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Let $G$ be a group. We say that $f : G \to G$ is a square root of $G$, if $f(x)^2=x$ for all $x \in G.$

Prove that a compact connected Lie group $G$ has no continuous square root.

What if, instead of a Lie group we consider a compact connected group with more that one element. Does the statement remain true?

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Are there any compact connected Hausdorff topological groups that aren't Lie groups? I thought not. – David Speyer Aug 6 '11 at 13:27
@David Speyer: what's your intuition about it? why it shouldn't exist? – Ehsan M. Kermani Aug 8 '11 at 7:04
The locally compact Hausdorff topological groups were classified by Yamabe. I don't understand the full classification, but my memory was that basically there were two types: Things that look like Lie groups, and things that look like the p-adics, plus mixes of the two. Things that look like the p-adics aren't connected, so that's why I think that the groups you are interested in are simply the Lie groups. But, since I don't know the details of Yamabe's result, don't rely on that. – David Speyer Aug 8 '11 at 14:46
Any idea for the main problem? – Ehsan M. Kermani Aug 23 '11 at 12:04
up vote 6 down vote accepted

Nudged by ehsanmo's comment, I returned to an idea I had a while ago and got it to work.

By hypothesis, $G$ is a compact connected manifold. Notice also that it is an oriented manifold: Choose any orientation on the tangent space at the identity and use left multiplication to translate it to the rest of the group. Let $\delta$ be the doubling map: $g \mapsto g^2$. For any continuous map of compact connected manifolds of the same dimension, we can define the degree of the map. The degree of a composition of two maps is the product of their degrees. So, if $\epsilon$ were a continuous square root of $\delta$, we would have $\delta \circ \epsilon = \mathrm{Id}$ and thus $\deg(\delta) \deg(\epsilon) = 1$.

I will compute that $\deg(\delta) = 2^r$, where $r$ is the dimension of a maximal torus of $G$. Hence, unless $G$ is trivial, $\deg(\delta)$ does not divide $1$ and there is no continuous square root.

Let $T$ be a maximal torus of $G$, with $T \cong (S^1)^r$, and let $W$ be the Weyl group. We will use the description of degree as the number of preimages of a generic point, counted with sign. Let $t$ be an element of $T$ such that the only way to write $t$ as $g u g^{-1}$, with $u \in T$ is to take $g \in W$. A generic element of $T$ will have this property. We claim that $g$ has $2^r$ square roots, and that $\delta$ is orientation preserving near each of them.

Consider a hypothetical square root of $t$. We can write it in the form $h u h^{-1}$ for $u \in T$, so $(h u h^{-1})^2 = h u^2 h^{-1} = t$. By our choice of $t$, we must have $h \in W$ and, replacing $(h,u)$ by $(\mathrm{Id}, h^{-1} u h)$, we may assume that $h = \mathrm{Id}$. In other words, we have shown that every square root of $t$ is in the torus. There are clearly $2^r$ square roots of $t$ in $T$.

Inside the tangent space to $G$ at $\mathrm{Id}$, choose a complement to $T$ and translate it around $T$ by left action. So the tangent bundle to $G$, restricted to $T$, splits as a direct sum of the tangent bundle to $T$ and a complementary trivial bundle. Using the formula $\delta(h u h^{-1}) = h u^2 h^{-1}$, we see that $\delta$ acts on the tangent bundle to $T$ by multiplication by $2$, and acts trivially on the chosen orthogonal complement. So the eigenvalues of the Jacobian of $\delta$ are all $2$'s and $1$'s, and we see that $\delta$ is orientation preserving. We have verified that the $\delta$-preimage of a generic point has $2^r$ elements, each counted with sign $+1$, as desired.

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I think the following toplogical argument will also work:

I will denote $\delta: G \to G$, $\delta(x) = x^2$ and the hypothetical continuous square root $\epsilon$. Since $\delta \circ \epsilon = \mathrm{id}$ we have that $\epsilon$ is injective. By invariance of domain this implies that $\epsilon$ is open. By compactness of $G$ it follows that $\epsilon$ is also closed; hence we see that, because of connectedness of $G$, the square root $\epsilon$ must be a homeomorphism with inverse $\delta$.

So, in particular, $\delta$ needs to be injective. But looking at tori in $G$, this is not the case.

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To conclude that $\delta$ is an inverse, I think you also have to show that $\epsilon \circ \delta = Id$. – user10676 Aug 25 '11 at 9:22
But this holds very generally: Any left inverse $g: G \to G$ of a bijective map $f:G\to G$ must also be a right inverse: Suppose we are given $x,y$ s.t. $g(x) = g(y)$. By surjectivity of $f$ there are $a,b$ such that $f(a) = x$, $f(b) = y$. Hence $a = g(f(a)) = g(f(b)) = b$ implies $x = f(a) = f(b) = y$. So $g$ is injective. Surjectivity is clear. – Sam Aug 25 '11 at 16:33
Nice! I believe this works. – David Speyer Aug 26 '11 at 15:00

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