# Derivation of the crhristoffel symbol

How can we derive the formula for the christoffel symbols

$$\Gamma_{ij}^k = \frac{f_{ij}f_k}{(1+f_1^2+f_2^2)}$$

for the monge patch $\mathbf{x}(x,y) = (x,y,f(x,y))$ ?

I got the normal, $L_{11}$, $L_{12}$, $L_{21}$, $L_{22}$, $g_{ij}$'s but can't see what's missing.

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you asked yesterday already. –  Bombyx mori Dec 9 '12 at 6:50
Did you ever tried compute the synbols by using the formula? –  Tomás Dec 11 '12 at 10:21
I tried it, but I dont understand what to do with it after getting the info above –  Buddy Holly Dec 11 '12 at 10:29
Why not show the work you did to get $L_{11}, L_{12}... g_{ij}$ etc? –  mck Dec 11 '12 at 15:32
I would, but it is really easy computation. I wanted help in area where I'm stuck and I'm not a pro in Latex. Hope you can help. Thanks –  Buddy Holly Dec 12 '12 at 0:55

First things first: if you're going to compute Christoffel symbols, you first have to have some coordinates chosen, since this will effect the computation. Of course, the Monge patch $M\subset \mathbb{R}^3$ has an obvious choice of coordinates, namely those given by the projection $\pi\colon M\to \mathbb{R}^2$ onto the first two coordinates. Instead of calling these $x$ and $y$, I will use $x_1$ and $x_2$ so that your indices make more sense.
You state in your question that you were able to compute the metric $g = (g_{ij})$ in these coordinates. For this reason, I will not compute them here, but instead note the final answer:$$g_{11} = 1 + f_1^2,\,\,\,\,\,g_{12} = g_{21} = f_1f_2,\,\,\,\,\,g_{22} = 1 + f_2^2.$$ From this one can compute the inverse $g^{-1} = (g^{ij})$, to obtain $$g^{11} = \frac{1 + f_2^2}{1 + f_1^2 + f_2^2},\,\,\,\,\,g^{12} = g^{21} = \frac{-f_1f_2}{1 + f_1^2 + f_2^2},\,\,\,\,\,g^{22} = \frac{1 + f_1^2}{1+f_1^2 + f_2^2}.$$ The Christoffel symbols $\Gamma_{ij}^k$ are then computed using the formula $$\Gamma_{ij}^k = \frac{1}{2}\sum_m\left\{\frac{\partial g_{jm}}{\partial x_i} + \frac{\partial g_{mi}}{\partial x_j} - \frac{\partial g_{ij}}{\partial x_m}\right\}g^{mk}.$$ To illustrate how this computation, I will compute $\Gamma_{11}^1$: $$\Gamma_{11}^1 = \frac{1}{2}\left\{\frac{\partial g_{11}}{\partial x_1} + \frac{\partial g_{11}}{\partial x_1} - \frac{\partial g_{11}}{\partial x_1}\right\}g^{11} + \frac{1}{2}\left\{\frac{\partial g_{12}}{\partial x_1} + \frac{\partial g_{21}}{\partial x_1} - \frac{\partial g_{11}}{\partial x_2}\right\}g^{21}$$$$= \frac{1}{2}\frac{\partial g_{11}}{\partial x_1}g^{11} + \frac{1}{2}\left\{2\frac{\partial g_{12}}{\partial x_1} - \frac{\partial g_{11}}{\partial x_2}\right\}g^{21}.$$ Now $$\frac{\partial g_{11}}{\partial x_1} = 2f_1f_{11},\,\,\,\,\,\frac{\partial g_{12}}{\partial x_1} = f_{11}f_2 + f_1f_{21},\,\,\,\,\,\frac{\partial g_{11}}{\partial x_2} = 2f_1f_{12},$$ so $$\Gamma_{11}^1 = \frac{1}{2}(2f_1f_{11})\frac{1 + f_2^2}{1 + f_1^2 + f_2^2} + \frac{1}{2}\left\{2(f_{11}f_2 + f_1f_{21})-2f_1f_{12} \right\}\frac{-f_1f_2}{1 + f_1^2 + f_2^2}$$$$= \frac{f_1f_{11}}{1 + f_1^2 + f_2^2}.$$ The other $\Gamma_{ij}^k$ are computed similarly, and I leave it to you.