0
$\begingroup$

Let $\pi:E\rightarrow B$ be a smooth vector bundle. I call this vector bundle to be an oriented vector bundle, if I can choose an orientation on $\pi^{-1}(b)$ for each $b\in B$ and a trivialization cover $\{U_\alpha\}$ of $B$ with trivializations $\psi_\alpha:\pi^{-1}(U_\alpha)\rightarrow U_\alpha\times \mathbb{R}^n$ such that, the isomorphisms $\pi^{-1}(x)\rightarrow \{x\}\times \mathbb{R}^n$ coming from $\psi_\alpha$ are orientation preserving taking standard orientation on $\mathbb{R}^n$.

This is ok for me. I do not see the reason of how/why one should think of some homology/cohomology class when talking about orientation. Any comments are welcome.

$\endgroup$
  • 1
    $\begingroup$ Usually one thinks of the orientation class of an oriented, compact $n$-dimensional manifold as being given by the fundamental class in $H_n(M,\Bbb Z)$ (i.e., the orientation gives a choice of generator). In the vector bundle case you need the Thom class of the bundle. $\endgroup$ – Ted Shifrin Mar 14 at 18:48
  • $\begingroup$ @TedShifrin Hello Sir. Can you help me to relate the notion of orientation I have mentioned.. $\endgroup$ – Praphulla Koushik Mar 14 at 19:17
  • 1
    $\begingroup$ The key thing to understand is the fact that, topologically, you can see the orientation class for a vector space $\Bbb R^n$ as a relative cohomology class, the generator of $H^n(\Bbb R^n,\Bbb R^n-\{0\})$. $\endgroup$ – Ted Shifrin Mar 14 at 19:45

Your Answer

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

Browse other questions tagged or ask your own question.