HOMOGENEOUS EINSTEIN METRICS

Transcription

1 HOMOGENEOUS EINSTEIN METRICS Megan M. Kerr A Dissertation in Mathematics Presented to the Faculties of the University of Pennsylvania in Partial Fulfillment of the Requirements for the Degree of Doctor of Philosophy 1995 Wolfgang Ziller, Thesis Advisor Wolfgang Ziller, Graduate Group Chairman 1

2 ACKNOWLEDGEMENTS I would like to thank some of the many people who have helped me along: I thank David Harbater, Steve Shatz, and Frank Warner for encouragement throughout my graduate studies. I also thank Chris Croke and Herman Gluck for generously sharing their expertise. I am especially indebted to my advisor, Wolfgang Ziller, for so many hours of his time, for his confidence in me, and most of all for introducing me to differential geometry. I thank Andy Hicks for sharing the daily ups and downs of research with me, and for reminding me that life is funny. The friendship of Seon-In Kwon and Kate Stevenson has been invaluable; their sense of humor, sympathy, and example inspire me. I thank Tom Powers for always listening. ii

3 ABSTRACT HOMOGENEOUS EINSTEIN METRICS Megan M. Kerr Wolfgang Ziller (Supervisor) We consider homogeneous Einstein metrics on symmetric spaces and we describe their geometry. For compact irreducible symmetric spaces with rank(m) > 1, not isometric to a Lie group, we classify all homogeneous Einstein metrics. Whenever there exists a closed proper subgroup G of Isom(M) acting transitively on M we find all G-homogeneous (nonsymmetric) Einstein metrics on M. We next examine three reducible symmetric spaces on which a simple Lie group G acts transitively, and we describe all G-invariant Einstein metrics. Finally, we discuss the SO(n + 1)-invariant Einstein metrics on the Stiefel manifold SO(n + 1)/ SO(n 1) = V (R n+1 ). iii

4 Contents Introduction 1 Chapter 0: Preliminaries 3 Chapter 1: Irreducible Symmetric Spaces SO(n)/U(n) SU(n)/Sp(n) G + (R7 ) G + 3 (R8 ) 3 Chapter : Products of Symmetric Spaces S 7 S S 6 S S 7 G + (R8 ) 44 Chapter 3: V (R n+1 ) 50 Appendix 54 References 56 iv

5 Introduction We study homogeneous Einstein metrics on compact symmetric spaces and Stiefel manifolds. In chapter 1 we look at compact symmetric spaces presented homogeneously, i.e. as M = G/H, where G = Isom 0 (M) is simple, and we consider the cases where there exists a closed subgroup G G which acts transitively on M. Denote by H the isotropy subgroup in G, then M = G /H. Since G is smaller than G, we expect more G -invariant metrics on M than G-invariant metrics, and thus we can hope for non-symmetric G -invariant Einstein metrics on our symmetric space M. We find the following. Lemma 1.1. Let M be a compact irreducible symmetric space of rank > 1, M not isometric to a compact Lie group with biinvariant metric. Let G = Isom 0 (M), and M = G/H. Then there exists a proper subgroup G G acting transitively on M (1) G = SO(n), H = U(n), G = SO(n 1), H = U(n 1) (n 4); () G = SU(n), H = Sp(n), G = SU(n 1), H = Sp(n 1) (n 3); (3) G = SO(7), H = SO() SO(5), G = G, H = U() (U() SU(3)); (4) G = SO(8), H = SO(3) SO(5), G = Spin(7), H = SO(4) (SO(4) G ). Theorem 1.. Among the compact irreducible symmetric spaces of rank > 1, not isometric to a Lie group with a biinvariant metric, only G + (R7 ), G + 3 (R8 ), and SO(n)/U(n), for n 4, carry non-symmetric homogeneous Einstein metrics (cases (3), (4), and (1), respectively). The Grassmannians G + (R7 ) and G + 3 (R8 ) each carry two and SO(n)/ U(n) carries one; the only homogeneous Einstein metric on SU(n)/ Sp(n) is the symmetric metric. The analogous results for symmetric spaces of rank 1 were studied in [Z] and for compact Lie groups with biinvariant metrics in [DA-Z]. 1

6 Remark. The following result should be true for all compact irreducible symmetric spaces M, up to diffeomorphism: If G is a compact connected Lie group acting transitively and effectively on M, then G is conjugate to a subgroup of Isom 0 (M). This would imply that Theorem 1. classifies all homogeneous Einstein metrics on compact irreducible symmetric spaces of rank > 1, not isometric to a Lie group. Such a result is well known for rank 1 symmetric spaces, but does not seem to be known for all symmetric spaces of rank > 1. Partial results can be found in [O], [O3], [S], [T]. For example, in [O3] Oniščik showed that if M is diffeomorphic to G k (R n ) for n even, n > 5, 1 < k < n n 3, or for n odd, < k <, and if a compact connected Lie group G acts transitively and effectively on M, then G is conjugate to SO(n) with the standard action. Tsukada proved in [T] that if M is diffeomorphic to G k+1 (R n ), and G is compact, connected, and simple, then if G acts transitively and effectively, the action of G is conjugate to the standard action. In chapter we consider three products of irreducible symmetric spaces on which simple compact Lie groups act transitively. We consider their homogeneous metrics to find the Einstein metrics among them. Proposition.1. (1) S 7 S 7 = Spin(8)/ G carries two distinct Spin(8)-invariant Einstein metrics: the product metric and the metric induced by the Killing form. () S 7 S 6 = Spin(7)/ SU(3) carries three distinct Spin(7)-invariant Einstein metrics: the product metric and two others. (3) S 7 G + (R8 ) = Spin(8)/ U(3) carries two distinct Spin(8)-invariant Einstein metrics: the product metric and one other.

7 For these examples, we could not consider diagonal metrics only; it was necessary to develop a scalar curvature formula which does not assume one has an orthonormal, or even orthogonal, basis. Few examples of this type, where diagonal metrics do not represent all homogeneous metrics, even up to isometry, have been previously examined. In chapter 3 we look at the Stiefel manifold V (R n+1 ) of two-flags in (n + 1)-dimensional Euclidean space. (One can also think of this space as the unit tangent bundle of the n- sphere S n.) This can be presented homogeneously as SO(n + 1)/ SO(n 1), and we classify the SO(n + 1)-invariant Einstein metrics. Chapter 0: Preliminaries A Riemannian manifold (M, g) is Einstein if Ric(X, Y ) = λg(x, Y ) for some constant λ, for all vector fields X, Y. We say a Riemannian manifold (M, g) is a symmetric space if for all p M there exists an isometry σ p : M M such that σ p (p) = p and (dσ p ) p = Id. Symmetric spaces make up a class of manifolds which includes spheres, projective spaces, and Grassmannians; their geometry is well understood. In fact, every symmetric space is homogeneous. A manifold M is defined to be G-homogeneous if we have a Riemannian metric g and a closed subgroup G Isom(M, g) such that for any p and q M, there exists a g G with g(p) = q. We write H p = {g G g(p) = p}, called the isotropy subgroup corresponding to p. Notice H p is compact, since H p O(T p M). Via the map g g(p) we identify the two manifolds G/H p and M. Any two isotropy subgroups H p and H q are conjugate: if q = g(p), then g 1 H q g = H p, hence we will usually suppress the point p. Given a homogeneous manifold G/H, where G is compact and H is closed, what metrics can we put on G/H so that G acts by isometries? 3

8 Just as a left-invariant metric on a Lie group is determined by any inner product on its Lie algebra, a G-invariant metric on G/H is determined by an inner product on g/h = T [H] (G/H), with the additional requirement that the inner product be Ad(H)-invariant. We identify the quotient g/h with an Ad(H)-invariant complement p to h in g; compactness of H guarantees such a p exists. If g is semisimple then the Killing form is Ad(H)-invariant and we can use it to define p = h. We want to consider all Ad(H)-invariant inner products on p. A homogeneous space M = G/H is said to be isotropy irreducible if the isotropy action, denoted χ : H GL(T p M), or equivalently Ad : H GL(p), is an irreducible representation of H. When this is the case, the G-invariant metric on G/H is unique, up to scaling, and it is Einstein. When G/H is a symmetric space with G simple and G = Isom 0 (G/H), then G/H is an irreducible symmetric space. (In fact, the only irreducible symmetric space with G not simple is (K K)/ K, for K a compact simple Lie group, and (K K)/ K with the symmetric metric is isometric to K with a biinvariant metric.) In 196 A.L. Oniščik classified all simple compact Lie algebras g with Lie subalgebras g and g, such that g = g + g. In terms of transitive group actions, let G be the simply connected compact Lie group corresponding to g and let G, G be Lie subgroups corresponding to g, g, respectively, then G/G = G /(G G ) and G/G = G /(G G ). When G/G or G/G is a symmetric space, Oniščik s list tells us when a subgroup of G still acts transitively. Comparing this with the classification of compact irreducible symmetric spaces one obtains Lemma 1.1. Here are the symmetric spaces on his list [O1]: (See the 4

9 appendix of this paper for the non-symmetric homogeneous spaces on his list.) SO(n)/ SO(n 1) = U(n)/ U(n 1) SO(n)/ SO(n 1) = SU(n)/ SU(n 1) SO(4n)/ SO(4n 1) = Sp(n)/ Sp(n 1) SO(4n)/ SO(4n 1) = Sp(n) U(1)/ Sp(n 1) U(1) = S n 1 = S n 1 = S 4n 1 = S 4n 1 SO(4n)/ SO(4n 1) = Sp(n) Sp(1)/ Sp(n 1) Sp(1) = S 4n 1 SO(7)/ SO(6) = G / SU(3) = S 6 SO(8)/ SO(7) = Spin(7)/ G = S 7 SO(16)/ SO(15) = Spin(9)/ Spin(7) = S 15 SU(n)/ U(n 1) = Sp(n)/ Sp(n 1) U(1) = CP n SO(n)/ U(n) = SO(n 1)/ U(n 1) = spec. orth. cx. str. on R n SU(n)/ Sp(n) = SU(n 1)/ Sp(n 1) = spec. orth. quat. str. on C n SO(7)/ SO() SO(5) = G / U() = G + (R7 ) SO(8)/ SO(3) SO(5) = Spin(7)/ SO(4) = G + 3 (R8 ). Each of these symmetric spaces in the left-hand presentation is an irreducible symmetric space. Up to scaling, each has exactly one Einstein metric, the symmetric metric, homogeneous with respect to the left-hand presentation. However, with respect to the right-hand presentation, only the sixth and seventh symmetric spaces are isotropy irreducible. The first nine examples are discussed in [Z]. In this paper we consider the last four examples in the table. Of these, the first is originally described in [W-Z]. On SU(n)/ Sp(n), 5

10 the only homogeneous Einstein metric is the original one. However, the last two spaces each carry two new Einstein metrics, homogeneous with respect to the smaller group. For any homogeneous space M = G/H, with g = h p on the Lie algebra level, we parametrize the space of G-invariant metrics on M by decomposing p into its Ad(H)- irreducible subspaces, p = p 1 p p k. If the p i s are pairwise inequivalent representations, a G-homogeneous metric is determined by an inner product on p of the form, = x 1 Q p1 x Q p x k Q pk, for Q an Ad(H)-invariant inner product and x i > 0 for all i. If p i and p j are equivalent for some i and j, then p i, p j does not necessarily vanish. We discuss examples of this type in chapters and 3. Assume G/H is compact, and let S(g) denote the scalar curvature of g. Einstein metrics are the critical points of the total scalar curvature functional T (g) = M S(g)dvol g on the space M of Riemannian metrics of volume one [Ber], [H]. Let M G denote the set of all G-invariant metrics of volume one on M. Notice that on M G, T (g) S(g). Furthermore, critical points of T MG are precisely G-invariant Einstein metrics of volume one [Bes, p.11]. If for our homogeneous space G/H, every homogeneous metric is diagonal, i.e.,, = x 1 Q p1 x Q p x k Q pk, with x i > 0 for all i, then we use equation (1.3) for the scalar curvature given in [W-Z]. S = 1 i d i b i x i i,j,k ( ) k xk. i j x i x j

11 In their formula, for each i, B pi = b i Q pi, where B denotes the Killing form, and d i = dim(p i ); the triple ( k i j) = Q([Xα, X β ], X γ ), summed over {X α }, {X β }, and {X γ }: Q- orthonormal bases for p i, p j, and p k, respectively. Notice ( k i j) is symmetric in all three entries. 7

12 Chapter 1: Irreducible Symmetric Spaces Let M G denote the space of G -invariant metrics of volume one. Our results are summarized in the following table. In the next four sections we discuss each of the spaces in turn, proving Theorem 1.. G /H G /H dim M G no. Einstein SO(n)/ U(n) SO(n 1)/ U(n 1) 1 SU(n)/ Sp(n) SU(n 1)/ Sp(n 1) 1 SO(7)/ SO() SO(5) G / U() 3 SO(8)/ SO(3) SO(5) Spin(7)/ SO(4) SO(n)/U(n) We begin with the symmetric space SO(n)/ U(n). Consider the space of orthogonal complex structures on R n, and let M 0 be the connected component containing J 0, the ( ) 0 Idn complex structure represented by with respect to the standard basis on R Id n 0 n. We will show that M 0 = SO(n)/ U(n). Let J M 0. Since J is an orthogonal complex structure Jv = v and J = Id. We construct an orthonormal basis {v α } for R n such that for 1 i n, Jv i = v n+i and Jv n+i = Jv i. Let v 1 = e 1, and let v n+1 = Jv 1. We have v 1, Jv 1 = Jv 1, J v 1 = v 1, Jv 1, hence {v 1, v n+1 } is an orthonormal basis for a J-invariant subspace. Let v be any unit vector in span{v 1, v n+1 }, and let v n+ = Jv. Continue up to v n, v n. With ( ) 0 Idn respect to the basis {v α }, J =. Let P be the change of basis transformation Id n 0 from the standard basis to {v α }, then J = P J 0 P 1. The hypothesis that J be in the connected component containing J 0 corresponds exactly to the fact that P must be in SO(n). Via conjugation, SO(n) acts transitively on M 0. 8

13 The isotropy subgroup of J 0 is the set of all P SO(n) such that P J 0 = J 0 P. If we identify R n R n = C n via (u, v) u + iv then J 0 is multiplication by i and hence P J 0 = J 0 P implies P SO(n) GL(n, C) = U(n). Thus M 0 = SO(n)/ U(n), where ( ) A B we have U(n) embedded in SO(n) in the following way: A + ib. It is well B A known that SO(n)/ U(n) is an irreducible symmetric space [W, p.87]. Let µ n denote the standard complex n-dimensional representation of U(n); the isotropy representation of U(n) is [ µ n ] R. Since µ n is unitary, [ µ n ] R is the irreducible real representation whose complexification is isomorphic to the direct sum of µ n and its dual. If we look at the low dimensional examples, for n 4, we find SO(4)/ U() = S, SO(6)/ U(3) = CP 3, and SO(8)/ U(4) = G + (R8 ). For n 4 the rank of the symmetric space is greater than one. Notice that in our description above we let v 1 = e 1, so the subgroup SO(n 1) SO(n) fixing span{e 1 } also acts transitively on M 0. The isotropy subgroup of SO(n 1) corresponding to J 0 is U(n 1) SO(n ), where SO(n ) is the subgroup fixing e 1 and e n+1. On the Lie algebra we have, for X, Y gl(n 1, R), u(n 1) X 0 Y = X = Xt, Y = Y t so(n 1); Y 0 X therefore, p X v Y = v t 0 w t X = Xt, Y = Y t, v, w R n 1. Y w X We find that p decomposes into the sum of two irreducible representations. In fact U(n 1) SO(n ) SO(n 1) gives rise to the following fibration: SO(n )/ U(n 1) SO(n 1)/ U(n 1) SO(n 1)/ SO(n ) = S n. 9

14 Both base and fibre are irreducible symmetric spaces. Let p 1 denote the Ad U(n 1)- invariant complement to u(n 1) in so(n ); p 1 corresponds to the tangent space of the fibre. This representation of U(n 1) on p 1 is the irreducible isotropy representation of the fibre symmetric space, [ µ n 1 ] R [W, p.87]. Let p denote the Ad SO(n )-invariant complement to so(n ) in so(n 1); in our fibration p corresponds to the tangent space of the base. The representation of U(n 1) on p is the restriction of the standard representation of SO(n ) on p = R n to U(n 1), which is [µ n ] R, again irreducible. We have p = p 1 p. The dimensions of p 1 and p are (n 1)(n ) and (n 1), respectively. We see that p 1 and p are clearly inequivalent representations of U(n 1) for n 4, and for n = 4, while µ 3 and µ 3 are equivalent representations on SU(3), they are inequivalent on the center of U(3). We apply Schur s lemma to know that p 1, p and Ric(p 1, p ) must vanish. Thus any SO(n 1)-homogeneous metric on M 0 must be of the form, = x 1 Q p1 x Q p, where Q(X, Y ) = 1 tr(xy ) and x 1, x > 0. We express the scalar curvature in terms of x 1 and x using [W-Z, (1.3)]. The equation is S = 1 i d i b i x i 1 4 i,j,k ( ) k xk. i j x i x j Since for so(k), B(X, Y ) = (k ) tr(xy ), we have b 1 = b = (n 3). From the fibration it follows that [p 1, p 1 ] u(n 1), [p, p ] u(n 1) p 1, hence the only nonzero triple (up to rearrangements) is ( 1 ). Let E ij denote the skew-symmetric matrix in so(n 1) with 1 in the ij th entry and 1 in the ji th entry, and zeros everywhere else. p 1 = span{e ij E n+i,n+j, E i,n+j E j,n+i 1 i < j n 1} p = span{ E in, E n,n+i 1 i n 1}. 10

15 We find ( 1 ) = (n 1)(n ), and then we substitute into the scalar curvature equation to find S. Let S be the equation for S with the boundary constraint that volume = 1: S = (n 1)(n ) (n 1)(n 3) + (n 1)(n )x 1 x 1 x x S = S λ(x 1 (n 1)(n ) x (n 1) 1). We find the partial derivatives of S: S (n 1)(n ) (n 1)(n ) = x 1 x 1 x (n 1)(n )λx (n 1)(n ) 1 1 x (n 1) S (n 1)(n 3) = x x + (n 1)(n )x 1 x 3 (n 1)λx (n 1)(n ) 1 x n 3. Setting both equations equal to zero is equivalent to the following equation: n 1 x 1 + (n ) x = (n 3)x. x 1 We find that the solutions are x = 1 x 1, and x = (n 1) (n ) x 1. The second solution is a (nonsymmetric) SO(n 1)-invariant Einstein metric, discovered earlier in [W-Z, Ex. 3.6]. The first solution is the SO(n)-invariant symmetric metric, but this is not obvious until we see how to compare them. Let p denote the Q-orthogonal complement to u(n) in so(n): p = span{e ij E n+i,n+j, E i,n+j E j,n+i 1 i < j n}. We must project p to p. We take a basis element of p 1 : 1 (E ij E n+i,n+j ). Under the embedding of so(n 1) in so(n), 1 (E ij E n+i,n+j ) 1 (E i+1,j+1 E n+i+1,n+j+1 ), already in p; hence an element of norm = x 1 is sent to an element of norm = 1. A basis element of p is E in, which the embedding sends to E i+1,n+1. Next we write E i+1,n+1 as the sum of an element in u(n) and an element in p: E i+1,n+1 = 1 (E i+1,n+1 + E 1,n+i+1 ) 1 (E i+1,n+1 E 1,n+i+1 ). 11

16 This shows that an element of norm = x is sent to an element of norm = 1. The symmetric metric on SO(n)/ U(n) is given by the restriction of Q to p. It corresponds to 1 = x 1 x, i.e., x = 1 x 1. To see that these metrics are distinct, we can compare the scale-invariant product (S) d (V ) 1, where S is the scalar curvature, V is the volume, and d is the dimension of M. The first metric has S = n(n 1) x 1 and V = x (n 1)(n ) 1 x (n 1), so that (S) n(n 1) (V ) 1 = (n 1)(n ) (n(n 1) ) n(n 1). The second metric has S = n(n )(n n 1) (n 1)x 1, and V = ( n 1 (n ) )(n 1) x n(n 1) 1, hence (S) n(n 1) ( (V ) 1 n(n )(n ) n(n 1) ( ) n 1) n 1 n 1 =. n 1 (n ) 1.. SU(n)/Sp(n) Our next example is the symmetric space M = SU(n)/ Sp(n), an analogue of the previous example. This is the set of special orthogonal quaternionic structures on C n. We identify R 4n = C n via a fixed orthogonal complex structure I on R 4n. An orthogonal quaternionic structure on C n is given by J SO(4n) such that J = Id and IJ = JI. As a map from C n to itself, J is complex anti-linear, i.e., J(λv) = λj(v). We show that the set of all orthogonal quaternionic structures can be written homogeneously as U(n)/ Sp(n), and we call the submanifold M = SU(n)/ Sp(n) the set of special orthogonal quaternionic ( ) 0 Idn structures. We first observe that if I = and if we identify C Id n 0 n C n = H via (u, v) u + jv, then multiplication by j on C n = R 4n becomes 0 Id n 0 0 J 0 = Id n Id n, 0 0 Id n 0 1

17 which is the standard orthogonal quaternionic structure. We want to show that U(n) acts transitively on {J SO(4n) J = Id and IJ = JI}, and the isotropy subgroup is Sp(n). Since U(n) = GL(n, C) SO(4n), A is unitary when A SO(4n) and AI = IA, and for A unitary, AJA 1 is a quaternionic structure if J is: (AJA 1 )(AJA 1 ) = Id and (AJA 1 )I = AJIA 1 = AIJA 1 = I(AJA 1 ). Furthermore, IJ is also a quaternionic structure: (IJ)(IJ) = I(IJ)J = ( Id), and (IJ)I = I(IJ). Given an orthogonal quaternionic structure J, we construct a unitary basis of C n in which J is represented by the matrix J 0. Let v 1 = e 1, the first element of the standard basis. Let v n+1 = Jv 1, v n+1 = Iv 1, and v 3n+1 = IJv 1. Clearly {v 1, Jv 1, Iv 1, IJv 1 } is an orthonormal basis for a 4-plane invariant under I, J, and IJ. Choose v to be any unit vector in the orthogonal complement, and repeat the process above, continuing up to v n, v n, v 3n, v 4n. Notice this is a unitary basis for R 4n, since v n+i = Iv i for all 1 i n. The isotropy subgroup of U(n) corresponding to J 0 is all A U(n) such that AJ 0 = J 0 A. I.e., A commutes with I, J, and IJ: A is quaternionic linear. We embed Sp(n) ( ) A B U(n) via A + jb. The image of this embedding is contained in SU(n), B A and we now restrict ourselves to the orbit of SU(n), which is the symmetric space of special orthogonal quaternionic structures on R 4n, or SU(n)/ Sp(n). This symmetric space is irreducible; up to scaling the symmetric metric is the unique SU(n)-invariant metric, and it is Einstein. Notice that when n =, SU(4)/ Sp() = S 5 ; for n 3 the rank of the symmetric space is greater than one. 13

18 Let p be the orthogonal complement to sp(n) in su(n) with respect to the inner product Q(X, Y ) = 1 tr(xy ). {( ) } X Y We have su(n) = Y t X, Z u(n), Y gl(n, C), tr Z = tr X Z {( ) and sp(n) X Y = X u(n), Y = Y t gl(n, C)}, Y X {( ) X Y hence p = X su(n), Y = Y t gl(n, C), and tr X = 0}. Y X The isotropy representation of Sp(n) on p is [ ν n Id] R, where ν n is the standard representation of Sp(n) on H n = C n. (The representation ν n is the sum of a complex (n + 1)(n 1)-dimensional irreducible representation and a one-dimensional trivial representation. We denote by ν n Id the non-trivial summand.) We write [ ν n Id] R for the real representation whose complexification is ν n Id. The subgroup SU(n 1) SU(n) fixing e 1 acts transitively on M, just as in the previous example. The isotropy subgroup of SU(n 1) corresponding to J 0 is A 0 B H = SU(n 1) B 0 A A + jb Sp(n 1, C) = Sp(n 1) SU(n ) fixing e n+1. In su(n 1), for X, Y gl(n 1, C), X 0 Y h = sp(n 1) = t X u(n 1), Y = Y. Y 0 X We denote by p the Q-orthogonal complement to sp(n 1) in su(n 1): X ū t Y p = u z v t X u(n 1), Y = Y t, u, v C n 1, z = tr X. Y v X We have the following fibration of our symmetric space, which tells us how to decompose p into irreducible Ad Sp(n 1)-invariant subrepresentations: SU(n )/ Sp(n 1) SU(n 1)/ Sp(n 1) SU(n 1)/ SU(n ) = S 4n 3. 14

19 From the fibration we see that p = p 1 p, where p 1 is tangent to the fibre and the Ad Sp(n 1) action on p 1 is [ ν n 1 Id] R, and dim(p 1 ) = (n 1)(n ). The subspace p is tangent to the base, and Ad SU(n ) acts on p by [µ n ] R Id, which when restricted to Sp(n 1) is [ν n 1 ] R Id. That is, p = p p 3 ; dim(p ) = 4(n 1) and dim(p 3 ) = 1 (the Ad Sp(n 1) action on p 3 is trivial). The set of elements listed below gives a Q-orthogonal basis for p. We write E ij for the skew-symmetric (n 1) (n 1) matrix with 1 in the ij th entry and 1 in the ji th entry, and zeros elsewhere. We denote by F ij the symmetric (n 1) (n 1) matrix with 1 in both the ij th and ji th entries. p 1 = span{(e kl E n+k,n+l ), i(f kl + F n+k,n+l ), (E k,n+l E l,n+k ), i(f k,n+l F l,n+k ) 1 k < l n 1} span{i(f kk F n 1,n 1 + F n+k,n+k F n 1,n 1 ) 1 k < n 1} p = span{e nk, if nk, E n,n+k, if n,n+k 1 k n 1} p 3 = span{diag(η,..., η, (n 1)η, η,..., η)}, where η = i (n 1)(n 1). Since p 1, p, and p 3 are inequivalent irreducible representations of Sp(n 1), any SU(n 1)-invariant metric on M must take the form, = x 1 Q p1 x Q p x 3 Q p3, with x i > 0 for i = 1,, 3. To find all SU(n 1)-invariant Einstein metrics on M, we solve for the critical points of the scalar curvature equation in terms of x 1, x, and x 3 (restricting to unit volume). As in the previous case we use the formula given in [W-Z, (1.3)]: S = 1 i d i b i x i 1 4 i,j,k 15 ( ) k xk. i j x i x j

20 In our example we have b i = 4(n 1) for all i, since for su(k), B(X, Y ) = k tr(xy ). And d 1 = (n 1)(n 1), d = 4(n 1), d 3 = 1. We find that [p 1, p 1 ] sp(n 1), [p 1, p ] p, [p, p ] sp(n 1) + p 1 + p 3, [p 1, p 3 ] = 0, [p 3, p 3 ] = 0, [p, p 3 ] p. Therefore ( 1 ) (, 3 ) ( (and rearrangements) are the only nonzero triples. We compute 1 ) = 4(n 1)(n ) and ( 3 ) = 4(n 1). We now have the equation for the scalar curvature of M in terms of x 1, x, and x 3. ( 4(n 1)(n ) 8(n 1) S = (n 1) + x 3 x 1 x x We normalize for volume 1 metrics: S = S λ(x d 1 1 xd x 3 1). (n )x ) 1 x. S 4(n 1)(n 1)(n ) (n 1)(n ) = x 1 x 1 x (n 1)(n )λx d x d x 3 S 8(n 1)(n 1) = x x + (n 1)((n )x 1 + x 3 ) x 3 4(n 1)λx d 1 1 xd 1 x 3 S (n 1) = x 3 x λx d 1 1 xd. Setting S x 1 = S x = S x 3 = 0 simultaneously is equivalent to 4(n 1)x + x 1 = (n 1)x 1 x 3 = (n 1)x 1 x n 1 n ((n )x 1 + x 1 x 3 ). There is only one solution; it is x = 1 x 1 and x 3 = n n 1 x 1, unique up to scaling. This is not a new metric, rather it is the symmetric metric, which we knew must solve our equations. (It is SU(n)-invariant, hence SU(n 1)-invariant.) Thus the only homogeneous Einstein metric on M = SU(n 1)/ Sp(n 1) = SU(n)/ Sp(n) is the symmetric metric. 16

21 1.3. G + (R7 ) The Grassmann manifold of oriented two-planes through the origin in R 7 is generally written homogeneously G + (R7 ) = SO(7)/ SO() SO(5). It is irreducible: the symmetric metric is not only Einstein, it is the only SO(7)-invariant metric. We will show that this Grassmannian manifold can also be written homogeneously as G / U() and we find it carries two non-symmetric G -invariant Einstein metrics. First we must see how G SO(7), following [M, p.190]. We identify R 8 with the Cayley numbers, or Octonians, the normed division algebra O = H Hε. Then G is the set of automorphisms of O. Any automorphism of the Cayley numbers must take 1 to itself and must preserve the inner product, so elements of G also preserve Im(O), the space of imaginary Cayley numbers, the orthogonal complement to 1. In this way we see G SO(7). To see that G acts transitively on G + (R7 ), we use the following observation [M, p.186]. Lemma 1.1. Given three imaginary orthogonal unit octonians: v 1, v, and v 3 {v 1, v, v 1 v }, there exists a unique automorphism A of O with A(i) = v 1, A(j) = v, and A(ε) = v 3. Using the lemma we take any P = span{v, w} to the oriented two-plane P 0 = span{i, j}, where we take v and w an orthonormal basis for P. We must find the isotropy subgroup fixing P 0. It will be useful to recall the following well known fact. Lemma 1.. The quotient G / SU(3) = S 6. Proof. By the previous lemma, G acts transitively on S 6 (1) Im(O). We need to show that the isotropy subgroup H v of G corresponding to any v S 6 is SU(3). We observe first 17

22 that A(v) = v implies the map L v is a complex structure on O and on V = span{1, v}. This shows that H v U(V ) = U(3). Furthermore, dim(h v ) = 8. Since G and S 6 are connected and simply connected, H v is connected. Consider the homomorphism det : H v S 1 ; it must be either trivial or onto. If it is trivial, H v = SU(3). If it is onto, then let H denote the kernel. Then H is a normal subgroup of H v of dimension seven, and if H 0 is the connected component of the identity, rank(h 0 ) rank(su(3)) =. But the only compact connected seven-dimensional Lie groups, up to finite cover, are T 7, S 3 T 4, and S 3 S 3 S 1, and each of these has rank > : rank(t 7 ) = 7, rank(s 3 T 4 ) = 5, rank(s 3 S 3 S 1 ) = 3. This shows H = H v, and thus H v = SU(3). We are now ready to describe the isotropy subgroup H corresponding to the oriented two-plane P 0. Since G and G + (R7 ) are connected and simply connected, we know H is connected. If we have A G such that A(P 0 ) = P 0 (with orientation), then A(i) = i cos θ j sin θ, A(j) = i sin θ + j cos θ, thus A(k) = A(i)A(j) = k. The isotropy subgroup H {A G A(k) = k} = SU(3), and since the ij-plane is a complex line with respect to our complex structure L k, H must preserve this complex line and the complex twoplane perpendicular to it. Hence H S(U(1) U()) SU(3). A dimension count shows H = S(U(1) U()) = U(). If we look on the Lie algebra level, and we take {i, j, k, ε, iε, jε, kε} as our basis for Im(O), then using the automorphism property we can write g so(7) in the following way: g = span{e 1 + E 56, E 47 + E 56, E 45 + E 67, E 46 E 57, E 46 + E 57 + E 13, E 67 E 45 + E 3, E 14 E 7 E 36, E 15 + E 6 E 37, E 16 E 5 + E 34, 18

23 E 17 + E 4 + E 35, E 14 + E 7, E 15 + E 6, E 16 + E 5, E 17 + E 4 }. We use the inner product on g given by Q(X, Y ) = 1 tr(xy ), in which the basis for g above is orthogonal. The subalgebra h corresponds to H = U(): u() = span{e 1 + E 56 E 47, E 47 + E 56, E 45 + E 67, E 46 E 57 }. The isotropy representation of U() is the action of Ad U() on p, the Q-orthogonal complement of u() in g. We can use the following fibrations to decompose p into its irreducible Ad U() representations: First, CP = SU(3)/ U() G / U() G / SU(3) = S 6. The tangent space to the base is isomorphic to [µ 3 ] R ; when restricted to U() it gives [(µ 1 ˆ Id) (Id ˆ µ )] R = p 1 p. Thus, p 1 = [µ 1 ˆ Id] R, p = [Id ˆ µ ] R. The tangent space to the fibre is p 3 = [µ 1 ˆ µ ] R. We have p = p 1 p p 3. p 1 = span{e 46 + E 57 + E 13, E 67 E 45 + E 3 }, p = span{e 14 E 7 E 36, E 15 + E 6 E 37, E 16 E 5 + E 34, E 17 + E 4 + E 35 }, p 3 = span{e 14 + E 7, E 6 E 15, E 16 + E 5, E 4 E 17 }. We have a second fibration of our manifold: we claim U() SO(4) G. Before showing this, we briefly discuss the embedding SO(4) G and the irreducible symmetric space G / SO(4). Lemma 1.3. The quotient G / SO(4) is the space of quaternionic subalgebras of the Cayley numbers, O. 19

24 Proof. We have SO(4) = (Sp(1) Sp(1))/{(q 1, q ) ( q 1, q )}, and it acts on O = H Hε by (q 1, q ) : a + bε q 1 a q 1 + (q b q 1 )ε. A calculation shows that SO(4) G, and this embedding of SO(4) in G can also be described as the subgroup of G which leaves the subalgebra H = span{1, i, j, k} invariant. Since U() is the subgroup of G preserving the plane spanned by i and j, elements of U() take 1 to itself and k to itself, hence they preserve span{1, i, j, k}. This also shows U() SO(4) SU(3). We also have SO(4) SU(3) U(): under G, 1 1; under SO(4), span{1, i, j, k} span{1, i, j, k}; under SU(3), k k. Thus SO(4) SU(3) = U(). Our second fibration is SO(4)/ U() G / U() G / SO(4). Here p 1 is tangent to the fibre, p and p 3 are tangent to the base: From the two fibrations we obtain the following Lie bracket relations among the p i s: [p 1, p 1 ] u(), [p 1, p ] p p 3, [p, p ] u() p 1, and [p 3, p 3 ] u(). Since p 1, p, and p 3 are mutually inequivalent, any G -invariant metric on our space is of the form, = x 1 Q p1 x Q p x 3 Q p3, with x i > 0, for i = 1,, 3. As in the previous cases we can write the scalar curvature on G + (R7 ) as function of x 1, x, and x 3 via the formula given in [W-Z]: S = 1 i d i b i x i 1 4 i,j,k ( ) k xk. i j x i x j When we compute the non-zero Lie bracket relations between the p i s we find that ( 3 1 ) = 4 and ( 1 ) = Also we compute (using the basis elements for g ) b i = 8 for each i, and of 0

25 course d 1 =, d = 4, and d 3 = 4. This gives us ( 1 S = ) 4 ( x1 x 1 x x 3 3 x + ) ( x1 + x + x ) 3. x 1 x x 3 x 1 x 3 x 1 x Then S = S λ(x 1 x4 x4 3 1) includes our boundary condition, volume = 1. Einstein metrics on G / U() will be critical points of S. S = 16 x 1 3x 4 ( x 1 3x + x 1 x 3 S = 16 x x + 8x 1 3x 3 + S = 16 ( x1 x 3 x + 3 x x 3 Now we look for all solutions to S x 1 find the following: ( x1 x x 3 + x 3 x 1 x 1 ) λx 1 x 4 x x x x 3 x 1 x 1 ) 4λx x 1 x 1x 3 x x x 1 x 1 ) 4λx 3 x 1 x 1x 4 x 3 3. = S x = S x 3 = 0. We solve these using Maple, and we either x = 1 x 1 and x 3 = 3 x 1 ( 7 or x = ζx 1 and x 3 = 10 ζ ζ ζ ζ 1 ) x 1, 40 where ζ is a root of 56ζ 5 53ζ ζ ζ + 776ζ 60 = 0. We need both a positive solution to this polynomial in order for x > 0, and also ( 7 10 ζ ζ ζ ζ 1 ) > 0, so that x 3 > There are exactly three real solutions to the quintic polynomial. We give the approximate values for x and x 3, setting x 1 = 1: x = x 3 =.155 x = x 3 = x = x 3 =

26 We must eliminate the first of the solutions from the quintic, since it gives a negative value for x 3. The solution x = 1 x 1 and x 3 = 3 x 1 is the symmetric metric; that is, we will see that it is SO(7)-invariant, after projection. If we denote by p the Q-orthogonal complement to so() so(5) in so(7), then p = {E ij 1 i, 3 j 7}. When we project p to p, we see that in p 1, the basis element 1 6 (E 13 + E 46 + E 57 ) projects to 6 E 13, i.e., an element of norm = x 1 projects to an element of norm = 3. In p, the basis element 1 6 (E 36 E 14 + E 7 ) projects to 1 6 ( E 13 + E 7 ), so norm = x is projected to norm = 1 3. In p 3, basis element 1 (E 14 + E 7 ) is already an element of p, so norm = x 3 is projected to norm = 1. This shows that the symmetric metric in our basis satisfies x x 1 = 1 and x 3 x 1 the Grassmannian G + (R7 ). = 3. Thus we end up with two new G -homogeneous Einstein metrics on Remark. Notice none of these is a fibration metric, since a metric of the first fibration would require x 1 = x and a metric of the second fibration would require x = x 3. Many examples of Einstein metrics have been obtained by using fibrations over Einstein spaces and with Einstein fibres [Bes, ch.9]. In our example we have two fibrations; in each the fibre and base space are isotropy irreducible, so they must be Einstein. However, in each fibration the isotropy representation of the base, when restricted to U(), decomposes into the sum of two irreducible subrepresentations. Recall Q(X, Y ) = 1 tr XY is our comparison metric. The Casimir constant corresponding to Q and to the restriction to U() of the isotropy representations for these homogeneous spaces differs on these two subrepresentations. This implies that the O Neill tensor restricted to horizontal vectors is not a scalar multiple of Q. Using Besse s Proposition 9.70 [Bes, p.53] we see that therefore our fibrations cannot give rise to Einstein metrics.

27 To see that we have three non-isometric solutions we compare the scale-invariant product: (S) d (V ) 1, where S is the scalar curvature and V is the volume, and d is the dimension of our homogeneous space, for our three metrics. If we let x = 1 x 1 and x 3 = 3 x 1, we obtain S = 100 3x 1, and V = 34 8 x 10 1, so (S)5 (V ) 1 = This is approximately The second solution gives (S) 5 (V ) 1 = , and the third solution gives (S) 5 (V ) 1 = We note that these have been found previously in [A] and [K]. They observe that one of the three metrics is Kähler and the other two are not Kähler for any complex structure on M. Neither author observes that the Kähler Einstein metric is the symmetric metric G + 3 (R8 ) We can write the Grassmannian manifold of oriented three-planes in R 8 homogeneously G + 3 (R8 ) = SO(8)/ SO(3) SO(5), but we can also write it as G + 3 (R8 ) = Spin(7)/ SO(4). With respect to SO(8), G + 3 (R8 ) is irreducible, and therefore the symmetric metric is Einstein and it is the unique SO(8)-invariant metric, up to scaling. However, we find there are two more Spin(7)-invariant Einstein metrics which are not symmetric. We first describe how Spin(7) sits inside SO(8). We again identify R 8 with the Cayley numbers O; from Murakami [M] we know Spin(7) = {A SO(8) B SO(8) such that B(x)A(y) = A(xy) x, y O}. In this definition notice B(1) = 1, so B SO(7) and if A corresponds to B, A corresponds to B as well, which shows Spin(7) is indeed a double cover of SO(7). We also remark that that {L a a Im(O), = } Spin( ). (The corresponding B is conjugation by a.) We need to show that C a (x)l a (y) = L a (xy) for any x, y O. Because a is a unit 3

28 imaginary octonian, a 1 = a, thus C a (x)l a (y) = (axa 1 )(ay) = (axa)(ay). Then the first Moufang identity tells us that (axa)(ay) = a(x(aay)), and using that aa = 1, we have a(x(aay)) = a(xy) = L a (xy). It is also convenient to identify two subgroups of Spin(7), they are G, the automorphisms of the Cayley numbers, and SU(4), complex linear maps with respect to L i. We see G Spin(7) by letting B = A in the definition of Spin(7). Murakami shows how to see that SU(4) Spin(7) with the following lemma [M]. Lemma 1.4. In SO(8), U(4) = {A SO(8) ia(x) = A(ix) x, y O} and U(4) Spin(7) = SU(4). Proof. Let p + : Spin(7) SO(7) be the homomorphism sending A B, for A and B in the definition of Spin(7). For every A U(4) Spin(7) we have B(i) = i, so p + (U(4) Spin(7)) SO(6). And for every B SO(6) (B(i) = i), the corresponding A must be in U(4) Spin(7), hence SO(6) p + (U(4) Spin(7)). Furthermore, p + is a local isomorphism, thus U(4) Spin(7) is a 15-dimensional connected Lie group with a simple Lie algebra. Observe that SU(4) is the commutator subgroup of U(4). Since its Lie algebra is simple, U(4) Spin(7) is its own commutator subgroup, thus U(4) Spin(7) SU(4), and a dimension count tells us that these subgroups are equal. We embed SU(4) SO(8) via A + ib ( ) A B. To embed SU(4) into SO(8) in B A this way, we want R 4 R 4 = C 4 with (u, v) u + iv; this restricts our choice of ordered bases for O: we choose {1, j, ε, jε, i, k, iε, kε}. The intersection of our two subgroups is G SU(4) = SU(3), and this time SU(3) in G fixes i instead of k. We check that Spin(7) acts transitively on G + 3 (R8 ): Let P 0 = span{i, j, k}, an oriented three-plane through the origin. Let P be given by span{v 1, v, v 3 }, where v 1, v and v 3 are 4

29 ordered orthonormal vectors. Without loss of generality we may assume v 1, v Im(O), since P is a three-plane, so dim(p Im(O)). We know from the previous example that we can find an element A G such that A(i) = v 1, A(j) = v. Let x = A 1 (v 3 ). Observe that x, i = A 1 (v 3 ), i = v 3, v 1 = 0 x, j = A 1 (v 3 ), j = v 3, v = 0. We claim there exists A Spin(7) such that A (i) = i, A (j) = j, and A (k) = x. This is because the subgroup of Spin(7) fixing one unit octonian is conjugate to G, and the subgroup of Spin(7) fixing two orthonormal octonians is conjugate to SU(3), which acts transitively on S 5 (1) span{i, j}. The composition A A Spin(7) is our map taking i v 1, j v, and k v 3, so that P 0 goes to P, and this shows Spin(7) acts transitively on G + 3 (R8 ). Next we must determine the isotropy subgroup H of P 0. We claim that H G. To see this, we note that if A(P 0 ) = P 0, then if B is the element of SO(7) in the definition of Spin(7) corresponding to A, since B(i)A(j) = A(k) we know that B(i) span{i, j, k}. Thus A(1) = B(i)A(i) span{1, i, j, k}, and furthermore A(1) P 0, so A(1) = ±1. Since Spin(7) and G + 3 (R8 ) are connected and simply connected, we know H is connected, hence A(1) = 1. From the definition of Spin(7) it follows that A = B and this implies H G. Furthermore, any element of H takes 1 to itself and preserves the standard quaternionic subalgebra span{1, i, j, k}. Thus H SO(4) G, and by a dimension count H = SO(4). 5

30 Now we are ready to find the isotropy representation. On the Lie algebra level we have spin(7) = span{e ij + E 4+i,4+j, E i,4+j + E j,4+i 1 i < j 4} span{e i,4+i E 48 1 i 3} span{e 7 E 45, E 3 + E 58, E 4 E 57, E 8 + E 35, E 56 E 78 E 5 E 38 + E 47 }. The subalgebra corresponding to the isotropy subgroup is h = su() su(): h = span{e 37 E 48, E 34 + E 78, E 38 + E 47 } span{e 56 + E 34 E 78, E 6 E 48 E 37, E 5 E 38 + E 47 }. Notice each copy of su() is an ideal in h and its basis vectors above are orthogonal with respect to the inner product on spin(7) given by Q(X, Y ) = 1 tr(xy ). As usual we denote by p the Q-orthogonal complement of h in spin(7). There are two fibrations of our symmetric space Spin(7)/ SO(4). The first is G / SO(4) Spin(7)/ SO(4) Spin(7)/ G = S 7. Let p be the subspace tangent to the fibre; let p be the subspace tangent to the base. Let θ k denote the unique irreducible complex representation of su() in k dimensions; the first fibration tells us that p = [θ ˆ θ 4 ] R, since this is the representation of the symmetric space G / SO(4) [W, p.87]. The isotropy representation of Spin(7)/ G is the seven-dimensional representation of G SO(7). We restrict this representation to SO(4), to see that p = [ρ 3 ˆ Id] [θ ˆ θ ] R, where ρ 3 denotes the standard representation of so(3) on R 3. We let p 1 = [ρ 3 ˆ Id], and p = [θ ˆ θ ] R, and p 3 = [θ ˆ θ 4 ] R. We have dim(p 1 ) = 3, dim(p ) = 4, and dim(p 3 ) = 8. For the second fibration, we first need some explanation. 6

31 Lemma 1.5. The compact group (Spin(4) Spin(3))/{±(Id, Id)} satisfies SO(4) (Spin(4) Spin(3))/{±(Id, Id)} Spin(7). Proof. Recall, for any n, Spin(n) is the simply connected double cover of SO(n); let π : Spin(n) SO(n) denote the two-fold homomorphism. Because π 1 (SO(k)) π 1 (SO(n)) is a surjection, π(spin(k)) = SO(k) for all k n. Observe that ker(π) = {±(Id, Id)} = Spin(k) Spin(n k), thus it is (Spin(k) Spin(n k))/{±(id, Id)} Spin(n). In Spin(7) we can consider the subgroup (Spin(3) Spin(4))/{±(Id, Id)}. We know Spin(4) = Spin(3) Spin(3) (and Spin(3) = SU()). Thus Spin(7) has a subgroup (Spin(3) Spin(3) Spin(3))/{±(Id, Id, Id)}. Our isotropy subgroup SO(4) Spin(7) is exactly ( Spin(3) Spin(3))/{±(Id, Id)}, where Spin(3) is the diagonal subgroup of Spin(3) Spin(3) isomorphic to Spin(3). This can be seen via the restriction to SO(4) of the homomorphism from Spin(7) to SO(7) taking A to B (A and B in the definition of Spin(7)). We obtain the following fibration: S 3 = Spin(4) Spin(3) {±(Id,Id)} / Spin(3) Spin(3) {±(Id,Id)} Spin(7)/ SO(4) = G + 3 (R 8 ) Spin(7) / Spin(4) Spin(3) {±(Id,Id)} = G + 3 (R7 ). In the second fibration, it is p 1 which is the subspace tangent to the fibre, while p p 3 is the subspace tangent to the base. p 1 = span{3e 1 E 34 + E 56 + E 78, 3E 15 E 6 E 37 E 48, 3E 16 + E 5 + E 38 E 47 } p = span{3e 13 + E 4 + E 57 E 68, 3E 14 E 3 + E 58 + E 67, 3E 17 E 8 + E 35 + E 46, 3E 18 + E 7 E 36 + E 45 } p 3 = span{e 3 + E 67, E 4 + E 68, E 7 + E 36, E 8 + E 46, E 57 E 4 + E 68, E 58 + E 3 E 67, E 35 + E 8 E 46, E 45 E 7 + E 36 }. 7

32 Since each of the p i s has a different dimension, they are inequivalent representations. This means any Spin(7)-invariant metric on G + 3 (R8 ) is determined by an inner product on p satisfying, = x 1 Q p1 x Q p x 3 Q p3, for x 1, x, x 3 > 0. From the fibrations we obtain the following Lie bracket relations: [p 1, p 1 ] h p 1, [p 1, p ] p p 3, [p, p ] h p 1, [p 3, p 3 ] h. Recall the scalar curvature formula from [W-Z]: S = 1 i d i b i x i 1 4 i,j,k ( ) k xk. i j x i x j On Spin(7) we find that b i = 10 for i = 1,, 3, and we know d 1 = 3, d = 4, and d 3 = 8. From the Lie bracket relations we know ( 1 ) ( 1 1, 1 ) (, 1 3) 0; all other triples (except rearrangements) are zero. We find that ( 1 ) ( 1 1 =, 1 ) ( = 4, and 3 1 ) = 8. We now have the scalar curvature function in x 1, x, x 3 : S = x ( 1 x1 x 1 x x 3 x 4 + x + x ) 3. x x 3 x 1 x 3 x 1 x We want the critical points of the scalar curvature function with the constraint equation of volume 1: S = S λ(x 3 1 x 4 x8 3 1). S = 5 x 1 x 1 1 x 4 + 4x x x 3 x 1 x + 4x 3 3 x 1 x 3λx 1x 4 x 8 3 S = 0 x x + x 1 x 3 + 4x 1 x x 4 + x 3 3 x 1 x 3 x 1 x 4λx 3 1x 3 x 8 3 S = 40 x 3 x + 4x 1 3 x x + 4x 3 x 1 x x 1 x 8λx 3 1x 4 x 7 3.

33 A solution to S x 1 two polynomials: = S x = S x 3 = 0 is equivalent to the simultaneous solution of the following 10x 1 x 10x 1 x x 3 + x 1x + x 1x 3 + 3x x 3 3x 3 = 0 11x 1x + 11x x 3 + 5x 3 5x x x 1 x x 1x 3 = 0. We obtain three solutions, using Maple. The first is the symmetric solution: x 1 = 3 4 x 3, x = 1 4 x 3. Let p denote the Q-orthogonal complement to so(3) so(5) in so(8); p = span{e ij i =, 5, 6, j = 1, 3, 4, 7, 8}. (Of course we take so(3) so(5) corresponding to P 0.) We must project p 1, p, and p 3 to p. In p 1, we take the basis element 1 3 (3E 1 E 34 + E 56 + E 78 ) of norm = x 1. It is projected to 3 E 1 in p, an element of norm = 3. In p we take the basis element 1 3 (3E 13 + E 4 + E 57 E 68 ) of norm = x, which projects to 1 (E E 57 E 68 ) in p, an element of norm = 1 1. Finally, the element (E 3 + E 67 ) is already in p, so norm = x 3 corresponds to norm = 1. Hence x 1 = 3 4 x 3, x = 1 4 x 3 is indeed the symmetric metric. The second and third solutions are x = ηx 3, and x 1 = for η a positive root of the polynomial ( η η η3 39 ) 33 η4 x 3, 4704t t t t 398t This polynomial has three real roots, of which two are positive and yield two positive values for x 1 and x in terms of x 3. We give the approximate values, setting x 3 = 1: x 1 = x = x 1 = x =.9019 x 1 = x =

34 These are two new Einstein metrics on G + 3 (R8 ). Remark. None of these is a fibration metric, since the first fibration required x 1 = x, and the second required x = x 3. Just as in the previous example in both fibrations the fibre and base space are isotropy irreducible, therefore Einstein. However, the isotropy representation of each base, when restricted to SO(4), again decomposes into the sum of two irreducible subrepresentations, where the Casimir constant corresponding to Q and to the restriction to SO(4) of the isotropy representations for these homogeneous spaces differs. Again this implies the O Neill tensor restricted to horizontal vectors is not a scalar multiple of Q. Using Besse s Proposition 9.70 [Bes, p.53] we know our fibrations cannot give rise to Einstein metrics. We verify that they are all distinct with the (scale-invariant) product: (S) 15 (V ) 1, where S is the scalar curvature of the metric and V is the volume of the metric. For the symmetric metric, S = 90 x 3 and V = 33 x , and so (S) 15 (V ) 1 = For the two new metrics, we find that (S) 15 (V ) 1 = , and (S) 15 (V ) 1 = , respectively. This shows that they are non-isometric. 30

35 Chapter : Products of Symmetric Spaces In this chapter we consider some products of two irreducible symmetric spaces on which there is a simple compact Lie group acting transitively. The same classification by Oniščik of simple compact Lie algebras g with Lie subalgebras g, g such that g = g + g (from page 5) also gives a list of simple groups acting on products. Let G be the simply connected compact Lie group corresponding to g and let G and G be the Lie subgroups corresponding to g and g, respectively. Then G/(G G ) = G/G G/G. Here are the products of compact irreducible symmetric spaces we obtain from Oniščik s result: SU(n)/ Sp(n 1) = S 4n 1 SU(n)/ Sp(n) SU(n)/ Sp(n 1) U(1) = CP n 1 SU(n)/ Sp(n) SO(n + )/ U(n) = S n+1 SO(n + )/ U(n + 1) Spin(8)/ G = S 7 S 7 Spin(7)/ SU(3) = S 7 S 6 Spin(8)/ U(3) = S 7 G + (R8 ) Spin(8)/ SO(4) = S 7 G + 3 (R8 ) Spin(7)/ U() = S 7 G + (R7 ). In our analysis of most of these spaces, we found that the equations were too complicated to solve, either by hand or with the help of Maple. We summarize our results in the following table, denoting by M = G/H the product of two irreducible symmetric spaces, and M G = {G-homogeneous metrics g on M with (}) = }. 31

36 M G/H dim M G no. Einstein S 7 S 7 Spin(8)/ G S 7 S 6 Spin + (7)/ SU(3) 4 3 S 7 G + (R8 ) Spin(8)/ U(3) 3.1. S 7 S 7 Just as S 7 is the unit sphere in R 8 = O, the Cayley numbers (or octonians), the product of two seven-spheres is a natural submanifold of O O. Since it is a product of symmetric spaces, S 7 S 7 is homogeneous; the simple Lie group Spin(8) acts transitively on S 7 S 7 with isotropy subgroup G. One expects at least two distinct Spin(8)-invariant Einstein metrics: one the product metric, and the other induced by the Killing form [W-Z1, p. 575]. We find that it carries exactly these, and no others. We begin by describing the homogeneous presentation of S 7 S 7, then we can determine M Spin( ), the space of invariant metrics, and consider it for Einstein metrics. We have a natural matrix group representation for Spin(8): Spin(8) = {(A, B, C) SO(8) 3 A(x)B(y) = C(xy) x, y O}. The Triality Principle tells us that Spin(8) is indeed a double cover of SO(8), since a choice of A SO(8) determines the corresponding B and C, up to sign [M]. The Moufang identities give us three families of triples in Spin(8): for each z Im(O) with z = 1, (R z, L z R z, R z ), (L z R z, L z, L z ), and (L z, R z, L z R z ) Spin(8). We note some of the subgroups of Spin(8): Spin + (7) = {(A, B, C) Spin(8) B = C} generated by {(L z R z, L z, L z )}, Spin (7) = {(A, B, C) Spin(8) A = C} generated by {(R z, L z R z, R z )}, and G = {(A, B, C) Spin(8) A = B = C} = Spin + (7) Spin (7). 3

37 For a triple (A, B, C) in Spin + (7) Spin(8), notice that B = C implies A(1) = 1, thus A SO(7). This shows that Spin + (7) is indeed a double cover of SO(7). For a triple (A, B, C) in Spin (7) Spin(8), notice that A = C implies B(1) = 1, so B SO(7), and Spin (7) double covers SO(7). We define the action of Spin(8) on S 7 S 7 via (A, B, C) : (x, y) (Ax, By). To show that this action is transitive we take any point (x, y) S 7 S 7 and construct a map from (x, y) to (1, 1). We can first find (A, B, C) : (x, y) (1, y ) for some y, since A can be any element of SO(8). Next, since Spin + (7) acts transitively on S 7, there exists an element (A, B, B ) of Spin + (7) mapping (1, y ) (1, 1). Next we determine the isotropy subgroup H Spin(8) corresponding to the point (1, 1). We know that Spin + (7) fixes the first component of (1, ), and Spin (7) fixes the second component of (, 1), hence the subgroup H fixing (1, 1) satisfies H Spin + (7) Spin (7) = G. Every element of G takes (1, 1) to itself, thus H = G. This shows that Spin(8)/ G = S 7 S 7. Under the double covering homomorphism from Spin(8) to SO(8) (A, B, C) C, the subgroups Spin + (7) and Spin (7) are isomorphic to their images in SO(8). We use the homomorphism to identify the Lie algebras spin(8) = so(8). If we order our basis for the octonians in the following way: {1, j, ε, jε, i, k, iε, kε}, then G is a subset of SO(7) = ( ) 1 0 SO(8). While g 0 SO(7) is invariant under the triality automorphism of so(8), the Lie subalgebras so(7), spin + (7), and spin (7) (subalgebras of so(8)), which all contain g, are interchanged. We decompose so(8) into g p, where p is the orthogonal complement to g in so(8) with respect to the inner product Q(X, Y ) = 1 tr(xy ). Using the following fibrations we 33