Problem: Show that $M:=\{ (x_1,x_2,x_3,x_4) \in \mathbb{R}^4: x_1^2+x_2^2 = x_3^2 + x_4^2 \}$ is not a smooth manifold.


Consider the map $f:\mathbb{R}^4 \rightarrow \mathbb{R}$ such that $(x_1,x_2,x_3,x_4) \mapsto x_1^2 + x_2^2 - x_3^2 - x_4^2$. Then, $M = f^{-1}(0)$.

First, I make the observation that if $g:=f|_{\mathbb{R}^4 \setminus \{ (0,0,0,0) \}}$ then $g^{-1}(0) = M \setminus \{ (0,0,0,0) \}$ is a submanifold of dimension $3$ because $0$ is a regular value: for all $(x_1,x_2, x_3, x_4) \in \mathbb{R}^4 \setminus \{ (0,0,0,0) \}$, $$dg_{(x_1,x_2,x_3,x_4)} = \begin{bmatrix} 2x_1 & 2x_2 & -2x_3 & -2x_4 \end{bmatrix}$$ has full rank. Since $M\setminus \{ (0,0,0,0) \}$ is a submanifold, it admits a smooth manifold structure. Therefore, if $M$ were to be a smooth manifold, the issue would arise from the point $(0,0,0,0) \in M$ so it remains to analyze this point.

Remark: I tried writing a proof that is analogous to Example 5.45. of Lee's Introduction to Smooth Manifolds showing that $S= \{ (x,y): y = |x| \} \subset \mathbb{R}^2$ has no smooth manifold structure but this relies on the fact that there is a global minimum for $y$ which we cannot assert here.

So what I did instead is say that if $M$ is a manifold of dimension $m$ then each tangent space at a point $T_pM$ must be a vector space of dimension $m$. Since we have seen that $M\setminus \{ (0,0,0,0) \}$ is a smooth manifold of dimension $3$, for the sake of contradiction, if $M$ also admits a smooth manifold structure, then $T_{(0,0,0,0)} M$ is a vector space of dimension $3$. However, $T_{(0,0,0,0)}M = \mathrm{ker}(df_{(0,0,0,0)}) \cong \mathbb{R}^4$ which is $4$-dimensional. Thus, we arrived at a contradiction.

However, this argument I just wrote makes me believe that I might have shown that $M$ is not a smooth submanifold of $\mathbb{R}^4$ as opposed to showing that it is not a smooth manifold. My skepticism arises from the fact that my argument leverages $M$ to obtain the smooth manifold structure from a regular level set of a smooth function as opposed to looking at all possible smooth structures.


1 Answer 1


Smooth submanifolds $X$ of dimension $d$ of $\mathbb{R}^n$ have this property: around every point $p\in X$ there exists a neighbourhood of $U$ $p$ in $\mathbb{R}^n$, and $d$ of the components of $\mathbb{R}^n$ ( say $x_1$, $\ldots$, $x_d$) such that $X\cap U= \{(x_1, \ldots x_d, \phi(x_1, \ldots, x_d))$ where $\phi$ is a function from $\mathbb{R}^d$ to $\mathbb{R}^{n-d}$. Basically, the other $n-d$ components are locally a (smooth) function of the first $d$.

Now, you can check that this is not so for $X$ around the point $(0,0,0,0)$. Not matter how you choose the $3$ components, arount $(0,0,0,0)$ a vertical line will intersect $X$ in at least $2$ points.

Note: it might be possible to see that $X$ with the induced topology is not a topological manifold ( around $(0,0,0,0)$) Indeed, we can parametrize the points of $X$ by $(r, \phi, \theta)$, $r\ge 0$, $\phi$, $\theta \in \mathbb{R}/2\pi \mathbb{Z}$ as

$$(r, \phi, \theta) \mapsto (r \cos \phi, r \sin \phi, r \cos \theta, r \sin \theta)$$

which is a bijection except at $r=0$. Therefore we have

$$X \simeq [0, \infty) \times T/ (\{0\} \times T)$$

that is a (positive) cone over the $2$ torus $T$. That does not seem to a a topological manifold around the vertex, but I lack an argument now.


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