3
$\begingroup$

Recently I go through obstruction class illustrated by Milnor. He defined $\mathfrak{o}_i$by an element in $H^i(M; \pi_{i-1}(V_{n-i+1}(F))$, which is cohomology with local coefficients.

But the 0th homotopy group has no group structure, and the definition for $\mathfrak{o}_1$ doesn’t work. So does 1st degree obstruction class exist and if exists, how do you define it?

$\endgroup$

1 Answer 1

3
$\begingroup$

When $i=1$, the Stiefel manifold is $V_n({\mathbb R})$ which is the bundle of $n$-frames in an $n$-dimensional vector space. As such it is homeomorphic to $GL(n,{\mathbb R})$ and hence $\pi_0$ has a group structure. It has two components and it is easy to see that ${\mathfrak o}_1$ measures orientability in the way that you might expect.

Probably this question is more suited to MSE than Mathoverflow.

$\endgroup$
2
  • $\begingroup$ Then $\pi_0 (V_n(R^n)) = Z/2$, right? $\endgroup$
    – XT Chen
    Apr 27, 2020 at 17:44
  • 1
    $\begingroup$ That's right; there are two components, distinguished by the sign of the determinant. $\endgroup$ Apr 28, 2020 at 18:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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