ON CHRISTOFFEL SYMBOLS AND TEOREMA EGREGIUM
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1 ON CHRISTOFFEL SYMBOLS AND TEOREMA EGREGIUM LISBETH FAJSTRUP 1. CHRISTOFFEL SYMBOLS Thisisasectiononatechnicaldevicewhichisindispensableboth in the proof of Gauss Theorema egregium and when handling geodesics and geodesic curvature. To compare with C. Bär: Elermentary Differential geometry, notice that a chart is denoted x(u, v), x:u R 3.Theentriesg ij inthegrammatrixofthefirstfundamentalformaredenotede =g 11.F =g 1 =g 1 andg =g 1.1.Definition. Inthefollowing,wefixaparametrizationx:U V SforthesurfaceS.Ateverypointp =x(q),thethreevectors (q), (q)andn(p)forma(moving)basisforr 3.Letusexpress thedoublederivativesu (q),v (q)andv (q)aslinearcombinationsofthesebasisvectors(weomitthepsandqsandregardthe derivatives etc. as vector fields): u = Γ Γ 11 + Γ 3 11 N v = Γ Γ 1 + Γ 3 1 N (1) v = Γ 1 + Γ + Γ 3 N. withunknownrealcoefficientfunctions Γ k ij : U R.Sincethe vectorfields, andnarelinearlyindependet,thesystem(1)has auniquesolutionateverypointp S. 1.. The metric determines the Christoffel symbols. We know alreadythecoefficients Γ 3 ij ofn: Lemma1.1.Thecoefficients Γ 3 ij coincidewiththecoefficientse,fandgof the second fundamental form: Γ 3 11 =e, Γ3 1 =f, Γ3 =g. 1
2 LISBETH FAJSTRUP Bevis. Perform the dot-product with the normal vector N to the three equations in(1). The result is: e =u N = Γ 3 11 f =v N = Γ 3 1 g =v N = Γ 3. Hence, we can reformulate(1) in matrix language as follows: Let Γdenotethe3 3-matrix Γ = Γ1 11 Γ 1 1 Γ 1 Γ 11 Γ 1 Γ e f g Then, the following matrix equation holds:. () [u v v ] = [ N]Γ. Multiplyingthisequation(fromtheleft)withthe3by3-matrix [,,N] T, we obtain: u v v u v v N u N v N v and hence (3) Γ = 1 EG F We conclude: = N E F 0 F G G F 0 F E EG F [u v v ] = Γ, N u v v u v v N u N v N v Proposition1..TheChristoffel-symbols Γij k,1 i,j,k areintrinsic entities, i.e., they depend only on the metric(first fundamental form) along the surface S. [ N]Γ = Bevis.SincetheChristoffelsymbolsarethefirsttworowsof Γ,and sincetheentries(13)and(3)intheinversematrixare0,itisenough toshowthatthefirsttworowsofthematrix x u u v v u v v N u N v N v.
3 ON CHRISTOFFEL SYMBOLS AND TEOREMA EGREGIUM 3 areexpressibleintermsofthecoefficientse,f,gofthefirstfundamentalform.tothisend,weformthepartialderivativesofthefunctionse,f,gwithrespecttothevariablesuandvandobtainthe following six equations: E u = u,e v = v,f u = u + v,f v = v + v, G u = v,g v = v. This system of equations has the following solutions: u = E u, v = E v, v =F v G u, u =F u E v, v = G u, v = G v.. TEOREMA EGREGIUM The Gauss curvature is defined in terms of the second fundamentalform,anditiscertainlynottrue,thatthecoefficientsh ij canbe expressedbyg ij.norisitthecasethattheprincipalcurvaturesare intrinsic. However, the remarkable theorem, Teorema Egregium says, thatthegausscurvatureisinfactintrinsic.sotheproduct κ 1 κ isintrinsic,evenif κ 1 and κ arenot. Theorem.1.LetS 1 ands beregularsurfacesandlet φ:s 1 S be a local isometry. Then the Gauss curvature is preserved; K(p) = K(φ(p)) forallp S 1. Bevis.Letp S 1.ByBärexercise4.3,itsufficestoseeforachart (U,F,V),thatK(p)maybeexpressedviathekoefficientsofthefirst fundamental form. From Corollary.3, we get d p N( ) d p N( ) =K(p). Onbothsidesofthisequation,taketheinnerproductwith toget K(g 11 g g 1 ) = (d pn( ) )(d p N( ) ) (d p N( ) )(d p N( ) ) WeusedLagrangesformula.4ontherighthandside.Thenormal vector is N x(u,v) = (u,v)
4 4 LISBETH FAJSTRUP and (u,v) = (g 11 g g 1 )(u,v)so d p N( ) = N x u and symmetrically: d p N( ) = N x v = u + x vu = v + x vv + u ( 1 )( ) + v ( 1 )( ) Substitute these in the equation for Gauss curvature to get: K(g 11 g g 1 ) = ((u ) )(( v ) ) (( v ) )((v ) ) Thedeterminantofa3x3matrixisexpressedviathecrossproduct andinnerproductofrows(orcolumns)andherethisgives =det(u )det( v ) det( v )det(v ) wherethevectorsx ij arecolumns.interchangecolumnsandtransposetoget =det((u ) x u v ) det(( v ) x uv ) =det((u ) v ) det((v ) v =det u v u u v g 11 g 1 v g 1 g det x uv v v v v g 11 g 1. v g 1 g We have already expresses the entrances of these matrices in terms of the first fundamental form(via the Christoffel symbols), except the upper left entrances. Since we take the determinant, it suffices to expressthedifferenceu v v v (Why?). Since v and v havealreadybeenexpressedintermsofthe first fundamental form, there are expressions of their derivatives. Now u ( v ) =u v + vu and v ( v ) =v v + vv.thechartxisc,sovu =vv.subtractthetwoexpressionsandconcludethatx uu v v v isexpressiblebyg ij and their derivatives. )
5 ON CHRISTOFFEL SYMBOLS AND TEOREMA EGREGIUM 5 Lemma..Let [v 1 v ]bea3xmatrixwithcolumnsv i.letabeax matrixanddefine [w 1 w ] = [v 1 v ]A.Thenthevectorproductsatisfies w 1 w =v 1 v det(a) Bevis. Left to the reader. Corollary.3.d p N( ) d p N( ) =K(p) Bevis.LetWbethematrixoftheWeingartenmapwrt.,.Then d p N( ) =w 11 +w 1 and d p N( ) =w 1 +w.hence thelemmaapplieswitha=w,andk(p) =det(w) Lemma.4.Lagrange sformula:forcrossproductofvectorsin R 3,we have (v 1 v ) (w 1 w ) = (v 1 w 1 )(v w ) (v 1 w )(v w 1 )
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