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I have to show that Lie groups have the rational homotopy type of a wedge of spheres. Unfortunately, my knowledge of Sullivan models is a bit shaky. The proof I come up with shows that they have the RHT of a product of spheres, which should be something else (I guess):

Let $G$ by a Lie group. Therefore, $G$ is an H-space and it's cohomology algebra $H^*(G,\mathbb Q)$ is a Hopf algebra (see Hatcher p283). By a theorem of Hopf, $H^*(G,\mathbb Q)\cong \Lambda V$ is isomorphic to a free graded-commutative algebra (see Hatcher 3C.4). Since $G$ is a manifold, it's higher cohomology-groups are trivial, so $V$ cannot have generators in even degree.

Now it is easy to see, that $\Lambda V$ with the zero-differential is a minimal Sullivan model of $G$.

Next, I want to factorize the sullivan model $\Lambda V \cong \bigotimes\limits_i \Lambda(e_i)$ , where all $e_i$ have odd degree. If I look at the product of spheres $P=\prod\limits_i S^{|e_i|}$, then I get a Sullivan model for $P$ by taking the tensor product of the Sullivan models for each single sphere (by Example 2 of chapter 12 in Félix/Halperin/Thomas' Rational Homotopy Theory). By Example 1 on the same page, the Sullivan model of $S^{|e_i|}$ is exactly $\Lambda(e_i)$. So I conclude $\Lambda V$ is a Sullivan model for $P$ and $G$, so $P$ and $G$ have the same rational homotopy type. But my aim was to show that $G$ has the RHT of a wedge of spheres.

Where am I mistaken? Where should I do something else to obtain the desired consequence?

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It's not true that a Lie group has the rational homotopy type of a wedge of spheres (for instance, this would imply the cup product on their rational cohomology is always trivial, which is certainly false). The correct statement is instead, as you have found, that a Lie group has the rational homotopy type of a product of spheres.

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  • $\begingroup$ Wow. That's not what I expected. Thank you very much. $\endgroup$ – Babelfish Mar 29 at 7:48

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