Local graph for $F(x_1,x_2,x_3,x_4)=\left(x_1^2+x_2^2+x_3^2-x_4^2,\sum x_i\right)=(0,0)$ when implicit function theorem fails? 
Let $X=\{(x_1,x_2,x_3,x_4)\in{\Bbb R^4}:x_1^2+x_2^2+x_3^2-x_4^2=2,\ \sum x_i=2\}$ and let $p=(1,1,1,-1)$. Then $p\in X$. Is it possible to find a product of open set $V$ containing $p$ such that $X\cap V$ is the graph of a function expressing two of the variables $x_1,x_2,x_3,x_4$ in terms of the other two?


A quick calculation shows that the implicit function theorem cannot be applied here since $F'(1,1,1,-1)=0$ where 
$$
F(x_1,x_2,x_3,x_4)=\left(x_1^2+x_2^2+x_3^2-x_4^2,\sum x_i\right).
$$
If $X=\{p\}$, then the answer would be trivial. But I don't see whether this is true or not. How can I go on?
 A: If the independent variables are $(x_1,x_2)$, then we must have
$x_3^2-x_4^2 = 2 -(x_1^2+x_2^2)$, $(x_3+x_4) = 2-(x_1+x_2)$. Now choose $(x_1,x_2) = (t,2-t)$, then these equations reduce to
$(x_3+x_4) = 0$, and $x_3^2-x_4^2 = (x_3+x_4) (x_3-x_4) = -2 (1-t)^2$. Hence for $t\neq 1$ these equations have no solution. Hence we cannot express $(x_3,x_4)$ as a local function of $(x_1,x_2)$.
By symmetry, any pair of variables from $x_1,x_2,x_3$ cannot be the independent variables.
It is also not possible with $x_3,x_4$ as the independent variables.
To see this, let $\phi_+(x_3,x_4) = {  2-(x_3+x_4) + \sqrt{ (x_3+x_4)( x_4-3 x_3 + 4 )   } \over 2} $, $\phi_-(x_3,x_4) = {  2-(x_3+x_4) - \sqrt{ (x_3+x_4)( x_4-3 x_3 + 4 )   } \over 2} $
Both of the following are local, distinct solutions that pass through $p$:
$(x_3,x_4) \mapsto (\phi_+(x_3,x_4), 2-(\phi_+(x_3,x_4)+x_3+x_4))$ and
$(x_3,x_4) \mapsto (\phi_-(x_3,x_4), 2-(\phi_-(x_3,x_4)+x_3+x_4))$.
By symmetry, the same applies to the pairs $(x_1,x_4)$, $(x_2,x_4)$ as well.
A: This is the same   what copper.hat did, but presented differently. First, shift the point to the origin by writing $x_i=y_i+1$ for $i=1,2,3$, and 
$x_4=y_4-1$. The equations simplify to 
$$\begin{cases}y_1^2+y_2^2+y_3^2-y_4^2 &= 0 \\  y_1+y_2+y_3+y_4&=0\end{cases}$$


*

*If $y_1=y_2=0$, we   have a bunch of solutions: entire line $y_3+y_4=0$. Similar for other pairs from $y_1,y_2,y_3$. 

*If $y_1=0$  and $y_4=\epsilon$, we have two solutions: intersection of the circle $y_2^2+y_3^2=\epsilon^2$ with the line 
$y_2+y_3=-\epsilon$. Similar for other pairs $(y_i,y_4)$. 

A: There is a generalization of the implicit function theorem called the Constant Rank Theorem.
To express two of the variables in terms of the other two you would need the derivative matrix to have rank two in a neighborhood of $p$.  But the derivative matrix is
$$
DF(x_1,x_2,x_3,x_4) = \begin{pmatrix} 2x_1 & 1 \\ 2x_2 & 1 \\ 2 x_3 & 1 \\ -2 x_4 & 1 \end{pmatrix}
$$
At $p$ the first column is twice the second, so the rank is only one.
