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I have the following exercise:

Let $f:\mathbb{R}^3\rightarrow \mathbb{R}$, and let $S$ be a surface. Let $p$ be a regular value of $f$, and $S=f^{-1}(p)$. Determine the coefficients of the first fundamental form.

I think that since I have to work with partial derivatives, I can use the Theorem of the Implicit Function. However, I am not sure how to approach this exercise. Any help would be appreciated.

Ted Shifrin
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user2699
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  • @whpowell96 no I do not., If I am not wrong the basis is the partial of f in all of the points x_i in p? – user2699 Oct 12 '23 at 01:34
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    The first coefficient of the first fundamental form depends on a coordinates. I suppose you need to find some local coordinates first. – Arctic Char Oct 12 '23 at 01:46
  • So use the implicit function theorem: Say the surface is locally $z=\phi(x,y)$. Relate the partial derivatives of $\phi$ to those of $f$ and finish. Post the answer when you’re done. – Ted Shifrin Oct 12 '23 at 04:31
  • @TedShifrin I was thinking of the first fundamental form as the matrix of inner products of a basis of the tangent space. These are obviously constructed from partial derivatives of $f$ via the IFT map, but I personally find it easier to interpret from a linear algebra POV. I think I may have accidentally implied that $f$ is a parameterization of the manifold instead of the $f^{-1}$ so I will delete my previou comment. – whpowell96 Oct 12 '23 at 14:22
  • @whpowell96 But neither gives a parametrization. Yes, linear algebra will construct a basis for the tangent plane given a basis for the normal space, but this doesn’t give local coordinates or parametrization. – Ted Shifrin Oct 12 '23 at 16:17
  • @TedShifrin Thank you for your comment. I see my mistake now. In my head I was equivocating "parameterization" and "local representation in some coordinates." – whpowell96 Oct 12 '23 at 16:35

1 Answers1

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By definition of the inverse image, $f^{-1}(p) = \{(x,y,z) : f(x,y,z) = p\}$, so $f(x,y,z) - p = 0$. Now, since p is a regular value, it means that one of its partial derivatives is nonzero. Without loss of generality, we can assume that the partial derivative with respect to z is nonzero.

And by the implicit function theorem, there exists $\phi: U \subset \mathbb{R}^2 \rightarrow \mathbb{R}^3$ such that $\phi(u,v) = (u,v,h(u,v))$, where h(u,v) is a local parameterization at a point.

If we consider the composition $f \circ \phi = f(u,v,h(u,v)) = p$,

Now, differentiating partially with respect to u and using the chain rule $f \circ \phi$ , we obtain

$\frac{\partial h}{\partial u} = \frac{-\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial z}}$

and

$\frac{\partial h}{\partial v} = \frac{-\frac{\partial f}{\partial y}}{\frac{\partial f}{\partial z}}$

and if I do the partial derivative of $\phi$

$\frac{\partial \phi}{\partial u} = (1,0,\frac{\partial h}{\partial u}) = (1,0,\frac{-\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial z}})$

and

$\frac{\partial \phi}{\partial v} = (0,1,\frac{\partial h}{\partial v}) = (0,1,\frac{-\frac{\partial f}{\partial y}}{\frac{\partial f}{\partial z}})$

And the coefficients would be equal to

$E = \langle \frac{\partial \phi}{\partial u} \cdot \frac{\partial \phi}{\partial u} \rangle = 1 + \left(\frac{-\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial z}}\right)^2$

$F = \langle \frac{\partial \phi}{\partial u} \cdot \frac{\partial \phi}{\partial v} \rangle = \frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial z}} \cdot \frac{\frac{\partial f}{\partial y}}{\frac{\partial f}{\partial z}}$

$G = \langle \frac{\partial \phi}{\partial v} \cdot \frac{\partial \phi}{\partial v} \rangle = 1 + \left(\frac{-\frac{\partial f}{\partial y}}{\frac{\partial f}{\partial z}}\right)^2$

user2699
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