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I am teaching a summer qualifying exam class, and a student was looking up practice questions online. I cannot figure one of them out.

I am trying to solve this problem: Let $M$ be an $n$-dimensional manifold embedded in $\mathbb{R}^{n + 1}$. Then almost every hyperplane in $\mathbb{R}^{n + 1}$ is not tangent to $M$ at any point.

The hint given is to consider the map $f: M \to S^n$ that takes $x \in M$ to the unit normal at $x$.

I first thought to use Sard's Theorem and analyze the critical values of this map. Then I found examples where there are points with tangent hyperplanes, but $f$ has no critical values. I've tried to define other maps and analyze them, but can't produce a map that has critical values precisely where I want them.

I'd like if the solution was via the hint, but any solution is welcome.

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    $\begingroup$ Hmm. What about looking at the map from $M$ to the Grassmanian of hyperplanes? (Diff Geo is not really my area, so forgive me if I have my names wrong). Then I think the dimension of the graddmanian will just be too large. $\endgroup$
    – Steven Gubkin
    Commented Jun 23, 2017 at 0:24
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    $\begingroup$ Any normal hyperplane is uniquely determined by its normal versor, that can be identified with a point is $S^n$. So the tangent hyperplanes form a submanifold of dimension at most $n$ in the space of hyperplanes of $R^{n+1}$, which has dimension $n+1$. This implies the result. $\endgroup$
    – Francesco Polizzi
    Commented Jun 23, 2017 at 0:44
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    $\begingroup$ Following up on previous comments, the space of hyperplanes of $\mathbb{R}^{n+1}$ (not exactly a Grassmannian) is $\mathbb{RP}^{n+1}\setminus \{pt\}$. For any given unit vector, there is a 1-parameter family of hyperplanes with that unit normal. $\endgroup$
    – macbeth
    Commented Jun 23, 2017 at 1:03
  • $\begingroup$ Added my previous comment as a answer. $\endgroup$
    – Francesco Polizzi
    Commented Jun 23, 2017 at 6:30
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    $\begingroup$ I don't see any explicit mention of the fact that there seems to be two different definitions of "hyperplane" being used here. The hint above seems to indicate that in the statement of the problem, it is assumed that "hyperplane" means a linear subspace of dimension $n$. In that case, the assertion is obviously incorrect. However, in all of the answers below, "hyperplane" is assumed to be "affine hyperplane", in which case the assertion is correct for the reasons given below. $\endgroup$
    – Deane Yang
    Commented Jun 23, 2017 at 21:18

2 Answers 2

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This should be a little easier than using Sard's theorem:

The manifold $H$ of hyperplanes in $(n+1)$-space has dimension $n+1$. The map from $M$ to $H$ that sends each point to its tangent hyperplane is a differentiable map from an $n$-dimensional manifold to an $(n+1)$-dimensional manifold. The image must have measure zero.

You just need a lemma saying that if $f: U \rightarrow \mathbb{R}^m$ is a $C^1$ map from an open subset of $\mathbb{R}^k$ to $\mathbb{R}^m$ with $k < m$, then the image has measure zero. This is also a lemma for the Morse-Sard theorem (see Hirsch's Differential Topology).

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Any tangent hyperplane is uniquely determined by its normal versor, that via the Gauss map can be identified to a point is $S^n$. So the tangent hyperplanes form a submanifold of dimension at most $n$ in the space of hyperplanes of $\mathbb{R}^{n+1}$. Such a space is the dual space $(\mathbb{R}^{n+1})^*$, whose dimension is $n+1$, so the result follows.

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  • $\begingroup$ did you mean "Any <tangent> hyperplane is uniquely determined by its normal versor" ? It seems also that you must keep track of the base point, no ? $\endgroup$
    – Duchamp Gérard H. E.
    Commented Jun 23, 2017 at 7:37
  • $\begingroup$ Yes, "tangent". Corrected, thanks. $\endgroup$
    – Francesco Polizzi
    Commented Jun 23, 2017 at 10:06
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    $\begingroup$ Well, my argument shows that in any case the dimension of the space of tangent hyperplanes is the same as the dimension of the image of the Gauss map, and this is at most n (and exactly n when the Gauss map is generically finite into its image). $\endgroup$
    – Francesco Polizzi
    Commented Jun 23, 2017 at 10:09
  • $\begingroup$ I don't feel that the above argument is correct. At least, it is not complete. Tangent planes at different points can be different but parallel. Then these parallel tangents are represented by the same point of the Gauss sphere. Along the line of the above argument, if Gauss sphere had dimension $0$ (instead of $n$) then this would make things still easier, and this is false--it would show that the presented logical argument were a complete miss. $\endgroup$
    – Wlod AA
    Commented Jun 23, 2017 at 14:18
  • $\begingroup$ I think that in the smooth case (when M is not a plane) there is at least one point where the differential of the Gauss map has maximal rank. This implies that it is generically finite onto its image. I will look for references. $\endgroup$
    – Francesco Polizzi
    Commented Jun 23, 2017 at 15:47

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