Given a $2$ dimensional riemannian manifold with a local orthonormal frame $e_1,e_2$ I want to evaluate the covariant derivatives $\nabla_{e_1}e_1$, $\nabla_{e_2}e_1$, $\nabla_{e_1}e_2$ and $\nabla_{e_2}e_2$, assuming the Levi-Civita connection. I am using two methods and finding different results, so I would like to understand where I am wrong.
If I express the covariant derivative with the connection form $\omega^i_j$ associated to the frame, defined by $\nabla_{X}e_j= \Sigma \omega^i_j (X)e_i$ I get the following equations
$$\nabla_{e_1}e_1 = \omega_1^1(e_1)e_1+\omega_1^2(e_1)e_2$$ $$\nabla_{e_2}e_1 = \omega_1^1(e_2)e_1+\omega_1^2(e_2)e_2$$ $$\nabla_{e_1}e_2 = \omega_2^1(e_1)e_1+\omega_2^2(e_1)e_2$$ $$\nabla_{e_2}e_2 = \omega_2^1(e_2)e_1+\omega_2^2(e_2)e_2$$
as the connection matrix $\omega_i^j$ with respect to an orthonormal frame is skew symmetric, the previous equations further simplify as
$$\nabla_{e_1}e_1 = -\omega^1_2(e_1)e_2$$ $$\nabla_{e_2}e_1 = -\omega^1_2(e_2)e_2$$ $$\nabla_{e_1}e_2 = \omega^1_2(e_1)e_1$$ $$\nabla_{e_2}e_2 = \omega^1_2(e_2)e_1$$
However, I am not fully conviced because if I instead use the formula for the covariant derivative in local coordinates I get a seemingly different result.
Assuming two general vector fields $v=v^je_j$ and $u=u^ie_j$ the formula says:
$$\nabla_v u= (v^ju^i\Gamma^k_{ij}+v^j\frac{\partial u^k}{\partial x^j})\frac{\partial}{\partial x^k}$$
To find a local expression for an orthonormal frame, I assume that in the coordinates $x^1,x^2$ the metric is expressed by the first fundamental form $[\begin{smallmatrix} E & F \\ F & G \end{smallmatrix}]$. Then an orthonormal frame is given by (D is the determinant of the matrix):
$$(\frac{1}{\sqrt{E}}\frac{\partial}{\partial x^1}, \frac{-F}{\sqrt{ED}}\frac{\partial}{\partial x^1}+\sqrt{\frac{E}{D}}\frac{\partial}{\partial x^2})$$.
In particular,$\frac{1}{\sqrt{E}}\frac{\partial}{\partial x^1}$ has norm $1$ for the metric in question.
My problem is that if now calculate $\nabla_{e_1}e_1$ I don't get an expression only in $\frac{\partial}{\partial x^2}$ as I would have expected, as I get
$$\nabla_{\frac{1}{\sqrt{E}}\frac{\partial}{\partial x^1}}(\frac{1}{\sqrt{E}}\frac{\partial}{\partial x^1})=\frac{1}{E}\Gamma^1_{11}\frac{\partial}{\partial x^1}+\frac{1}{E}\Gamma^2_{11}\frac{\partial}{\partial x^2}-\frac{1}{2}\frac{E_1}{E^2}\frac{\partial}{\partial x^1}$$
Further expanding the coefficient for $\frac{\partial}{\partial x^1}$ and using the expressions for the Christoffel symbols in coefficients of the first fundamental form I get $$\frac{GE_1-2FF_1+FE_2}{2E(EG-F^2)}\frac{\partial}{\partial x^1}-\frac{1}{2}\frac{E_1}{E^2}\frac{\partial}{\partial x^1}$$ Which does not seem to vanish to me, unless I am mistaken.
Where am I wrong?
- applying the connection form and then assuming that $\nabla_{e_1}e_1$ is a multiple of $e_2$ only
- assuming that the vector fields above are an orthonormal frame
- applying the formula for the covariant derivative in local coordinates
thanks!
Your set-up is correct. Your $e_1,e_2$ do in fact form an orthonormal frame. The computation is a messy bookkeeping job, but here is most of it. As you pointed out, we will need the classical formulas for the Christoffel symbols. Note, also, that we can compute $\nabla X$ rather than $\nabla_Y X$ just by using basis $1$-form "placeholders"; I find this clearer.
I start with the equation $$\nabla e_1 = \nabla \left(\frac1{\sqrt E}\frac{\partial}{\partial x^1}\right) = -\frac{dE}{2E^{3/2}}\frac{\partial}{\partial x^1} + \frac1{\sqrt E}\nabla\left(\frac{\partial}{\partial x^1}\right).$$ Now, using $f_i = \partial f/\partial x^i$, we have \begin{align*} \nabla\left(\frac{\partial}{\partial x^1}\right) &= \sum\Gamma^j_{1k} dx^k \frac{\partial}{\partial x^j} \\ &= \frac1{2D}\left((GE_1-2FF_1+FE_2) dx^1\frac{\partial}{\partial x^1} + (GE_2-FG_1)dx^2\frac{\partial}{\partial x^1} +\\ (-FE_1+2EF_1-EE_2)dx^1\frac{\partial}{\partial x^2} + (-FE_1+EG_2)dx^2\frac{\partial}{\partial x^2}\right). \end{align*} What you want to check is that $\nabla e_1 \cdot \frac{\partial}{\partial x^1} = 0$. Separate this into the $dx^1$ and $dx^2$ pieces. I'll write just one of these out. The coefficient of $dx^1$ in that dot product will be $$E\Big({-}\frac{E_1}{2E^{3/2}} + \frac1{2D\sqrt E}(GE_1-2FF_1+FE_2)\ \Big) + F\frac1{2D\sqrt E}(-FE_1+2EF_1-EE_2).$$ Multiplying through by $2D\sqrt E$ and simplifying, we get $$-E_1(EG-F^2)+ EGE_1-2EFF_1+EFE_2 - F^2 E_1 + 2EFF_1-EFE_2 = 0,$$ as desired.
I hope you can learn from this and feel ... um ... reassured that it works. As a final remark, I'll say that I've spent my whole mathematical career (and graduate but not undergraduate teaching career) using orthonormal/unitary moving frames and it's a beautiful and powerful tool. However, to use them in coordinates, one really wants orthogonal coordinates, as this mess shows :P