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So we want to...

i.) Calculate the Levi-Civita Connection of two vector fields X,Y: $$\nabla_{X}Y$$ for the standard euclidean metric in n-dimensions: $$ds^{2} = dx_{1}^{2} + ... + dx_{n}^{2}$$

ii.) Calculate the curvature function of this metric for n = 2.


So for...

i.) I have for the covariant derivative of two vector fields, $V = v^{i}e_{i}$ and $U = u^{j}e_{j}$ that $\nabla_{V}U$ = $v^{i}u^{j}\Gamma^{k}_{ij}e_{k} + v^{i}\frac{du^{j}}{dx^{i}}e_{j}$

  • However looking online I found an example where it looked like all he was doing was calculating the christoffel symbols. I was going to calculate $\Gamma^{i}_{ii}, \Gamma^{i}_{ij} = \Gamma^{i}_{ji}, \Gamma^{i}_{jj}, \Gamma^{j}_{ii}, \Gamma^{j}_{ij} = \Gamma^{j}_{ji}$ and $\Gamma^{j}_{jj}$ and since for the euclidean metric only the diagonal (i.e., i=j) elements are 1 with all else 0 and since the symbols all involve derivatives of metric elements, thus making all the symbols equal to zero right?

So should I just calculate those Christoffel Symbols for the Levi-Civita Connection (Bothered because doesn't involve Vector Fields at all)? Or should I use that equation for $\nabla_{V}U$ (in which case the term with christoffel symbols vanish leaving only second term and then I don't know how to make a meaningful answer out of components of arbitrary vector fields)?

ii.) For this I just need to know what the curvature function is. I know theres a normal curvature which looks to be defined as: $$ K_{N} = <c''(0), N(p)>$$ for a curve c with c(0) = p but I've seen it as $$K_{N} = II(c'(0), c'(0)) where II is 2nd fundamental form.

  • I'm not exactly sure how to calculate something with the Second fundamental form if anyone can elucidate this I would appreciate that.

    Basically I just need to know what formula I should be using to calculate the curvature in n = 2.

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1 Answer 1

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Yes, since all the $g_{ij}$ are constant, all the Christoffel symbols will be $0$. Your formula for normal curvature is not what is going on here (for starters, we have an abstract Riemannian metric, not a surface in $\Bbb R^3$; next, we want Gaussian curvature, not normal curvature, even in the setting of a surface in $\Bbb R^3$). Do you have a formula for the curvature tensor in terms of things like $\nabla_X\nabla_Y Z$ ... or, perhaps, in terms of the Christoffel symbols and their derivatives?

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  • $\begingroup$ So your saying for the Levi-Civita calculation I can basically just calculate the various 'types' of Christoffel Symbols? And so I looked up the Curvature Tensor and it appears that $$R(X,Y)Z = \nabla_{X}\nabla_{Y}Z - \nabla_{Y}\nabla_{X}Z - \nabla_{[X,Y]}Z$$ So given what I've done in the first part would I take Z to be the metric g? And then calculate the chistoffel symbols? What concerns me then would be that for the 3 terms the various vector spaces X, Y and Z and their placement doesn't seem to make a difference in the calculation of $\nabla$ given that in part 1 all that was needed were $\endgroup$ May 2, 2017 at 0:08
  • $\begingroup$ the elements of the metric, $g_{ij}$ and not the elements of the vector spaces. $\endgroup$ May 2, 2017 at 0:08
  • $\begingroup$ What does your course expect you to do for the curvature? I know 50 ways of approaching this, and I can't mind-read. But I guarantee nothing to do with second fundamental form. This is for an abstract Riemannian surface. $\endgroup$ May 2, 2017 at 0:12
  • $\begingroup$ Well we are asked to "calculate the Curvature Function of this metric for n = 2" and then later on in reference to the "Curvature function" the symbol $K_g$ is used, seemingly indicating Gaussian Curvature? $\endgroup$ May 2, 2017 at 0:17
  • $\begingroup$ Also "nothing to do with second fundamental form" is the nicest concatenation of words I've heard all day :) $\endgroup$ May 2, 2017 at 0:18

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