Let $c \neq 0$ be a real number. We define $M_c$ as $$M_c = \{ (x_1,x_2,x_3,x_4) \in \mathbb R^4|x_1x_2 + x_2x_3+ x_3x_4 =c \}.$$
(1) Show $M_c$ is submanifold of $\mathbb R^4$.
(2) Show $M_c$ is diffeomorphism to $\mathbb R^2 \times \mathbb S^1$.
My approach (I cannot solve neither):
(1) I consider $f:\mathbb R^4 \to \mathbb R$, $f(x_1,x_2,x_3,x_4)= x_1x_2 + x_2x_3+ x_3x_4$. Also $df = (x_2,x_1+x_3,x_2+x_4,x_3)$.
(2) I tried to construct smooth map from $M_c$ to $\mathbb R^2 \times \mathbb S^1$ and inverse function of it. But I cannot.
[Edited]
From Mike F's approach, $M_c$ is diffeomorphic to Special linear group $SL(2,R)$. Generally $SL(n,R) \simeq SO(n,R) \times R^{(n+2)(n-1)/2}$ and $SO(2,R) \simeq S^1$. Hence $M_c \simeq SL(2,R) \simeq S^1 \times R^2$.
Here's a suggestion, though perhaps not the most efficient approach. Define new coordinates \begin{align*} y_1 = x_1 + x_3 && y_2 = \frac{1}{c} x_2 && y_3 = -x_3 && y_4=\frac{1}{c} x_4 \end{align*} Note this coordinate change is mediated by an invertible 4-by-4 matrix. In the new coordinate system, the equation under consideration becomes $$y_1 y_2 - y_3y_4 = 1$$ So, the problem is tantamount to showing that the set of matrices $\begin{bmatrix} y_1 & y_3 \\ y_4 & y_2\end{bmatrix}$ whose determinant is $1$ is a submanifold of the space of all 2-by-2 matrices which is moreover diffeomorphic to $\mathbb{R}^2 \times \mathbb{S}^1$. In my opinion, this translated form of the problem is a better use of your time.