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Let $M$ be a smooth manifold. I am trying to prove that $M$ admits a vector field with only finitely many zeros.

This will follow if we can find a function $f : M\rightarrow \mathbb R$ such that $df$ has only finitely many zeros, but I cannot find such a function with this property either. My initial idea was to try to embed $M$ in $\mathbb R^N$ for some $N$ and look at $x\mapsto u \cdot x$ for fixed $u\in \mathbb R^N$, but I could not find a way to prove that there must be a $u$ such that the differential of this map has only finitely many zeros.

Does anyone have an elementary construction of such a vector field (or function)?

user15464
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    Are you assuming $M$ is compact? Even then, it is not a trivial result. Most height functions (projection on $u$, as you were considering) will end up being Morse functions, and hence will give you what you want, but it takes something like Sard's Theorem to prove that. So ... what tools do you have?! – Ted Shifrin May 12 '13 at 21:52
  • I can use Sard's theorem, but I would prefer to see a proof which doesn't require much beyond that. We can assume $M$ is compact if that makes the argument easier. – user15464 May 12 '13 at 21:55
  • If $M$ is open this post gives an outline how to construct a vector field without zeros, so it suffices to consider the compact case. – Martin May 12 '13 at 21:56
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    A Morse function has isolated critical points; thus, on a compact manifold, you are guaranteed that its gradient has only finitely many zeroes. Sard's theorem will show that for almost every $u$ (in your notation), the vector field $\text{grad}(f_u)$ is transverse to the zero section of the tangent bundle of $M$. Again, I don't know what you know and what you don't. – Ted Shifrin May 12 '13 at 22:03
  • Can you explain how to get the transversality from Sard's theorem? – user15464 May 12 '13 at 22:05
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    The usual game is to represent the submanifold $Z$ to which one wants to be transverse as the zero-set of a submersion $g$. Then $f\pitchfork Z$ if and only if $0$ is a regular value of $g\circ f$. Sard's Theorem comes in if you have parameters $s\in S$. The theorem is that if $F\colon X\times S\to Y$ is transverse to $Z$, then $F(\cdot,s)$ is transverse to $Z$ for almost all $s$. Appropriately set up, this is an application of Sard. – Ted Shifrin May 12 '13 at 22:58
  • Obtaining a vector field with transverse zeros is (slightly) easier than obtaining the existence of a Morse function. As Ted explained, you want to use Sard's theorem. The idea is much more general -- it allows you to say that you can always arrange the section of a vector bundle to be transverse to the 0 section. Any book on differential topology should explain this -- for instance Hirsch does it. – Sam Lisi May 14 '13 at 00:26

2 Answers2

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Here's a fun proof that employs the Transversality theorem to show that on any smooth manifold $M$, there is a vector field that vanishes only on a $0$-dimensional submanifold of $M$. Of course, when $M$ is compact, every $0$-dimensional submanifold is finite, which gives you your desired result.

Assume without loss of generality that $M^n$ is embedded in $\mathbb{R}^N$ with $N>n$. Define a map $F:M\times \mathbb{R}^N\to TM$ by $F(p,v)=\text{proj}_{T_pM}v$. Then $F$ is a smooth submersion. In particular, $F$ is transverse to $Z=M\times \{0\}$. So, by the transversality theorem, there exists some $v\in \mathbb{R}^N$ so that $f_v(x):M\to TM$ is transverse to $Z$. Now, $f_v(x)$ is a smooth section of $TM$, and so $f_v$ is a vector field. So $f_v^{-1}(M\times \{0\})$-the zeros of $f_v$-is a submanifold of $M$ of codimension $\dim TM-\dim M\times \{0\}= \dim M$, as claimed.

one potato two potato
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Adam Azzam
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    One does not need the manifold to be embedded in Euclidean space. It is already embedded as the zero section of its tangent bundle. Use the transversality theorem to produce a map of the manifold into the tangent bundle that is transverse to the zero section. This is a vector field with finitely many zeros. – lavinia Dec 12 '13 at 15:35
  • Why is $F$ a submersion? – Eduardo Longa Apr 27 '16 at 03:04
  • @EduardoLonga use local charts for a nbhd of a point of $TX$ in order to have $TX\cong U \times \Bbb R^{\dim X}$. the jacobian matrix of the map (locally) looks like a matrix $2\dim(X) \times x+n$ which has $1$'s on the "main diagonal". (in fact first component will be the identity of $U$ and the second is the derivative of the projection, which -after choosing suitable basis) has $1$'s on the diagonal and 0 elsewhere) – Luigi M Aug 22 '17 at 18:07
  • @lavinia That idea would use Thom’s transversality theorem, instead of the (weaker) parametric one. – Bcpicao Oct 29 '24 at 02:40
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A classic method - I think of Steenrod - is to triangulate the manifold then form the vector field whose singularities are the barycenters of the triangulation.For instance on a triangle the field flows away from the barycenter of the triangle towards the vertices and the centers of the edges. along the edges the field flows away from the centers towards the vertices. Draw a picture. It is easy to see.

lavinia
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