# How to show that this manifold has dimension 2?

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.

• Hint: consider the determinants formed by the first two and the last two columns. Can they both be zero? – Wojowu Jan 8 at 16:36
• I do not understand. To show that rank is 2, you are suggesting that every 2x2 determinant should be non-zero ? – me10240 Jan 8 at 16:41
• just one of them is enough – Carlos Campos Jan 8 at 16:42
• A matrix has rank at least $r$ iff you can find some $r\times r$ submatrix which has nonzero determinant. – Wojowu Jan 8 at 16:45
• 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. – 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 $$M$$.