The mean curvature of a surface is defined as $$H = \frac{1}{2}(k_1+k_2)$$ where $k_1$, $k_2 $ are the principal curvatures.
More abstractly, it is the trace of the shape operator $$H = tr(S)=\frac{eG-2fF+gE}{2(EG-F^2)},$$
where $E$, $F$, $G$ and $e$, $f$, $g$ are the coefficients of the first/second fundamental form.
How do those two definitions/formulas coincide?
Best Answer
The mean curvature $H$ is equal to half of the trace of the shape operator $S$: $$ H = \frac{1}{2}\mathrm{trace}(S). $$ By definition, the principle curvatures $k_1$, $k_2$ are the eigenvalues of $S$, hence $H=\frac{1}{2}(k_1 + k_2)$. The argument goes as follows. Since $S$ is a symmetric operator, at each point $p$ of the surface, there exist two orthonormal eigenvectors $e_1$ and $e_2$, the principal directions: $Se_1 = k_1 e_1$ and $S e_2 = k_2 e_2$. So with respect to these eigenvectors the matrix of the shape operator is $\mathrm{diag}(k_1, k_2)$, whose trace is clearly $k_1 + k_2$.
The same argument, but explained a little differently: since the shape operator is diagonalisable, there exists at each point an orthogonal matrix $U$ such that $U^T S U = \mathrm{diag}(k_1, k_2)$. Then $$ k_1 + k_2 = \mathrm{trace}(U^T S U) = \mathrm{trace}(S U U^T) = \mathrm{trace}(S). $$
Here we used the fact that $\mathrm{trace}(AB) = \mathrm{trace}(BA)$ for any two square matrices.
When we have coordinates $\mathbf{x}(u,v)$ on the surface, we can calculate the matrix of $S$. One can show that the matrix is $$ \begin{bmatrix} E & F \\ F & G \end{bmatrix}^{-1} \begin{bmatrix} e & f \\ f & g \end{bmatrix} = \frac{1}{EG-F^2} \begin{bmatrix} e G - f F & f G - g F \\ f E - e F & g E- f F \\ \end{bmatrix}, $$ where $$ \begin{align*} E = \mathbf{x}_u\cdot \mathbf{x}_u, \quad F = \mathbf{x}_u\cdot \mathbf{x}_v, \quad G = \mathbf{x}_v\cdot \mathbf{x}_v \\ e = \mathbf{x}_{uu}\cdot N, \quad f = \mathbf{x}_{uv}\cdot N, \quad g = \mathbf{x}_{vv}\cdot N. \end{align*} $$ Taking half the trace of this matrix gives the second expression for $H$.