Let $p:M\to \mathbb{R}^{n+1}$ be the closed immersed hypersurface. Is the following thing right? If there exists a point $x_0$ in $\mathbb{R}^{n+1}$ such that $\langle x(p)-x_0,\nu(p)\rangle>0$ for all $p\in M$, then $M$ is diffeomorphic to a sphere. Here $\nu(p)$ is the outward normal vector.
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1$\begingroup$ A cylinder in $\mathbb{R}^3$ satisfies your condition (with $x_0=0$ for the cylinder $\{(x,y,z)|x^2+y^2=1\}$). Maybe you want to assume compactness of $M$ ? $\endgroup$– Thomas RichardCommented Jul 22, 2014 at 12:38
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8$\begingroup$ Please use $\langle a,b\rangle$ (with \langle and \rangle) rather than $<a,b>$ for the inner product. Besides making the formulas more readable, it makes fewer people furious. (I took the liberty of making this change.) $\endgroup$– Joonas IlmavirtaCommented Jul 22, 2014 at 12:41
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2 Answers
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Assuming that $M$ is connected compact without boundary the answer is YES.
Consider a map $\pi: p \mapsto \frac{x(p)-x_0}{||x(p)-x_0||}$ from $M$ to $S^n$.
$\pi$ is an immersion since $\langle x(p)-x_0,\nu(p)\rangle >0$.
$\pi$ is a covering since $\pi$ is an immersion, $M$ is a closed $n-$manifold, and $S^n$ is connected.
$\pi$ is a homeomorphism since it is a covering over the simply connected $S^n$ . So $M$ is diffeomorphic to $S^n$.
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$\begingroup$ Can you tell me the details of step 2? Why $\pi$ is an immersion? $\endgroup$ Commented Jul 23, 2014 at 20:36
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$\begingroup$ Consider a map from $\mathbb{R}^{n+1}\setminus x_0$ to $S^n$, $\pi':x\mapsto\frac{x-x_0}{\|x-x_0\|}$. It is clear that $D\pi':T_x\mathbb{R}^{n+1}\to T_{\pi'(x)}S^n$ is the orthogonal projection along the vector $x-x_0$ which generates the kernel of $D\pi'$. We have $\pi=\pi'\circ p$ where $p$ is the embedding of $M$. So we have $D\pi=D\pi'\circ Dp$. Condition $\langle x(p)-x_0,\nu(p)\rangle>0$ implies that the image of $Dp$ does not contain $x-x_0$ - the kernel of $D\pi'$. Hence $D\pi$ is an injection, and more over, an isomorphism. $\endgroup$ Commented Jul 24, 2014 at 8:34
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Let me suppose that $M$ is in addition compact, and for simplicity that $x_0=0$. Let $\phi : M \to S^n$ be the radial projection, which is a smooth map.
If $\phi$ is not an immersion, there is a tangent vector $v \in T_x M$ such that $Dp(v)$ is parallel to the vector $p(x) \in \mathbb{R}^{n+1}$. But $$\langle v, \nu(x)\rangle=0$$ as the tangent and normal spaces are orthogonal, so $\langle p(x), \nu(x)\rangle=0$, a contradiction. Thus $\phi$ is an immersion.
But $\phi$ is also proper, and a proper codimension 0 immersion is a covering map. As any compact covering space of $S^n$ is a finite union of copies of $S^n$ (this is phrased strangely so that it still holds for $n=1$), it follows that $M$ is diffeomorphic to a finite union of copies of $S^n$.
One clearly can't insist that it is a single copy of $S^n$ in general, just by taking e.g. two spheres of different radii. So assume $M$ is connected if this is what you want.
answered Jul 22, 2014 at 12:54
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$\begingroup$ Here you use $Dp(v)$ is parallel to $p(x)$ and $\langle v,\nu(x)\rangle=0$ to conclude that $\langle p(x),\nu(x)\rangle =0$. But I don't see how to prove this unless you have $Dp(v)$ is parallel to $v$. $\endgroup$ Commented Jul 23, 2014 at 20:33