Warped Products by Peter Petersen De nitions We shall de ne as few concepts as possible. A tangent vector always has the local coordinate expansion a function the di erential v = dx i (v) df = f dxi We start with a Riemannian metric (M; g) : In local coordinates we obtain the metric coe cients g ij g = g ij dx i dx j = g ; dx i dx j. The Gradient The gradient of a function f : M! R is a vector eld dual to the di erential df In local coordinates this reads showing that Using g ij for the inverse of g ij we then obtain g (rf; v) = df (v) g ij dx i (rf) dx j (v) = f dxj (v) g ij dx i (rf) = f One can easily show that: Note that since we also have dx i ij f (rf) = g ij f rf = g : g ij = g dx i ; dx j = g rx i ; rx j : g ik g kj = i j g ik x l g kj + g ik g kj x l = 0; g ij g ik x l g kj = g ik g kj x l x s = g gik kl x s glj : A slightly more involved calculation shows that det g ij x s = g kl g kl x s det g ij
. The Divergence To motivate our de nition of the Hessian of a function we rst de ne the divergence of a vector eld using a dynamical approach. Speci cally, by checking how the volume form changes along the ow of the vector eld: L X dvol = div (X) dvol The volume form is a metric concept just like the metric in positively oriented local coordinates it has the form dvol = p det g kl dx ^ ^ dx n Thus L X dvol = p L X det gkl dx ^ ^ dx n = p D X det gkl dx ^ ^ dx n + p X det g kl dx ^ ^ L X dx i ^ ^ dx n = D p X det gkl p dvol det gkl + p X det g kl dx ^ ^ dd X x i ^ ^ dx n = D p X det gkl p dvol + p X det g kl dx ^ ^ d dx i (X) ^ ^ dx n det gkl = D p X det gkl p dvol + p X det g kl dx ^ ^ dxi (X) det gkl dx i ^ ^ dx n p! D X det gkl = p + dxi (X) det gkl dvol p det g kl dx i (X) = p det gkl dvol The Laplacian is now naturally de ned by.3 The Hessian f = div (rf) A similar approach can also be used to de ne the Hessian: = f gkl x k L Hessf = L rf g The local coordinate calculation is a bit worse, but yields something nice in the end Hessf ; = (L rf g) ; = L g kl f g x k x l ; kl f g x k g x l ; + = f g ij gkl x k x l + kl f g x k g lj + = g kl f g lj x k + f gkl g li x k + = f + gkl gij x l g lj g li g lj + kl f g x k x j g kl g ij x l f x k g li kl f g x k g li + gkl g lj + gkl g li f x k
This usually rewritten by introdocing the Christo el symbols of the rst second kind ijl = glj + g li g ij x l ; : k ij = glj gkl + g li Observe that this formula gives the expected answer in Cartesian coordinates, that at a critical point the Hessian does not depend on the metric. Next we should tie this in with our de nition of the Laplacian. Computing the trace in a coordinate system uses the inverse of the metric coe cients tr (Hessf) = g ij Hessf ; On the other h Examples f = div (rf) = = g ij f + gij g kl p det gkl pdet ij f gkl g gij x l g ij x l g lj g li f x k = g ij f + gij f f p det g kl + gij p xj det gkl = g ij f = g ij f = g ij f g ik g lj g kl = g ij f + gij g kl = g ij f + gij g kl = tr (Hessf) g ik g lj g kl f + g ij g kl g jl f x k + f + f (det g kl ) gij det g kl g ij x l gij x l f g gij gkl kl gkl f g jl g lj g x k gij ij x l f x k g li f x k Starting with a Riemannian metric (N; h) a warped product is de ned as a metric on I N; where I R is an open interval, by g = dr + (r) h where > 0 on all of I: One could also more generally consider (r) dr + (r) h However a change of coordinates de ned by relating the di erentials d = (r) dr allows us to rewrite this as d + (r ()) h: Important special cases are the basic product polar coordinates on (0; ) S n representing the Euclidean metric. g = dr + h dr + r ds n 3
. Conformal Representations The basic examples are still su ciently broad that we can reduce almost all problems to these cases by a simple conformal change: or dr + (r) h = () d + h ; dr = () d; (r) = () dr + (r) h = () d + h ; dr = () d; (r) = () The rst of these changes has been studied since the time of Mercator. The sphere of radius R can be written as R ds n = R dt + sin (t) ds n = dr + R sin r R ds n The conformal change envisioned by Mercator then takes the form R ds n = R () d + ds n As this means that we have () d = dt; () = sin (t) d = dt sin (t) Thus is determined by one of these obnoxious integrals one has to look up. Mercator had no access to calculus making this problem even more di cult it took hundreds of years to settle this issue! Switching the spherical metric to being conformal to the polar coordinate representation of Euclidean space took even longer probably wasn t studied much until the time of Riemann R ds n = R () d + ds n = R 4 ( + ) d + ds n Hyperbolic space is also conveniently de ned using warped products: g = dr + sinh (r) ds n ; r (0; ) it too has a number of interesting other forms related to warped products: dr + sinh (r) ds 4 n = ( ) d + ds n ; (0; ) And after using inversions to change the unit ball into a half space we can rewrite this as x dx + can R n = ds + e s can R n 4
This works as follows. Let the half space model be H = ( ; 0)R n with the metric x dx + can R n : De ne (x ; z) F (x; z) = (; 0) + jx j + jzj (x ) = + jx j + jzj ; z jx j + jzj (x ) = + r ; z r as the inversion in the sphere of radius p centered at (; 0) R R n jf (x; z)j = + 4x r = : Next we check what F does to the metric. We have F (x ) = + r ; = 4 df X + ( ) r 4x F i = zi r ; i > : k> dx r (x ) rdr (r )! df k!! + X r dz k z k rdr 4x r (r k> ) = (x ) rdr dx x r + X x k> = x dx + X! dz k k> x dx(x ) rdr X r x (x x ) rdr r dx dz k + x (x ) rdr r k> X x k> = rdr x dx + can R n + x r r rdr rdr x r rdr x r rdr = x dx + can R n dz k zk rdr r z k rdr r dz k! : This maps H to the unit ball since z k rdr r + x X k> z k rdr Showing that F also transforms the conformal unit ball model to the conformal half space model.. Singular Points The polar coordinate conformal model dr + (r) h = () d + h r 5
can be used to study smoothness of the metric as we approach a point r 0 I where (r 0 ) = 0: We assume that (r 0 ) = 0 in the reparametrization. When h = ds n we then note that smoothness on the right h side () d + ds n only depends on () being smooth. Thinking of as being Euclidean distance indicates that this is not entirely trivial. In fact we must assume that (0) > 0 Translating back to we see that (odd) (0) = 0 0 (0) = ; (even) (0) = 0: Better yet we can also prove that when h is not ds n ; then it isn t possible to remove the singularity. Assume that g is smooth at the point p M corresponding to = 0; i.e., we assume that on some neighborhood U of p we have U fpg = (0; b) N; g = d + h The = level correponds naturally to N: Now observe that each curve in M emmanating from p has a unique unit tangent vector at p also intersects N in a unique point. Thus unit vectors in T p M points in N can be identi ed by moving along curves. This gives the isometry from S n ; the unit sphere in T p M; to N: 3 Characterizations We start by o ering yet another version of the warped product representation. Rather than using r ; the goal is to use only one function f: Starting with a warped product we construct the function f = R dr on M: Since dr + (r) h df = dr we immediately see that The gradient of f is dr + (r) h = (r) df + (r) h rf = rr = r so the Hessian becomes (Hessf) (X; Y ) = L g (X; Y ) r = L g (X; Y ) + r (D X) g (rr; Y ) + (D Y ) g (rr; X) = L g r = L g r (X; Y ) + 0 (D X r) g (rr; Y ) + 0 (D Y r) g (rr; X) (X; Y ) + 0 dr (X; Y ) 6
This is further reduced Hessf = L g + 0 dr r = L r dr + (r) h + 0 dr = D r = 0 h + 0 dr = 0 g (r) h + 0 dr In other words: it is possible to nd a function f whose Hessian is conformal to the metric. Using 0 = d dr = d df df dr = d df = d jrfj df we have obtained a representation that only depends on f jrfj g = jrfj df + jrfj h; Hessf = d jrfj g df Before stating the main theorem we need to establish a general result. Lemma If f is a smooth function on a Riemannian manifold, then Hessf (rf; X) = D X jrfj : Proof. At points where rf vanishes this is obvious. At other points we can the assume that f = x is the rst coordinate in a coordiate system. Then rf = g i With that informaltion the calculation becomes Hessf rf; jrfj = g i g ij g j = g = g i Hessx ; = gi g kl gij x l The formula now follows by exping X in coordinates. We can now state prove our main result. g lj g lj g li x x k = gi g l gij g li x l = gi g l g li + gi g l g ij x l gi g l g lj = g + gi g l g ij x l gl g i g ij x l = jrfj 7
Theorem If there are smooth functions f; on a Riemannian manifold so that Hessf = g then the Riemannian structure is a warped product around any point where rf 6= 0: Proof. We rst need to show that only depends on f: This uses the previous lemma: D X jrfj = Hessf (rf; X) = g (rf; X) So we see that jrfj is locally constant on level sets of f by taking X? rf: Thus we can locally assume that jrfj = b (f) : If we let X = rf; then we obtain or We now claim that b = D rf b = g (rb; rf) = db df b = db df g = b df + bh where h is de ned such that b (c) h is simply the restriction of g to the level set f = c: In particular, we have that gj p = b df + bh j p ; when f (p) = c Next we observe that L rf g = g L rf b df + bh = db df b df + bh = b df + bh Thus g b df +bh both solve the same di erential equation with the same initial values at f = c; showing that the two metrics must agree. We can also hle singular points if we assume they are isolated. Corollary 3 If there are smooth functions f; on a Riemannian manifold so that Hessf = g; rfj p = 0; (p) 6= 0 then the Riemannian structure is a warped product around p: Proof. As we ve assumed the Hessian to be nondegenerate at p it follows that there are coordinates in a neighborhood around p such that f x ; :::; x n = f (p) + (p) x + + (x n ) : Thus the level sets near p are spheres, we know as before that g = b df + bh; b (f) = jrfj Since g is a smooth metric at p we can then also conclude that h = ds n : The nal goal is to characterize the three stard geometries at least locally. 8
Corollary 4 Assume that there is a function f on a Riemannian manifold such that () If f (p) = 0; rfj p = 0: Hessf = g then the metric is Euclidean in a neighborhood of p: () If Hessf = ( f) g then the metric is the unit sphere metric in a neighborhood of p: (3) If Hessf = ( + f) g then the metric is hyperbolic in a neighborhood of p: Proof. Note that in each of three cases is already a function of f: Thus we can nd by solving the initial value problem b (f) = jrfj db df = (f) ; b (0) = 0 The solution is obviously unique so it actually only remains to be checked that the three stard geometries have functions with the described properties. This works as follows using the stard warped product representations: Euclidean space The unit sphere Hyperbolic space f (r) = r f (r) = cos r dr + r ds n ; dr + sin rds n ; dr + sinh rds n ; f (r) = + cosh r In all three cases r = 0 corresponds to the point p: 4 Generalizations In view of what we saw above it is interesting to investigate what generalizations are possible. Transnormal functions simply satisfy: jrfj = b (f) Note that the equation Hessf (rf; X) = D X jrfj shows that this is locally equivalent to saying that rf is an eigenvector for Hessf. Wang proved that such functions further have the property that the only possible critical values are the maximum minimum 9
values, moreover the corresponding max/min level sets are submanifolds. Such functions can be reparametrized as f = f (r) ; where r is a distance function, i.e., jrrj : Note that it is easy to de ne r as a function of f Z df r = p b It is important that b is di erentiable for these properties to hold as any strictly increasing function on R is transnormal for some continuous b: In this case there can certainly be any number of critical points that are merely in ection points. Isoparametric functions are transnormal functions where in addition f = a (f) : These functions in addition have the property that the min/max sets for r are minimal submanifolds. These functions were introduced studied by E. Cartan. Even stricter conditions would be that the function is transnormal the eigenvalues of Hessf are functions of f: A very good general model case for this situation is a cohomogeneity manifold with f = f (r) r : M! M=G = I R: For this class of functions it is a possibility that the eigenspace distributions for the Hessian are not integrable. Speci cally select M 4 ; g with U () symmetry, e.g., C bundles over S ; R 4 with the Taub-NUT metric, or CP. The generic isotropy of U () is U () : The subspace corresponding to U () is then a -dim distribution that must lie inside an eigenspace for Hessf: This distribution is however not integrable as it corresponds to the horizontal space for the Hopf bration. As long as the metric isn t invariant under the larger group SO (4) the Hopf ber direction, i.e., the direction tangent to the orbits but perpendicular to this distribution, will generically correspond to an eigenvector with a di erent eigenvalue. Finally one can assume that the function is transnormal satis es R (X; Y ) rf = 0 for all X; Y? rf: The curvature condition is automatic in the three model cases with constant curvature so it doesn t imply that f is isoparametric. But it is quite restrictive as it forces eigenspaces to be integrable distributions. It is also equivalent to saying that Hess(r) is a Codazzi tensor where r is the distance function so that f = f (r) : A good model case is a doubly warped product M = I N N g = dr + (r) h + (r) h ; f = f (r) : Note that the functions f coming from warped products satisfy all of the conditions mentioned above. Another interesting observation that might aid in a few calculations is that a general warped product g B + g F = g B + g F on a Riemannian submersion M! B; where g B is a metric on B g F represents the family of metrics on the bers, ; functions on B: Here the metric g B + g F is again a Riemannian submersion with the conformally changed metric g B on the base. Thus general warped products are always conformally changed on the total space of a Riemannian submersion, where the conformal factor has a gradient that is basic horizontal. 5 Geodesics We obtain formulas for geodesics on warped products that mirror those of geodesics on surfaces of revolution. 0