, connected Lie groups, but any connected Lie group has the homotopy type of its maximal compact subgroup. (Everything here is for finite-dimensional Lie groups, of course.)

Edit: Maybe I should add, given that two of the other answers mention similar proofs, that the one in this book does not use Morse theory. It uses only basic covering space techniques once it is shown that $\pi_2(G)$ is isomorphic to $\pi_2(G_r)$, where $G_r$ are the regular elements, itself not a difficult lemma.

Edit: The following is wrong! It is only true rationally, which is why I didn't remember having seen a proof of the general case :)

Also if you believe that simply-connected compact Lie groups have the homotopy type of a product of odd spheres, then this follows. It is fairly easy to see that this is the case rationally, but do not remember whether the general statement is hard to prove.

This is proved in the book Representations of compact Lie groups by Bröcker and tom Dieck and reviewed here. It is Proposition 7.5 in Chapter V. The proof is for compact, connected Lie groups, but any connected Lie group has the homotopy type of its maximal compact subgroup. (Everything here is for finite-dimensional Lie groups, of course.)

Edit: Maybe I should add, given that two of the other answers mention similar proofs, that the one in this book does not use Morse theory. It uses only basic covering space techniques once it is shown that $\pi_2(G)$ is isomorphic to $\pi_2(G_r)$, where $G_r$ are the regular elements, itself not a difficult lemma.

Edit: The following is wrong! It is only true rationally, which is why I didn't remember having seen a proof of the general case :)

Also if you believe that simply-connected compact Lie groups have the homotopy type of a product of odd spheres, then this follows. It is fairly easy to see that this is the case rationally, but do not remember whether the general statement is hard to prove.

This is proved in the book Representations of compact Lie groups by Bröcker and tom Dieck and reviewed here. It is Proposition 7.5 in Chapter V. The proof is for compact, connected Lie groups, but any connected Lie group has the homotopy type of its maximal compact subgroup. (Everything here is for finite-dimensional Lie groups, of course.)

Edit: Maybe I should add, given that two of the other answers mention similar proofs, that the one in this book does not use Morse theory. It uses only basic covering space techniques once it is shown that $\pi_2(G)$ is isomorphic to $\pi_2(G_r)$, where $G_r$ are the regular elements, itself not a difficult lemma.

Edit: The following is wrong! It is only true rationally, which is why I didn't remember having seen a proof of the general case :)

Also if you believe that simply-connected compact Lie groups have the homotopy type of a product of odd spheres, then this follows. It is fairly easy to see that this is the case rationally, but do not remember whether the general statement is hard to prove.

This is proved in the book Representations of compact Lie groups by Bröcker and tom Dieck and reviewed here. It is Proposition 7.5 in Chapter V. The proof is for compact, connected Lie groups, but any connected Lie group has the homotopy type of its maximal compact subgroup. (Everything here is for finite-dimensional Lie groups, of course.)

Edit: Maybe I should add, given that two of the other answers mention similar proofs, that the one in this book does not use Morse theory. It uses only basic covering space techniques once it is shown that $\pi_2(G)$ is isomorphic to $\pi_2(G_r)$, where $G_r$ are the regular elements, itself not a difficult lemma.

Also if you believe that simply-connected compact Lie groups have the homotopy type of a product of odd spheres, then this follows. It is fairly easy to see that this is the case rationally, but do not remember whether the general statement is hard to prove.

This is proved in the book Representations of compact Lie groups by Bröcker and tom Dieck and reviewed here. It is Proposition 7.5 in Chapter V. The proof is for compact, connected Lie groups, but any connected Lie group has the homotopy type of its maximal compact subgroup. (Everything here is for finite-dimensional Lie groups, of course.)

Edit: Maybe I should add, given that two of the other answers mention similar proofs, that the one in this book does not use Morse theory. It uses only basic covering space techniques once it is shown that $\pi_2(G)$ is isomorphic to $\pi_2(G_r)$, where $G_r$ are the regular elements, itself not a difficult lemma.

Edit: The following is wrong! It is only true rationally, which is why I didn't remember having seen a proof of the general case :)

Also if you believe that simply-connected compact Lie groups have the homotopy type of a product of odd spheres, then this follows. It is fairly easy to see that this is the case rationally, but do not remember whether the general statement is hard to prove.