How to compute the metric tensor for a surface defined by an equation?

Question

How to compute the Riemannian metric tensor exactly for a generic surface $$f\left(x_{1},x_{2},x_{3}\right)=0$$ in $\mathbb{R}^{3}$? Thanks.

Answer 1

Welcome to Mathematics StackExchange, phobe!

Since the ambient space is $ \mathbb R ^ 3 $, you can start with the metric there, which is $ { \mathrm d x _ 1 } ^ { \! 2 } + { \mathrm d x _ 2 } ^ { \! 2 } + { \mathrm d x _ 3 } ^ { \! 2 } $, and then perform a substitution. You need to decide what variables you're going to use. (When you have a parametrized surface, then you can put everything in terms of the parameters, but now you don't have any parameters.) I'll assume that you're going to use $ x _ 1 $ and $ x _ 2 $, but you could make a different choice. (Indeed, different choices are often best at different places on the surface.)

So given $ f ( x _ 1 , x _ 2 , x _ 3 ) = 0 $, differentiate this to get $$ \mathrm D _ 1 f ( x _ 1 , x _ 2 , x _ 3 ) \, \mathrm d x _ 1 + \mathrm D _ 2 f ( x _ 1 , x _ 2 , x _ 3 ) \, \mathrm d x _ 2 + \mathrm D _ 3 f ( x _ 1 , x _ 2 , x _ 3 ) \, \mathrm d x _ 3 = 0 \text . $$ (The function that I'm calling $ \mathrm D _ 1 f $ here is the derivative of $ f $ with respect to its first variable, also called $ f _ 1 $, $ f _ { x _ 1 } $, $ \partial f / \partial x _ 1 $, $ \partial _ 1 f $, etc.) Since I'm using $ x _ 1 $ and $ x _ 2 $ as the variables, solve this for the differential to be eliminated: $$ \mathrm d x _ 3 = - \frac { \mathrm D _ 1 f ( x _ 1 , x _ 2 , x _ 3 ) \, \mathrm d x _ 1 + \mathrm D _ 2 f ( x _ 1 , x _ 2 , x _ 3 ) \, \mathrm d x _ 2 } { \mathrm D _ 3 f ( x _ 1 , x _ 2 , x _ 3 ) } \text . $$ Then substitute into the ambient metric: $$ \eqalign { { \mathrm d x _ 1 } ^ { \! 2 } + { \mathrm d x _ 2 } ^ { \! 2 } + { \mathrm d x _ 3 } ^ { \! 2 } & = { \mathrm d x _ 1 } ^ { \! 2 } + { \mathrm d x _ 2 } ^ { \! 2 } + \bigg ( { - \frac { \mathrm D _ 1 f ( x _ 1 , x _ 2 , x _ 3 ) \, \mathrm d x _ 1 + \mathrm D _ 2 f ( x _ 1 , x _ 2 , x _ 3 ) \, \mathrm d x _ 2 } { \mathrm D _ 3 f ( x _ 1 , x _ 2 , x _ 3 ) } } \bigg ) ^ 2 \\ & = { \mathrm d x _ 1 } ^ { \! 2 } + { \mathrm d x _ 2 } ^ { \! 2 } + \frac { \mathrm D _ 1 f ( x _ 1 , x _ 2 , x _ 3 ) ^ 2 } { \mathrm D _ 3 f ( x _ 1 , x _ 2 , x _ 3 ) ^ 2 } \, { \mathrm d x _ 1 } ^ { \! 2 } \\ & \qquad { } + \frac { 2 \mathrm D _ 1 f ( x _ 1 , x _ 2 , x _ 3 ) \mathrm D _ 2 f ( x _ 1 , x _ 2 , x _ 3 ) } { \mathrm D _ 3 f ( x _ 1 , x _ 2 , x _ 3 ) ^ 2 } \, \mathrm d x _ 1 \, \mathrm d x _ 2 + \frac { \mathrm D _ 2 f ( x _ 1 , x _ 2 , x _ 3 ) ^ 2 } { \mathrm D _ 3 f ( x _ 1 , x _ 2 , x _ 3 ) ^ 2 } \, { \mathrm d x _ 2 } ^ { \! 2 } \\ & = \frac { \mathrm D _ 1 f ( x _ 1 , x _ 2 , x _ 3 ) ^ 2 + \mathrm D _ 3 f ( x _ 1 , x _ 2 , x _ 3 ) ^ 2 } { \mathrm D _ 3 f ( x _ 1 , x _ 2 , x _ 3 ) ^ 2 } \, { \mathrm d x _ 1 } ^ { \! 2 } \\ & \qquad { } + \frac { 2 \mathrm D _ 1 f ( x _ 1 , x _ 2 , x _ 3 ) \mathrm D _ 2 f ( x _ 1 , x _ 2 , x _ 3 ) } { \mathrm D _ 3 f ( x _ 1 , x _ 2 , x _ 3 ) ^ 2 } \, \mathrm d x _ 1 \, \mathrm d x _ 2 \\ & \qquad \quad { } + \frac { \mathrm D _ 1 f ( x _ 1 , x _ 2 , x _ 3 ) ^ 2 + \mathrm D _ 3 f ( x _ 1 , x _ 2 , x _ 3 ) ^ 2 } { \mathrm D _ 3 f ( x _ 1 , x _ 2 , x _ 3 ) ^ 2 } \, { \mathrm d x _ 2 } ^ { \! 2 } \text . } $$ Notice that this requires division by $ D _ 3 f ( x _ 1 , x _ 2 , x _ 3 ) $. Where this is zero, $ x _ 1 $ and $ x _ 2 $ are not an appropriate pair of coordinates, and you will need to use a different pair of variables. (Where the entire gradient of $ f $ is zero, it will take some more work to find suitable coordinates; indeed, there is no guarantee in that case that the surface is a manifold at all!)

If you don't want to write your answer as a quadratic differential form, then you'll need to convert this into your desired notation. For example, if you want to express the metric as a matrix (relative to the coordinate system $ ( x _ 1 , x _ 2 ) $), then the answer is $$ \left [ \matrix { \displaystyle \frac { \mathrm D _ 1 f ( x _ 1 , x _ 2 , x _ 3 ) ^ 2 + \mathrm D _ 3 f ( x _ 1 , x _ 2 , x _ 3 ) ^ 2 } { \mathrm D _ 3 f ( x _ 1 , x _ 2 , x _ 3 ) ^ 2 } & \displaystyle \frac { \mathrm D _ 1 f ( x _ 1 , x _ 2 , x _ 3 ) \mathrm D _ 2 f ( x _ 1 , x _ 2 , x _ 3 ) } { \mathrm D _ 3 f ( x _ 1 , x _ 2 , x _ 3 ) ^ 2 } \\ \displaystyle \frac { \mathrm D _ 1 f ( x _ 1 , x _ 2 , x _ 3 ) \mathrm D _ 2 f ( x _ 1 , x _ 2 , x _ 3 ) } { \mathrm D _ 3 f ( x _ 1 , x _ 2 , x _ 3 ) ^ 2 } & \displaystyle \frac { \mathrm D _ 2 f ( x _ 1 , x _ 2 , x _ 3 ) ^ 2 + \mathrm D _ 3 f ( x _ 1 , x _ 2 , x _ 3 ) ^ 2 } { \mathrm D _ 3 f ( x _ 1 , x _ 2 , x _ 3 ) ^ 2 } } \right ] \text . $$ Notice that the $ \mathrm d x _ 1 \, \mathrm d x _ 2 $ term has been split evenly between the two off-diagonal entries in this symmetric matrix. You should be able to fit this into any other system of notation for a Riemannian metric on a surface; if you're looking for four terms, then the two that mix the two coordinates will be equal, but if you're looking for three terms, then the one that mixes the two coordinates will have a factor of $ 2 $.

You can work out the partial derivatives for your example and substitute these into the formulas above. Or you can you take your specific equation and adapt the derivation technique that I used; I generally find that more convenient. (Of course, there are other derivation techniques that you could use instead, and you may get other answers that demonstrate these.)