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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?

  1. applying the connection form and then assuming that $\nabla_{e_1}e_1$ is a multiple of $e_2$ only
  2. assuming that the vector fields above are an orthonormal frame
  3. applying the formula for the covariant derivative in local coordinates

thanks!

ebenezer
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    The expression involving the Christoffel symbols is valid for a coordinate frame, not an orthonormal frame – Didier May 12 '23 at 18:38
  • thanks @Didier for your comment. however I did not understand it. I am using a coordinate frame when I apply the Christoffel symbol, that is $x^1,x^2$. Indeed, in that coordinate frame a possible orthonormal frame (if my calculations are not wrong) is $(\frac{1}{\sqrt{E}}\frac{\partial}{\partial x^1}, \frac{-F}{\sqrt{ED}}\frac{\partial}{\partial x^1}+\sqrt{\frac{E}{D}}\frac{\partial}{\partial x^2})$ and when I apply the formula with Christoffel symbols to $\frac{1}{\sqrt{E}}\frac{\partial}{\partial x^1}$ I consider it as a generic vector field... – ebenezer May 12 '23 at 19:00
  • The main thing I don't see is the Lie bracket. This is zero for the $\partial x^j$ from local coordinates, so the mixed "partials" commute. Not so for orthonormal frames – Will Jagy May 12 '23 at 19:08
  • I was referring to the equation $\nabla_vu = (\cdots) e_k$. This should be $\nabla_vu=(\cdots)\frac{\partial}{\partial x^k}$ instead – Didier May 12 '23 at 19:09
  • @Didier thanks I have amended the formula – ebenezer May 12 '23 at 19:18
  • @Will Jagy where is the bracket missing in your opinion? – ebenezer May 12 '23 at 19:18
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    If you’re going to use upper and lower indices, then please look for consistency. You should have $\nabla e_i = \sum \omega_i^j \otimes e_j$. – Ted Shifrin May 12 '23 at 19:21
  • https://en.wikipedia.org/wiki/Fundamental_theorem_of_Riemannian_geometry – Will Jagy May 12 '23 at 19:21
  • @TedShifrin thanks I have amended the formula, even if the text I have does not use tensor product. but I guess it is because I have indicated $X$ explicitely in my formula so in my case is vector = vector, while in your case is tensor = tensor, as $X$ becomes an input. do you have also an idea of wnere my calculations are wrong? – ebenezer May 12 '23 at 19:37
  • @Will Jagy yes but here I am calculating $\nabla_{e_1}e_1 = -\omega^1_2(e_1)e_2$ so at least in this case the bracket should vanish, no? it would be very helpful for me if you could be a bit more explicit, thanks, I am in a very early learning stage :) – ebenezer May 12 '23 at 19:41
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    If $e_1,e_2$ is an orthonormal moving frame, then, yes, $\nabla e_1$ must be in the $e_2$ direction. You actually do not need to do $\nabla_{e_1}e_1$. But you do need to rewrite your "classical" answer as a linear combination of $e_1$ and $e_2$. Brackets do not vanish when you're working with orthonormal frames; they vanish only when you're computing with coordinate frames. Perhaps you should work with the structure equations for a $GL(2)$ connection, rather than reducing to $O(2)$ in the general case? – Ted Shifrin May 12 '23 at 19:42
  • @ebenezer If the Lie bracket $[e_1,e_2]$ vanishes and ${e_1,e_2}$ is orthonormal, then ${e_1,e_2}$ is both a coordinate frame and an orthonormal frame, and your surface is flat. – Didier May 12 '23 at 19:43
  • @TedShifrin your comment is interesting but I fear I don't have enough knowledge of the matter to understand it. what are the structure equations that you are mentioning? and why do groups O(2) and GL(2) come into picture? I only know of affine connections, and here I am working with the Levi-Civita one. also, what is the "classical" answer that I need to review? – ebenezer May 12 '23 at 19:59
  • @ebenezer What exactly are you trying to show actually? – Didier May 12 '23 at 20:32
  • $GL(2)$ means that you're working with general frames (such as coordinate frames), and then the connection matrix is no longer skew-symmetric. You get the standard $\nabla v_i = \sum \omega^j_i v_j$, with $\omega^j_i = \sum \Gamma^j_{ik} dx^k$. But I want to echo Didier's last question: What are you trying to do here? – Ted Shifrin May 12 '23 at 20:43
  • @TedShifrin well as exercise I derived expression for $\nabla_{e_i}e_j$ when $(e_i,e_j)$ is an orthonormal frame using connection forms. I have found as result the second display in my question. Wasn't convinced, tried to verify this with another method (using the orthonormal frame of my 4th display). Found a different result and wanted to understand where is wrong. Note that I am not using a coordinate frame in the expression with Christoffel symbols, but an ad hoc orthonormal frame for the metric, and I am taking into account its coefficients in the formula of the second to last display. – ebenezer May 12 '23 at 21:10
  • @Didier I explained what I am trying to do in the comment above. I have also edited the question to clarify what I am after. thank you all for the help! I may not be 100% clear because I am new to the subject. – ebenezer May 12 '23 at 21:33

1 Answers1

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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

Ted Shifrin
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  • thanks a lot for your answer. it clarifies. basically I was expecting $\nabla e_1$ to be a multiple of $\frac{\partial}{\partial x^2}$ which is wrong as it is indeed a multiple of $e_2$. The coordinate frames are not ortogonal in this setup. – ebenezer May 13 '23 at 08:16