1
$\begingroup$

The following theorem is proved in Milnor's famous book "Morse theory".

Theorem 21.7 (Bott). Let $G$ be a compact, simply connected Lie group. Then the loop space of $G$ has the homotopy type of a CW-complex with no odd dimensional cells.

It is not clear to me where the author uses the simply connectedness of $G$. Is it a necessary condition? Can someone please illuminate?

$\endgroup$
  • $\begingroup$ No. Milnor shows that the space of maps $Map(X,Y)$ (compact-open topology) is CW whenever $X,Y$ are CW, $X$ is finite and $Y$ is countable and locally finite. I think this can be improved somewhat, but I don't have a reference off the top of my head. (In any case it applies fo $\Omega G=Map((S^1,\ast),(G,\ast))$ for all compact Lie groups $G$, no simple connectedness needed). $\endgroup$ – Tyrone Dec 8 '19 at 16:56
  • $\begingroup$ Thanks! @Tyrone. $\endgroup$ – Arun Dec 9 '19 at 5:33
1
$\begingroup$

You can easily drop the connectivity condition as long as each (equivalently, one) component is simply-connected. But the assumption that $\pi_1(X)=1$, is a necessary condition. If $X$ is a complex without odd-dimensional cells then $\pi_1(X)=1$: Indeed, by the cellular approximation theorem, every loop $c$ in $X$ is homotopic to a loop $c'$ in $X^1$. If $X^1$ contains no 1-cells, then $X^1=X^0$, implying that $c'$ is constant. Thus, $\pi_1(X)=1$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.