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I can show that the sphere $x_1^2+x_2^2+\ldots+x_n^2=1$ is an $(n-1)$-dimensional manifold by considering the map $f(x_1,\ldots,x_n)=x_1^2+\ldots+x_n^2$, and noticing that $1$ is a regular value of $f$, i.e. $Df(p)$ is surjective for all $p\in f^{-1}(1)$.

What about the quadric $x_1^2+x_2^2+\ldots+x_{n-1}^2=x_n^2$? I want to use the same trick, so I consider the map $g(x_1,\ldots,x_n)=x_1^2+\ldots+x_{n-1}^2-x_n^2$. Now, if $p=(p_1,\ldots,p_n)$, $Dg(p)$ is the $1\times n$ matrix $$\begin{bmatrix}2p_1 & 2p_2 & \ldots & -2p_n \end{bmatrix}.$$

Unfortunately, $Dg(p)$ is not surjective when $p=(0,0,\ldots,0)$.

How can I show that the quadric is a ($(n-1)$-dimensional?) manifold?

Mika H.
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  • Well, that quadric isn't. The origin is a point that has no neighbourhood homeomorphic to an $n-1$-dimensional ball. – Daniel Fischer Dec 09 '13 at 22:58
  • @DanielFischer That's interesting. There's an exercise that asks to compute the tangent space of this quadric. But the definition of a tangent space that I know is based on the fact that we have a manifold. Any ideas how to do it in this case? – Mika H. Dec 10 '13 at 00:23
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    Well, if you remove the offending point, you get a manifold. The origin is the only point where things go wrong. Roughly, the quadric, the double cone, is two $\mathbb{R}^{n-1}$ glued together at one point. If you punch a hole removing that point, you get two separate disjoint copies of $\mathbb{R}^{n-1}\setminus {0}$, a perfectly fine (disconnected) manifold. – Daniel Fischer Dec 10 '13 at 00:27

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The property of being a manifold is (if we assume second countability from the beginning) a local property: it is something that fails or holds on small scale, in a neighborhood of every point. Thus, one can speak of manifold points of a set that is not a manifold itself. A manifold point is one that has a neighborhood homeomorphic to a Euclidean space. Or diffeomorphic, if we want to talk about smooth manifolds and have a notion of differentiability (which we do for submanifolds of a Euclidean space, as here).

The cone $x_1^2+x_2^2+\ldots+x_{n-1}^2=x_n^2$ is not a manifold, only because of the vertex. Every other point is a manifold point, with a well-defined tangent. Another way to put it: cone minus its vertex is a manifold (as Daniel Fischer said), because the implicit function theorem applies to the defining function $x_1^2+x_2^2+\ldots+x_{n-1}^2-x_n^2$ on the set $\mathbb R^n\setminus\{0\}$.

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