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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.

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  • $\begingroup$ It looks to me like the surfaces are related by a rotation around the $z$-axis through an angle $\pi/4$, if so that rotation should be a (global) isometry between the surfaces. $\endgroup$ – mcwiggler Jan 8 at 10:10
  • $\begingroup$ @mcwiggler How did you observe this? It's not too bad if you realise this is what to go for, but the way the question is phrased suggests it should be done without the aid of a graph plotting program and I certainly wouldn't be able to recognise this without looking at it for a while so was wondering if there are any alternative approaches $\endgroup$ – backstrapp Jan 8 at 16:54
  • $\begingroup$ What made me observe that was that I recognized the numerator of the second surface as a product of $x_1 - x_2$ and $x_1 + x_2$, and since the first equation is just the product of the coordinates this made me think that they are basically the same thing! I can’t explain too much in this space but maybe look into surfaces defined by quadratic forms, especially the parts where we diagonalize the quadratic form, and express the surface equation in an eigenbasis. $\endgroup$ – mcwiggler Jan 9 at 4:28
  • $\begingroup$ @mcwiggler I managed to get the isometry you suggested to work since rotations preserve the length of all vectors, in particular the ones of the tangent and that the differential of a rotation is just the rotation itself (since it's a linear map). We haven't gone through the topics you suggested at any point in my course though which leads me to think there should be an alternative method that doesn't require it $\endgroup$ – backstrapp Jan 10 at 11:34

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