RIEMANNIAN MANIFOLDS WHOSE SKEW-SYMMETRIC
CURVATURE OPERATOR HAS CONSTANT EIGENVALUES
Peter B. Gilkeyy, John V. Leahy, Hal Sadofsky y
Abstract. A Riemannian metric on a manifold is said to be IP if the eigen*
*values
of the skew-symmetric curvature operator are pointwise constant, i.e. the*
*y depend
upon the point of the manifold but not upon the particular 2 plane in the*
* tangent
bundle at that point. We classify the IP metrics in dimensions m = 5, m =*
* 6, and
m 9.
x0 Introduction
Let R be the curvature of a connected Riemannian manifold Mm of dimension
m. If X is a unit tangent vector in the tangent space TP Mm to Mm at a point
P , let J(X) : TP Mm ! TP Mm be the Jacobi operator; J(X) : Y 7! R(Y; X)X.
If Mm is a local 2-point homogeneous space, then the local isometries of Mm a*
*ct
transitively on the bundle of unit tangent vectors S(Mm ) so that the eigenvalu*
*es
of J(X) are constant as X varies in S(Mm ). Osserman conjectured [12] that the
converse might hold. Chi [5] showed this to be the case if m is odd, if m 2 mod
4, or if m = 4; the case m = 4k + 4 for k 1 remains open in this conjecture. C*
*hi's
proof is a lovely blend of algebraic topology and differential geometry and use*
*s in
an essential fashion work of Adams [1] concerning vector fields on spheres.
If ss is an oriented 2 plane in the tangent bundle, let R(ss) := R(X; Y ) be*
* the
skew-symmetric curvature operator; here {X; Y } is any oriented orthonormal bas*
*is
for ss and R(ss) is independent of the particular {X; Y } chosen. In this pape*
*r,
we study when the eigenvalues of R(ss) are independent of ss; we are motivated
at least partially by the results cited above for the Osserman conjecture. Let
Rijkl:= (R(ei; ej)ek; el) be the components of the curvature tensor relative to*
* a
local orthonormal frame ei where (.; .) denotes the inner product. Then we have
the following relations; these are the curvature symmetries and the first Bianc*
*hi
identity:
(0.1) Rijkl= -Rjikl; Rijkl= Rklij; and Rijkl+ Riklj+ Riljk= 0:
Definition. A 4 tensor R defined at a point P of Mm is an algebraic curvature
tensor if the identities of equation (0.1) hold at P .
Note that if R is an algebraic curvature tensor, then there exists a metric "ge*
*xtending
the metric on TP Mm so that R is the curvature tensor of "gat P .
______________
1991 Mathematics Subject Classification. Primary 53B20.
Key words and phrases. Curvature operator, K theory.
yResearch partially supported by the NSF (USA).
Gilkey-Leahy-Sadofsky W09v7 printed 19 January 1998
Typeset by AM S-T*
*EX
1
Definition. Let Gr+2(TP Mm ) be the Grassmanian of oriented 2 planes in TP Mm .
We say that an algebraic curvature tensor R is IP if the eigenvalues of R(ss) a*
*re
constant on Gr+2(TP Mm ); let the rank of R be dim Range (R(ss)).
We have chosen this notation as the fundamental papers in this subject are due *
*to
Ivanov and Petrova [9]; see also related work in [3, 7, 8].
Definition. We say a metric g is IP if the associated curvature tensor R is IP*
* at
every point; such a metric has rank at most k if rank(R) k everywhere.
Note that if g is IP, then the eigenvalues of R(ss) are constant on Gr+2(TP Mm )
for each P but can vary with P . If rank (R) = 0, then R = 0; if g has rank 0
everywhere, then g is flat. Ivanov and Petrova [9] give the following examples *
*of IP
metrics:
Example 0.2.
(1) Let g be a metric of constant sectional curvature C on a manifold Mm . *
*The
curvature tensor is CR for R(X;pY;_Z; W ) := (X; W )(Y; Z) - (X; Z)(Y; *
*W ).
The eigenvalues of R(ss) are { -1 C(t); 0; :::; 0}. Let {X; Y; Z} be a*
*n or-
thonormal set. Then
R(X; Y )X = -CY; R(X; Y )Y = CX; R(X; Y )Z = 0:
(2) Let M = I x N be a product manifold where I is a subinterval of R and
where ds2N is a metric of constant sectional curvature K on N. Give Mm *
* the
2+At+B
metric ds2M :=pdt2_+ f(t)ds2N where f(t) := Kt________2> 0. The eigenva*
*lues
2
of R(ss) are { -1 C(t); 0; :::; 0} for C(t) := 4KB-A___4f(t)2. If {@t*
*; X; Y; Z} is
an orthonormal set, then
R(X; Y )X = -C(t)Y; R(X; Y )Y = C(t)X; R(X; Y )Z = 0; R(X; Y )@t = 0
R(@t; X)X = -C(t)@t; R(@t; X)Y = 0; R(@t; X)Z = 0; R(@t; X)@t = C(t)X:
If Mm has constant sectional curvature, the local isometries of Mm act transi*
*tively
on the set of all 2 planes in the tangent bundle of Mm so the metric is global*
*ly IP.
In Lemma 3.3, we will show the metrics in (2) are IP; note that if A2 - 4BK 6= *
*0,
then the metric in (2) does not have constant sectional curvature. We have the
following examples of algebraic IP curvature tensors.
Example 0.3. Let R be the algebraic curvature tensor associated to a metric of
constant sectional curvature +1 given above. If OE is an isometry of Rm with O*
*E2 = Id
and if C 6= 0, set ROE;C(X; Y ) = CR(OEX; OEY ).
We will show these algebraic curvature tensors are IP in Lemma 2.3. We can give*
* a
geometric realization of the algebraic curvature tensor ROE;Cas follows. Decomp*
*ose
Rm = Rp x Rq into the 1 eigenvalues of OE. Let ds2xand ds2ybe the flat metrics*
* on
Rp and Rq. Then R is the curvature tensor of the following metric at the origin*
*; as
we shall not need this fact, we omit the details:
ds2 := {1 + 1_2C(|y|2 - |x|2)}ds2x+ {1 + 1_2C(|x|2 - |y|2)}ds2y:
Note that the curvature tensor R given in Example 0.2 (2) corresponds to ROE;C
where OE has eigenvalue -1 with multiplicity 1 and where C(t) is as given above.
2
The following Theorem is the main result of this paper; it shows that the IP
metrics of Example 0.2 and the algebraic IP curvature tensors of Example 0.3 are
the only examples if m = 5, m = 6, or m 9.
Theorem A. Let m 5.
(1) Let R be an algebraic IP curvature tensor. If m 6= 7; 8, then rank(R) *
*2.
(2) An algebraic curvature tensor R is IP with rank (R) = 2 if and only if
R = ROE;Cas in Example 0.3. Furthermore, ROE;C= R"OE;C"if and only if
C = "Cand OE = O"E.
(3) A metric g is IP of rank 2 everywhere if and only if g is locally isome*
*tric
to one of the metrics given in Example 0.2.
(4) If g is an IP metric of rank at most 2, then either g is flat or g has *
*rank 2
everywhere.
Ivanov and Petrova [9] constructed IP metrics in dimension m = 3 which are
not of the form given Example 0.2 and classified the IP metrics in dimensions 4.
We therefore assume m 5 henceforth. Gilkey and Petrova [6] have shown that
Theorem A (1) holds if m = 8; thus only the case m = 7 is still open.
In contrast to the methods of [9] used in dimension 4 which are purely diffe*
*rential
geometric, we will use topological methods to prove Theorem A (1). Let so() be
the Lie algebra of the orthogonal group; so() is the vector space of skew-symme*
*tric
x real matrices. Let Gr+2(m) be the Grassmanian of oriented 2 planes in Rm .
Definition. We say that R : Gr+2(m) ! so() is admissible if R is continuous,
if R(-ss) = -R(ss), and if dim ker R(ss) is constant. If R is admissible, then*
* let
rank (R) := dim Range (R(ss)).
Theorem A (1) will follow from the following result:
Theorem B. Let m 5 and let R : Gr+2(m) ! so() be admissible.
(1) If = m and if m 6= 7, 8, then rank(R) 2.
(2) If < m and if m 6= 8 then R(ss) = 0.
(3) There exists an admissible R : Gr+2(8) ! so(8) so that rank(R) = 8.
(4) There exists an admissible R : Gr+2(8) ! so(7) so that rank(R) = 6.
(5) There exists an admissible R : Gr+2(7) ! so(7) so that rank(R) = 6.
There are similar investigations in the literature dealing with an analogous
problem. Let L(m; n; k) be the dimension of the largest linear subspace V in
Hom (Rm ; Rn) so that the rank of 0 6= v 2 V is k. Various authors have trie*
*d to
bound L for given k where k is relatively large. Adams [1] determines L(m; m; m*
*).
Lam and Yiu [10] determine L(m; m; m-1), L(m; m-1; m-2), and L(m; m; m-2).
There is an extensive literature on the subject; see the bibliographies in Lam *
*and
Yiu [10] and in Meshulam [11] for further references.
Here is a brief guide to the paper. In x1, we will give the proof of Theo-
rem B. If R is admissible, dim ker(R(ss)) is constant so W0(ss) := ker R(ss) and
W1(ss) := W0(ss)? = Range (R(ss)) define vector bundles over Gr+2(m). Since
R(-ss) = -R(ss), the Wi descend to define vector bundles Vi over the unoriented
3
Grassmanian Gr2(m). The map which sends to span{; em } defines the canonical
embedding of RP m-2 in Gr2(m); let Ui be the restriction of Vi to RP m-2 . The
restriction of the non-trivial real line bundle
L := Gr+2(m) x R=(ss; ) ~ (-ss; -)
over Gr2(m) to RP m-2 is the non-trivial real line bundle over RP m-2 ; thus we*
* may
use L to denote both line bundles without fear of confusion. Note that we have
W0 W1 = m . 1. We also have R(-ss) = -R(ss). This equivariance property
implies that R descends to induce an isomorphism between V1 and V1 L over
Gr2(m). We have:
W0(ss) := kerR and W1 := W0? = Range (R) over Gr+2(m);
(0.4) V0 V1 = m . 1 and V1 = V1 L over Gr2(m);
U0 U1 = m . 1 and U1 = U1 L over RP m-1 :
We will use the Stieffel-Whitney classes and these observations to prove assert*
*ion
(1) of Theorem B. If m 10, we use the cohomology ring of RP m-2 ; if m = 5, if
m = 6, or if m = 9, we use the cohomology ring of Gr2(m). We will derive assert*
*ion
(2) similarly. We will use the spin representations to prove the remaining asse*
*rtions
of Theorem B by constructing suitable examples.
In x2, we will prove Theorem A (2). In x3, we will use the second Bianchi id*
*entity
(0.5) Rijkl;n+ Rijln;k+ Rijnk;l= 0
to prove Theorem A (3). In x4, we prove Theorem A (4). Theorem A classifies the
IP metrics in dimensions m = 5, m = 6, and m 9; it also classifies the IP metr*
*ics
of rank at most 2 in dimensions m = 7 and m = 8. The cases m = 7 and m = 8 are
exceptional in Theorem B; we refer to [6] for a proof of Theorem A (1) if m = 8;
we do not know if Theorem A (1) fails if m = 7.
We summarize below for the convenience of the reader some notational conven-
tions we shall use. We do not adopt the Einstein convention; we do not sum over
repeated indices in this paper.
(1) Let A2(m) be the set of the algebraic IP curvature tensors of rank 2.
(2) Let G2(m) be the set of the IP metrics of rank 2.
(3) Let R(X; Y; Z; W ) := (X; W )(Y; Z) - (X; Z)(Y; W ).
(4) Let I(m) be the set of isometries OE of Rm so OE2 is the identity.
(5) For OE 2 I(m), let p(OE) and q(OE) be the dimensions of the +1 and -1
eigenspaces.
(6) Let W0(T ) := ker(T ) and W1(T ) := ker(T )? = Range (T ) for T 2 so(m).
(7) Let so2(m) = {T 2 so(m) : rank(T ) = 2}.
4
x1 The proof of Theorem B
If U is a real vector bundle over a topological space X, let w(U) be the tot*
*al
Stiefel-Whitney class of U. We have:
a) w(U) = 1 + w1(U) + w2(U) + ::: for wi 2 Hi(X; Z2).
b) wi(U) = 0 for i > dim (U).
c) w(U V ) = w(U)w(V ).
d) If U is a trivial bundle, then w(U) = 1.
e) w is natural with respect to restriction.
Let [U] denote the corresponding element in the K theory group KO(X). The
work of Adams [1] shows the elements [1] and [L] generate KO(RP m-2 ) and that
[L] - [1] is an element of order 2OE(m-2)where OE(m - 2) is given below.
If R : Gr+2(m) ! so(m) is admissible, let Wi, Vi, and Ui be as given in equa*
*tion
(0.4). We will show dim (U1) 2 if m = 5, m = 6, or m 9. Decompose
[Ui] = ai([L] - [1]) + dim(Ui)[1] in KO(RP m-2 )
where ai is well defined modulo 2OE(m-2).
1.1 Lemma. Let m = 5, let m = 6, or let m 9. Choose j so 2j m - 2 < 2j+1.
Let R be admissible.
(1) If x = w1(L), then H*(RP m-2 ; Z2) = Z2[x]=xm-1 .
(2) We have OE(0) = 0, OE(1) = 1, OE(2) = 2, OE(3) = 2, OE(4) = 3, OE(5) = *
*3,
OE(6) = 3, OE(7) = 3, and OE(8k + j) = 4k + OE(j).
(3) We have w(Ui) = (1 + x)ai and a0 + a1 0 mod 2OE(m-2).
(4) We have dim (U0) 6= 0 and 2a1 dim (U1) mod 2OE(m-2).
(5) If 2a1 dim (U1) mod 2j+2, then rank(R) 2.
Proof. Assertion (1) is well known. Assertion (2) follows from the work of Adams
[1, Theorem 7.4]. Assertion (3) follows from the fact that U0 U1 = m . 1 is tr*
*ivial.
If dim (U0) = 0, then m([1] - [L]) = [U1] - [U1 L] = 0 in KO(RP m-2 ) since
U1 = U1 L by equation (0.4). This implies 2OE(m-2) divides m. The following
table is immediate from the definitions which we have given:
Table 1.2
_________________________________________
| | | | || | | | | |
|_________m_|5_|6_|68||9_|10_|11_|12_|13_|
| | | | || | | | | |
|_____m_-_2_|3_|4_|56||7_|_8_|_9_|10_|11_|
| || | | | | | | | |
|_OE(m_-_2)_2||3_|3_|3_|3_|4_|_5_|6_|__6_|
| | | | || | | | | |
|_____j(m)__|1_|2_|22||2_|_3_|_3_|3_|__3_|
It is clear that 2OE(m-2)does not divide m in this range for m 6= 8. Since OE(m*
* - 2)
grows roughly linearly with slope 1_2, 2OE(m-2)> m for m 9 and hence 2OE(m-2)d*
*oes
5
not divide m. This shows that dim (U0) > 0. We complete the proof of assertion
(4) by equating coefficients of ([L] - [1]) mod 2OE(m-2)in the equation:
[U1] =a1([L] - 1) + dim(U1)[1]
= [U1 L] = a1([1] - [L]) + dim(U1)[L]
= (dim (U1) - a1)([L] - 1) + dim(U1)[1]:
Assume R 6= 0. Decompose m - 2 = 2j1 + ::: + 2jr + ffi in a 2-adic expansion
where ffi = 0; 1 and j = j1 > ::: > jr > 0. Recall that ai is defined mod 2OE(*
*m-2)
and since 2j m - 2 < 2j+1, we have 2OE(m-2) 2j+1. So we can take ai with
0 ai< 2j+1 so ai ai mod 2j+1; then w(Ui) = (1 + x)ai. If 2a1 dim (U1) mod
2j+2, then a1 1_2dim(U1) mod 2j+1. As dim (U1) is even and as 1 dim (U0),
2 dim (U1) 2j+1. Then
1 a1 2j; a0 = 2j+1 - a1; and 2j a0 2j+1 - 1:
Choose s maximal with 1 s r so that the powers 2j1, ..., 2js appear in the
2-adic expansion of a0. Let := 2j1 + ::: + 2js m - 2. Then the coefficient *
*of
x is non-trivial in w(U0) = (1 + x)a0. Thus for dimensional reasons dim (U0).
Suppose that s < r. As dim U1 is even, ffi appears in the 2 adic expansion of
dim (U0). Then
dim U1 = m - dim U0 (m - 2) - - ffi + 2
= 2js+1+ . .+.2jr+ 2 2 . 2js+1 and
a0 = 2j+1 - 1_2dim(U1) 2j + 2j-1 + . .+.2js+1
contradicting the maximality of s. Thus s = r, = m - 2 - ffi dim (U0), and
dim (U1) 2 + ffi. Since dim (U1) is even, dim (U1) 2.
Let m 11. We use Table 1.2 to see that OE(m - 2) j(m) + 2; the function
OE grows linearly and the function j grows logarithmically. Consequently we ha*
*ve
2a1 dim (U1) mod 2j+2 and we can apply Lemma 1.1 to see rank (R) 2. To
complete the proof of Theorem B (1) if m = 5, m = 6, m = 9, or if m = 10, we
must eliminate the possibility that 2a1 dim (U1) + 2j+1 mod 2j+2.
If m = 10, 2j = 8, H*(RP 8) = Z2[x]=x9 and OE(m - 2) = 4. There are 3 cases:
(1) dim(U1) = 8, dim (U0) = 2, a1 = 12. Then w(U1) = (1 + x)12 so that
w(U0) = (1 + x)4. This contains x4 and is impossible since dim (U0) = 2.
(2) dim(U1) = 6, dim (U0) = 4, a1 = 11. Then w(U1) = (1+x)11. This contains
x8 and is impossible since dim (U1) = 6.
(3) dim(U1) = 4, dim (U0) = 6, a1 = 10. Then w(U1) = (1+x)10. This contains
x8 and is impossible since dim (U1) = 4.
We will use the cohomology of the Grassmanian Gr2(m) to study the cases
m = 5, 6, and 9. We decompose m . 1 = E2 E?2 where E2 is the classifying
2 plane bundle over Gr2(m); E2 := {(ss; ) 2 Gr2(m) x Rm : 2 ss}. We define
w := w(E2) = 1 + w1 + w2 and w := w(E?2) = 1 + w1 + w2 + ::::. We compute that:
6
Table 1.3
____________________________________________________
| | 2 |
|_w1_=_w1__________|w2_=_w1_+w2___________________|_
| 3 | 4 2 2 |
|_w3_=_w1__________|w4_=_w1_+w1w2_+_w2___________|__
| 5 2 | 6 4 3 |
|_w5_=_w1_+_w1w2__|_w6_=_w1_+_w1w2_+_w2___________|_
| w = w7 |w = w8 + w6w + w4w2 + w4 |
|__7____1__________|_8____1____1_2____1_2____2___|__
We refer to Borel [4] for the proof of the following result:
1.4 Lemma. H*(Gr2(m); Z2) = Z2[w1; w2]=(wi = 0) for m - 1 i.
If dim (V0) = 1, then V0 = 1 or V0 = L; this is determined by w(V0) or equiv*
*a-
lently by w(U0). There are 5 remaining cases to rule out to prove Theorem B (1);
we use the calculations of Table 1.3.
(1) m = 5, 2j = 2, dim V1 = 4, dim V0 = 1, and a1 = 0. Then w(U1) = 1 so
w(U0) = 1 and U0 is the trivial line bundle; thus V0 is the trivial lin*
*e bundle.
5 = V0 V1, so 5 . L = V0 L V1. Then we have (1 + w1)w(V1) = (1 + w1)5
in H*(Gr2(5); Z2) so w(V1) = (1+w1)4. This implies 1 = w(V1) = (1+w1)4
so w41belongs to the span of w41+ w21w2 + w22in Z2[w1; w2] which is fal*
*se.
(2) m = 6, 2j = 4, dim V1 = 4, dim V0 = 2, and a1 = 6. Then we have
w(U1) = (1 + x)6 so that w(U0) = (1 + x)2. Thus U0 and hence V0 are
orientable. Note that dim Gr2(6) = 8. We have w(V0) = 1 + w21+ ffw2 and
w(V1) = w(V0)-1 = (1 + w21+ ffw2)7 so (w21+ ffw2)3 = 0 in H6(Gr2(6); Z2)
since dim (V1) = 4. Thus w61+ ff(w41w2 + w21w22+ w32) belongs to the sp*
*an of
w1w5 = w61+ w21w22and w6 = w61+ w41w2 + w32in Z2[w1; w2]; this is false.
(3) m = 9, 2j = 4, dim (V1) = 8, dim (V0) = 1, a1 = 0. Then w(U1) = 1 so
w(U0) = 1; Thus U0 and V0 are trivial. Since 9 . L = V0 L V1, we have
(1 + w1)w(V1) = (1 + w1)9 in H*(Gr2(9); Z2) so w(V1) = (1 + w1)8. This
implies 1 = w(V0)w(V1) = (1 + w1)8 so w81= 1. This implies w81belongs to
the span of w8 = w81+ w61w2 + w41w22+ w42in Z2[w1; w2] which is false.
(4) m = 9, 2j = 4, dim (V1) = 6, dim (V0) = 3, a1 = 7. Then w(U1) = (1 + x)7
in Z2[x]=x8. This contains x7 and is impossible since dim (U1) = 6.
(5) m = 9, 2j = 4, dim (V1) = 4, dim (V0) = 5, a1 = 6. Then w(V1) = (1 + x)6
in Z2[x]=x7. This contains x6 and is impossible since dim (V1) = 4.
This completes the proof of Theorem B (1). We now prove Theorem B (2). Let
m 5 and m 6= 8. Let R : Gr+2(m) ! so() be admissible and non-trivial for
< m. By replacing R by R 0 if necessary we may suppose = m - 1. Suppose
first m 6= 7. Apply Theorem B (1) to R 0 to see dim U1 = 2, where we decompose
[m] = W0 W1 using R 0 as in (0.4). Thus a1 = 1 or a1 = 1 + 2j. If a1 = 1, then
a0 = 2j+1 - 1 and w(U0) = (1 + x)a0 = 1 + x + x2 + ::: + xm-2 ; write U0 = "U0 *
*[1]
since the decomposition given in equation (0.4) uses R 0. Then w(U"0) = w(U0),
which implies dim U"0 m - 2. This is false since dimjU"0 - 2 < m - 2. If
a1 = 1 + 2j, then w(U1) = (1 + x)a1. This contains x2 so 2j dim (U1) = 2.
Thus j = 1 and m = 5. Since m - 2 = 3, x2j+1 = x3 survives in w(U1). Thus
3 dim (U1) = 2 which is false.
7
Suppose next m = 7. We have 5 cases to eliminate to complete the proof of
Theorem B (2); note that dim (V0) + dim(V1) = 6.
(1) dim(V1) = 6, dim (V0) = 0. Then 6([L] - [1]) = 0 in gKO (RP 5) which is*
* false
since [L] - [1] has order 8.
(2) dim(V1) = 4, dim (V0) = 2, and a1 = 2. Then w(U1) = (1 + x)2 so that
w(U0) = (1 + x)6 in Z2[x]=x6. This contains x4 and is impossible since
dim(U0) = 2.
(3) dim(V1) = 4, dim (V0) = 2, and a(U1) = 6. Then w(U1) = (1 + x)6 so
w(U0) = (1 + x)2 in Z2[x]=x6. This shows that w(V0) = 1 + w21+ ffw2 and
that w(V1) = (1 + w21+ ffw2)7. Thus 0 = w61+ ff(w21w42+ w41w2 + w32) in
H6(Gr2(7)) since dim (V1) = 4. Thus w61+ ff(w21w42+ w41w2 + w32) belongs
to the span of w6 = w61+ w41w2 + w32in Z2[w1; w2] which is false.
(4) dim(V1) = 2, dim (V0) = 4, and a(U1) = 1. Then w(U1) = 1 + x so we have
w(U0) = (1 + x)7. This contains x5 and is impossible since dim (U0) = 4:
(5) dim(V1) = 2, dim (V0) = 4, and a(U1) = 5. Then w(U1) = (1 + x)5. This
contains x5 and is impossible since dim (U1) = 2.
We now prove assertions (3), (4), and (5) of Theorem B. We refer to Atiyah, *
*Bott,
and Shapiro [2] for further details concerning Clifford algebras and spinors. R*
*ecall
that the Clifford algebra Clif(m) is the universal real unital algebra generate*
*d by
Rm subject to the commutation relations v * w + w * v = -2(v; w)1. Let Spin(m)*
* be
the subset of Clif(m) generated by all products of the form ! = v1*:::*vk where*
* the
vi are unit vectors in Rm and k is even. This is a smooth submanifold of Clif(*
*m)
which has the structure of a Lie group under Clifford multiplication. If {v1; *
*v2}
is an oriented orthonormal basis for a 2 plane ss, let oe(ss) := v1 * v2 2 Spin*
*(m);
this is independent of the particular oriented orthonormal basis for ss and def*
*ines
a smooth embedding oe : Gr+2(m) 7! Spin(m). There is a natural representation
ae : Spin(m) 7! SO (m) that exhibits Spin(m) as the universal cover of SO (m) f*
*or
m 3; ae(oe(ss)) is -1 on ss and +1 on ss? .
Suppose m = 8. The spin representation c gives a representation of the Clif-
ford algebra Clif(8) on R16. There is a decomposition R16 = R8+ R8- so that
Clifford multiplication by a vector v 2 R8 interchanges the two summands, i.e.
c(v) : R8 ! R8 . Clifford multiplication by an element ! 2 Spin(m) preserves the
two summands; thus c restricts to define representations, called the half spin *
*repre-
sentations, c of Spin(8) on R8 . Let R(ss) = c+ (oe(ss)); R is an admissible m*
*ap from
Gr+2(8) to so(8) which has constant rank 8 since R(ss)2 = -1 and R(-ss) = -R(ss*
*).
Fix e8 2 R8+. Let E(ss) := span {e8; R(ss)e8}. This is invariant under R(s*
*s) so
we may decompose R = R0 R1 where R0 is the restriction of R to E and R1 is
the restriction of R to E? . We have E? e?8 = R7; the map ss 7! R1(ss) is an
admissible map from Gr+2(8) to so(7) which has constant rank 6; the restriction*
* of
R1 to Gr+2(7) Gr+2(8) defines an admissible map from Gr+2(7) to so(7) which has
rank 6. This completes the proof of Theorem B.
8
x2 The proof of Theorem A (2)
If T 2 so(m), define !(T ) 2 2(Rm ) by !(T )(; j) = (T ; j); ! is an isomor-
phism from so(m) to 2(Rm ). Let (T1; T2) := - 1_2Tr(T1T2) be the Killing metric
on so(m). Let Tiej:= R(ei; ej) = !-1 (dej ^ dei) where e is an orthonormal bas*
*is
for Rm . Then {Tiej}i 0. Then !(T ) = 1arank(T) adua^dva. Consequently
T has rank 2 if and only if !(T )2 = 0 which proves assertion (1).
If the hypotheses of assertion (2) are satisfied, expand T (f2) = k 3. By Lemma 2*
*.1,
ae13ae2i= ae1iae23. Thus ae1i= 0 implies ae2i= 0 for i > 3 and T2 = ae13T1e3+ a*
*e23T2e3.
Thus dim (W1(T1) + W1(T2)) 3 and W1(T1) \ W1(T2) 6= 0.
Choose the orthonormal basis e for Rm so e1 2 W (T1) \ W (T2), T1 = T1e2,
and so that T2 = T1e3. Let f be a unit vector with f ? f1 and f ? f2. Expand
T (f) = ijaeij;fTiej. By Lemma 2.1, ae12;f= 0 and aeij;f= 0 for (i; j) disjoint*
* from
(1; 2); similarly ae13;f= 0 and aeij;f= 0 for (i; j) disjoint from (1; 3). Cons*
*equently
T (f) = ae23;fT2e3+ i>3ae1i;fT1ei:
If i > 3, then ae23;fae1i;f= ae13;fae2i;f; thus ae23;fae1i;f= 0 since ae2i;f= 0*
*. Since T
is an isometry, i3ae1i;fT1ei, and e1 2 W1(T (f*
* ))
for any .
We note that Lemma 2.2 can fail if k = 3. Let T () = 1T23 - 2T13 + 3T12;
T is an isometry from R3 to so2(3) with no common non-trivial eigenvector. Thus
the restriction that k 4 is an essential one. We establish the first equivalen*
*ce in
assertion (2) of Theorem A by proving:
2.3 Lemma. Let m 5. We have R 2 A2(m) if and only if R = ROE;Cfor some
OE 2 I(m) and for some C 6= 0.
Proof. Suppose R = ROE;C. Since OE2 is the identity, R(X; Y ) = OE-1 CR(X; Y )O*
*E so
R has rank 2. Choose an orthonormal basis for Rm so that OEea = ea for a p and
so that OEeff= -efffor ff > p. The non-zero curvatures are
R(ea; eb)eb = Cea and R(ea; eb)ea = -Ceb fora < b p;
(2.4) R(eff; efi)efi= Ceffand R(eff; efi)eff= -Cefifor p < ff < fi;
R(ea; eff)eff= -Cea and R(ea; eff)ea = Cefffor a p < ff:
We note that Rijkl = 0 unless (i; j) = (k; l) or (i; j) = (l; k). The curvature
symmetries of equation (0.1) are now immediate so R is an algebraic curvature
tensor.
Conversely, let R 2 A2(m). By rescaling R, we may assume the eigenvalues of
R2 are 0 and -1; this means that |R(ss)| = 1. Let f = {fi} be an orthonormal
basis for Rm . Let T (f) := R(f1; f) for f ? f1. Then |T (f)| = |R(f1; f)| = |f*
*| so T
is a linear isometry. We apply Lemma 2.2 to choose a unit vector 2 Rm so that
T (f) = R(-T (f)) for all f 2 f?1. Let e1 = . For i > 1, let T (fi)e1 = ei; e1 *
*? ei.
Since T is an isometry, e := {e1; :::; em } is an orthonormal basis for Rm . Wi*
*th this
normalization, we have
R(f1; fi) = T (fi) = -T1eifori > 1:
Expand R(f ; f ) = i 1. Then R(fi; fj) = -Ti"ejfor all i and j.
(3) We have ae; = 0 for all 1 < < . Then R(f ; f ) = i>1;i6= aei; T1ei.
Consequently e1 2 W1(R(ss)) for all ss. Let {e1; f2; f3} be an orthono*
*r-
mal set and let e4 := R(f2; f3)e1. Since (R(f2; f3)e1; e4) = 1, the cu*
*rva-
ture symmetries given in equation (0.1) imply (R(e1; e4)f2; f3) = 1. T*
*hus
f2 2 W1(R(e1; e4)) and f3 2 W1(R(e1; e4)) so e1 62 W1(R(e1; e4)); this
contradiction eliminates this case from consideration.
Let OE(fi) := ei define an isometry of Rm . Then R(X; Y ) = R(OEX; OEY ). *
*We
complete the proof by showing OE2 is the identity. If || = 1, let
P() := \|j|=1;j?W1(R(; j)):
By Lemma 2.2, dim (P()) = 1. Furthermore OE() 2 P(). Note that {OE(); OE(j)}
is an orthonormal basis for W1(R(; j)). The curvature symmetry
(R(; j)OE(); OE(j)) = (R(OE(); OE(j)); j)
shows that {; j} is an orthonormal basis for W1(R(OE(); OE(j))) so 2 P(OE()).
Consequently = ffiOE(OE()) for ffi = 1. To complete the proof, we must show
ffi = 1. We assume ffi = -1 and argue for a contradiction. If ffi = -1, then OE*
* defines
a complex structure on Rm . Choose an orthonormal basis for Rm so OE(ffa) = f*
*ia
and so OE(fia) = -ffa. As R satisfies the Bianchi identities given in equation *
*(0.1),
0 = R(ffa; fib)ffb + R(fib; ffb)ffa + R(ffb; ffa)fib
= {-R(fia; ffb)ffb - R(ffb; fib)ffa + R(fib; fia)fib}
= {-fia + 0 - fia} 6= 0:
The second equivalence in assertion (2) of Theorem A will follow from the fo*
*l-
lowing Lemma:
11
2.5 Lemma. Let m 5.
(1) We have ROE;C= R"OE;C"if and only if C = "Cand OE = O"E.
(2) Let Ri 2 A2. Then R1 and R2 are orthogonally equivalent if and only if
C1 = C2 and p(OE1) = p(OE2) or p(OE1) = q(OE2).
(3) We have A2(m) is a smooth submanifold of 4Rm .
(4) Let R be a smooth map from a simply connected manifold to A2(m). Then
there exists (OE; C) smooth so R = ROE;C.
Proof. If R 2 A2(m), let V1R = Range (R(ss)) be the 2 plane bundle over Gr2(m)
discussed in the introduction. We define a continuous map ffR from Gr2(m) to
Gr2(m) by setting ffR (ss) = V1R(ss). Let -C2 6= 0 be the non-trivial eigenvalu*
*e of
R(ss)2.
If {X; Y } is an orthonormal basis for ss, then {Z; W } is an orthonormal ba*
*sis
for ffR (ss) if and only if (R(X; Y )Z; W ) = C. Thus the curvature symmetry
(R(X; Y )Z; W ) = (R(Z; W )X; Y ) shows that V1R(V1R(ss)) = ss so ff2Ris the id*
*entity.
We use equation (2.4) to see that the fixed point set of ffR is the disjoint un*
*ion of
Gr2(p), Gr2(q), and RP p-1xRP q-1 . We use these two Grassmanians to decompose
Rm into complementary orthogonal subspaces of dimensions p and q. If R = ROE;C,
these subspaces are the 1 eigenspaces of OE. Thus R determines OE up to sign.
Let aeR be the Ricci tensor of R; aeR is diagonal with respect to any basis *
*which
diagonalizes OE. By equation (2.4), aeR (ea; ea) = (p - q - 1)C for a p *
*and
aeR (eff; eff) = (q - p - 1)C for p < ff. If p = q or p = 0 or q = 0, then -C
is the only eigenvalue of aeR . Otherwise, there are two distinct eigenvalues;*
* the
eigenvalue with the greater multiplicity is (|q - p| - 1)C and the eigenvalue w*
*ith
the lesser multiplicity is (-|q - p| - 1)C. Thus R also determines C; this pro*
*ves
assertion (1). Two isometries OEi 2 I(m) are orthogonally equivalent if and onl*
*y if
p(OE1) = p(OE2); assertion (2) now follows.
We normalize the choice of OE so p(OE) q(OE); distinct values of p define d*
*ifferent
components Ap2(m) of A2(m) so we may fix p in proving assertion (3). Let
Ip(m) := {OE 2 I(m) : p(OE) = p}:
The orthogonal group acts transitively on Ip(m) by conjugation so Ip(m) is a
smooth homogeneous manifold. The parameter C ranges over R* := R - 0. If
p < q, then Ap2(m) = Ip(m) x R*. If p = q, the map OE 7! -OE defines a fixed
point free action of Z2 on Ip(m) and Ap2(m) = (Ip(m)=Z2) x R*. Assertion (3) now
follows.
Let R(P ) satisfy the hypothesis of assertion (4). The function C(P ) is uni*
*quely
defined and is smooth. If p < q, we can define OE uniquely by requiring that
p(OE(P )) < q(OE(P )). If p = q, we must define a lifting from Ap2(m) to the d*
*ouble
cover Ip(m) x R*; this is possible as the domain of R is simply connected.
12
x3 The proof of Theorem A (3)
In this section, we adapt arguments of Ivanov and Petrova [9]. Let g 2 A2(m)
for m 5 be an IP metric of rank 2. Let R be the curvature tensor of g. We
are interested in local questions so we may assume Mm is a ball in Rm and hen*
*ce
simply connected. Thus by Theorem A (2) and Lemma 2.5, R = ROE;Cfor smooth
OE and C. Let indices i, j etc. range from 1 through m, let indices a, b, etc. *
*range
from 1 through p, and let indices ff, fi range from p + 1 through m. Choose a l*
*ocal
frame e diagonalizing OE; this means OEea = ea and OEeff= -eff. Let OEij, OEij;*
*k, Rijkl,
and Rijkl;nbe the components of OE, rOE, R, and rR. Let F be the distributions
defined by the 1 eigenspaces of OE; the ea span F+ and the effspan F- . We begin
with a technical Lemma:
3.1 Lemma. Let m 5, let g 2 G2(m), and let R = ROE;C.
(1) Rijkl;n= C;n(OEilOEjk - OEikOEjl) + C(OEil;nOEjk + OEilOEjk;n- OEik;nOE*
*jl- OEikOEjl;n).
(2) We have OEij;k= OEji;kfor any i, j, and k.
(3) We have OEab;i= 0 and OEfffi;i= 0 for any a, b, ff, fi, and k.
(4) If i, j, and k are distinct, then OEij;k= OEik;j.
(5) If a 6= b, then OEaff;b= 0; if ff 6= fi, then OEaff;fi= 0.
(6) The Christoffel symbols iaff= - 1_2OEaff;i.
(7) The distributions F are integrable.
(8) If there exists ff 6= fi, then C;a= -COEfia;fi- COEffa;ffand C;fi= COEa*
*fi;a.
(9) If there exists a 6= b, then C;ff= COEffa;a+ COEffb;b, and C;b= -COEffb*
*;ff.
(10) If q 3, then C;ff= 0. If p 3, then C;a= 0.
(11) Either p 1 or q 1.
Proof. We covariantly differentiate the identity Rijkl= C(OEilOEjk - OEikOEjl) *
*to es-
tablish assertion (1). Since OE 2 I(m), OEij= OEji and assertion (2) follows.
We covariantly differentiate the relation ffiij = lOEilOElj to establish the*
* identity
0 = l{OEil;kOElj+OEilOElj;k}. By our hypotheses, OEab = ffiab, OEfffi= -ffifffi*
*, and OEaff= 0.
Thus we may set i = a and j = b in the previous identity to see 0 = 2OEab;k. Si*
*milarly
we may set i = ff and j = fi to see that 0 = -2OEfffi;k. This proves assertion *
*(3). (If
i = a and if j = fi, then the two terms cancel and we gain no new information).
Let i, j, and k be distinct. Since m 5, we may choose l distinct from i, j,*
* and
k. By assertion (1) Rillj;k= COEij;kOEll, Riljk;l= 0, and Rilkl;j= -COEik;jOEl*
*l. By
the second Bianchi identity of equation (0.5), C(OEij;k- OEik;j)OEll= 0; assert*
*ion (4)
follows as C 6= 0 and OEll= 1. If a 6= b, then OEaff;b= OEab;ff= 0 by assertion*
*s (3)
and (4). Similarly, if ff 6= fi, then OEaff;fi= OEfffi;a= 0. This proves assert*
*ion (5).
Let := 1_2(1 OE) be orthogonal projection on F ; eff= - effand ea = + ea.
We prove assertion (6) by computing:
iaff= (reiea; eff) = (reiea; - eff) = (- reiea; eff)
=(- reiea - rei- ea; eff) = -;affi= - 1_2OEaff;i:
Since abff= - 1_2OEaff;b= 0 for a 6= b, - ([ea; eb]) = - (reaeb - rebea) = 0.
This shows F+ is integrable; similarly F- is integrable and assertion (7) follo*
*ws.
13
Let ff 6= fi. We use assertion (1) to compute
Rfffififf;a= C;a; Rfffiffa;fi= COEfia;fi; Rfffiafi;ff= COEffa;ff;
Rffaaff;fi= -C;fi; Rffafffi;a= COEafi;a; and Rffafia;ff= 0:
Assertion (8) now follows from the second Bianchi identity and from assertion (*
*1).
We replace OE by -OE and interchange the roles of the greek and roman indices to
derive assertion (9) from assertion (8). If q 3, we may choose ff, fi, and fl *
*distinct
and compute:
Rflfififl;ff= C;ff; and Rflfiflff;fi= Rflfifffi;fl= 0:
The second Bianchi identity now implies C;ff= 0; similarly C;a = 0 if p 3. If
p 2 and q 2, we use assertions (2) through (9) to show rC = 0, rOE = 0, and
iaff= 0. Thus the distribution F+ is parallel and 0 = (R(eff; ea)ea; eff) = -*
*C;
this is false.
By replacing OE by -OE if necessary, we may suppose that p(OE) q(OE). Thus
p(OE) = 0 or p(OE) = 1. If p = 0, then Mm has constant sectional curvature C. *
*We
therefore suppose p(OE) = 1. Let y = (y1; :::; ym-1 ) be local coordinates on a*
* leaf of
the foliation F- . Let T (t; y) := expy(te1(y)) define local coordinates on Mm .
3.2 Lemma. Let m 5, let g 2 G2(m), let R = ROE;C, and let p(OE) = 1.
(1) We have C;ff= 0, C;1= -2COE1ff;fffor any ff, and ff1fi= 1_4ffifffiC-1 C*
*;1.
(2) For fixed y0, the curves t ! T (t; y0) are unit speed geodesics in Mm *
*which
are leaves of the foliation F+ .
(3) For fixed t0, the surfaces T (t0; y) are leaves of the foliation F- an*
*d inherit
metrics of constant sectional curvature.
(4) Locally ds2 = dt2 + fds2K where f(t) is a positive smooth function and *
*ds2K
is a metric of constant sectional curvature K.
Proof. We apply Lemma 3.1. Since p = 1, a = 1. Since q 3, C;ff= 0 and
C;1 = -C(OE1ff;ff+ OE1fi;fi) = -C(OE1ff;ff+ OE1fl;fl) so C;1 = -2COE1ff;fffor a*
*ny ff and
ff1fi= 1_4ffifffiC-1 C;1. Since 111 = 0 and 11ff= - 1_2OE1ff;1= - 1_4C-1 C;ff*
*= 0,
the integral curves for e1 are unit speed geodesics; assertion (2) now follows.*
* We
compute
@t(@t; @yff) = (@t; r@t@yff) = (@t; r@yff@t) = 1_2@yff(@t; @t) = 0:
This shows (@t; @yff) = 0 so the @yffspan the perpendicular distribution F- an*
*d the
manifolds T (t0; y) are leaves of the foliation F- . Since C;ff= 0, we have tha*
*t C and
ff1;fi= 1_4ffifffiC-1 C;1are constant on the leaves of F- . Let R- be the curva*
*ture of
the induced metric on the leaves of F- . Assertion (3) now follows from the ide*
*ntity
R- (eff; efi; efl; eoe) = R(eff; efi; efl; eoe) - fifl1ff1oe- fffl1fi1oe. Let *
*@yff= flaeffflefl.
Since C-1 C;1only depends on the parameter t, we show that the metric is a twis*
*ted
product by computing:
(r@t@yff; @yfi) = (r@yff@t; @yfi) = floeaeffflaefioe(refl@t; eoe)
= 1_4floeaeffflaefioeffifffiC-1 C;1= 1_4C-1 C;1gfffi;
@tgfffi= 1_2C-1 C;1gfffi:
14
Theorem A (3) will follow from Lemma 3.1, from Lemma 3.2, and from the
following Lemma that determines which warping functions give IP metrics. Let
f_= @tf and "f= @2tf.
3.3 Lemma. Let ds2 := dt2 + f(t)ds2Kwhere ds2Kis a metric of constant sectional
curvature K. Then ds2 is an IP metric of rank 2 if and only if f" = 2K i.e.
f = Kt2 + At + B where A and B are constants. Let C = _1_4f2(4KB - A2). Let
OE(@t) = @t and OE(X) = -X for X ? @t. Then R = ROE;C. The metric has constant
sectional curvature if and only if 4KB - A2 = 0.
Proof. Let y = (y2; :::; ym ) be geodesic polar coordinates on a leaf of the fo*
*liation
F- near some point y0; gfffi= ffifffi+ O(|y|2). Let " be the Christoffelpsym*
*bols_
relative to a coordinate frame @t, @yff; 2 ff m. We have eff(y0) = (@yff= f)*
*(y0)
and
(1) "ff1fi= "1fffi= 1_2ffifffi_f+ O(y2), "ff1fi= "1fffi= _1_2fffififf_f+ O(*
*y2),
(2) "fffi1= - 1_2ffifffi_f+ O(y2), "fffi1= - 1_2ffifffi_f+ O(y2),
(3) R(@yff; @yfi; @yfl; @yoe)(y0) = f{R- (@yff; @yfi; @yfl; @yoe) + "ff1oe"*
*fifl1- "fi1oe"fffl1}
= f(K - _1_4f_f2)(ffiffoeffififl- ffifioeffifffl),
(4) R(eff; efi; efl; effi)(y0) = 1_f(K - _1_4f_f2)(ffiffoeffififl- ffifioef*
*fifffl).
(5) R(@yff; @t; @t; @yfi)(y0) = f{-@t"ff1ff- "ff1ff"1ffff}ffifffi
= f{- 1_2f"f+ _1_2f2_f2- _1_4f2_f2}ffifffi.
(6) R(eff; e1; e1; efi)(y0) = {- 1_2f"f+ _1_4f2_f2}ffifffi.
The remaining curvatures vanish. Suppose g is an IP metric with p = 1. Then
K_- __f2_= -(- f"_+ __f2_):
f 4f2 2f 4f2
This shows K = 1_2"fso we may expand f = Kt2 + At + B where A and B are
constant. Conversely, if ds2 has this form, we use equation (2.4) to see R = RO*
*E;C
where OE(@t) = @t, OE(@yff) = -@yff, and
_f2 4K(t2K + At + B) - (2Kt + A)2 4KB - A2
C = K_f- ____4f2= ________________________________4f2= ___________4f2:
It is now immediate that ds2 has constant sectional curvature if and only if C *
*is
constant or equivalently if 4KB - A2 = 0; furthermore, the metric is flat in th*
*is
setting.
15
x4 The proof of Theorem A (4)
We suppose that Theorem A (4) fails. Let g be an IP metric of rank at most
2. Assume thatpg_has rank 0 at some point and that g has rank 2 at some other
point. Let -1 C for C 0 be the non-zero eigenvalues of R. As the manifold
is connected and C is continuous, there exists a unit speed geodesic fl defined*
* for
s 2 [a; b] so that C(s) := C(R(fl(s))) satisfies C(a) = 0 and C(b) 6= 0. The se*
*t of s
where C(s) = 0 is closed; let s0 be the least upper bound of this set. Then C(s*
*) 6= 0
for s 2 (s0; b] while C(s0) = 0.
Let p = p(R(fl(s))); this is independent of s 2 (s0; b]. If p = 0, the manif*
*old has
constant sectional curvature near fl(s) so C(s) 6= 0 is constant and C(s) does *
*not
tend to zero as s # s0. Thus p = 1.
Let t(s) measure the distance along the foliation F+ for s in the range (s0;*
* b];
Z b
t(s) := - (f_l(u); e1(fl(u))du
s
Note that |t(s)| |b - a| so t is uniformly bounded. Let R(fl(s)) = ROE(s);C(s)*
*. The
parameter t determines C(s) = C(t(s)); C(s) = |C(s)|. The numerator of C(t) in
Lemma 3.3 is constant. The denominator 4f2 can have zeros; near these zeros, C
tends to infinity. Although 4f2 ! 1 as t ! 1, the denominator is bounded
since |t(s)| |b - a| is uniformly bounded. Thus C is uniformly bounded away
from 0 for s 2 (s0; b] and thus C does not tend to 0 as s # s0. This provides *
*the
desired contradiction and completes the proof of Theorem A.
16
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Mathematics Department, University of Oregon, Eugene Or 97403 USA
E-mail address: gilkey@math.uoregon.edu, leahy@math.uoregon.edu,
& sadofsky@math.uoregon.edu
17