Q. Given that $M=\{\mathbf{x} \in \mathbb{R}^4: x_1^2 + x_2^2 + x_3^2 + x_4^2 =1, x_1x_2 = x_3x_4\}$. Show that $M$ is a smooth manifold of dimension 2.

I write $M = \mathbf{F}^{-1}(\{\mathbf{0}\})$, where $\mathbf{F}: \mathbb{R}^4 \rightarrow \mathbb{R}^2$ is given by $\mathbf{F} = \begin{pmatrix} x_1^2 + x_2^2 + x_3^2 + x_4^2 -1 \\ x_1x_2 - x_3x_4 \end{pmatrix}$. I calculate the derivative $D\mathbf{F} = \begin{pmatrix} 2x_1 & 2x_2 & 2x_3 & 2x_4 \\ x_2 & x_1 & -x_4 & -x_3\end{pmatrix}$.

If I can show that $D\mathbf{F}$ has rank 2, then that will imply that $M$ has dimension 4-2 =2. But I am unable to show the rank is 2.

I could manipulate the constraints to get $(x_1 \pm x_2)^2 + (x_3 \mp x_4)^2 =1$, but am stuck after that.

  • $\begingroup$ Hint: consider the determinants formed by the first two and the last two columns. Can they both be zero? $\endgroup$ – Wojowu Jan 8 at 16:36
  • $\begingroup$ I do not understand. To show that rank is 2, you are suggesting that every 2x2 determinant should be non-zero ? $\endgroup$ – me10240 Jan 8 at 16:41
  • 1
    $\begingroup$ just one of them is enough $\endgroup$ – Carlos Campos Jan 8 at 16:42
  • $\begingroup$ A matrix has rank at least $r$ iff you can find some $r\times r$ submatrix which has nonzero determinant. $\endgroup$ – Wojowu Jan 8 at 16:45
  • $\begingroup$ So if I consider the first determinant, it will be 0 if $x_1 + x_2 =0 $ or $x_1 - x_2 =0$. If $x_1 =x_2$, then I get $ (x_3 + x_4)^2 =1$. Otherwise I get $(x_3 -x_4)^2 =1$. I am still unable to see what it means. $\endgroup$ – me10240 Jan 8 at 16:54

For every $x$, $rank(DF_x)\leq 2$. Now $rank(dF_x)\leq 1$ iff there is $\lambda$ s.t.

$(*)$ $x_2=2\lambda x_1,x_1=2\lambda x_2,x_4=-2\lambda x_3,x_3=-2\lambda x_4$ (since $x\in M$, $x\not=0$).

Consider the cases when $\lambda=\pm 1/2$ and $\lambda\not=\pm 1/2$ and deduce that there are no solutions of $(*)$ in $M$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.