0
$\begingroup$

Could anyone give a non-trivial example of a bundle-mapping over $S^4$, i.e. find two complex rank 2 vector bundles $E_0,E_1$ over $S^4$ and a bundle mapping $$0\to E_0\overset{v}{\to}E_1\to0$$ such that the singularity set of $v$(where $v$ is not an isomorphic mapping) is equal to the embedding sphere $S^2$ of $S^4$?

$\endgroup$
1
  • $\begingroup$ For the point singularity case, there is an example. Since outside the equator $S^3$, $S^4\setminus S^3$ can shrink to points,we can use the standard identification, but near S3S3 we use the coordinate of the point by $(\alpha,\beta)\to\rho(\sqrt{|\alpha|^2+|\beta|^2})\left(\begin{array}{cc}\alpha&-\bar\beta\\ \beta&\bar\alpha \end{array}\right)$, where we use the complex coordinate $(\alpha,\beta)$ and $\rho$ is the cut off function near the point, by the partition of unity, we can have a global mapping. But I do not how to give the sphere singularity example. $\endgroup$
    – DLIN
    Jul 22, 2016 at 11:20

1 Answer 1

3
$\begingroup$

The answer is YES. First of all, notice the following lemma :

Let $M$ be a smooth manifold and let $K$ be a closed subset in $M$. Then, there exist a smooth function $f$ on $M$ such that $f^{-1}(0)=K$.

We can deduce this lemma by using a partition of unity.

Then, for $M=S^4$ and $K=S^2$, you can construct such an example by setting $E_0=E_1=E$ (where $E$ is an arbitrary complex vector bundle) and $v=f\cdot\text{id}$.

$\endgroup$
2
  • $\begingroup$ Could you give the reference of that Lemma. Well, I want to find two different vector bundles $E_0,E_1$. But is is also usefull. Thank you for your sharing.@nsmath $\endgroup$
    – DLIN
    Jul 26, 2016 at 3:44
  • $\begingroup$ @DLIN For example, see the Theorem 2.29 in a Lee's textbook Introduction to Smooth Manifolds Second Edition (available free online). Be careful of the Edition number. $\endgroup$ Jul 26, 2016 at 8:07

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.