# Can I lift the maps from $S^2 \times S^3 \rightarrow \frac{U(n)}{O(n)}$ to $S^2 \times S^3 \rightarrow U(n)$

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 $$\tilde{f}$$?

• Why are you interested in this question? – Arctic Char Oct 15 at 18:35
• It's for some physics application. – Yen-Ta Huang Oct 15 at 22:32
• 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 $n\geq 3$. – Jason DeVito Oct 16 at 0:12

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.

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