# Prove that certain subset $M$ of $\mathbb R^4$ is a smooth manifold

differential-geometryjacobiansmooth-manifoldssubmanifold

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?

$$J_{14}(f)=\left(\begin{array}{cc} 2x &-2t\\ -t&-x \end{array}\right)$$
$$J_{23}(f)=\left(\begin{array}{cc} 2y&-2z\\ z&y \end{array}\right)$$
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 $$F$$.