# Christoffel Symbols on a Surface

In Do Carmo's Differential Geometry of Curves and Surfaces he does the following:

Let $\vec r$ be a parametrization of a surface $S\subset\mathbb{R}^3$ so that $\vec r_u,\vec r_v$ forms a basis for $T_pS$ at each point in the image of $\vec r$. We can then define $N=\frac{\vec r_u\times\vec r_v}{|\vec r_u\times \vec r_v|}\in (T_pS)^\perp$ so that $\{\vec r_u,\vec r_v,N\}$ is a basis for $\mathbb{R}^3$.

We can express $\vec r_{uu},\vec r_{uv}=\vec r_{vu},\vec r_{vv}$ in terms of this basis. We set $$\vec r_{uu}=\Gamma_{11}^1\vec r_u+\Gamma_{11}^2\vec r_v+L_1 N$$ $$\vec r_{uv}=\Gamma_{12}^1\vec r_u+\Gamma_{12}^2\vec r_v+L_2N$$ $$\vec r_{vv}=\Gamma_{22}^1\vec r_u+\Gamma_{22}^2\vec r_v+L_3N$$

I am failing to recover this definition of the Christoffel symbols from that on an arbitrary Riemannian manifold as $\Gamma_{ij}^k=(\nabla_{E_i}E_j)^k$. There's obviously something simple I am missing. If $\nabla$ is the Euclidean connection on $\mathbb{R}^3$, and $\vec r=(x,y,z)$ then $\nabla_{\vec r_u}\vec r_u=\vec r_u(x_u)+\vec r_u(y_u)+\vec r_u(z_u)$, right? Where is $\vec r_{uu}$ coming in?

• do you know the Gauss equation? Between the standard connection, $D$, and the induced on the tangent of the surface, $\nabla$, we have $\nabla_XY=D_XY-(D_XY\bullet N)N$ – janmarqz Nov 18 '14 at 0:56
• Yes, I believe you are referring to the second fundamental form? So on an arbitrary manifold $\nabla_XY=D_XY+B(X,Y)$, where $B$ is the second fundamental form. In what you wrote I think you have the orthogonal projection $h(X,Y)=\langle B(X,Y),N\rangle$, and $h(X,Y)=\langle sX,Y\rangle$ where $s$ is the shape operator. I also know that $s=-dN_p$, the derivative of the Gauss map on surfaces. – JonHerman Nov 18 '14 at 1:38
• this argument works for a codimension one submanifold only – janmarqz Nov 18 '14 at 2:51
• I was thinking you could just fix a normal vector. Sorry, how do I get the above relations from the Gauss equation? – JonHerman Nov 18 '14 at 4:02
• $D_XY$ usually is not tangent, but its projection on the tangent, $\nabla_XY$ is, so $D_XY=\nabla_XY+\alpha N$ for some scalar $\alpha$. Then $\alpha=D_XY\bullet N$ – janmarqz Nov 18 '14 at 4:18