Consider the subset $M\subset\mathbb R^4$ given by the equations:
$$M\equiv \left\{\begin{array}{ll}
x^2+y^2-z^2-t^2=1\\
-xt+yz=1
\end{array}\right.$$
Prove that $M$ is a smooth manifold.
My idea is to define the smooth map $f:\mathbb R^4\to\mathbb R^2$ given by:
$$f(x,y,z,t)=(x^2+y^2-z^2-t^2-1,-xt+yz-1)$$
It is well know that if $J(f)$ has rank $2$ in $M=F^{-1}(0,0)$ then $M$ must be a 2-dimensional smooth manifold.
Unfortunately, I am not able to find a regular submatrix of:
$$J(f)=\left(\begin{array}{cccc}
2x & 2y&-2z&-2t\\
-t&z&y&-x
\end{array}\right)$$
taking in count that $(x,y,z,t)\in M$.
Any help?
Best Answer
Consider the two minors
$$J_{14}(f)=\left(\begin{array}{cc} 2x &-2t\\ -t&-x \end{array}\right)$$
and
$$J_{23}(f)=\left(\begin{array}{cc} 2y&-2z\\ z&y \end{array}\right)$$
Which has determinant $-2(x^2 + t^2)$ and $2(y^2+z^2)$ respectively. If both are zero, then $x=y=z=t = 0$, but $(0,0,0,0)$ is not in $M$. Thus either one of the above minors is invertible for all $(x, y, z, t)\in M=F^{-1}(0,0)$. Hence $(0,0)$ is a regular value of $F$.