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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.

J126
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    One way is to show the space of hyperplanes in $\Bbb R^{n+1}$ is a manifold of dim $n+1$. This can be done rather easily by realizing it as a homogeneous space quotient of the Euclidean group (it should be $\operatorname {Euc} (n+1)/ \operatorname {Euc} (n)$) – PVAL-inactive Jun 23 '17 at 04:06
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    The set ${f(x)}$ has measure zero in $S^n$. Perhaps you could use this to show that the set of hyperplanes tangent to $M$ has measure zero in the set of hyperplanes passing through points in $M$? Choosing $v\in S^n$ determines a hyperplane passing through $x$; only two choices of $v$ will give a hyperplane that is tangent to $M$. – 211792 Jun 24 '17 at 15:48

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EDITED

We can give the space of affine hyperplanes in $\Bbb R^{n+1}$ explicitly as an $(n+1)$-dimensional manifold as follows: To the hyperplane $\nu\cdot (x-a) = 0$ with $\nu\in S^n$ and $x,a\in\Bbb R^{n+1}$, we associate the point $(\nu,a\cdot\nu)\in S^n\times\Bbb R = \mathscr H$. (Geometrically, we're taking the point on the hyperplane closest to the origin when we look at $(a\cdot\nu)\nu$.) Indeed, with the natural differentiable structure on the space of (oriented) affine hyperplanes, it is diffeomorphic to $S^n\times\Bbb R$.

As requested, we now work with the Gauss map $f\colon M\to S^n$, which assigns to each $x\in M$ the unit outward-pointing normal vector at $x$. We get the corresponding mapping $F\colon M\to\mathscr H$, representing the set of tangent hyperplanes to $M$, as follows: Set $F(x) = (f(x),x\cdot f(x))$. Since $M$ is an $n$-dimensional manifold and $F$ is smooth, the image in $\mathscr H=S^n\times\Bbb R$, an $(n+1)$-dimensional manifold, has measure zero. That is, almost every (affine) hyperplane in $\Bbb R^{n+1}$ is nowhere tangent to $M$.

Ted Shifrin
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