Graph Geodesics via Christoffel Symbols

The purely intrinsic approach, using only the induced metric.

Given a real valued function f(x,y)f(x,y), its graph defines a surface which inherits an intrinsic geometry from the ambient E3\mathbb{E}^3. In this note, we compute the geodesic equations for this surface.

TheoremGeodesics on a Graph

If γ(t)=(x(t),y(t))\gamma(t)=(x(t),y(t)) is a geodesic on the graph (x,y,f(x,y))(x,y,f(x,y)) of a function f ⁣:R2Rf\colon\mathbb{R}^2\to\mathbb{R}, then x¨=fxfxxx˙2+2fxyx˙y˙+fyyy˙21+fx2+fy2\ddot{x}=-f_x\frac{f_{xx}\dot{x}^2+2f_{xy}\dot{x}\dot{y}+f_{yy}\dot{y}^2}{1+f_x^2+f_y^2} y¨=fyfxxx˙2+2fxyx˙y˙+fyyy˙21+fx2+fy2\ddot{y}=-f_y\frac{f_{xx}\dot{x}^2+2f_{xy}\dot{x}\dot{y}+f_{yy}\dot{y}^2}{1+f_x^2+f_y^2}

In this note we approach the calculation intrinsically, by expanding out the defining equation γ˙γ˙=0\nabla_{\dot{\gamma}}\dot{\gamma}=0 in coordinates. This involves some preliminary calculation of Christoffel symbols, which are recorded in generality in the note ‘2D Geometry Cheatsheet’. Here we focus on specializing this generality to the case of a surface (x,y,f(x,y))(x,y,f(x,y)) in R3\mathbb{R}^3.

The Metric & Christoffel Symbols

From general calculations for a 2D Riemannian metric, we have the following form for the Christoffel symbols.

TheoremGeneral Christoffel Symbols in 2 Dimensions

The Christoffel symbols of a general metric g=(EFFG)g=\left(\begin{smallmatrix}E&F\\ F&G\end{smallmatrix}\right) are Γxxx=GExF(2FxEy)2DΓxxy=E(2FxEy)FEx2D\Gamma_{xx}^x=\frac{G E_x - F \left( 2 F_x - E_y \right)}{2D}\hspace{1cm}\Gamma_{xx}^y= \frac{E\left( 2 F_x - E_y \right)-FE_x}{2D} Γyyx=G(2FyGx)FGy2DΓyyy=EGyF(2FyGx)2D\Gamma_{yy}^x=\frac{G\left( 2 F_y - G_x \right)-FG_y}{2D}\hspace{1cm}\Gamma_{yy}^y=\frac{E G_y - F \left( 2 F_y - G_x \right)}{2D} Γxyx=Γyxx=GEyFGx2DΓxyy=Γyxy=EGxFEy2D\Gamma_{xy}^x=\Gamma_{yx}^x=\frac{G E_y -F G_x}{2D}\hspace{1cm}\Gamma_{xy}^y=\Gamma_{yx}^y=\frac{E G_x -F E_y}{2D} For D=detg=EGF2D=\det g = EG-F^2

The Metric Components

Specializing to a surface parametrized by the function F ⁣:(x,y)(x,y,f(x,y))F\colon (x,y)\mapsto (x,y,f(x,y)), the metric is computed in coordinates (x,y)(x,y) by pullback. Specifically, let TxT_x and TyT_y be the xx- and yy- partial derivatives of the parameterization

Tx=1,0,fxTy=0,1,fyT_x = \langle 1,0, f_x\rangle\hspace{1cm}T_y=\langle 0,1, f_y\rangle

Then the components of the metric g=(EFFG)g=\left(\begin{smallmatrix}E&F\\ F&G\end{smallmatrix}\right) are the pairwise dot-products of these

E=TxTxF=TxTyG=TyTyE=T_x\cdot T_x\hspace{1cm} F=T_x\cdot T_y\hspace{1cm} G=T_y\cdot T_y

Below we compute these, and their derivatives (which show up in the Christoffel formulas referenced above).

E=1+fx2F=fxfyG=1+fy2Ex=2fxfxxFx=fxxfy+fxfxyGx=2fyfxyEy=2fxfxyFy=fxyfy+fxfyyGy=2fyfyy\begin{align*} E &= 1 + f_x^2 & F &= f_x f_y & G &= 1 + f_y^2 \\ E_x &= 2 f_x f_{xx} & F_x &= f_{xx} f_y + f_x f_{xy} & G_x &= 2 f_y f_{xy} \\ E_y &= 2 f_x f_{xy} & F_y &= f_{xy} f_y + f_x f_{yy} & G_y &= 2 f_y f_{yy} \end{align*}

These three numbers are the only thing about the surface that the rest of this note will use, so it is worth seeing what they say. Written as a matrix, g=I+ffTg=I+\nabla f\,\nabla f^{\mathsf T}, whose eigenvectors are the gradient direction, with eigenvalue 1+f21+|\nabla f|^2, and the direction across it, with eigenvalue 11. So if we draw at each point the vectors of length one in this metric, we get a circle squashed by 1/1+f21/\sqrt{1+|\nabla f|^2} along the fall line and left alone across it.

With these in hand, we compute the three independent Christoffel symbols with upper index xx. Under the symmetry exchanging variables xx and yy (and thus simultaneously exchanging EE and GG) these determine all Christoffel symbols, which then determine the geodesic equations. All Christoffel symbols have the same denominator 2D2D, so its helpful to compute this quickly here:

D=EGF2=(1+fx2)(1+fy2)(fxfy)2=1+fx2+fy2+fx2fy2fx2fy2=1+fx2+fy2\begin{align*} D &= EG-F^2\\ &=(1+f_x^2)(1+f_y^2)-(f_xf_y)^2\\ &=1+f_x^2+f_y^2+f_x^2f_y^2-f_x^2f_y^2\\ &=1+f_x^2+f_y^2 \end{align*}

Computing Γxxx\Gamma_{xx}^x

The numerator of this Christoffel symbol is GExF(2FxEy)G E_x - F \left(2 F_x - E_y\right). Simplifying with what we know,

GExF(2FxEy)=(1+fy2)2fxfxxfxfy(2fxxfy+2fxfxy2fxfxy)=(1+fy2)2fxfxxfxfy(2fxxfy)=2fxfxx+2fxfy2fxx2fxfy2fxx=2fxfxx\begin{align*} G E_x - F \left(2 F_x - E_y\right) &= \left(1 + f_y^2\right) 2 f_x f_{xx} - f_x f_y \left(2 f_{xx} f_y + 2 f_x f_{xy} - 2 f_x f_{xy}\right) \\ &= \left(1 + f_y^2\right) 2 f_x f_{xx} - f_x f_y \left(2 f_{xx} f_y\right) \\ &= 2 f_x f_{xx} + 2 f_x f_y^2 f_{xx} - 2 f_x f_y^2 f_{xx} \\ &= 2 f_x f_{xx} \end{align*}

This gives the full Christoffel symbol upon division by 2D2D:

Γxxx=2fxfxx2D=fxfxx1+fx2+fy2\Gamma_{xx}^x = \frac{2f_xf_{xx}}{2 D}=\frac{f_xf_{xx}}{1+f_x^2+f_y^2}

Computing Γxyx\Gamma_{xy}^x

The numerator of this Christoffel symbol is GEyFGxG E_y - F G_x . Simplifying with what we know,

GEyFGx=(1+fy2)2fxfxyfxfy2fyfxy=2fxfxy+2fxfy2fxy2fxfy2fxy=2fxfxy\begin{align*} G E_y - F G_x &= \left(1 + f_y^2\right) 2 f_x f_{xy} - f_x f_y \cdot 2 f_y f_{xy} \\ &= 2 f_x f_{xy} + 2 f_x f_y^2 f_{xy} - 2 f_x f_y^2 f_{xy} \\ &= 2 f_x f_{xy} \end{align*}

This gives the full Christoffel symbol upon division by 2D2D:

Γxyx=2fxfxy2D=fxfxy1+fx2+fy2\Gamma_{xy}^x = \frac{ 2 f_x f_{xy}}{2 D}=\frac{f_xf_{xy}}{1+f_x^2+f_y^2}

Computing Γxxy\Gamma_{xx}^y

The numerator of this Christoffel symbol is G(2FyGx)FGyG\left(2 F_y - G_x\right) - F G_y . Simplifying with what we know,

G(2FyGx)FGy=(1+fy2)(2fxyfy+2fxfyy2fyfxy)fxfy2fyfyy=(1+fy2)(2fxfyy)2fxfy2fyy=2fxfyy+2fxfy2fyy2fxfy2fyy=2fxfyy\begin{align*} G\left(2 F_y - G_x\right) - F G_y &= \left(1 + f_y^2\right) \left(2 f_{xy} f_y + 2 f_x f_{yy} - 2 f_y f_{xy}\right) - f_x f_y \cdot 2 f_y f_{yy} \\ &= \left(1 + f_y^2\right)(2 f_x f_{yy}) - 2 f_x f_y^2 f_{yy} \\ &= 2 f_x f_{yy} + 2 f_x f_y^2 f_{yy} - 2 f_x f_y^2 f_{yy} \\ &= 2 f_x f_{yy} \end{align*}

This gives the full Christoffel symbol upon division by 2D2D:

Γyyx=2fxfyy2D=fxfyy1+fx2+fy2\Gamma_{yy}^x = \frac{ 2 f_x f_{yy}}{2 D}=\frac{f_xf_{yy}}{1+f_x^2+f_y^2}

The Geodesic Equations

We again reference the 2D geometry ‘cheatsheet’ where the geodesics of a general metric are calculated to satisfy the following ODEs:

TheoremGeodesics

A curve γ(t)=(x(t),y(t))\gamma(t)=(x(t),y(t)) is a geodesic if it satisfies the following ODEs:

{x¨+x˙2Γxxx+2x˙y˙Γxyx+y˙2Γyyx=0y¨+x˙2Γxxy+2x˙y˙Γxyy+y˙2Γyyy=0}\left\{\begin{matrix} \ddot{x} + \dot{x}^2 \Gamma_{xx}^x + 2\dot{x} \dot{y} \Gamma_{xy}^x + \dot{y}^2 \Gamma_{yy}^x=0\\ \ddot{y} + \dot{x}^2 \Gamma_{xx}^y + 2\dot{x} \dot{y} \Gamma_{xy}^y + \dot{y}^2 \Gamma_{yy}^y =0 \end{matrix}\right\}

With our calculation of the Christoffel symbols Γxxx,Γxyx\Gamma_{xx}^x,\Gamma_{xy}^x and Γyyx\Gamma_{yy}^x, we have everything to write down the first equation. Noting that all share the same denominator 1+fx2+fy21+f_x^2+f_y^2 we combine fractions to

x¨+x˙2(fxfxx)+2x˙y˙(fxfxy)+y˙2(fxfyy)1+fx2+fy2=0\ddot{x}+\frac{\dot{x}^2\left(f_xf_{xx}\right)+ 2\dot{x} \dot{y} \left(f_xf_{xy}\right) + \dot{y}^2 \left(f_xf_{yy}\right)}{1+f_x^2+f_y^2}=0

Each term of the fraction contains a multiple of fxf_x, so factoring this out and moving to the other side gives

x¨=fxfxxx˙2+2fxyx˙y˙+fyyy˙21+fx2+fy2\ddot{x}=-f_x\frac{f_{xx}\dot{x}^2+2f_{xy}\dot{x}\dot{y}+f_{yy}\dot{y}^2}{1+f_x^2+f_y^2}

This entire situation is symmetric under the exchanges of xx and yy, and performing this swap gives the second equation. Of interest is that the entire fraction is itself invariant under this interchange, so the only changes are to the second derivative and the single-partial-derivative prefactor.

y¨=fyfxxx˙2+2fxyx˙y˙+fyyy˙21+fx2+fy2\ddot{y}=-f_y\frac{f_{xx}\dot{x}^2+2f_{xy}\dot{x}\dot{y}+f_{yy}\dot{y}^2}{1+f_x^2+f_y^2}

Integrating this pair from a point, in every direction at once, gives a fan of geodesics. Everything below happens in the coordinate plane — these are curves (x(t),y(t))(x(t),y(t)), and no surface has been drawn:

They bend, and they bend with no force in sight. All of that turning is the Christoffel symbols: it is what the coordinates cost, not anything the curves are doing. Evaluating the very same solutions at (x,y,f(x,y))(x,y,f(x,y)) puts them back where they came from, and there the bending is gone:

Sight along one of these curves edge-on and it runs straight. It never turns left or right within the surface; it only follows the surface down.

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