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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.

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  • $\begingroup$ 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
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    $\begingroup$ 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
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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 $M$.

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