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The Milnor-Moore theorem states that for path-connected H-spaces $X$, $H_*(X;\Bbb Q)$ is a free graded-commutative algebra on $\pi_*(X)\otimes \Bbb Q$. I am asked to use this to compute $\pi_i(S^n)\otimes \Bbb Q$.

I know another way of computing $\pi_i(S^n)\otimes \Bbb Q$ which does not use the Milnor-Moore theorem. I'm not sure how one would use this theorem since the only spheres which are H-spaces are in dimensions 1,3 and 7.

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  • $\begingroup$ Or consider the $H$-space $\Omega S^n$. $\endgroup$ – Justin Young Jan 15 '17 at 21:00
  • $\begingroup$ Indeed $\Omega S^n$ works, I wish I could delete the question now. $\endgroup$ – iwriteonbananas Jan 16 '17 at 8:51
  • $\begingroup$ Not sure I agree with the statement of the Milnor-Moore theorem, though. I think you get the universal enveloping algebra, which should not be the same as the free commuative algebra. $\endgroup$ – Justin Young Jan 19 '17 at 15:58
  • $\begingroup$ Or maybe you mean cohomology? $\endgroup$ – Justin Young Jan 20 '17 at 11:36
  • $\begingroup$ Hm, no I mean homology @JustinYoung, but I agree, I think there is a mistake in the statement of the theorem. $\endgroup$ – iwriteonbananas Jan 20 '17 at 12:38
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I thought I would post an answer anyway since it's a nice problem. The Milnor-Moore theorem (in this context) states that if $X$ is a path-connected $H$-space then $H_*(X) = U(\pi_*(X)\otimes \mathbb Q)$ is isomorphic as a Hopf algebra to the universal enveloping algebra over the Lie algebra $\pi_*(X)\otimes \mathbb Q$ (the Lie bracket is the Whitehead product, and the inclusion of the primitives is induced by the Hurewicz map). Applying this to $X =\Omega S^n$ for $n\ge 2$ we calculate (using spectral sequences for example) that $H_*(X) = \mathbb Q[x]$ as an algebra on a class $x$ in degree $n-1$. For degree reasons, $x$ is primitive. Using the Hopf algebra property we get that $\Delta (x^k)$ is a signed sum of terms of the form $x^i \otimes x^j$ where $i+j = k$.

If $n$ is odd, then $n-1$ is even and therefore $x^k$ is not primitive for $k>1$.

But if $n$ is even, then $n-1$ is odd and we get $\Delta(x^2) = 1\otimes x^2 + x^2 \otimes 1$, the middle terms cancel. For $k>2$, we have $\Delta(x^k) = \Delta(x^2 x^{k-2}) = \Delta(x^2) \Delta(x^{k-2})$, since $x^2$ has even degree, none of the terms cancel, and so we have, at least, a nonzero term of the form $\alpha x^2 \otimes x^{k-2}$. It follows that $x^k$ is not primitive for $k>2$.

Thus, the primitive Lie algebra is just $x$ in degree $n-1$ for $n$ odd and $x, x^2$ in degrees $n-1$ and $2(n-1)$ for $n$ even. Since the Lie algebra is also $\pi_*(\Omega S^n)\otimes \mathbb Q = \pi_{*+1}(S^n)\otimes \mathbb Q$, we get the standard result.

For $n$ odd $\pi_*(S^n)\otimes \mathbb Q = \mathbb Q$ for $* = n$ and zero otherwise.

For $n$ even $\pi_*(S^n)\otimes \mathbb Q = \mathbb Q$ for $* = n, 2n-1$ and zero otherwise.

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Indeed, you can only use this theorem for $S^1$, $S^3$ and $S^7$ since these are the only three that are connected H-spaces (don't forget that $S^0 = \mathbb{Z}/2\mathbb{Z}$ is an H-space too, but it's not path-connected). You have (let $n \in \{1,3,7\}$): $$H_i(S^n; \mathbb{Q}) = \begin{cases} \mathbb{Q} & i = 0,n; \\ 0 & \text{otherwise.} \end{cases}$$

The theorem tells you that this is a free graded-commutative algebra on some graded vector space $V$. Clearly you must have a generator $e \in V_n$ of degree $V$ corresponding to the fundamental class, it cannot be written as a product of lower-degree elements. Since $n$ is odd, $e^2 = 0$. There is nothing else in the homology, so you get that $V$ is one-dimensional spanned by an element of degree $n$, and thus for $n \in \{1,3,7\}$: $$\pi_i(S^n) \otimes \mathbb{Q} = \begin{cases} \mathbb{Q} & i = n; \\ 0 & \text{otherwise.} \end{cases}$$

I don't know what the author of the problem had in mind precisely, but note that you can't directly apply the theorem to even-dimensional spheres in any case. Indeed if $H_*(S^{2n})$ were a free graded-commutative algebra, there would have to be a generator $e$ in degree $2n$ corresponding to the fundamental class. But then $e^2 \in H_{4n}(S^{2n})$ would be a nonzero element... Which is not possible. For odd-dimensional spheres you get the same result as for the ones that are H-spaces ($\pi_{2n+1}(S^{2n+1}) \otimes \mathbb{Q} = \mathbb{Q}$ and zero otherwise).

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  • $\begingroup$ Yes, thank you. I had figured out how to deduce the cases $n\in\{1,3,7\}$ from the Milnor-Moore theorem but it's a mystery to me how one would use this theorem for the case $n\notin \{1,3,7\}$. $\endgroup$ – iwriteonbananas Jan 7 '17 at 15:25

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