
\vspace*{.2in}
\noindent{\bf Supplementary problem 3: }
The set $C$ is just the $(1,0)$ level set of the function $f:\R^2\to\R^4$.
Therefore $C$ will be a submanifold of $\R^4$ with both dimension and 
codimension equal to two except at points where
$$
f'(x,y,z,w) = \left(\begin{array}{cccc}
2x & -2y & -w & -z \\
w & z & y & x
\end{array}\right)
$$
has rank less than two.  Having rank less than two means that \emph{all}
$2\times 2$ submatrices of $f'$ have zero determinant.  There are six of
these, and their determinants are
$$
2xz + 2yw, \quad 2xy + w^2, \quad 2x^2 + wz, \quad -2y^2 + wz, 
\quad -2xy + z^2, \quad -xw+yz.
$$ 
If all six vanish, then comparing the second and fourth expression shows that
$w^2 = -z^2$, which can only happen if $z=w=0$; and the third and fifth
combine to imply that $x^2 = -y^2$, so that again $x=y=0$.  Hence the
only trouble occurs at $(0,0,0,0)$ which is not actually in $C$.  So
$C$ is a submanifold of $\R^4$ at all points.

Now the tangent space to $C$ at $(1,1,0,0)$ is just the $(1,0)$ level
set of the linear approximation of $f$ at that point.  That is, we
seek all vectors $(a,b,c,d)$ satisfying
$$
\left(\begin{array}{c} 
1 \\ 0
\end{array}\right)
= 
\left(\begin{array}{c} 
1 \\ 0
\end{array}\right)
+
f'(1,1,0,0)
\left(\begin{array}{c} 
a \\ b \\ c \\ d
\end{array}\right).
$$
From this it is clear that we're just after the nullspace of 
$$
f'(1,1,0,0)
=
\left(\begin{array}{cccc} 
2 & -2 & 0 & 0 \\
0 & 0 & 1 & 1
\end{array}\right).
$$
After dividing the matrix by two, it's in reduced row echelon form with
free variables $b$ and $d$.  The nullspace is therefore the two dimensional
subspace of $\R^4$ generated by the vectors $(1,1,0,0)$ and $(0,0,-1,1)$
obtained by setting one free variable equal to one and the other equal to
$0$.
