[Differential Geometry] Comparing the Christoffel Symbols for two Parametrisations of the Plane

We can parametrise an open set of the plane using Cartesian coordinates, φ:UR2R3\varphi: U \subset \mathbb{R}^2\rightarrow \mathbb{R}^3, where φ(u,v)=(u,v,0)\varphi(u,v) = (u,v,0). Then φu=(1,0,0)\varphi_u = (1,0,0) and φv=(0,1,0).\varphi_v = (0,1,0).     φuu=(0,0,0)=φuv=φvv.\implies \varphi_{uu} = (0,0,0) = \varphi_{uv} = \varphi_{vv}.

So Γijk=0,i,j,k=1,2\Gamma_{ij}^k =0, \forall i,j,k =1,2.

How about in polar coordinates? We can parametrise an open set of the plane with φ:VR2R3\varphi: V \subset \mathbb{R}^2 \rightarrow \mathbb{R}^3 where φ(u,v)=(ucosv,usinv,0)\varphi(u,v) = (u\cos v, u\sin v, 0) this time.

Then φu=(cosv,sinv,0),\varphi_u = (\cos v, \sin v, 0), and φv=(usinv,ucosv,0),\varphi_v = (-u \sin v, u\cos v, 0), and the coefficients of the first fundamental form are E=cos2v+sin2v=1,E= \cos^2v+sin^2v=1, F=usinvcosv+usinvcosv=0,F= -u\sin v\cos v+ u\sin v\cos v= 0, G=u2.G=u^2. Considering the trihedron given by the vectors Xu,Xv,NX_u, X_v, N, taking inner product with XuX_u, XvX_v, we have the following system of equations:

(for reference, see fifth note on computing Gaussian curvature)

Γ111E+Γ112F=12Eu,\Gamma^1_{11}E + \Gamma^2_{11}F = \frac{1}{2} E_u , Γ111F+Γ112G=Fu12Ev.\Gamma^1_{11}F + \Gamma^2_{11}G = F_u-\frac{1}{2}E_v. Γ121E+Γ122F=12Ev,\Gamma^1_{12}E + \Gamma^2_{12}F = \frac{1}{2}E_v, Γ121F+Γ122G=12Gu.\Gamma^1_{12}F+ \Gamma^2_{12}G = \frac{1}{2} G_u. Γ221E+Γ222F=Fv12Gu,\Gamma^1_{22}E + \Gamma^2_{22}F = F_v - \frac{1}{2}G_u, Γ221F+Γ222G=12Gv.\Gamma^1_{22}F + \Gamma^2_{22}G = \frac{1}{2}G_v. To find the Christoffel symbols, we let

[EFFG][Γ111Γ121Γ221Γ112Γ122Γ222]=[12Eu12EvFv12GuFu12Ev12Gu12Gv]. \begin{bmatrix} E & F \\ \\ F & G \\ \end{bmatrix} \begin{bmatrix} \Gamma^1_{11} & \Gamma^1_{12} & \Gamma^1_{22}\\ \\ \Gamma^2_{11} & \Gamma^2_{12} &\Gamma^2_{22}\\ \end{bmatrix} = \begin{bmatrix} \frac{1}{2} E_u & \frac{1}{2}E_v & F_v - \frac{1}{2}G_u\\ \\ F_u-\frac{1}{2}E_v & \frac{1}{2} G_u & \frac{1}{2}G_v\\ \end{bmatrix}.

Hence, [Γ111Γ121Γ221Γ112Γ122Γ222]=[EFFG]1[12Eu12EvFv12GuFu12Ev12Gu12Gv]. \begin{bmatrix} \Gamma^1_{11} & \Gamma^1_{12} & \Gamma^1_{22}\\ \\ \Gamma^2_{11} & \Gamma^2_{12} &\Gamma^2_{22}\\ \end{bmatrix} = \begin{bmatrix} E & F \\ \\ F & G\\ \end{bmatrix}^{-1} \begin{bmatrix} \frac{1}{2} E_u & \frac{1}{2}E_v & F_v - \frac{1}{2}G_u\\ \\ F_u-\frac{1}{2}E_v & \frac{1}{2} G_u & \frac{1}{2}G_v\\ \end{bmatrix}.

And since Eu=0,Ev=0,Fu=0,Fv=0,Gu=2u,Gv=0E_u =0, E_v =0, F_u =0, F_v=0, G_u=2u, G_v=0, (fromE=1,F=0,G=u2)(from E=1, F=0, G=u^2) RHS=[ 10 0u2 ]1[ 00u  0u0 ]RHS = \begin{bmatrix} ~1 & 0 ~ \\ \\ 0 & u^2 ~\\ \end{bmatrix}^{-1} \cdot \begin{bmatrix} ~0 & 0 & -u ~\\ \\ ~0 & u & 0 ~\\ \end{bmatrix}

=[ 00u  01u0 ].= \begin{bmatrix} ~0 & 0 & -u ~\\ \\ ~0 & \frac{1}{u} & 0 ~\\ \end{bmatrix}.

Therefore, Γ111=0,Γ121=0,Γ222=u,Γ112=0,Γ122=1u,Γ222=0\Gamma_{11}^1=0, \Gamma_{12}^1=0, \Gamma_{22}^2=-u, \Gamma_{11}^2=0, \Gamma_{12}^2= \frac{1}{u}, \Gamma_{22}^2=0.

In fact, we can check that the Gaussian curvature of the open set of the plane is zero (K0)(K \equiv 0), by substituting the Christoffel symbols above in the formula for Gauss Curvature:

EK=(Γ122)u(Γ112)v+Γ121Γ112+Γ122Γ122Γ112Γ222Γ111Γ122. -EK = (\Gamma_{12}^2)_u - (\Gamma_{11}^2)_v + \Gamma_{12}^1\Gamma_{11}^2+\Gamma_{12}^2\Gamma_{12}^2-\Gamma_{11}^2\Gamma_{22}^2-\Gamma_{11}^1\Gamma_{12}^2 .

EK=(12GuG)u(12EvG)v14EuEGuG14EvEEvG+14EvGGvG+14GuGGuG-EK = \left(\frac{1}{2}\frac{G_u}{G}\right)_u - \left(-\frac{1}{2}\frac{E_v}{G}\right)_v - \frac{1}{4}\frac{E_u}{E}\frac{G_u}{G} - \frac{1}{4}\frac{E_v}{E}\frac{E_v}{G} +\frac{1}{4}\frac{E_v}{G}\frac{G_v}{G} +\frac{1}{4}\frac{G_u}{G}\frac{G_u}{G}     K=(1u)u0+0+(1u)(1u)00 \implies -K = \left(\frac{1}{u}\right)_u -0+0+ \left(\frac{1}{u}\right)\left(\frac{1}{u}\right)-0-0     K=1u2+1u2=0,  K=0 \implies -K = -\frac{1}{u^2} + \frac{1}{u^2} =0,~~\therefore K = 0

Note by Tasha Kim
1 year, 5 months ago

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