Math 396. Metrc tensor on hypersurfaces 1. Motvaton Let U R n be a non-empty open subset and f : U R a C -functon. Let Γ U R be the graph of f. The closed subset Γ n U R proects homeomorphcally onto U wth nverse (x 1,..., x n ) (x 1,..., x n, f(x 1,..., x n )) that s a smooth mappng from U to U R. In fact, ths latter mappng s trvally an mmerson, so n ths way ts mage Γ acqures a (necessarly unque) structure of closed smooth submanfold of U R. In partcular, the coordnate functons x 1,..., x n on R n+1 restrct to a global C coordnate system on Γ. In what follows, we let x = x Γ for 1 n. Example 1.1. The most classcal example s n = 2, whch are the old z = f(x, y) parametrc surfaces n R 3 over open parts of the xy-plane. Whenever meetng a new concept n dfferental geometry, t s always a good dea to fgure out how t works n the case of such surfaces n R 3, or more generally for hypersurface graphs Γ n R n+1 as above (as the mmerson theorem tells us that all hypersurfaces n a manfold locally have ths form). Usng the standard Remannan metrc on U R R n+1, we wsh to compute the nduced metrc on Γ, whch s to say the nduced nner product on each T p (Γ) T p (R n+1 ) for p Γ. We want to express ths metrc tensor n terms of the coordnate system {x 1,..., x n} on Γ. The metrc tensor at p Γ s x p, x p p dx (p) dx (p),, wth the -coeffcent computed as an nner product n T p (R n+1 ) (snce the nduced metrc on Γ s gven pontwse by the ncluson of T p (Γ) nto T p (R n+1 ) for each p Γ). Snce the x p s are an orthonormal bass of T p (R n+1 ), the problem s therefore largely that of fgurng out the mage of each x p T p (Γ) n T p (R n+1 ) as a lnear combnaton n { x1 p,..., xn+1 p }. An easy begnner s mstake s to thnk that snce x = x Γ, x p T p (Γ) vewed n T p (R n+1 ) should equal x p for each n. That ths s nonsense s easy to recognze geometrcally as follows: the x -coordnate lne through p n R n+1 s usually not tangent to Γ at p and so ts tangent lne at p s generally not contaned n the tangent hyperplane T p (Γ) (so the nonzero tangent vector x p n ths lne cannot belong to the hyperplane T p (Γ) T p (R n+1 )). To better apprecate what s happenng, consder a smple example: Example 1.2. Let p = (a, b, c) be a pont on the open upper unt hemsphere H = {z = 1 x 2 y 2 } n R 3 (wth (x, y) nsde the open unt dsc centered at the orgn n R 2 ). Geometrc ntuton suggests that away from the ntersecton of H wth the yz-plane, the vector x p T p (R 3 ) s not tangent to any curve n the sphere at p and so t certanly cannot be a velocty vector along a coordnate lne for a local coordnate system on H near p. In partcular, t generally cannot equal x p when the latter s vewed n T p (R 3 ). What s the mage of x p T p (H) n T p (R 3 ) as a lnear combnaton of x p, y p, z p? There s a smple geometrc procedure to fgure ths out. On any manfold, the partal dervatve operator wth respect to a coordnate parameter s ust the velocty vector feld for the coordnate lne wth tme gven by the chosen coordnate. Thus, we have to take the x-coordnate lne y = b n passng through (a, b), consder t as parameterzed va x, and map ths parametrc curve 1
2 nto H usng the {x, y} parameterzaton of H. That s, we take the embedded parametrc curve σ(t) = (t, b, 1 t 2 b 2 ) n H for t near a, and ts velocty vector n T σ(t) (H) T σ(t) (R 3 ) at tme t s σ (t) = x σ(t) 2t(1 t 2 b 2 ) 1/2 z σ(t). At t = a, snce c = c 2 (as c > 0) we get x p = x p 2(a/c) z p n T p (R 3 ). Note that ths s equal to x p f and only f a = 0, whch s to say along the yz-plane (exactly as we see physcally). At all other ponts of H, the vectors x p and x p n T p (R 3 ) are lnearly ndependent. 2. Some tangent vectors and normal vectors The precedng example wth the sphere suggests a general method for computng the mage of x p T p (Γ) n T p (R n+1 ) n terms of the bass of x p s. But before we do that, we want to address another possble begnner s mstake, whch s to thnk that maybe x p should be the orthogonal proecton of x p onto T p (Γ). A moment of geometrc thought shows that ths s generally false, for two reasons: () the problem of computng the mage of x p n T p (R n+1 ) has absolutely nothng to do wth Remannan metrcs, and so the answer cannot possbly nvolve the crutch of a metrc (as s mplct wth notons such as orthogonal proecton ), and () vsualzng the case of the sphere shows that the angle between x p and x p n T p (R n+1 ) can be pretty much anythng n (0, π/2), so agan orthogonal proecton should have nothng to do wth the answer. We shall sort out the exact relatonshp wth orthogonal proecton from T p (R n+1 ) onto T p (Γ), as an applcaton of solvng our frst problem (to compute the mage of each x p n T p (R n+1 ) n terms of the bass of x p s). Let us now gve the formula for x p n T p (R n+1 ): x p = x p + x f(p) xn+1 p. To ustfy ths, consder the parametrc path σ (t) n Γ whch (n the x -coordnate system) has th coordnate fxed at p for, and has th coordnate p + t. In R n+1 ths s the path wth the same frst n coordnates, and wth last coordnate t f(p 1,..., p + t,..., p n ). Thus, we readly compute σ (t) much as we dd for the hemsphere example above, and for t = 0 ths gves the asserted formula. (In the case of the upper unt hemsphere n R 3, we take f(x, y) = 1 x 2 y 2 and recover the formula found n Example 1.2.) Theorem 2.1. The nonzero vector N p = n x f(p) x p xn+1 p T p (U R) = T p (R n+1 ) (.e., (..., x f(p),..., xn f(p), 1) f we dentfy T p (U R) wth R n+1 n the usual way) s a non-zero normal vector to T p (Γ), and the vectors N p, x1 p,..., xn p T p (U R) are a bass. Proof. A normal vector to T p (Γ) n T p (R n+1 ) s one that s normal to each of the n vectors x p (as these span T p (Γ)). The mages of these n vectors n T p (R n+1 ) are gven above, and N p as defned n the theorem s clearly perpendcular to all of them. Snce N p has nonzero coeffcent for xn+1 p, the fnal part of the theorem s clear. Snce orthogonal proecton to T p (Γ) klls N p, t follows from Theorem 2.1 that the orthogonal proectons {v (p)} of the x p s ( n) are a bass of T p (Γ) (so n partcular, they are all nonzero!).
3 Explctly, the orthogonal proecton v (p) s gven by v (p) = x p x p, N p p N p, N p p N p nsde of T p (R n+1 ). What s ths n terms of the bass of x p s for T p (Γ)? We know that v (p) must be a unque lnear combnaton of the x p s, and n T p (R n+1 ) the expanson of x p nvolves x p wth a coeffcent of 1 and no xk p s for k n wth k. Hence, by consderng coeffcents of x p s for n the only possblty s v (p) = x f(p) x f(p) 1 + r n ( x r f(p)) 2 x p + 1 + r,r n ( x r f(p)) 2 1 + r n ( x r f(p)) 2 x p. (The reader may verfy as a safety check that the coeffcents for xn+1 p mplct on both sdes are ndeed equal, as they must be.) Ths s qute dfferent from x p n general! Usng the determnaton of the mage of x p T p (Γ) n T p (R n+1 ) as a lnear combnaton of the x p s, we also see mmedately that x p, x p p = δ + x f(p) x f(p), where δ = 1 when = and δ = 0 when. Hence, we arrve at the desred formula for the metrc tensor of Γ n the x -coordnates: n + ( x f(p)) =1(1 2 ) dx dx + ( x f(p) x f(p)) dx dx. Example 2.2. Consder the classcal case of a surface z = f(x, y) n R 3. In ths case, the metrc tensor on the surface n {x, y} coordnates s (1 + f 2 x) dx 2 + (1 + f 2 y ) dy 2 + f x f y dx dy + f x f y dy dx, where we ndulge n the standard shorthand f x = x f and f y = y f. 3. Surfaces of revoluton In addton to the graph surfaces z = f(x, y), another nterestng class of surfaces n R 3 s the surfaces of revoluton. We shall focus on a specal subclass. Let I (0, ) be a nontrval nterval and let f : I R be a postve smooth functon such that f s nowhere zero. Let S I R 2 R 3 be the surface of revoluton obtaned from revolvng the graph of f (n I R R 2, the xy-plane) about the x-axs. Ths surface does not touch the x-axs (snce f s postve). By Exercse 3 n Homework 4 (whch used an nterval (a, b), but works the same wth (a, b) replaced by an arbtrary non-trval n R), S s a smooth submanfold of R 3 f I s open, and f I has endponts then S s a smooth submanfold wth boundary n R 3. Snce f has constant sgn (as t s non-vanshng and contnuous over the nterval I) the growth of f s strctly monotone. Hence, the surface ether unformly approaches the central axs or dverges from t; t does not wggle. Hence, the surface S naturally parameterzed by polar coordnates n the yz-plane. More rgorously, by the nverse functon theorem from calculus (and a moment of thought for the endponts) f must be a C somorphsm onto a nontrval nterval J = f(i) n (0, ) and so t has an nverse functon g : J I R. (Classcally, g s x as a functon of y.) We then have a smooth map J S 1 R 3 gven by h : (r, θ) (g(r), r cos θ, r sn θ)
4 that s a becton onto S. It must therefore be a C somorphsm. Ths can also be verfed by hand, as follows. The tangent mappngs for h are nectve (look at the y and z parts), so as maps nto the 2-dmensonal tangent spaces of the embedded submanfold S the tangent mappngs of the C map J S 1 S are nectve. Thus, these latter tangent maps are somorphsms for dmenson reasons, so J S 1 S s a bectve local C somorphsm. It s therefore a C somorphsm. What s the metrc tensor on S n terms of ts {r, θ} parameterzaton? The smplest way to solve ths s to recognze that the parameterzaton of S s better adapted to cylndrcal coordnates (wth x-axs as the axs of symmetry) rather than rectangular coordnates. Thus, we wll fnd our task easer f we pull back the standard metrc tensor from R 3 when t s expressed n cylndrcal coordnates rather than n rectangular coordnates. Let us therefore dgress to compute the tensor on R 3 n such coordnates, adapted to the x-axs as the axs of symmetry ( tradtonally t s the z-axs that s taken as the axs of symmetry, but the dstncton s a trvalty): Lemma 3.1. The C mappng R (0, ) S 1 R 3 defned by (x, r, θ) (x, r cos θ, r sn θ) s a C somorphsm onto the complement U of the x-axs n R 3, and on U the standard metrc tensor of R 3 has restrcton dx 2 + dr 2 + r 2 dθ 2. The absence of cross-terms n ths formula reflects the mutual orthogonalty (n tangent spaces to R 3 ) of the tangent lnes along each of the cylndrcal coordnate drectons through any ponts of R 3 away from the axs of symmetry. Proof. Polar coordnates provde the C somorphsm (0, ) S 1 R 2 {(0, 0)}, so takng the product aganst R on both sdes gves the frst part of the lemma (as U = R (R 2 {(0, 0)})). On a product of Remannan manfolds, endowed wth the product Remannan metrc (va the dentfcaton of the tangent bundle to a product manfold wth the drect sum of the pullbacks of the tangent bundles of the factors), the metrc tensor s bult fberwse as the orthogonal sum of the nner products on the tangent spaces to the factor manfolds; the same goes for manfolds wth corners. Snce R 3 equpped wth ts standard metrc tensor s the product of the Remannan manfolds R (the x-axs) and R 2 (the yz-plane) equpped wth ther standard metrc tensors, we get the asserted formula for the metrc tensor n cylndrcal coordnates by usng the result from the theory of polar coordnates that on R 2 {(0, 0)} the restrcton of the standard metrc tensor on R 2 s dr 2 + r 2 dθ 2. Returnng to the stuaton wth our surface of revoluton the nduced metrc tensor s the pullback of dx 2 +dr 2 +r 2 dθ 2 under the mappng (r, θ) (g(r), r, θ) n terms of the cylndrcal coordnate system on R (R 2 {(0, 0)}). In other words, the metrc tensor on S (dentfed wth J S 1 ) s ds 2 = d(g(r)) 2 + dr 2 + r 2 dθ 2 = (g (r)dr) 2 + dr 2 + r 2 dθ 2 = (1 + g (r) 2 )dr 2 + r 2 dθ 2. Recall that g : J I R s the nverse functon to f : I J (0, ) (so classcally one would wrte (1 + x (r) 2 ) for the coeffcent of dr 2 n ths formula). Ths s the most general formula for the nduced metrc tensor on a surface of revoluton about the x-axs when the surface admts a parameterzaton by polar coordnates n the yz-plane (.e., the graph beng rotated s strctly monotone). Example 3.2. A very famous example of such a surface of revoluton s the Beltram surface that s gven explctly n nverse functon form by x(y) = ( a 2 y 2 + a 2 log a + ) a 2 y 2 a a 2 y 2
for 0 < y < a. Ths s an antdervatve to a 2 y 2 /y < 0 and t approaches as y 0 + and approach 0 as y a, so the graph n the frst quadrant for the nverse y as a functon of x s strctly decreasng from the pont (0, a) on the postve y-axs asymptotcally down toward the x-axs. In partcular, t satsfes the above requrements. Ths surface s nterestng because (n terms of concepts to be ntroduced later) t has constant negatve curvature (equal to 1/a 2 ). Let us compute the metrc tensor on the Beltram surface n the polar coordnate system from the yz-plane. Snce x (r) = a 2 r 2 /r, clearly 1 + x (r) 2 = a 2 /r 2. Thus, by the general formula gven above, the metrc tensor s (a 2 /r 2 )dr 2 + r 2 dθ 2. 4. The torus We conclude by studyng a very classcal example, the donought (or torus) T R 3 wth nner radus r a and outer radus r + a, 0 < a < r. Explctly, T s the mage of the smooth closed embeddng S 1 S 1 R 3 defned by (θ, ψ) ((a + r cos θ) cos ψ, (a + r cos θ) sn ψ, r sn θ). Geometrcally, θ s the angle measure for the crcles of radus r that go through the hole and ψ s the angle measure for the other famly of crcles that have rad varyng from r a to r + a. There s a global trvalzaton of the tangent bundle specfed by the ordered par of vector felds { θ, ψ } (that are globally well-defned, even though θ and ψ are not), and under the embeddng of S 1 S 1 nto R 3 and hence of T (S 1 S 1 ) nto (T (R 3 )) = (S 1 S 1 ) R 3 these go over to the vector felds θ = ( r sn θ cos ψ, r sn θ sn ψ, r cos θ), ψ = ( (a + r cos θ) sn ψ, (a + r cos θ) cos ψ, 0). That s, for a pont ξ S 1 S 1 wth angle coordnates (θ 0, ψ 0 ), the necton of T ξ (S 1 S 1 ) nto T (ξ) (R 3 ) satsfes θ ξ = r sn θ 0 cos ψ 0 x (ξ) r sn θ 0 sn ψ 0 y (ξ) + r cos θ 0 z (ξ) and ψ ξ = (a + r cos θ 0 ) sn ψ 0 x (ξ) (a + r cos θ 0 ) cos ψ 0 y (ξ). The lack of a z -term n ths fnal expresson s suggested by the pcture: the coordnate lnes for ψ are parallel to the xy-plane. As a pcture suggests and a calculaton wth the orthonormal frame { x, y, z } at (ξ) confrms, the vector felds θ and ψ are parwse orthogonal wth respect to the nduced metrc. By drect calculaton, the self nner-products are r 2 for θ ξ and (a + r cos θ 0 ) 2 for ψ ξ. Hence, the nduced metrc tensor s r 2 dθ 2 + (a + r cos θ) 2 dψ 2. The constancy of the frst coeffcent and the varyng nature of the second coeffcent reflect some basc geometrc propertes of the surface of the torus as t sts n R 3. That s, ths geometry s very senstve to the chosen embeddng, nsofar as ths s what determned the metrc tensor. The nherent asymmetry n the roles of θ and ψ s not apparent when consderng the bare manfold S 1 S 1, but the chosen embeddng nto R 3 sngles out dfferent roles for these two factors and so leads to the asymmetrc nature of ther appearance n the metrc tensor. Can you see a geometrc explanaton (wthn R 3 ) for the varyng length of ψ as we vary θ, but the constant length of θ as we wander across the surface? 5