Consider the following surfaces in $ \mathbb R ^3 $:
$$ \Sigma _ 1 = \{ (x_1, x_2, x_3) \in \mathbb R ^3 : x_3 = x_1x_2 \} \\ \Sigma_2 = \{ (x_1, x_2, x_3) \in \mathbb R ^3 : x_3 = \dfrac{x_1^2-x_2^2}{2} \; \}. $$ Construct a local isometry between the given surfaces.
Some thoughts on the question:
We take the definition of a local isometry to be a map between regular surfaces that preserves the length of the tangent vectors. More formally, a map $\Phi : \Sigma_1 \to \Sigma_2 $ is said to be a local isometry if $\forall p\in \Sigma_1, X \in T_p \Sigma _1$ it holds that $ | D_p \Phi (X) | = |X|, $ where $X$ is an element of the tangent space of the first surface and $D_p \Phi$ denotes the differential of the given map at a point $p \in \Sigma_1 $.
I've looked at a graph of the two surfaces and they both appear to be hyperbolic, also the expression in the numerator of the second surface seems to suggest hyperbolic trig functions? I've attempted to calculate the tangent space to $ \Sigma_1 $ at a general point by expressing it as the preimage of a suitable function, namely if we define $\varphi : \mathbb R^3 \to \mathbb R, \varphi(x_1, x_2, x_3 ) = x_1x_2 - x_3 $ then we can observe $\Sigma_1 = \varphi^{-1}(0) $.
Claim: the tangent space is precisely $ \text{ker} (D_p \varphi) $. Applying this statement and explicitly calculating the differential, we find that at a point $ p = (x_1, x_2, x_3 ) $, we have that $ T_p \Sigma_1 = \{ (v_1, v_2, v_3 ) \in \mathbb R^3 : x_2 v_1 + x_1 v_2 - v_3 = 0 \}$. This is pretty much as far as I've gotten, there may be some mistakes but I'm not sure where to proceed from here.