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 $\mu:M \to \mathbb{R}$ be a fixed surjective smooth function on a smooth manifold $M$. Let $N$ be a smooth compact manifold that embeds smoothly into $M$ via $\iota:N \to M$.

What conditions on $\iota$ -- if any -- guarantee that $\mu\circ\iota:N \to \mathbb{R}$ is a Morse function on $N$?

Of course, one can always use Atlases on $M$ and $N$ to compute Hessians on each patch containing a critical point of $\mu \circ \iota$ and so forth, but is there an equivalent characterization which does not involve local coordinate computations?

If this is elementary, I will delete it.

share|cite|improve this question
up vote 7 down vote accepted

Saying a function $f : M \to \mathbb R$ is Morse amounts to saying that $D^* f : M \to T^* M$ is transverse to the zero section, where $D^*f$ is the derivative of $f$, though of as a section of the cotangent bundle, alternatively that the Hessian $Hf_p$ is non-degenerate at the critical points $p$ of $f$.

A critical point $p$ of $\mu \circ i$ is a place where $Di$ maps into the kernel of $D \mu$. The Hessian $H(\mu \circ i)_p$ (as a function of two input vectors $v$ and $w$) at such a point can be computed by the Hessian chain rule

$$H(\mu \circ i)_p(v,w) = H\mu_{i(p)}(Di_p(v),Di_p(w)) + D\mu_{i(p)}(Hi_p(v,w)) $$

This has to be non-degenerate. So this is something that can be achieved in a variety of ways since it's the sum of two quadratic functions. Generically you could break this up into two situations, ones where $i$ passes through critical points of $\mu$ and ones where it does not. In the former case the left-hand-side plays a dominant role, in the latter, you can use the right hand side. If you're okay perturbing the embedding and your embedding is not co-dimension zero, you can always ensure $i$ does not pass through the critical points of $\mu$, since they're isolated.

share|cite|improve this answer
Great, thank you! – Vidit Nanda Jul 23 '12 at 21:57

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.