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A proof of the Chern-Gauss-Bonnet theorem forindefinite signature metrics using analytic
continuation (arXiv:1405.7613)
P. Gilkey and J. H. Park
22 Aug. 2014 ICM Satellite Conference on Geometric Analysis
Peter B Gilkey The Gauss Bonnet Theorem
Introduction
The use of analytic continuation to pass from positive definitemetrics to indefinite metrics or to pass between the spacelike andthe timelike settings in pseudo-Riemannian geometry has proven tobe a very fruitful technical tool; results holding in the Riemanniancontext can often be extended to the pseudo-Riemannian contextwith relatively little additional fuss and results in the spacelike(resp. timelike) settings often give results in the timelike (resp.spacelike) setting. In all cases, one first extends the relevantnotions to complex “metrics” (i.e. sections g to S2(T ∗M)⊗R Csatisfying det(g) 6= 0) or complex “tangent vectors”, appliesanalytic continuation, and thereby passes from one signature toanother.
Peter B Gilkey The Gauss Bonnet Theorem
Examples
Osserman Geometry
In the study of Osserman geometry, Garcıa-Rıo et al. [1] examinedcomplex “tangent vectors” (i.e. elements of TM ⊗R C) to showthat spacelike Osserman and timelike Osserman were equivalentconcepts.
[1] E. Garcıa-Rıo, D. N. Kupeli, M. E. Vazquez-Abal, and R.Vazquez-Lorenzo, “Affine Osserman connections and their Riemannextensions”, Differential Geom. Appl. 11 (1999), 145–153.
E. Garcıa-Rıo M. E. Vazquez-Abal R. Vazquez-Lorenzo
Peter B Gilkey The Gauss Bonnet Theorem
Examples
Ricci solitons
If (M, g , f ) is a gradient Ricci solution and if Hf is the Hessian,then one has the identity
(∇∇f Ric) + Ric ◦Hf = R(∇f , ·)∇f + 12∇∇τ .
This was proved in the Riemannian setting; one can use analyticcontinuation to extend this identity to the indefinite setting
The Gauss-Bonnet theorem
We shall now derive the Gauss-Bonnet theorem in the indefinitesignature (p, q) setting from the corresponding theorem in thedefinite case; this gives an explanation of the factor of (−1)p/2
that appears.
Peter B Gilkey The Gauss Bonnet Theorem
Examples
Ricci solitons
If (M, g , f ) is a gradient Ricci solution and if Hf is the Hessian,then one has the identity
(∇∇f Ric) + Ric ◦Hf = R(∇f , ·)∇f + 12∇∇τ .
This was proved in the Riemannian setting; one can use analyticcontinuation to extend this identity to the indefinite setting
The Gauss-Bonnet theorem
We shall now derive the Gauss-Bonnet theorem in the indefinitesignature (p, q) setting from the corresponding theorem in thedefinite case; this gives an explanation of the factor of (−1)p/2
that appears.
Peter B Gilkey The Gauss Bonnet Theorem
Gaussian Curvature
Intrinsic Geometry
Let M be a 2-dimensional Riemannian manifold. If ~x = (x1, x2) isa system of local coordinates on M, let gij := g(∂xi , ∂xj ). TheGaussian curvature is K := det(gij)
−1R1221 and the volumeelement is dvolg :=
√det(gij)dx1dx2. Let Vol(Bε(P)) be the
volume of the geodesic ball in M of radius ε about P ∈ M. ThenVol(Bε)(P) = πr2{1− K(P)
12 r2 + O(r4)}.We refer to Gray [2] for further terms in the expansion.
[2] A. Gray, “The volume of a small geodesic ball of a Riemannianmanifold”, Michigan Math. J. 20 (1974), 329–344.
A. Gray
Peter B Gilkey The Gauss Bonnet Theorem
Geodesic curvature
Let γ be a curve in M and let ν be a unit normal along γ. Thegeodesic curvature κg := g(∇γ γ, ν).
Euler characteristic
Let χ be the Euler-characteristic of M. If M is triangulated, thenχ(M) = #vertices −#edges + #triangles + . . . .
Extrinsic geometry
If T (x1, x2) is an isometric immersion of M into R3, then
gij = ∂x i T · ∂x j T . If ν :=∂x1T×∂x2T‖∂x1T×∂x2‖
is a unit normal, then
II (∂xi , ∂xj ) := ∂xi∂xj T · ν is the second fundamental form andK = det(II )/ det(g).
Peter B Gilkey The Gauss Bonnet Theorem
Geodesic curvature
Let γ be a curve in M and let ν be a unit normal along γ. Thegeodesic curvature κg := g(∇γ γ, ν).
Euler characteristic
Let χ be the Euler-characteristic of M. If M is triangulated, thenχ(M) = #vertices −#edges + #triangles + . . . .
Extrinsic geometry
If T (x1, x2) is an isometric immersion of M into R3, then
gij = ∂x i T · ∂x j T . If ν :=∂x1T×∂x2T‖∂x1T×∂x2‖
is a unit normal, then
II (∂xi , ∂xj ) := ∂xi∂xj T · ν is the second fundamental form andK = det(II )/ det(g).
Peter B Gilkey The Gauss Bonnet Theorem
Geodesic curvature
Let γ be a curve in M and let ν be a unit normal along γ. Thegeodesic curvature κg := g(∇γ γ, ν).
Euler characteristic
Let χ be the Euler-characteristic of M. If M is triangulated, thenχ(M) = #vertices −#edges + #triangles + . . . .
Extrinsic geometry
If T (x1, x2) is an isometric immersion of M into R3, then
gij = ∂x i T · ∂x j T . If ν :=∂x1T×∂x2T‖∂x1T×∂x2‖
is a unit normal, then
II (∂xi , ∂xj ) := ∂xi∂xj T · ν is the second fundamental form andK = det(II )/ det(g).
Peter B Gilkey The Gauss Bonnet Theorem
The Gauss-Bonnet Theorem
Theorem [Gauss-Bonnet (1848)-von Dyck (1888)]
Let (M, g) be a 2-dimensional Riemannian manifold with piecewisesmooth boundary ∂M. Let κ = R1221 be the Gaussian curvature,let κg be the geodesic curvature, and let αi be the interior anglesat which consecutive smooth boundary components meet. Then
2π · χ(M) =
∫Mκ+
∫∂M
κg +∑i
(π − αi ).
Gauss Bonnet von Dyck
Peter B Gilkey The Gauss Bonnet Theorem
The Chern-Gauss-Bonnet Theorem in general
The Euler form (or Pfaffian) is defined by setting:
E2`(g) :=Ri1i2j2j1 ...Ri2`−1i2`j2`j2`−1
g(e i1 ∧ ... ∧ e i2` , e j1 ∧ ... ∧ e j2`)
(8π)``!.
Let ρkl := g ilRijkl be the Ricci tensor and let τ = gklρkl the scalarcurvature. One has:
E2 = τ4π
E4 = τ2−4|ρ|2+|R|232π2 ,
E6 = 1384π3 {τ3 − 12τ |ρ|2 + 3τ |R|2 + 16ρabρacρbc − 24ρabρcdRacbd
−24ρuvRuabcRvabc + 8RabcdRaucvRbvdu − 2RabcdRabuvRcduv}Set dvolg = det(gij)
1/2dx1 . . . dxm.
Peter B Gilkey The Gauss Bonnet Theorem
Theorem [Chern [4] — see also Allendoerfer and Weil [3]]Let M be a compact Riemannian manifold of dimension m = 2`without boundary. Then
∫M E2`(g) dvolg = χ(M).
[3] C. Allendoerfer and A. Weil, “The Gauss-Bonnet theorem forRiemannian polyhedra”, Trans. Am. Math. Soc. 53 (1943),101–129.[4] S. Chern, “A simple intrinsic proof of the Gauss-Bonnetformula for closed Riemannian manifolds”, Ann. of Math. 45,(1944). 747–752.
C. Allendoerfer S. Chern A. Weil
Peter B Gilkey The Gauss Bonnet Theorem
Extension to the pseudo-Riemannian setting
Definition (M, g) is a pseudo-Riemannian manifold of signature(p, q). Let
E2`(g) :=Ri1i2j2j1 ...Ri2`−1i2`j2`j2`−1
g(e i1 ∧ ... ∧ e i2` , e j1 ∧ ... ∧ e j2`)
(8π)``!.
Theorem Let (M, g) be a compact pseudo-Riemannian manifoldof signature (p, q) without boundary. If dim(M) ≡ 1 mod 2 or ifp ≡ 1 mod 2, then χ(M) vanishes. If dim(M) ≡ p ≡ 0 mod 2,
χ(M) = (−1)p/2∫M
Em| dvolg |.
[5] A. Avez, “Formule de Gauss-Bonnet-Chern en metrique designature quelconque”, C. R. Acad. Sci. Paris 255 (1962),2049–2051.[6] S. Chern, “Pseudo-Riemannian geometry and theGauss-Bonnet formula”, An. Acad. Brasil. Ci. 35 (1963), 17–26.(There is a small mistake in the paper of Chern as the factor of(−1)p is missing).
Peter B Gilkey The Gauss Bonnet Theorem
Examples
1 Let (M, g) be a Riemannian manifold with of dimensionm = 2`. Let g = −g be a pseudo-Riemannian metric ofsignature (m, 0). Then the Levi-Civita connection isunchanged and the curvature operator is not changed. AsRijkl = glnRijk
n, we have that Rijkl(g) = −Rijkl(g) and
E2`(g) =Ri1 i2j2j1
...Ri2`−1 i2`j2`j2`−1g(e i1∧...∧e i` ,e j1∧...∧e j` )
(8π)``!
= (−1)`E2`(g). Since the underlying volume form is notchanged, it is necessary to introduce a factor of (−1)` tocorrect the Chern-Gauss-Bonnet formula.
2 Let (Mi , gi ) be Riemannian manifolds withdim(Mi ) = mi = 2`i . Let (M = M1 ×M2, g = −g1 + g2)be a pseudo-Riemannian manifold of signature (m1,m2).As R = −R1 + R2,
Em(g) = Em1(−g1)Em2(g2) = (−1)`1Em(g1 + g2).
Peter B Gilkey The Gauss Bonnet Theorem
Proof - complex metrics
We follow Gilkey-Park [14]. Let g ∈ S2(T ∗M)⊗ C be a complexvalued symmetric bilinear form on the tangent space. We assumedet(g) 6= 0 as a non-degeneracy condition. The Levi-Civitaconnection and curvature tensor may then be defined. Setdvol(g) := det(gij)
1/2dx1 ∧ · · · ∧ dxm. We do not take theabsolute value. There is a subtlety here since, of course, there are2 branches of the square root function. We shall ignore this for themoment in the interests of simplicity and return to this presently.
[14] P. Gilkey and J.H. Park, “A proof of the Chern-Gauss-Bonnettheorem for indefinite signature metrics using analyticcontinuation”, arXiv:1405.7613.
P. Gilkey J.H. ParkPeter B Gilkey The Gauss Bonnet Theorem
Proof - Euler-Lagrange Equations
Consider∫M Em(g) dvol(g). Let h ∈ S2(T ∗M)⊗ C be a
perturbation of the metric. We form gε := g + εh and differentiatewith respect to the parameter ε. We then integrate by parts todefine the Euler-Lagrange equations:
∂ε
{∫M
Em(g + εh) dvol(g + εh)
}∣∣∣∣ε=0
=
∫M
g((TEm)(g), h) dvol(g)
where TEm(g) ∈ S2(T ∗M)⊗ C is the transgression of the PfaffianEm. The Chern-Gauss-Bonnet theorem shows that TEm(g) = 0 ifdim(M) = m and if g is Riemannian, i.e. if g and h are real since∫M Em(g + εh) dvol(g + εh) = χ(M) is independent of h.
Peter B Gilkey The Gauss Bonnet Theorem
Proof - Analytic continuation
The invariant TEm is locally computable. In any coordinatesystem, it is polynomial in the derivatives of the metric, thecomponents of the metric, and det(gij)
−1/2. Thus it is holomorphicin the metric. Thus vanishing identically when g is real implies itvanishes identically when g is complex. Consequently, theEuler-Lagrange equations vanish. In particular, if gε is a smooth1-parameter family of complex metrics starting with a real metric,then we can define a branch of the square root function along thisfamily so that the following integral is independent of ε:∫
MEm(gε) dvol(gε) .
Peter B Gilkey The Gauss Bonnet Theorem
Proof - Decomposing the tangent bundle
We may find an orthogonal direct sum decompositionTM = V− ⊕ V+ where V− is timelike with dim(V−) = p andwhere V+ is spacelike with dim(V+) = q. This decomposesg = g− ⊕ g+ where g− is negative definite on V− and g+ ispositive definite on V+. For ε ∈ [0, π], define:
gε = −eε√−1g− ⊕ g+.
Then g0 = −g− ⊕ g+ is a Riemannian metric while gπ = g− ⊕ g+is the given pseudo-Riemannian metric. We have
det(gε) = eε√−1p det(−g−) det(g+) = eε
√−1p det(g0)
so the branch of the square root function along the deformationfrom g0 to gπ is given by eε
√−1p/2. Thus if p is odd, this is purely
imaginary for gπ and hence the Euler characteristic vanishes. If p iseven, det(g0) = det(gπ) so (−1)p/2
√det(g0) = (−1)p/2
√det(g).
This completes the proof.
Peter B Gilkey The Gauss Bonnet Theorem
The Chern-Gauss-Bonnet Theorem for Riemannian manifolds with bd
Definition: Let {e1, e2, . . . , em} be a local orthonormal frame fieldwhere em is the inward unit normal. Let Lab := g(∇eaeb, em) bethe second fundamental form. We sum over indices ai and bi
ranging from 1 to m − 1 to define:
Fm,ν :={
Ra1a2b2b1...Ra2ν−1a2νb2νb2ν−1
La2ν+1b2ν+1...Lam−1bm−1
(8π)νν! Vol(Sm−1−2ν)(m−1−2ν)!
}×g(ea1 ∧ · · · ∧ eam−1 , eb1 ∧ · · · ∧ ebm−1).
Theorem Let (M, g) be a compact smooth Riemannian manifoldof dimension m. If m is odd, χ(M) = 1
2χ(∂M). If m is even,
χ(M) =
∫M
Em(g) vol(g) +∑ν
∫∂M
Fm,ν vol(g |∂M).
[4] S. Chern, “A simple intrinsic proof of the Gauss-Bonnetformula for closed Riemannian manifolds”, Ann. of Math. 45,(1944). 747–752.
Peter B Gilkey The Gauss Bonnet Theorem
Examples
Let M be a compact smooth Riemannian manifold with boundaryof dimension m.
1 If m = 2, then χ(M) = 14π
∫M τ + 2
∫∂M Laa.
2 If m = 3, thenχ(M) = 1
8π
∫∂M Ra1a2a2a1 + La1a1La2a2 − La1a2La1a2
= 18π
∫∂M τ∂M .
3 If m = 4, then χ(M) = 132π2
∫M4(τ2 − 4|ρ|2 + |R|2)
+ 124π2
∫∂M(3τLaa + 6RamamLbb + 6RacbcLab
+2LaaLbbLcc − 6LabLabLcc + 4LabLbcLac).
Peter B Gilkey The Gauss Bonnet Theorem
Extension to the pseudo-Riemannian setting
Theorem: Let (M, g) be a compact smooth manifoldpseudo-Riemannian manifold of even dimension m and signature(p, q) which has smooth non-degenerate boundary ∂M. If m isodd, then χ(M) = 1
2χ(∂M). If p is odd, then χ(M) = 0.Otherwise
χ(M) = (−1)p/2
{∫M
Em(g) vol(g) +∑ν
∫∂M
Fm,ν vol(g |∂M)
}.
[7] L. J. Alty, “The generalized Gauss-Bonnet-Chern theorem”, J.Math. Phys. 36 (1995), 3094–3105. doi: 10.1063/1.531015.
[8] P. Gilkey and J.H. Park, “A proof of the Chern-Gauss-Bonnettheorem for indefinite signature metrics using analyticcontinuation”, arXiv:1405.7613
Peter B Gilkey The Gauss Bonnet Theorem
Proof
If (M, g) is Riemannian, this follows from the previous result ofChern. So the trick is to extend it to the pseudo-Riemanniansetting. Again, we will use analytic continuation. But there is animportant difference. We fix a non-zero vector field X which pointsin on the boundary and consider a smooth 1-parameter of complexvariations gε so that gε(X ,X ) 6= 0 and so that X ⊥ T (∂M) withrespect to gε. We may then set ν = X · gε(X ,X )−1/2. In theexpression for Fm,ν , there are an odd number of terms whichcontain L and hence gε(X ,X )−1/2. Given a pseudo-Riemannianmetric g , we construct g0 as above. Let g∂M0,ab and g∂Mab be therestriction of the metrics g0 and g , respectively, to the boundary. IfX is space-like, then det(g∂Mab ) = (−1)p det(g∂M0,ab) and the analysisproceeds as previously; the whole subtlety arising from the squareroot of the determinant.
Peter B Gilkey The Gauss Bonnet Theorem
Proof continued
If X is time-like, then det(g∂Mab ) = (−1)p−1 det(g∂M0,ab). Howeverg(X ,X ) = −g0(X ,X ) and thus once again, we must take thesquare root of (−1)p in the analytic continuation. Apart from this,the remainder of the argument is the same as that used to considerthe case of a manifold without boundary.
Peter B Gilkey The Gauss Bonnet Theorem
Euler Lagrange Equations
Let (M, g) be a Riemannian manifold of dimension m > 2`. Weconsider ∫
ME2`(g) vol(g) .
This is no longer a topological invariant but is still of interest. Leth be a symmetric 2-tensor field. One integrates by parts to definethe associated Euler-Lagrange equations:
∂ε
{∫M
E2`(g + εh) vol(g + εh)
}∣∣∣∣ε=0
=
∫M
g(TE2`(g), h) vol(g) .
The transgression TE2` is canonically defined symmetric 2-tensor.It vanishes identically if 2n = m but in general is non-zero form > 2n.
Peter B Gilkey The Gauss Bonnet Theorem
Euler-Lagrange Equations II
A-priori since E2` involves the 2nd derivatives of the metric TE2`
could involve one 4th derivative of the metric or two 3rd derivativesof the metric. Berger [9] conjectured that this was not in fact thecase; that TE2` only involved the curvature tensor.
[9] M. Berger, “Quelques formulas de variation pour une structureriemanniene”, Ann. Sci. Ec. Norm. Super. 3 (1970), 285–294.
M. Berger
Peter B Gilkey The Gauss Bonnet Theorem
Euler-Lagrange Identities III
Berger’s conjecture was established by Kuz’mina [10] (see relatedwork by Labbi [11], Gilkey, Park, and Sekigawa [12], and A and JNavarro [13]).
Theorem
TE2` = c(m)Ri1i2j2j1 ...Ri2`−1i2`j2`j2`−1e i2`+1 ◦ e j2`+1
×g(e i1 ∧ ... ∧ e i2`+1 , e j1 ∧ ... ∧ e j2`+1).
[10] G. M. Kuz’mina, “Some generalizations of the Riemannspaces of Einstein”, Math. Notes 16 (1974), 961–963.[11] M.-L. Labbi, “Double forms, curvature structures and the(p, q)-curvatures”, Trans. Am. Math. Soc. 357 (2005),3971–3992.[12] P. Gilkey, J.H. Park, and K. Sekigawa, “Universal curvatureidentities”, Diff. Geom. Appl., 29 (2011) 770–778.[13] A. Navarro and J. Navarro, “Dimensional curvature identitieson pseudo-Riemannian geometry”, arXiv:1310.2878.
Peter B Gilkey The Gauss Bonnet Theorem
Euler-Lagrange Identities III
Berger’s conjecture was established by Kuz’mina [10] (see relatedwork by Labbi [11], Gilkey, Park, and Sekigawa [12], and A and JNavarro [13]).
Theorem
TE2` = c(m)Ri1i2j2j1 ...Ri2`−1i2`j2`j2`−1e i2`+1 ◦ e j2`+1
×g(e i1 ∧ ... ∧ e i2`+1 , e j1 ∧ ... ∧ e j2`+1).
[10] G. M. Kuz’mina, “Some generalizations of the Riemannspaces of Einstein”, Math. Notes 16 (1974), 961–963.[11] M.-L. Labbi, “Double forms, curvature structures and the(p, q)-curvatures”, Trans. Am. Math. Soc. 357 (2005),3971–3992.[12] P. Gilkey, J.H. Park, and K. Sekigawa, “Universal curvatureidentities”, Diff. Geom. Appl., 29 (2011) 770–778.[13] A. Navarro and J. Navarro, “Dimensional curvature identitieson pseudo-Riemannian geometry”, arXiv:1310.2878.
Peter B Gilkey The Gauss Bonnet Theorem
Dimension m = 2
One has, for example, that
TE2 = c2{Rijjiek ◦ ek − 2Rijkie
j ◦ ek}
In dimension 2, TE2 = 0; this yields the familiar identity thatρ = 1
2τg . Furthermore,∫τ dvol has a critical point in higher
dimensions if and only if ρ = λg so (M, g) is Einstein.
Peter B Gilkey The Gauss Bonnet Theorem
Dimension m = 4
If m = 4, then
TE4 = c4{(RijjiRkllk − 4RijkiRljkl + RijklRijkl)en ◦ en
+{RklniRklnj − 2RknikRlnjl − 2RikljRnkln + RkllkRnijn}e i ◦ e j .
Since this vanishes in dimension m = 4, this yields the4-dimensional Eun-Park-Sekigawa identity [15] which hasimportant applications in studying the Euler Lagrange equationsfor the Chern-Gauss-Bonnet action in gravity.
[15] Y. Euh, J. H. Park, K. Sekigawa, “A curvature identity on a4-dim. Riemannian manifold”, Results Math. 63 (2013), 107–114.
J.H. Park K. Sekigawa Y. Euh
Peter B Gilkey The Gauss Bonnet Theorem
Dimension 6
If m = 6, one obtains the 6-dimensional Euh-Park-Sekigawaidentity [17] obtained by setting TE6 = 0 where
TE6 = ( τ3
2 + 32τ |R|
2 − 6τ |ρ|2 + 8ρabρbcρca − 12ρabρcdRacbd
−12ρuvRabcuRabcv + 4RabcdRaucvRubvd − 2RabuvRuvcdRabcd)gij−3τ2ρij − 3|R|2ρij + 12|ρ|2ρij + 12τρiaρja + 12τρabRiabj
−6τRiabcRjabc − 24ρiaρjbρab − 24ρacρbcRiabj
+24ρajρcdRacid + 24ρaiρcdRacjd + 24ρabRicdjRacbd
+48ρcdRiabcRjabd + 6ρjdRabciRabcd + 6ρidRabcjRabcd
+12RiuvjRabcuRabcv + 12RibacRjbuvRacuv − 24RiabcRjubvRaucv .
[17] Y. Euh, J. H. Park and K. Sekigawa, “A curvature identity on6-dimensional Riemannian Manifolds and its applications”, (privatecommunication).
Peter B Gilkey The Gauss Bonnet Theorem
Euler Lagrange Equations in indefinite signature
One can define E2` and TE2` in indefinite signature setting:
E2`(g) :=Ri1i2j2j1 ...Ri2`−1i2`j2`j2`−1
g(e i1 ∧ ... ∧ e i2` , e j1 ∧ ... ∧ e j2`)
(8π)``!.
TE2` = c(m)Ri1i2j2j1 ...Ri2`−1i2`j2`j2`−1e i2`+1 ◦ e j2`+1
×g(e i1 ∧ ... ∧ e i2`+1 , e j1 ∧ ... ∧ e j2`+1).
dvol(g) = (det(gij))1/2dx1 . . . dxm.
Theorem ∂ε
{∫M
Em(g + εh) dvol(g + εh)
}∣∣∣∣ε=0
=
∫M
g((TEm)(g), h) dvol(g).
[18] P. Gilkey, J.H. Park, and K. Sekigawa, “Universal curvatureidentities II”, J. Geom and Physics 62 (2012), 814–825.
Peter B Gilkey The Gauss Bonnet Theorem
Proof
Rather than proving this result a new as was done in [18], we canderive it from the result in the Riemannian setting using analyticcontination. We note that it holds for (g , h) real and then permit(g , h) to be complex. The same argument given previously usingthe decomposition of TM = V+ ⊕ V− then shows any twonon-degenerate pseudo-Riemannian metrics lie in the same arccomponent and provides a path between the metric g = g+ ⊕ g−and a positive definite metric g+ ⊕−g−; it also gives us a way ofanalytically continuing the square root function. Appropriatepowers of
√−1 will arise, of course, in comparing | dvol | with dvol,
but these cancel out on both sides of the equation.
Peter B Gilkey The Gauss Bonnet Theorem
Generalizations
1. An appropriate extension of the Euler-Lagrange equations in theRiemannian setting to manifolds with boundary (as was done in[19]) then yields the corresponding formulas in the pseudoRiemannian setting with non-degenerate boundary.
2. There are analogous results for the Euler-Lagrange equations inthe Kahler setting defined by Chern forms that extend work ofGilkey-Park-Sekigawa [20] for Kahler geometry to pseudo-Kahlergeometry.
[19] P. Gilkey, J.H. Park, and K. Sekigawa, “Universal curvatureidentities III”, International Journal of Geometric Methods inModern Physics 106 (2013), 1350025.[20] P. Gilkey, J.H. Park, and K. Sekigawa, “Universal curvatureidentities and Euler Lagrange Formulas for Kaehler manifolds”, toappear Joint with J. H. Park and K. Sekigawa, to appear J. Math.Soc. of Japan. http://arxiv.org/abs/1311.2622
Peter B Gilkey The Gauss Bonnet Theorem