# Gauss curvature using metric and Riemannian curvature

I learnt that the Gauss curvature is given by:

$$K = \frac {eg - f^2}{EG - F^2}$$

where $E, F, G$ are coefficients of the first fundamental form and $e,f, g$ are coefficients of the second fundamental form.

However, in a proof that I am reading, I saw the equation:

$$K = \langle R(E_1,E_2)E_2, E_1 \rangle$$

for an orthonormal frame $(E_1, E_2)$ on a surface $M$, where $R$ is the Riemannian curvature tensor.

Wikipedia says that this is a definition for $K$ but I was wondering if there is a proof for this?

• A proof for the equivalence of the two? – user122283 Apr 27 '14 at 3:40
• It's called the Gauss equation in classical surface theory. – Ted Shifrin Apr 27 '14 at 3:41

## 1 Answer

We can define a new quantity: $$R(E_1,E_2,E_2,E_1):=\langle R(E_1,E_2)E_2,E_1\rangle$$ The Gauss equation allows us to say (since $E_1,E_2$ are orthonormal): $$R(E_1,E_2,E_2,E_1)=\det(S)=K$$ Here, $S$ is the shape operator, and the Gauss equation is $$R(E_1,E_2)E_2=\langle S(E_2),E_2\rangle S(E_1)-\langle S(E_1),E_2\rangle S(E_2)$$

• How did you get $det(S)$ from your last equation? – user145817 Apr 27 '14 at 4:06
• @user145817 Read this lecture notes. – user122283 Apr 27 '14 at 4:08