Let $f_i\colon \mathbb{R}^4\rightarrow \mathbb{R},\ i=1,2,3$ be defined by

$f_1(x)=x_1x_3-x_2^2 \\ f_2(x)=x_2x_4 -x_3^2\\ f_3(x)=x_1x_4-x_2x_3$

I want to show that $M=\{x\in\mathbb{R}^4 -\{0\} \colon f_1(x)=f_2(x)=f_3(x)=0\}$ is a 2-dimensional submanifold of $\mathbb{R}^4$.

My problem is, that I can only show that it is a 1 dimensional submanfold.

It is clear, that every $f_i\in\mathcal{C}^1(\mathbb{R}^4)$ and that the functional matrix

$\frac{\partial (f_1,f_2,f_3)}{\partial( x_1,x_2,x_3,x_4)}= \begin{pmatrix} x_3 & -2x_2 & x_1 & 0 \\ 0 & x_4 & -2x_3 & x_2 \\ x_4 & -x_3 & -x_2 & x_1 \end{pmatrix}$

has rank 3, therefore it is surjective. This should show that $M$ is a 1 dimensional submanifold. At this point I have no idea, since if I remember correctly, the dimension of a submanifold is unique (think it follows from the uniqueness of the atlas for $M$).

  • $\begingroup$ It has rank $3$ generically, but what happens when the polynomials vanish? $\endgroup$ – Michael Burr Sep 8 '18 at 14:06
  • $\begingroup$ Can you specify what you mean by generically? The point $x=0$ is excluded by requirement. The polynomial can vanish, but why would that be a problem for the rank? $\endgroup$ – EpsilonDelta Sep 8 '18 at 14:29

The set of points that satisfy the equations $f_1(x)=f_2(x)=f_3(x)=0$ can be described as the image of the map $$ \mathbb{R}^2\rightarrow\mathbb{R}^4, $$ where $$ (s,t)\mapsto(s^3,s^2t,st^2,t^3). $$ Plugging these points into the Jacobian matrix that you wrote above gives $$ \begin{bmatrix} st^2&-2s^2t&s^3&0\\ 0&t^3&-2st^2&s^2t\\ t^3&-st^2&-s^2t&s^3 \end{bmatrix} $$

Now, observe that $t$ times row $1$ minus $s$ times row $3$ is $$ \begin{bmatrix}st^3&-2s^2t^2&s^3t&0\end{bmatrix}-\begin{bmatrix}st^3&-s^2t^2&-s^3t&s^4\end{bmatrix}=\begin{bmatrix}0&-s^2t^2&2s^3t&-s^4\end{bmatrix}, $$ which is $-\frac{s^2}{t}$ times row $2$. So, the rank of this image is $2$. At all (most?) other points, the rank is $3$ (this is what is meant by generically).

  • $\begingroup$ I understand, my mistake was simply missing that for some points e.g. $(1,0,0,1)$ the jacobi matrix does not have rank 3. Is this correct? Is there a specific reason you used a parametrisation, because you would additionally need to show that it is homeomorphic into its image (which is not hard, but still more work)? $\endgroup$ – EpsilonDelta Sep 8 '18 at 15:13
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    $\begingroup$ See: Twisted cubic for some background on the parametrization. $\endgroup$ – Michael Burr Sep 8 '18 at 15:36
  • $\begingroup$ At $(1,0,0,1)$, the Jacobian matrix has rank $3$, perhaps you meant $(1,0,0,0)$? $\endgroup$ – Michael Burr Sep 8 '18 at 15:37
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    $\begingroup$ Hint: Most points that solve $f_1$ and $f_2$ also solve $f_3$. Which ones don't? $\endgroup$ – Michael Burr Sep 8 '18 at 16:15
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    $\begingroup$ Note that $x\not=(0,0,0,0)$ is not the same as every coordinate being nonzero. It means that one coordinate is nonzero. $\endgroup$ – Michael Burr Sep 9 '18 at 19:04

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