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Let $a$ in $\mathbb S(\mathbb C^{n})$, the unit sphere in $\mathbb C^n$. Does there exists a continuous map $x\mapsto u_x$, from $\mathbb S(\mathbb C^{n})$ to $U(n)$, the group of unitary endomorphisms of $\mathbb C^{n}$, such that $u_x(a)$ equals $x$ for all $x$?

As far as I can tell (that is not very far), the fact that the composition of $x\mapsto u_x$ and $v\in U(n) \mapsto v(a)$ is the identity on $\mathbb S(\mathbb C^{n})$ does not seem to lead to an obvious contraction about the cohomology spaces. But still, I am unable to define such a map.

And what if we consider the analogous question but with $\mathbb P^{n-1} \to PU(n)$ rather than $\mathbb S(\mathbb C^{n}) \to U(n)$?

Lierre
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  • Two examples: for $n=1$ identify $SU(1)$ and $S^1$ through $x\rightarrow (u_x:y\rightarrow xy)$ and define the continuous map $u:S^1\rightarrow U(1)=S^1$ sending $x\rightarrow u_{x/a}$; $u$ satisfy $u(x)a=x$ for every $x\in S^1$. For $n=2$ identify $S^3$ with $SU(2)$ and define a similar $u:S^3\rightarrow SU(2)$ with $u:x\rightarrow (u_x:y\rightarrow xa^{-1}y)$. – Gabriele Oct 22 '14 at 19:14

1 Answers1

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Note that $U(n)$ acts smoothly on $S(\mathbb{C}^n) = S^{2n-1}$ by left multiplication. The action is transitive with isotropy subgroup $U(n-1)$. So $p : U(n) \to S^{2n-1}$ given by $p(M) = Ma$ is a smooth principal $U(n-1)$-bundle.

If $u : S(\mathbb{C}^n) \to U(n)$, $x \mapsto u_x$, then $p(u_x) = u_xa$, so a map $u$ such that $u_xa = x$ for all $x \in S(\mathbb{C}^n)$ is a section of $p$. Note that a principal bundle admits a section if and only if it is trivial.

If the bundle $p : U(n) \to S^{2n-1}$ where trivial, then $U(n)$ would be diffeomorphic to $U(n-1)\times S^{2n-1}$. If this were true, then

$$\pi_{2n-2}(U(n)) = \pi_{2n-2}(U(n-1))\oplus\pi_{2n-2}(S^{2n-1}) = \pi_{2n-2}(U(n-1)).$$

By Bott periodicity, the left hand side is zero, while the right hand side is isomorphic to $\mathbb{Z}_{(n-1)!}$ (this was also proved by Bott). So we see that the bundle can only be trivial if $n =1$ or $2$.

For $n = 1$, $a \in S^1 = U(1)$, while for $n = 2$, $a \in S^3 = SU(2) \subset U(2)$. In both cases, we can take $a^{-1}$ and therefore define the section $u(x) = xa^{-1}$.

Michael Albanese
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