# Show that there is no smooth manifold structure

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.

Attempt:

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.

• Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Mathematics Meta, or in Mathematics Chat. Comments continuing discussion may be removed. Aug 18, 2023 at 19:45

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.