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?

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.
