warped products - petersen

7/22/2019 Warped Products - Petersen

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WARPED PRODUCTS

PETER PETERSEN

1. Definitions

We shall dene as few concepts as possible. A tangent vector always has thelocal coordinate expansion

v = dx i (v)@

@ x iand a function the di ff erential

df = @ f @ x i dxi

We start with a Riemannian metric (M, g) . In local coordinates we obtain themetric coeffi cients gij

g = gij dx i dx j = g@

@ x i,

@ @ x jdx

i dx j

1.1. The Gradient. The gradient of a function f : M ! R is a vector eld dualto the di ff erential df g (r f, v ) = df (v)

In local coordinates this reads

gij dxi

(r f ) dxj

(v) =@ f @ x j dx

j

(v)showing that

gij dx i (r f ) =@ f @ x j

Using gij for the inverse of gij we then obtain

dx i (r f ) = gij@ f @ x j

and

r f = gij@ f @ x j

@ @ x i

.

One can easily show that:

gij = g dx i , dx j = g

rx i ,

rx j .

Note thatgik gkj = ij

implies@ gik

@ x lgkj + gik

@ gkj@ x l

= 0 ,

@ gik

@ x lgkj = gik

@ gkj@ x l

1


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WARPED PRODUCTS 2

and@ gij

@ xs= gik

@ gkl

@ xsglj .

A slightly more involved calculation shows that

@ det gij@ xs

= gkl@ gkl@ xs

det gij

1.2. The Divergence. To motivate our denition of the Hessian of a function werst dene the divergence of a vector eld using a dynamic approach. Specically,by checking how the volume form changes along the ow of the vector eld:

LX dvol = div ( X ) dvol

The volume form is a metric concept just like the metric and in positively orientedlocal coordinates it has the form

dvol = p det gkl dx1 ^ ^dxn

Thus

LX dvol = LXp det gkl dx1 ^ ^dx

n

= DXp det gkldx

1 ^ ^dxn+p det gkl Xdx

1 ^ ^LX dx i ^ ^dxn

=DX p det gkl

p det gkl dvol+p det gkl Xdx1 ^ ^dD X x

i ^ ^dxn

=DX p det gkl

p det gkl dvol + p det gkl Xdx1 ^ ^d dx i (X ) ^ ^dxn

=DX p det gkl

p det gkl dvol + p det gkl Xdx1 ^ ^

@ dx i (X )@ x i

dx i ^ ^dxn

=DX p det gkl

p det gkl +@ dx i (X )

@ x i !dvol=

1p det gkl

@ p det gkl dx i (X )@ x i

dvol

The Laplacian is now naturally dened by

f = div ( r f )

1.3. The Hessian. A similar approach can also be used to dene the Hessian :

Hessf =12

L r f g


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WARPED PRODUCTS 3

The local coordinate calculation is a bit worse, but yields something nice in the end

Hessf @

@ x i,

@ @ x j=

12

(L r f g)@

@ x i,

@ @ x j

=12

Lg kl @ f @ x k

@

@ x lg

@ @ x i

,@

@ x j=

12

gkl@ f @ xk

L @@ x l

g

@ @ x i

,@

@ x j+12

@

gkl @ f @ x k@ x i glj +12

@

gkl @ f @ x k@ x j gli

=12

gkl@ f @ xk

@ gij@ x l

+12

@

gkl @ f @ x k@ x i glj +12

@

gkl @ f @ x k@ x j gli= 1

2gkl glj @

2f

@ x i @ xk+ gkl gli @

2f

@ x j @ xk+ 1

2gkl @ gij

@ x l+ @ g

kl

@ x iglj + @ g

kl

@ x jgli

@ f

@ xk

=@ 2f

@ x i @ x j+

12

gkl@ gij@ x l

@ glj@ x i

@ gli@ x j

@ f @ xk

This is usually rewritten by introducing the Christo ff el symbols of the rst andsecond kind

ijl =12

@ glj@ x i

+@ gli@ x j

@ gij@ x l,

kij =

12

gkl@ glj@ x i

+@ gli@ x j

@ gij@ x l.

Observe that this formula gives the expected answer in Cartesian coordinates, andthat at a critical point the Hessian does not depend on the metric.

Next we should tie this in with our denition of the Laplacian. Computing thetrace in a coordinate system uses the inverse of the metric coe ffi cients

tr (Hess f ) = gij Hessf @

@ x i,

@ @ x j

= gij@ 2f

@ x i @ x j+

12

gij gkl@ gij@ x l

@ glj@ x i

@ gli@ x j

@ f @ xk


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WARPED PRODUCTS 4

On the other hand

f = div (

rf )

=1

p det gkl@ p det gkl gij @ f @ x j@ x i

= gij@ 2f

@ x i @ x j+

@ gij

@ x i@ f @ x j

+ gij@ f @ x j

1p det gkl

@ p det gkl@ x i

= gij@ 2f

@ x i @ x jgik glj

@ gkl@ x i

@ f @ x j

+12

gij@ f @ x j

1det gkl

@ (det gkl )@ x i

= gij@ 2f

@ x i @ x jgik glj

@ gkl@ x i

@ f @ x j

+12

gij@ f @ x j

gkl@ gkl@ x i

= gij@ 2f

@ x i @ x jgij gkl

@ gjl@ x i

@ f @ xk

+12

gkl@ f @ xk

gij@ gij@ x l

= gij@ 2f

@ x i @ x j + gij gkl

12

@ gij

@ x l@ g

jl@ x i

@ f @ xk

= gij@ 2f

@ x i @ x j+

12

gij gkl@ gij@ x l

@ glj@ x i

@ gli@ x j

@ f @ xk

= tr (Hess f )

2. Examples

Starting with a Riemannian metric (N, h ) a warped product is dened as a metricon I N, where I

R is an open interval, by

g = dr 2 + 2 (r ) h

where > 0 on all of I. One could also more generally consider2 (r ) dr 2 + 2 (r ) h

However a change of coordinates dened by relating the di ff erentials d= (r ) drallows us to rewrite this as

d2 + 2 (r ()) h.

Important special cases are the basic product

g = dr 2 + h

and polar coordinatesdr 2 + r 2ds2n 1

on (0, 1 )S n 1

representing the Euclidean metric.

2.1. Conformal Representations. The basic examples are still su ffi ciently broadthat we can reduce almost all problems to these cases by a simple conformal change:

dr 2 + 2 (r ) h = 2 () d2 + h ,dr = () d,

(r ) = ()


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WARPED PRODUCTS 5

or

dr 2 + 2 (r ) h = 2 () d2 + 2h ,dr = () d,(r ) = ()

The rst of these changes has been studied since the time of Mercator. Thesphere of radius R can be written as

R2ds2n = R2 dt2 + sin2 (t) ds2n 1

= dr 2 + R2 sin2

rR

ds2n 1

The conformal change envisioned by Mercator then takes the form

R2ds2n = R2 2 () d2 + ds2n1

As

() d = dt,() = sin ( t)

this means that we have

d=dt

sin(t)

Thus is determined by one of these obnoxious integrals one has to look up. Mer-cator had no access to calculus making this problem even more di ffi cult and it tookhundreds of years to settle this issue!

Switching the spherical metric to being conformal to the polar coordinate rep-resentation of Euclidean space took even longer and probably wasnt studied muchuntil the time of Riemann

R2ds2n = R2 2 () d2 + 2ds2n 1

= R24

(1 + 2)2d2 + 2ds2n 1

Hyperbolic space is also conveniently dened using warped products:

g = dr 2 + sinh 2 (r ) ds2n 1 , r 2 (0, 1 )it too has a number of interesting other forms related to warped products:

dr 2 + sinh 2 (r ) ds2n 1 =4

(1 2)2d2 + 2ds2n 1 , 2 (0, 1)

And after using inversions to change the unit ball into a half space we can rewritethis as

1x2

dx2 + can R n 1 = ds2 + e2s can R n 1


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WARPED PRODUCTS 6

This works as follows. Let the half space model be H = ( 1, 0)R n 1 with themetric 1x 2 dx2 + can R n 1 . DeneF (x, z ) = (1 , 0) + 2 (x 1, z)

|x 1|2 + |z |2

= 1 +2 (x 1)

|x 1|2 + |z|2,

z|x 1|2 + |z |2!

= 1 +2 (x 1)

r 2,

zr 2

as the inversion in the sphere of radius p 2 centered at (1, 0) 2RR n 1 . This mapsH to the unit ball since|F (x, z )|2 = 1 +

4xr 2

= 2 .

Next we check what F does to the metric. We have

F 1 = 1 + 2 (x 1)r 2 ,

F i =z i

r 2, i > 1.

and4

(1 2)2dF 1

2+ Xk> 1 dF k

2!=

r 2 2

4x22dxr 2

2 (x 1) 2rdr(r 2)2 !

2

+

Xk> 1

r 2 2

4x2

dzk

r2

zk 2rdr

(r2)

2

!2

=1

x2dx(x 1) 2rdr

r 2 2

+1

x2 Xk> 1dzkzk 2rdr

r 2 2

=1

x2dx2 + Xk> 1 dzk

2!+ 1x2(x 1) 2rdrr 2 2

+1

x2 Xk> 1zk 2rdr

r 2 2

1x2

dx(x 1) 2rdr

r 21

x2 Xk> 1 dzkzk 2rdr

r 2

1x2

(x 1) 2rdrr 2

dx1

x2 Xk> 1zk 2rdr

r 2dzk

= 1x2

dx2 + can R n 1 + 1x2

r 2

2rdr

r 2

2

1x2

rdr2rdr

r 21

x22rdr

r 2rdr

=1

x2dx2 + can R n 1


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WARPED PRODUCTS 7

Showing that F also transforms the conformal unit ball model to the conformalhalf space model.

2.2. Singular Points. The polar coordinate conformal modeldr 2 + 2 (r ) h = 2 () d2 + 2h

can be used to study smoothness of the metric as we approach a point r 0 2@ I where(r 0) = 0 . We assume that (r 0) = 0 in the reparametrization. When h = ds2n 1we then note that smoothness on the right hand side

2 () d2 + 2ds2n 1

only depends on 2 () being smooth. Thinking of as being Euclidean distanceindicates that this is not entirely trivial. In fact we must assume that

(0) > 0

and(odd ) (0) = 0

Translating back to we see that0 (0) = 1,

(even ) (0) = 0 .

Better yet we can also prove that when h is not ds2n 1 , then it isnt possible toremove the singularity. Assume that g is smooth at the point p 2M correspondingto = 0 , i.e., we assume that on some neighborhood U of p we have

U { p} = (0 , b)N,g = d2 + 2h

The = 1 level corresponds naturally to N. Now observe that each curve in M emanating from p has a unique unit tangent vector at p and also intersects N ina unique point. Thus unit vectors in T p M and points in N can be identied bymoving along curves. This gives the isometry from S n 1 , the unit sphere in T pM,to N.

3. Characterizations

We start by o ff ering yet another version of the warped product representation.Rather than using r and , the goal is to use only one function f. Starting with awarped product

dr 2 + 2 (r ) hwe construct the function f =

dr on M. Since

df = drwe immediately see that

dr 2 + 2 (r ) h =1

2 (r )df 2 + 2 (r ) h

The gradient of f is

r f = r r =@ @ r


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WARPED PRODUCTS 8

so the Hessian becomes

(Hessf ) (X, Y ) =

1

2

L

@@ r g

(X, Y )

=12

L @@ r

g

(X, Y ) +12

(DX ) g (r r, Y ) +12

(DY ) g (r r, X )=

12

L @@ r

g

(X, Y ) +12

0 (DX r ) g (r r, Y ) +12

0 (DY r ) g (r r, X )=

12

L @@ r

g

(X, Y ) + 0dr 2 (X, Y )

This is further reduced

Hessf =12

L @@ r

g + 0dr 2

=12

L @@ r

dr 2 + 2 (r ) h + 0dr 2

=12

D @@ r

2 (r ) h + 0dr 2

= 0 2h + 0dr 2

= 0g

In other words: it is possible to nd a function f whose Hessian is conformal to themetric. Using

0 =ddr

=ddf

df dr

=ddf

=12

d |r f |2

df

we have obtained a representation that only depends on f and |r f |g = 1

|r f |2 df 2 + |r f |2 h,

Hessf =12

d |r f |2

df g

Before stating the main theorem we need to establish a general result.

Lemma 3.1. If f is a smooth function on a Riemannian manifold, then

Hessf (r f, X ) =12

DX |r f |2 .

Proof. At points where r f vanishes this is obvious. At other points we can theassume that f = x1 is the rst coordinate in a coordinate system. Then

r f = g1 i@

@ x i

and

|r f |2 = g1i gij g1j

= g11


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WARPED PRODUCTS 9

With that information the calculation becomes

Hessf

r

f,@

@ x j

= g1i Hessx1

@

@ x i,

@

@ x j

= 12g1i gkl

@ gij@ x l

@ glj@ x i

@ gli@ x j

@ x1

@ xk

=12

g1i g1 l@ gij@ x l

@ glj@ x i

@ gli@ x j

=12

gi 1g1 l@ gli@ x j

+12

g1i g1 l@ gij@ x l

12

g1i g1l@ glj@ x i

=12

@ g11

@ x j+

12

g1i g1l@ gij@ x l

12

g1l g1 i@ gij@ x l

=12

@ |r f |2

@ x j

The formula now follows by expanding X in coordinates.

We can now state and prove our main result.

Theorem 3.2. If there are smooth functions f, on a Riemannian manifold sothat

Hessf = g

then the Riemannian structure is a warped product around any point where r f 6= 0 .Proof. We rst need to show that only depends on f. This uses the previouslemma:

12

DX |r f |2 = Hess f (r f, X ) = g (r f, X )

So we see that |r f |2

is locally constant on level sets of f by taking X ? r f. Thuswe can locally assume that|r f |

2 = b(f ) .

If we let X = r f, then we obtainb =

12

D r f b =12

g (r b,r f ) =12

dbdf

b

or

2 =dbdf

We now claim that

g =1

bdf 2 + bh

where h is dened such that b(c) h is simply the restriction of g to the level setf = c. In particular, we have that

g| p =1b

df 2 + bh| p , when f ( p) = cNext we observe that

L r f g = 2 g


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WARPED PRODUCTS 10

and

L r f

1

bdf 2 + bh

=db

df

1

bdf 2 + bh

= 2 1b

df 2 + bhThus g and 1b df

2 + bh both solve the same di ff erential equation with the same initialvalues at f = c, showing that the two metrics must agree.

We can also handle singular points if we assume they are isolated.

Corollary 3.3. If there are smooth functions f, on a Riemannian manifold sothat

Hessf = g,

rf | p = 0 ,( p) 6= 0

then the Riemannian structure is a warped product around p.

Proof. As weve assumed the Hessian to be nondegenerate at p it follows that thereare coordinates in a neighborhood around p such that

f x1 ,...,x n = f ( p) + ( p)

x12

+ + ( xn )2

.

Thus the level sets near p are spheres, and we know as before that

g =1b

df 2 + bh,

b(f ) = |

rf |2

Since g is a smooth metric at p we can then also conclude that h = ds2n 1 .

The nal goal is to characterize the three standard geometries at least locally.

Corollary 3.4. Assume that there is a function f on a Riemannian manifold such that

f ( p) = 0 ,

r f | p = 0 .(1) If

Hessf = g

then the metric is Euclidean in a neighborhood of p.(2) If Hessf = (1 f ) g

then the metric is the unit sphere metric in a neighborhood of p.(3) If

Hessf = (1 + f ) g

then the metric is hyperbolic in a neighborhood of p.


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Proof. Note that in each of three cases is already a function of f. Thus we cannd

b(f ) =|r

f |2

by solving the initial value problemdbdf

= 2 (f ) ,

b(0) = 0

The solution is obviously unique so it actually only remains to be checked that thethree standard geometries have functions with the described properties. This worksas follows using the standard warped product representations:

Euclidean space

dr 2 + r 2ds2n 1 ,

f (r ) =12

r 2

The unit sphere

dr 2 + sin 2 rds 2n 1 ,f (r ) = 1 cos r

Hyperbolic space

dr 2 + sinh 2 rds 2n 1 ,f (r ) = 1 + 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 generalizationsare possible.

Transnormal functions simply satisfy:

|r f |2 = b(f )

Note that the equation

Hessf (r f, X ) =12

DX |r f |2

shows that this is locally equivalent to saying that r f is an eigenvector for Hessf .Wang proved that such functions further have the property that the only possiblecritical values are the maximum and minimum values, moreover the correspondingmax/min level sets are submanifolds. Such functions can be reparametrized asf = f (r ) , where r is a distance function, i.e., |r r |

21. Note that it is easy to

dene r as a function of f r = df p b

It is important that b is diff erentiable for these properties to hold as any strictlyincreasing function on R is transnormal for some continuous b. In this case therecan certainly be any number of critical points that are merely inection points.

Isoparametric functions are transnormal functions where in addition

f = a (f ) .


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WARPED PRODUCTS 12

These functions in addition have the property that the min/max sets for r areminimal submanifolds. These functions were introduced and studied by E. Cartan.

Even stricter conditions would be that the function is transnormal and the eigen-values of Hessf are functions of f. A very good and general model case for this sit-uation is a cohomogeneity 1 manifold with f = f (r ) and r : M ! M/G = I R .For this class of functions it is a possibility that the eigenspace distributions for theHessian are not integrable. Specically select M 4 , g with U (2) symmetry, e.g.,C bundles over S 2 , R 4 with the Taub-NUT metric, or C P 2 . The generic isotropyof U (2) is U (1) . The subspace corresponding to U (1) is then a 2-dim distributionthat must lie inside an eigenspace for Hessf. This distribution is however not inte-grable as it corresponds to the horizontal space for the Hopf bration. As long asthe metric isnt invariant under the larger group SO (4) the Hopf ber direction,i.e., the direction tangent to the orbits but perpendicular to this distribution, willgenerically correspond to an eigenvector with a di ff erent eigenvalue.

Finally one can assume that the function is transnormal and satises

R (X, Y ) r f = 0for all X, Y ? r f. The curvature condition is automatic in the three model caseswith constant curvature so it doesnt imply that f is isoparametric. But it is quiterestrictive as it forces eigenspaces to be integrable distributions. It is also equivalentto saying that Hess(r ) is a Codazzi tensor where r is the distance function so thatf = f (r ) . A good model case is a doubly warped product

M = I N 1N 2g = dr 2 + 21 (r ) h1 +

22 (r ) h2 ,

f = f (r ) .Note that the functions f coming from warped products satisfy all of the condi-

tions mentioned above.

Another interesting observation that might aid in a few calculations is that ageneral warped product2gB + 2gF = 2 2 2gB + gF

on a Riemannian submersion M ! B, where gB is a metric on B and gF repre-sents the family of metrics on the bers, and , functions on B. Here the metric2 2gB + gF is again a Riemannian submersion with the conformally changed met-

ric 2 2gB on the base. Thus general warped products are always conformallychanged on the total space of a Riemannian submersion, where the conformal factorhas 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.

warped products - petersen

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