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For $n\geq 3$, I am considering all maps

$S^2 \times S^3 \xrightarrow{f} \frac{U(n)}{O(n)}$

I wish to know whether there exists a lift to

$S^2 \times S^3 \xrightarrow{\tilde{f}} U(n)$

where the map $U(n) \rightarrow \frac{U(n)}{O(n)}$ is the standard projection in the fibration

$O(n) \rightarrow U(n) \rightarrow \frac{U(n)}{O(n)}$

Revised Question:

Thanks to @Jason DeVito. I realized that I didn't post the question as precise as I should. I am only interested in the subclass of map $f$ which are trivial when restricted to $S^3$ and $S^2$. With this restriction, can the map $f$ be lifted to $\tilde{f}$?

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  • $\begingroup$ Why are you interested in this question? $\endgroup$ – Arctic Char Oct 15 at 18:35
  • $\begingroup$ It's for some physics application. $\endgroup$ – Yen-Ta Huang Oct 15 at 22:32
  • $\begingroup$ Well, when $n=2$, using the fact that $U(2)\rightarrow U(2)/SO(2)$ is a principal $S^1$-bundle, it's not too hard to show the answer is "yes" under the weaker hypothesis that $f$ is trivial on $H^2$. However, I don't know what happens when $n\geq 3$. $\endgroup$ – Jason DeVito Oct 16 at 0:12
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For $n=1$, yes, such a map lifts. This is because $S^2\times S^3$ is simply connected, so any map to $U(1)/O(1)\cong S^1$ lifts to any cover of $U(1)/O(1)$, including $U(1)$.

On the other hand, for each $n\geq 2$, there is a map $f:S^2\times S^3\rightarrow U(n)/O(n)$ which does not lift.

To see this, first note that the identity component of $O(n)$, $SO(n)$ is a subgroup of $SU(n)\subseteq U(n)$. So, the inclusion map $SO(n)\rightarrow U(n)$ is trivial on $\pi_1$ (with base point the identity), because it factors through $SU(n)$.

Now, from the long exact sequence in homotopy groups associated to $O(n)\rightarrow U(n)\rightarrow U(n)/O(n)$, we get a portion of the form $$...\rightarrow \pi_2(U(n))\rightarrow \pi_2(U(n)/O(n))\rightarrow \pi_1(O(n))\rightarrow \pi_1(U(n)) \rightarrow ...$$

Now, $\pi_2(U(n)) = 0$ as it is for every Lie group (see, e.g., this MO question), and the map $\pi_1(O(n))\rightarrow \pi_1(U(n))$ is trivial. So, this because $$0\rightarrow \pi_2(U(n)/O(n)) \rightarrow \pi_1(O(n))\rightarrow 0.$$

In other words, $\pi_2(U(n)/O(n))\cong \pi_1(O(n))\neq 0$ for $n\geq 2$.

Now, let $f:S^2\times S^3\rightarrow U(n)/O(n)$ be the composition $S^2\times S^3\rightarrow S^2\rightarrow U(n)/O(n)$, where the first map is the obvious projection an the second map is non-trivial on $\pi_2$. Then $f$ is non-trivial on $\pi_2$.

On the other hand, if $f$ were to lift, then $f$ would factor through $U(n)$. As $\pi_2(U(n)) = 0$, this would force $f$ to be trivial on $\pi_2$. Thus, $f$ cannot lift.

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  • $\begingroup$ Thanks to your answer. I realized that I didn't post the question as precise as I should. I just edited the question. $\endgroup$ – Yen-Ta Huang Oct 15 at 22:21

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