# Is this set a manifold?

For which $( \alpha , \beta ) \in \Bbb R^2$ set:
$\{ (x_1,x_2,x_3,x_4) \in \Bbb R^4 | x_1+x_4= \alpha, x_1 x_4 - x_2x_3 = \beta \}$
is a manifold?

I made a Jacobian matrix: $\begin{bmatrix} 1 & 0 & 0& 1 \\ x_4 & -x_3 & -x_2 & x_1 \\ \end{bmatrix}$

Now I think that something must happen with this matrix for my set to be a manifold, but don't know what it is..

Can anybody help me?

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You forgot the minus sign in the matrix, should be $-x_3$ – Dror Feb 17 '14 at 21:55
@Dror thanks ;) – Kuba Feb 17 '14 at 21:57
NP. ah and it also should be $-x_2$..! :P – Dror Feb 17 '14 at 21:58
I know, saw it already:) – Kuba Feb 17 '14 at 22:00
I know you did, it was j/k. – Dror Feb 17 '14 at 22:01

The theorem is that $(\alpha,\beta)$ has to be a regular value for $(x_1+x_4,x_1x_4-x_2x_3)$. That is, at every $(x_1,x_2,x_3,x_4)$ mapping to $(\alpha,\beta)$ the differential (Jacobian matrix) has to be surjective. If you remember your linear algebra, you know that your differential is surjective except when $x_3=x_2=0$ and $x_1=x_4$. Then the equations read as $2x_1=\alpha, x_1^2=\beta$, so the non-regular values occur on the parabola $\beta=\alpha^2/4$.
Your jacobian need to be of maximal rank (2) on every point in $f^{-1}(0)$, where $f$ is the function you give the jacobian. (You want this function to be a submersion everywhere.)
@Kuba: (& at Léo:) I think that a good example to understand what going on when $(\alpha, \beta)$ is a singular value is the set $$M=\{(x, y)\in \mathbb{R}^2\ :\ xy=0\}.$$Here $0$ is a singular value of the mapping $f\colon (x, y)\to xy$, and its level set is the union of the two coordinate axes, which ceases to be a manifold near $(0, 0)$. Indeed, if we exclude a neighborhood of $(0, 0)$, then $M$ is a one-dimensional manifold, meaning that it can be locally described in terms of a single coordinate. But clearly this breaks up near the origin, where we have a cross. [...] – Giuseppe Negro Feb 17 '14 at 22:23