MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $M^n$ be a differentiable manifold and $\pi\colon E\to M$ is $n$-dimensional vector bundle over $M$.

We have a zero section $s\colon M\to E$ of $\pi$.

How can I make a section $s'$ which is trnasversal to $s$? (i.e., $s'$ vanishes $s$ finitely many times.)

(In some text, it seems even possible to make $s$ and $s'$ are isotopic.)

I need this to interpret the euler class of $\pi$, $\chi(\pi)$ as an algebraic intersection number of $s$ and $s'$.

Are there anybody who can give me any references?

share|cite|improve this question
I'll answer the easy one: $s$ and $s'$ are always isotopic, since you can consider the isotopy $s_t = t\cdot s'$ (using the linear structure on each fiber, so $t\cdot s'$ is well-defined). – Marco Golla Apr 25 '11 at 14:50

This is a straightforward application of transversality theorem, which roughly speaking states that we can make a map transverse to a submanifold with an arbitrary small perturbation. It is a consequence of Morse-Sard theorem.

The statement you need is the following:

Theorem. Let $A$, $B$ be $\mathcal{C}^r$-submanifolds of $M$, $1 \leq r \leq \infty$. Then every neighborhood of the inclusion $i_B \colon B \to M$ in $\mathcal{C}^r(B, M)$ contains an embedding which is transverse to $A$.

For a proof, see [Hirsch, Differential Topology, Thm. 2.4 pag. 78].

share|cite|improve this answer
As for the second part of the question, it doesn't look like Hirsch covers the theorem "Euler number = self-intersection of the zero-section". That's covered in Bott and Tu's, Differential forms in algebraic topology, Theorem 11.17. That requires lots of technology, though (at least, the book does). – Marco Golla Apr 25 '11 at 14:56

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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