# Homotopy type of the loop space of a compact Lie group

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?

• 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). – Tyrone Dec 8 '19 at 16:56
• Thanks! @Tyrone. – Arun Dec 9 '19 at 5:33

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, $$\pi_1(X)=1$$.