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It is well known that a smooth vector field on a 2-sphere must vanish twice.

What is the general technique for counting singularities of a smooth map between manifolds? For example, how many singularities must a smooth, surjective map $\phi: V \rightarrow SU(n)$ possess? Here $V$ is a finite dim vector space over $\mathbb{R}$. Can this be established via any analogous method to the vector field on the sphere.

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  • $\begingroup$ The first sentence isn't a special case of the second sentence. Anyway, for maps to $\mathbb{R}$ what you want is called Morse theory, and for vector fields what you want is called the Poincare-Hopf theorem. I don't know what the general situation is. $\endgroup$ – Qiaochu Yuan May 28 '15 at 17:52
  • $\begingroup$ I realize that it's not exactly a special case. I just wondered if there was something in the same spirit using the Euler characteristic. $\endgroup$ – Benjamin May 28 '15 at 18:08
  • $\begingroup$ It also depends on what you mean by "singularities." In the case of maps to $\mathbb{R}$ I mean points at which the derivative vanishes, but in the case of vector fields I mean points at which the vector field vanishes. I don't know what you mean for a map to $SU(n)$. $\endgroup$ – Qiaochu Yuan May 28 '15 at 18:16
  • $\begingroup$ Points where the Jacobian has rank less than the largest possible rank. $\endgroup$ – Benjamin May 28 '15 at 18:17
  • $\begingroup$ The first sentence is false. Wikipedia kindly provides a counterexample: upload.wikimedia.org/wikipedia/commons/1/14/… $\endgroup$ – Neal May 29 '15 at 2:25
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Concerning your first case. Section 9 of Heinz Hopf's lecture notes on differential geometry is called: "the role of the euler characteristic in the theory vector fields" which generalizes your first statements to surfaces:

Theorem: Summing up the singularity of any vector field on a surface, gives the Euler characteristic.

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Well here's maybe a somewhat enlightening example. Any smooth map $f:M\rightarrow \Bbb R^n$ where $M$ is an $m$-dimensional closed compact manifold has infinitely many singularities (for $n\geq 2$ and $m\geq n$). If $f$ only had finitely many singularities, then $f(M)\subset \Bbb R^n$ would be compact with non-empty interior and hence have infinite topological boundary (for $n\geq 2$). By the inverse function theorem, a topological boundary point must be a critical value.

In fact using a little covering space theory, we can conclude from this that any map from $S^2$ to a closed orientable surface of genus $\geq 1$ also has infinitely many singularities.

Generically, I do not think one should expect there to even exist maps where singularities are isolated, so the case where the target is $\Bbb R$ (as in Morse theory) is quite special.

Edit: Here's a further generalization of the above that should be easy for anyone interested to fill in the details. Let $M$, $N$ be smooth compact manifolds (without boundary) with $\operatorname{dim}M \geq \operatorname{dim}N$. If we have $p(\pi_1(M))$ has infinite index in $\pi_1(N)$ for any homomorphism $p: \pi_1(M) \rightarrow \pi_1(N)$, then any map $f:M \rightarrow N$ has infinitely many singularities.

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  • $\begingroup$ I tried to up vote this, but it won't let me! Sorry +1 $\endgroup$ – Benjamin May 29 '15 at 5:02

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