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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.

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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.

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Great, thank you! –  Vidit Nanda Jul 23 '12 at 21:57
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