Map induced by $O(n)\hookrightarrow U(n)$ on homotopy groups There is an inclusion $O(n)\hookrightarrow U(n)$ which views an $n\times n$ orthogonal matrix as a unitary matrix. It is also a theorem, sometimes called Bott periodicity, that we have the following homotopy groups:
$\pi_{4i-1}(O(n)) = \mathbb{Z}$ and $\pi_{4i-1}(U(n))=\mathbb{Z}$
for $n$ large, say $n>2(4i +1)$. My question is: What is the induced map $\pi_{4i-1}(O(n))\rightarrow \pi_{4i-1}(U(n))$? It is a map from $\mathbb{Z}$ to $\mathbb{Z}$ so must be multiplication by some integer, but which integer is it?
The only thing I know that might help is that the map $O(n)\hookrightarrow U(n)$ induces a map on classifying spaces $BO(n)\rightarrow BU(n)$. One can think of this map as an inclusion of a real grassmannian into a complex grassmannian, but the map on homotopy groups seems just as mysterious.
Thanks for any help.

I had this question myself a few days ago and found this question, but I now have an answer with a hint from my supervisor.
The key seems to be in a theorem (3.2) in 'Homotopy theory of Lie Groups' by Mimura, which can be found in the Handbook of Algebraic Topology. The theorem gives a weak homotopy equivalence
$$BO\rightarrow \Omega (SU/SO)  .$$
This is enough for our requirement since there are isomorphisms
$$\pi_i(O)\cong\pi_i(SO) \mbox{ and } \pi_i(U)\cong\pi_i(SU)$$
for $i>2$. Notice that these homotopy groups are the same as $U(n)$ or $O(n)$ in the ranges you specified.   
Now there is a fibration
$$ SO\overset{f}{\rightarrow} SU \rightarrow SU/SO$$
and $f$ will induce the same map on $\pi_i$ for $i>2$.  Due to the resulting long exact sequence of homotopy groups, we are required to work out $\pi_i(SU/SO)$, but by the theorem, we have 
$$\pi_i(SU/SO)\cong \pi_{i-2}(O).$$
I shall leave the details to you, but for all $k$, we get $\pi_{3+8k}(O) \rightarrow \pi_{3+8k}(U) $ is multiplication by 2 and $\pi_{7+8k}(O)\rightarrow \pi_{7+8k}(U)$ is an isomorphism. 
