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Let $M\subset \mathbb{R}^2$ be a surface. The geodesic equation can be written:

$$\dfrac{D\gamma'}{dt}=0.$$

I want to show that the same equation, together with the Christoffel symbols can be obtained as the Euler Lagrange equation for the Lagrangian

$$L(u,u',v,v')=E(u,v)u'^2+2F(u,v)u'v'+G(u,v)v'^2,$$

where $E,F,G$ are the coefficients of the first fundamental form.

Now, I computed

$$\dfrac{\partial L}{\partial u}=\dfrac{\partial E}{\partial u}u'^2+2\dfrac{\partial F}{\partial u}u'v'+\dfrac{\partial G}{\partial u}v'^2,$$

$$\dfrac{\partial L}{\partial u'}=2Eu'+2Fv'.$$

In that case we have

$$\dfrac{d}{dt}\dfrac{\partial L}{\partial u'}=2\left(\dfrac{\partial E}{\partial u}u'+\dfrac{\partial E}{\partial v}v'\right)u'+2Eu''+2\left(\dfrac{\partial F}{\partial u}u'+\dfrac{\partial F}{\partial v}v'\right)v'+2Fv''.$$

In that case writing down the Euler-Lagrange equation for the $u$-coordinate we have

$$-\dfrac{\partial E}{\partial u}u'^2+\left(\dfrac{\partial G}{\partial u}-2\dfrac{\partial F}{\partial v}\right)v'^2+2\dfrac{\partial E}{\partial v}u'v'+2Eu''+2Fv''=0.$$

The problem is that the $u$-th coordinate of the first equation I wrote is:

$$u''+\Gamma^1_{11}u'^2+\Gamma^{1}_{12}u'v'+\Gamma^1_{22}v'^2=0.$$

The problem here is that in my equation there appears $2Fv''$. In the original equation there is no $v''$ term, so something is quite wrong here.

I should be able to recover from the Lagrangian the geodesic equation together with all Christoffel symbols.

What is wrong here? Why I'm not getting the geodesic equation as expected?

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Since your first fundamental form is not diagonal, the two equations you get from Euler-Lagrange will not be individually equivalent to the two geodesic equations, despite the fact that the two systems are equivalent. The relationship is via the first fundamental form matrix $g$ - if you view the geodesic equation as the vector ODE

$$ \gamma'' - \Gamma(\gamma',\gamma') = 0$$

then the E-L equation is simply

$$ g\gamma'' - g\Gamma(\gamma',\gamma')= 0.$$

Since $g$ is not diagonal, this transformation mixes the two equations; but it does not change their solutions (since $g$ is invertible).

If you compute both E-L equations, you should see that they are (independent) linear combinations of the two geodesic equations.

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