0
$\begingroup$

If $M$ is an orientable compact 3-manifold with boundary such that there is defined $s\in\Gamma(\partial M,V_2(TM))$ a section of the 2-frame bundle of $TM$, then $s$ extends to $\tilde s\in\Gamma(M,V_2(TM))$

Proof:

Step 1

Triangulate $M$. It gives rise to a cell structure on $M$.

Step 2

I define $\tilde s=s$ on $\partial M^{(1)}$, the 1-skeleton of $\partial M$.

Step 3

I extend $\tilde s$ over $M^{(1)}$, because $\pi_0(SO_3)=0$.

Step 4

I then extend it over $M^{(2)}$ (because the obstruction, which is the second Steifel-Whitney class $w_2(M)\in H^2(M,\pi_1(SO_3))$ vanishes for any orientable compact 3-manifold, due to the Wu formula and the vanishing of $w_1(M)$ as $M$ is orientable.)

Step 5

Now, having obtained $\tilde s$ in particular on $\partial M^{(2)}$, I can homotope $\tilde s$ to $s$ over $\partial M^{(2)}$, because they already agree on $\partial M^{(1)}$ and the obstruction to homotoping over the 2-skeleton is just $\pi_2(SO_3)$ which is the trivial group.

Step 6

Now, having obtained $\tilde s$ over $M^{(2)}$, we extend it to all of $M$, because again, the obstruction to extending over the 3-skeleton is the trivial group $\pi_2(SO_3)$.

I need this reviewed.

$\endgroup$
2
  • $\begingroup$ I don't understand Step 3. As the tangential bundle of $M$ is trivial, the datum of a $2$-frame is equivalent to a map to $SO(3)$. Obstructions to extending such a map from the $1$-skeleton to the $2$-skeleton lie in $\pi_1SO(3)$, which is non-trivial. $\endgroup$ Aug 9, 2013 at 9:59
  • $\begingroup$ Edited and changed numbering. Steifel-Whitney classes and other such characteristic classes help us when we at times encounter such situations as $\pi_1(SO_3)$ being non-zero, and the cohomology group being non-zero. $\endgroup$
    – Karthik C
    Aug 9, 2013 at 12:59

0

Your Answer

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