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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!

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What do you mean by $L_{11}, L_{12}, L_{22}$? I assume they're the coefficients of the second fundamental form, but I want to make sure. – Jesse Madnick Oct 10 '12 at 6:10
Yes, that is correct. Sorry about my wrong thought, still learning the material – mary Oct 10 '12 at 6:47
Can you maybe state all the definitions of the various symbols? Not only for the reader, but this might already help you see where you could begin a proof! – Sam Oct 12 '12 at 17:08
Hi! I'd like to know what does $H$ mean? where are the Christoffel symbols here? – c.p. Oct 14 '12 at 18:26
not very sure, still trying to learn that – mary Oct 14 '12 at 22:31

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 $-2g_{12}L_{12} =0$

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