Eliminating a parameter when intersecting a manifold with a hyperplane In the Euclidean space $\mathbb R^4$ we look at the intersection of the equations$$x^2 + y^2 = 1 \\ z^2 + w^2 = 1$$
sometimes known as the Clifford torus. This is known to be a 2-dimensional manifold, with global parameterization in 2 parameters, given by:
$$x = \cos(2\pi t_1) \\ y = \sin(2\pi t_1) \\ z = \cos(2\pi t_2) \\ w = \sin(2\pi t_2)$$
Now I want to intersect this with an affine hyperplane, given by the equation $$\vec v\cdot(x,y,z,w)=\gamma$$
($\vec v$ is a unit normal vector and $\gamma\in \mathbb R$ is an affine offset.)
Let's assume the intersection is a 1-dimensional manifold - i.e. the Jacobian matrix has rank $3$ at any point in the intersection. (I can show this happens for almost all $\vec v, \gamma$.)
Questions.


*

*Since the intersection is a manifold, there is a 1-parameter local parameterization. But can we say that there is a 1-parameter global parameterization of the intersection manifold?

*Can we say the parameterization is given by $x = \cos(2\pi t)$, $y = \sin(2\pi t)$ and $z,w$ some implicit functions of $t$?


I want to avoid using the quadratic formula and taking square roots because this requires care of when the value is positive and negative... Ideally, I want to use the implicit function theorem to solve this, but I tried to consider it a few times and it's been pretty confusing.
Also - can elimination theory be used to address this question?
 A: Consider the case where your hyperplane is given by $w=0$. Then your manifold (lets call it $V$) consists of two disjoint circles (one at the level $z=1$ resp. $z=-1$). In particular, $V$ is not connected, so there can not be any surjective continous map from $\mathbb{R}$ (or from an interval) to $V$ (which I guess is what you mean by "a 1-parameter global parametrization").
Of course you can always get defining equations by pluggin in a parametrization of the hyperplane.
A: I think that algebraic methods give better result. 
Rewrite this system of equations a little differently.
$\left\{\begin{aligned}&G^2+E^2=K^2\\&D^2+F^2=K^2\end{aligned}\right.$
Solutions of this system are of the form:
$G=2XY$
$E=X^2-Y^2$
$D=2ZR$
$F=Z^2-R^2$
$K=X^2+Y^2=Z^2+R^2$
We need solutions will be taken from the following formulas.
This equation is quite symmetrical so formulas making too much can be written:
So for the equation:    $X^2+Y^2=Z^2+R^2$  
solution:  
$X=a(p^2+s^2)$   
$Y=b(p^2+s^2)$   
$Z=a(p^2-s^2)+2psb$   
$R=2psa+(s^2-p^2)b$   
solution:   
$X=p^2-2(a-2b)ps+(2a^2-4ab+3b^2)s^2$   
$Y=2p^2-4(a-b)ps+(4a^2-6ab+2b^2)s^2$   
$Z=2p^2-2(a-2b)ps+2(b^2-a^2)s^2$   
$R=p^2-2(3a-2b)ps+(4a^2-8ab+3b^2)s^2$   
solution:   
$X=p^2+2(a-2b)ps+(10a^2-4ab-5b^2)s^2$   
$Y=2p^2+4(a+b)ps+(20a^2-14ab+2b^2)s^2$   
$Z=-2p^2+2(a-2b)ps+(22a^2-16ab-2b^2)s^2$   
$R=p^2+2(7a-2b)ps+(4a^2+8ab-5b^2)s^2$   
solution:   
$X=2(a+b)p^2+2(a+b)ps+(5a-4b)s^2$   
$Y=2((2a-b)p^2+2(a+b)ps+(5a-b)s^2)$  
$Z=2((a+b)p^2+(7a-2b)ps+(a+b)s^2)$   
$R=2(b-2a)p^2+2(a+b)ps+(11a-4b)s^2$   
solution:   
$X=2(b-a)p^2+2(a-b)ps-as^2$    
$Y=2((b-2a)p^2+2(a-b)ps+(b-a)s^2)$    
$Z=2((b-a)p^2+(3a-2b)ps-(a-b)s^2)$   
$R=2(b-2a)p^2+2(a-b)ps+as^2$  
solution:  
$X=(p^2-s^2)b^2+a^2s^2$  
$Y=b^2(p-s)^2-2abs^2+a^2s^2$ 
$Z=b^2(p-s)^2+2abps-a^2s^2$  
$R=s^2(a-b)^2+2abps-p^2b^2$   
number $a,b,p,s$ integers and sets us, and may be of any sign.
