Volume 2010, Article ID 187232,7pages doi:10.1155/2010/187232
Research Article
Heterogeneous Riemannian Manifolds
James J. Hebda
Department of Mathematics, Saint Louis University, St. Louis, MO 63103, USA
Correspondence should be addressed to James J. Hebda,[email protected]
Received 6 July 2009; Accepted 23 March 2010
Academic Editor: Wolfgang Kuehnel
Copyrightq2010 James J. Hebda. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
We solve Ambrose’s Problem for a generic class of Riemannian metrics on a smooth manifold, namely, the class of heterogeneous metrics.
1. Introduction
We define a Riemannian metricg on a manifoldMto be heterogeneous if no two distinct points ofMhave isometric neighborhoods. Intuitively, a heterogeneous metric is as far as possible from being homogeneous. Heterogeneity can be reformulated in terms of a multijet transversality condition so that by an application of the standard transversality theorems, the genericity of heterogeneous metrics is established.
Theorem 1.1. The set of heterogeneous metrics on a smooth manifold M of dimension n ≥ 2 is
residual in the space of Riemannian metrics onMwith the strongC∞topology.
A version ofTheorem 1.1for compact manifolds was stated and proved by Sunada1, Proposition 1.
Ambrose2asked whether or not a complete simply connected Riemannian manifold is determined up to isometry by the behavior of curvature under parallel transport along geodesics emanating from a point. For heterogeneous metrics the answer is always yes.
Proposition 1.2. IfM, gis a complete, connected, simply connected, heterogeneous Riemannian manifold of dimension n ≥ 2, then, for everyp ∈ M,M, gis determined up to isometry by the behavior of curvature under parallel transport along geodesics emanating fromp.
Theorem 1.3. The set of complete metrics on a connected, simply connected smooth manifold M of dimensionn ≥ 2 which, for every pointp inM, are determined up to isometry by the behavior of curvature under parallel transport along geodesics emanating from p, is residual in the space of complete Riemannian metrics onMwith the strongC∞topology.
Although Ambrose’s Problem has been completely settled in dimension 2 3, 4,
Theorem 1.3 does give a significant advance on the problem in higher dimensions, since earlier partial results in5,6apply only to metrics with rather special properties.
2. Ambrose’s Problem and Heterogeneity
Let us recall what it means for a complete, connected, simply connected, n-dimensional Riemannian manifoldM, gto be determined up to isometry by the behavior of curvature under parallel transport along geodesics emanating from a point p in M. Let M, g be another complete, connected, simply connected,n-dimensional Riemannian manifold. Let
p ∈M, and letI :TpM → TpMbe a linear isometry between the tangent spaces. For each
geodesicγ:0,1 → Msatisfyingγ0 p, there is a corresponding geodesicγ:0,1 → M
satisfying γ0 p characterized by the initial condition γ0 Iγ0. Given such a geodesic γ,Iγ Pγ ◦I ◦Pγ−1 defines a linear isometry from Tγ1M ontoTγ1M where Pγ
andPγdenote parallel transport alongγandγ, respectively. Consider the hypothesis
IγRX, YZ R
IγX, IγY
IγZ ∗
for every such geodesic γ and for all vectors X, Y, Z ∈ Tγ1M, where R and R are the
respective Riemann curvature tensors. Then M, g is determined up to isometry by the behavior of curvature under parallel transport along geodesics emanating from p if the hypothesis∗implies that there exists an isometryΦ:M → MwithΦp panddΦ I
atpcf.,2,5,6.
In order to proveProposition 1.2, suppose thatM, gis heterogeneous, and assume the hypothesis∗. We will first find an isometryΦ:M → MwithΦp panddΦ I−1at p. Thus we obtain the desired isometryΦby settingΦ Φ−1.
To prove the existence ofΦ, let qin Mbe any nonconjugate cut point topof order two, and letγ1andγ2be the two minimizing geodesics joiningptoq, whose respective initial tangent vectors areX1 γ10and X2 γ20. Thus exppX1 exppX2 q. Sinceqis
not conjugate topalongγ1orγ2, there are neighborhoodsV1ofX1andV2ofX2inTpMfor
which exppcarries bothV1 andV2 diffeomorphically onto the same neighborhoodW ofq.
As noted in6, page 561, fori ∈ {1,2}, the mapfi expp◦I−1◦exp
p |Vi−1 : W → M
is an isometryafter possibly cutting downWonto a neighborhoodUi ofγi1. It follows
thatγ11 γ21, because otherwisef2◦f−11 : U1 → U2 would be an isometry between
neighborhoods of two distinct points ofM, contradicting the assumed heterogeneity ofM. This verifies the hypothesis of Lemma 2.1 in6, with the roles ofMandMinterchanged. Hence there exists an isometric immersionΦ:M → MwithΦp panddΦ I−1. Actually Φ is an isometry because 1 an isometric immersion between two complete Riemannian manifolds of the same dimension is a covering map by Theorem IV.4.6 in7, page 176, and
3. Heterogeneity and Transversality
Let us say that twok-jets,Jk
pg andJqkh, of germs of Riemannian metricsg andhat points
p andqinM are equivalent if there is a germ of a diffeomorphism f with fp qsuch that Jk
pf∗h Jpkg. We then define a Riemannian metric g on M to be heterogeneous
of order k if thek-jets ofg at any two distinct points ofMare not equivalent. Obviously, being heterogeneous of orderk for some k ≥ 2 is a stronger condition than being simply heterogeneous. Let us proceed to explain how to express heterogeneity of orderkin terms of transversality whenkis sufficiently large.
Letπ : X → Mdenote the the bundle of positive definite symmetric covariant 2-tensors overM. Thus sections ofX are just Riemannian metrics onM. Letπk : Xk → M
denote the bundle ofk-jets of Riemannian metrics onM. Following8,9, let2πk : 2Xk → 2Mdenote the bundle of multijets of Riemannian metrics of orderkand multiplicity 2. Thus
2Xk
Jpkg, Jqkh∈Xk×Xk:p /q, 2M
p, q∈M×M:p /q , 3.1
and2πkJpkg, Jqkh πkJpkg, πkJqkh p, q. Given a Riemannian metricg, the multijet
extension
2Jkg: 2M−→ 2Xr 3.2
is defined by the formula2Jkgp, q Jpkg, Jqkgforp, q∈ 2M.
Since2Xkis just the set of ordered pairs ofk-jets of Riemannian metrics over distinct
points ofM, the equivalence relation onk-jets of Riemannian metrics onMdefines a subset
S ⊂ 2Xkconsisting of pairs of equivalentk-jets. Obviously,g is heterogeneous of orderkif
and only if the image of its multijet extension2Jkg2Mmisses the setS. In the next section
we investigate the structure of the setSand prove the following proposition.
Proposition 3.1. Letnbe the dimension ofM. Then the setSis a union of finitely many submanifolds of codimension at leastn1
2
nk
k
−nnk1
k1
n. In particular codimS>2nin any of the three casesin2 andk≥4,iin3 andk≥3, oriiin≥4 andk≥2.
As a corollary we obtain the following stronger version ofTheorem 1.1.
Theorem 3.2. The set of Riemannian metrics on a manifold M of dimension n, which are heterogeneous of orderk, is residual in the space of Riemannian metrics onMwith the strongC∞ topology as long any one of the three cases listed inProposition 3.1for which codimS>2nholds.
Proof. SinceSis a union of submanifolds, application of the multijet transversality theorem
Corollary 3.4 8or9, page 739, shows that the set ofg for which2Jkg is transverse to Sforms a residual set in the space of Riemannian metrics with the strongC∞topology. But because dim2M 2nand codimS>2n, it follows that2Jkgis transverse toSif and only
if its image missesS, that is, if and onlygis heterogeneous of orderk.
4. The Structure of
S
Consider the collectionMknofk-jets of Riemannian metrics onRnat 0 and the so-called jet
groupGk1n, consisting of thek1-jets of diffeomorphismsf :Rn → Rnsatisfyingf0
012, page 128. The jet group acts uponMknon the left by the formulaf·gJk 0f−1
∗
g
forf∈ Gk1nandg∈ Mkn. For any subgroupHofGk1n, letMkHndenote the set of points on orbits of typeGk1n/H, that is,
Mk Hn
g∈ Mkn: The isotropy group ofg is conjugate toH. 4.1
Finally, letMkHndenote the orbit spaceGk1n\ MkHn, and letQdenote the quotient projectionQ:Mk
Hn → M k Hn.
Proposition 4.1. Mk
HnandM k
Hnare smooth manifolds, and Qis a smooth fibration with
fibersGk1n/H. Moreover, there are only finitely many distinct orbit types, and these finitely many
submanifoldsMkHnstratifyMkn.
Proof. If Gk1n was compact, this proposition would follow from Theorem IV.3.3 in 10,
page 182. Although Gk1n is not compact, its action on Mkn reduces to a compatible
action of the orthogonal groupOn, which canonically includes as a subgroup ofGk1n,
on the subsetNknofMkn. HereNkndenotes the set ofk-jets of Riemannian metrics
Rnat 0 for which the standard coordinates onRnform a normal coordinate system. These are
the jets that satisfy the conditions2.5.1-3in11for 1≤ r ≤ k. If one carries out the proof of Theorem 2.6 in11fork-jets, rather than for∞-jets, one concludes that eachGk1norbit
inMknmeetsNknin anOnorbit. Thus the inclusion of the orbit spaceOn\ Nkn
intoGk1n\ Mknis a one-to-one correspondence. This also implies that every isotropy
subgroup for theGk1naction is conjugate to an isotropy subgroup of theOnaction. Thus
there is a one-to-one correspondence between the orbit types of the two actions. In addition, we see that the isotropy subgroups of the Gk1naction are compact. It also follows that
every slice for theOnaction onNk
nis a slice for theGk1naction onMknwhich proves
that slices exist for the latter action. Because of these observations, the conclusions of Theorem IV.3.3 in10which apply to the action ofOnonNknalso apply to the action ofGk1n
onMkn. That there are only finitely many of orbit types follows in the same way from the
well-known finiteness of orbit types for orthogonal actions10, page 112. This completes the proof ofProposition 4.1.
LetPk1Mdenote the principalGk1nbundle of thek1th order frames on the n-dimensional manifoldM12, page 122. Clearly, the bundle Xk ofk-jets of Riemannian
metrics onMis the associated bundlePk1M×Gk1nMkn. Since the stratification by orbit
types ofMknis invariant underGk1n, it induces a stratification ofXkwhere the typical
stratum takes the form of the associated bundle
XkHPk1M×Gk1nMkHn. 4.2
BecauseS⊂Xk×Xk, we may set
SHS∩Xk
H×XkH
. 4.3
Since two equivalent jets automatically have the same orbit type, we have
S
H
SH. 4.4
Clearly,SHis just the inverse image of the diagonal inMkHn×M k
Hnunder the product
submersionXk
H×XkH → M k
Hn× M k
Hninduced byQ. Therefore,SH is a smooth
submanifold ofXk
H×XkHwhose codimension satisfies
codimSH⊂XkH×XkHdim
MkHn
. 4.5
We may now compute the codimension ofSHin2Xk:
codimSHcodim
SH⊂Xk×Xk
codimSH⊂XkH×XkH
2codimXkH⊂Xk
dim
MkHn
2 codimMk
Hn⊂ Mkn
dimMkHn−dimGk1n/H 2dimMkn−dimMkHn dimMkn−dimGk1ndimH
dimMkn−dimMkHn
≥dimMkn−dimGk1n
n1
2
nk
k − n
nk1
k1
−n
.
4.6
5. The Space of Complete Metrics
Since every Riemannian metric on a compact manifold is complete, Theorem 1.3 is an immediate consequence of Theorem 1.1 and Proposition 1.2 when M is compact. If M is not compact, the space of complete metrics onMis a proper subspace of the space of all Riemannian metrics onM. Thus to proveTheorem 1.3in general, we need to show that the subspace of complete metrics inherits the property of being a Baire space from the space of all metrics, and that an open dense subset of the space of all metrics intersects the subspace of complete metrics in an open dense subset of the subspace. Both of these statements are immediate consequences of the next proposition.
Proposition 5.1. The set of all complete Riemannian metrics on a given smooth manifoldMis both open and closed in the space of Riemannian metrics onMwith the strongC∞topology.
Proof. Fix a Riemannian metricg0onM. It suffices to show that there is a neighborhoodUof g0 in the strongC∞topology consisting either entirely of complete metrics ifg0is complete
or entirely of incomplete metrics ifg0is incomplete.
To this end, letUbe the set of all Riemannian metricsgsuch that 1
2g0X, X< gX, X<2g0X, X 5.1
for allX /0 inTM. Obviously,Uis an open neighborhood ofg0in the strongC∞topology.
Clearly, ifg ∈ U, and if dist and dist0are the associated distance functions corresponding to gandg0, respectively, then we have
1 2dist0
p, q≤distp, q≤2 dist0
p, q 5.2
for allpandqinM. It follows that dist and dist0have the same Cauchy sequences and the
same convergent sequences with the same limits. Thusg andg0are either both complete or
both incomplete. This finishes the proof ofProposition 5.1.
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