1
$\begingroup$

I have read (Theorem 29.3, line 7, pg 144) the following statement:

$\pi_i(U(n-k)) \rightarrow \pi_i(U(n))$ is surjective for $i \le 2(n-k)+1$ and injective for $i \le 2(n-k)$.

and

Therefore $\pi_i(St_k(U(n))=0$ for $i \le 2(n-k)$.


What results are used here and where can I find a reference of this?

$\endgroup$
  • 1
    $\begingroup$ This follows from my answer to your question math.stackexchange.com/questions/3095144/…, since $Gl(\mathbb{C}^n)$ deformation retracts onto $U_n$ by polar decomposition of matrices. $\endgroup$ – Tyrone Feb 21 at 9:42
  • $\begingroup$ Oh my... I have completely forgotten to look back at the question! Thank you so much. $\endgroup$ – CL. Feb 21 at 10:38
1
$\begingroup$

I think they have made an off-by-$1$ error, they might have been thinking about the map $BU(n-k) \to BU(n)$. (The notes you're reading are unpublished and there is a disclaimer at the start about not being proofread, so an off-by-$1$ error isn't so bad.)

A great reference for a lot of results like this is Mimura and Toda's "Topology of Lie Groups " I and II. They compute a lot of homotopy and homology groups for classical Lie groups and their classifying spaces. For this particular example, look at Corollary 3.17 on page 68. The statement includes:

... For $i < 2n$, $\pi_i U(n) \cong \pi_i U(n+1)$, ...

Changing variables we see $\pi_iU(n-1)\cong \pi_i U(n)$ for $i < 2(n-1)$, and their argument actually also shows surjectivity in degree $2(n-1)$.

Typically you use the long exact sequence of the fibration

$$ U(n - 1) \to U(n) \to S^{2n - 1}$$

to compute the connectivity of $U(n - 1)\to U(n)$. The degree $i = 2n - 2$ is the last degree where $\pi_i S^{2n-1} = 0$ so our map induces surjective homomorphisms in all degrees $i \leq 2(n-1)$, and moreover they are injective if $i < 2(n-1)$. The (correct) connectivity of your map, and hence the Stiefel manifold, follows.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.