Christoffel symbols in Differential geometry iff proof I need help in proving that $H = 0$ for a surface iff $g_{11}L_{22} - 2g_{12}L_{12} + g_{22}L_{11} = 0.$
I think that these are the Christoffel symbols exploited in some manner and normally, I'm not sure what matrix representation this was derived from.
Thanks in advance!
 A: Usually when talking about surfaces $H$ refers to the mean curvature. Mean curvature is defined to be $$H = \frac{L_{11}}{g_{11}} + \frac{L_{22}}{g_{22}}$$ If you know some linear algebra, then you will recognize that this is the trace of the bilinear form $L$. Let me give you a hint on how to do your problem. Divide your equation by $g_{11}g_{22}$ (why can't $g_{11}g_{22}$ be zero?) to obtain $$\frac{L_{11}}{g_{11}} + \frac{L_{22}}{g_{22}} = \frac{2g_{12}L_{12}}{g_{11}g_{22}}$$ Since $H$ does not depend on the parametrization, it follows that $$\frac{2g_{12}L_{12}}{g_{11}g_{22}}$$ doesn't depend on the parametrization either. Thus, we may as well choose a parametrization in which $g_{12} = 0$ (it is always possible to do this- such parametrizations are called ``orthogonal parametrizations''). Thus $H = 0$.
Now let's consider the converse statement. Suppose that $H = 0$. Thus 
$$\frac{L_{11}}{g_{11}} + \frac{L_{22}}{g_{22}} = 0$$
To finish the problem, you want to show that $-2g_{12}L_{12} =0$. But this does not have to be true! (For a counterexample, consider the catenoid and choose any system of coordinates whose coordinate vectors are not orthogonal. Then it will turn out that $g_{12}$ and $L_{12}$ are nonzero). On the other hand, the converse is true if we assume that we are working in an orthogonal parametrization (that is, we choose coordinates that are orthogonal to each other). For if $g_{12} = 0$, then it is clear that $-2g_{12}L_{12} =0$
