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I am looking to calculate the Riemann curvature on an orthonormal frame for the tangent bundle of the surface parametrized by the equation $\phi: \mathbb{R}^2\to \mathbb{R}^3$ given by $$ (x,y) \mapsto (x,y,f(x,y))$$ for some continuous $f:\mathbb{R}^2 \to \mathbb{R}$. I would like to do this by calculating the Christoffel symbols with respect to the orthogonal tangent vectors $$ \begin{align} \frac{\partial}{\partial \varphi_1} &= -f_y\frac{\partial}{\partial x} + f_x\frac{\partial}{\partial y} \\ \frac{\partial}{\partial \varphi_2} &= f_x\frac{\partial}{\partial x} + f_y\frac{\partial}{\partial y} + (f_x^2 + f_y^2)\frac{\partial}{\partial z} \end{align} $$ (following the approach of Orthonormal basis for a tangent plane) where the metric is the restriction of the Euclidean metric from $\mathbb{R}^3$. I would then calculate $\nabla_\frac{\partial}{\partial \varphi_i}\frac{\partial}{\partial \varphi_j}$ for all $i,j$ and hence the curvature on the normalised frame. However, I am unclear on a few points:

  1. If $z = f(x,y)$, then should I calculate $\frac{\partial}{\partial z} = f_x \frac{\partial}{\partial x} + f_y \frac{\partial}{\partial y}$ (where $f_x = \frac{\partial f}{\partial x}$, etc.)? Doesn't this mess up orthogonality?
  2. Should I somehow try to rewrite $x,y$ in terms of $\varphi_1, \varphi_2$ and then do the Christoffel symbol calculation? My only previous example was calculating the Riemann curvature for a sphere where the parametrization was in terms of spherical polar coordinates, which made the calculus cleaner.
  3. There is an expression for Christoffel symbols in terms of $\phi_x, \phi_y, \phi_{xx}$, etc. as per Christoffel Symbols on a Surface. However, here $\phi_x$ and $\phi_y$ are not orthogonal, so would it still be possible for me to calculate Christoffel symbols using that method and then apply those to the orthogonal frame I wish to use?

EDIT MADE: Corrected the choice of tangent vectors in the orthonormal frame, as the original frame was not tangent.

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  • $\begingroup$ If you see your definition as a pullback from $\mathbb{R}^3$ to $\mathbb{R}^2$ you can start from the eucledian metric of $\mathbb{R}^3$ $ds^2=dx^2+dy^2+dz^3=\left(1+\left(\frac{df}{dx}\right)^2\right) dx^2+\left(1+\left(\frac{df}{dy}\right)^2\right) dy^2 + 2\frac{df}{dy}\frac{df}{dx}dxdy$ to get the metric on the surface f. Thats all you need to calculate the Christoffel symbols. $\endgroup$
    – Bort
    Commented Feb 15, 2018 at 18:07
  • $\begingroup$ Hello @Bort, thank you for your comment. My confusion is arising from the fact that I want to calculate the Christoffel symbols to derive the curvature tensor and then apply that tensor to an orthonormal frame. However, if I calculate the $\Gamma^{i}_{jk}$ with respect to one frame, will the curvature matrix need to be transformed as well? $\endgroup$
    – An Coileanach
    Commented Feb 16, 2018 at 19:31
  • $\begingroup$ To be more specific, can I say something like $ds^2 = d\varphi_1^2 + d\varphi_2^2$ and then compute the corresponding Christoffel symbols just in terms of those? $\endgroup$
    – An Coileanach
    Commented Feb 17, 2018 at 11:00

1 Answer 1

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I am currently struggling with surfaces as well at the moment, so I'm no expert, take this answer with a grain of salt. As you're correclty stating $\phi_x , \phi_y$ are NOT an orthonormal frame for $T_p S$, though I think that it is much easier to express Christoffel symbols in terms of those basis vectors. Using the system $$\phi_{xx} = \Gamma_{11}^1 \phi_x + \Gamma_{11}^2 \phi_y + L^1 N$$ $$\phi_{xy} = \Gamma_{12}^1 \phi_x + \Gamma_{12}^2 \phi_y + L^2 N$$ $$\phi_{yy} = \Gamma_{22}^1 \phi_x + \Gamma_{22}^2 \phi_y + L^3 N$$ and multiplying each equation scalarly by $\phi_x$ and $\phi_y$ the normal vector $N$ always cancels out (being orthgonal to $T_p S$), and you're left with a linear system of six equations from which you can derive the six Christoffel Symbols.

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  • $\begingroup$ In simpler terms, in my opinion you should follow the third route you suggested. The fact that $\phi_x,\phi_y$ are not orthogonal just means that you won't be able to extract $\Gamma_{ij}^k$ right away, and you'll have to solve a linear system. No big deal. $\endgroup$
    – barmanthewise
    Commented Feb 17, 2018 at 11:23
  • $\begingroup$ Thank you for your comment @barmanthewise. The thing that is confusing me here is whether the $\Gamma^{k}_{ij}$ coming from this will be the same as the $\Gamma^{k}_{ij}$ from the orthonormal coordinate system, or will I need to apply some linear transformation? I appreciate that you're in the same boat as me, as someone who is just learning this stuff, but you seem to have a better intuition for it than I do. $\endgroup$
    – An Coileanach
    Commented Feb 17, 2018 at 11:33
  • $\begingroup$ If by 'the' orthonormal coordinate system you mean the standard basis of $\Bbb R^3$ as the tangent space to $S$, that is $\partial_x,\partial_y,\partial_z$, then, yes, you'd need to apply a linear transformation, for exapmle the one that takes $\phi_x \mapsto \partial_x$, $\phi_y \mapsto \partial_y$, $N \mapsto \partial_z$. $\endgroup$
    – barmanthewise
    Commented Feb 17, 2018 at 14:04
  • $\begingroup$ I should have been clearer: by "the orthonormal coordinate system", I meant the normalised version of the frame corresponding to the tangent vectors $\frac{\partial}{\partial\varphi_1}$, $\frac{\partial}{\partial\varphi_2}$. I don't think $\partial_x$, $\partial_y$, $\partial_z$ would be a basis for the tangent space to $S$ in general. $\endgroup$
    – An Coileanach
    Commented Feb 17, 2018 at 19:45

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