# Mean curvature of a surface - Why does it equal this formula?

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?

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 $$H$$.
• Yes thanks, but why is $\frac{1}{2}tr(S)=\frac{1}{2}(k_1+k_2)$? May 3, 2020 at 12:58