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How do one prove (or disprove) that $\Omega S^{2}$ and $\Omega(S^{2} \vee S^{2})$ are non commutative Hopf spaces?

I thought this is a question for math.stackexchange, but not many people even viewed that question. Seems like it should be easy but right now I have no idea how to do it. Any hint?

Definition: If $X$ is commutative Hopf space if, for the multiplication $ \mu $ there is a homotopy between $ H: \mu \simeq \mu \circ T $, where $T : X \times X \to X \times X$ is the switch map.

Clarification: When I say multiplication, I mean multiplication induced by the loop structure. On $\Omega S^{2}$ there is another multiplication which is induced by multiplication on $S^{1}$ as $\Omega S^{2} = \Omega \Sigma S^{1} $ ( This is of course commutative).

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  • $\begingroup$ math.stackexchange.com/questions/310824/… Here is my question in math.stackexchange.com. One may put the answer wherever it is appropriate $\endgroup$
    – Prasit
    Commented Feb 22, 2013 at 22:47
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    $\begingroup$ Your non-commutative Hopf algebras look like topological spaces. $\endgroup$
    – Fernando Muro
    Commented Feb 22, 2013 at 22:52
  • $\begingroup$ Please use a tag with a two-letter prefix (instead of creating one without) if (and only if!) they already exist. [The complete list is given right below the field where you enter the tag.] $\endgroup$
    – user9072
    Commented Feb 22, 2013 at 23:16
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    $\begingroup$ On MSE you talk of H-spaces, here you talk of Hopf algebras. These are not strictly speaking the same thing. What is it you actually mean to ask? $\endgroup$
    – Yemon Choi
    Commented Feb 22, 2013 at 23:23
  • $\begingroup$ Sorry, I meant spaces and I changed the title! $\endgroup$
    – Prasit
    Commented Feb 22, 2013 at 23:31

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Let us run the Serre SS for homology with $\mathbb{Z}/2$ coefficients for the fibration

$\Omega(S^{2} \vee S^{2}) \to * \to S^{2} \vee S^{2}$

Let $x,y \in H_{2}(S^{2} \vee S^{2})$ be the generators. Since these guys has to die, there exists $x_{1}, y_{1} \in H_{1}(\Omega(S^{2} \vee S^{2}))$ which are hit by $x$ and $y$ respectively.

Notice that there are $4$ terms at $E^{2}_{2,1}$ namely $x_{1}x, y_{1}x, x_{1}y, y_{1}y$.

$d_{2}(x_{1}y) = x_{1}y_{1}$ and $ d_{2}(y_{1}x) =y_{1}x_{1} $.

So $x_{1}y + y_{1}x$ survives if $x_{1}y_{1} = y_{1}x_{1}$ and never dies. Which is a contradiction.

this shows that in homology the elements $x_{1},y_{1}$ do not commute. So $\Omega(S^{2} \vee S^{2})$ is not a commutative Hopf space.

I think this is correct. But the case of $\Omega S^{2}$ still remains unanswered.

Edit: The complete answer is given by Justin in the comments below.

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  • $\begingroup$ If $X$ is a space, the multiplication in $\Omega(X)$ is the concatenation of paths, and this is a commutative operation iff $\Omega(X)$ has exactly one point. Probably you were not asking something which has this as answer, but then I do not understand your question. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Feb 23, 2013 at 1:05
  • $\begingroup$ To start with I have to admit that the situation is very blurry to me at this moment. But I do mean multiplication by concatenation, the usual multiplication on a loop space, at least thats the multiplication I was thinking of. I do not understand what you mean when you say "exactly one point". $\endgroup$
    – Prasit
    Commented Feb 23, 2013 at 1:22
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    $\begingroup$ It would then be useful to add that information to the body of the question. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Feb 23, 2013 at 3:11
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    $\begingroup$ In general, if $\tilde H_*(X, R)$ is free then $H_*(\Omega \Sigma X, R)$ is a tensor algebra on $\tilde H_*(X,R)$, and tensor algebras are not commutative. You can prove this using your Serre SS. With $X = S^1$ and $X = S^1 \vee S^1$ you get your result. Hi Prasit! $\endgroup$
    – Justin Young
    Commented Feb 23, 2013 at 7:15
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    $\begingroup$ Edit: in the above I should specify that tensor algebras are not commutative provided the generators are more than one dimensional, or in odd degree. Remember, to be commutative, odd degree elements must satisfy $2x^2 = 0$. $\endgroup$
    – Justin Young
    Commented Feb 23, 2013 at 7:25

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