Ricci flow on locally homogeneous closed 4-manifolds

Transcription

1 communications in analysis and geometry Volume 14, Number 2, , 2006 Ricci flow on locally homogeneous closed 4-manifolds James Isenberg, Martin Jackson and Peng Lu We discuss the Ricci flow on homogeneous 4-manifolds. After classifying these manifolds, we note that there are families of initial metrics such that we can diagonalize them and the Ricci flow preserves the diagonalization. We analyze the long time behavior of these families. We find that if a solution exists for all time, then the flow exhibits a type III singularity in the sense of Hamilton. 1. Introduction. It is well-known that there are eight maximal, simply connected geometries (X, G) with compact quotient in dimension three ([17], p.474). In Thurston s geometrization conjecture, any closed three-manifold can be cut into pieces each of which admits one of these geometries. To explore the relation between the Ricci flow and the model geometries, the first two named authors analyze the long time behavior of the Ricci flow on locally homogeneous three-manifolds in [8]. In later work ([11]), using the notion of quasi-convergence, Knopf and McLeod identify the equivalence classes of all such flows except the case X = ŜL(2, R). Ricci flow has proven to be very successful in studying the geometric and topological properties of three manifolds ([4], [15], [16]), and there are indications ([3], [5], [7]) that it could be useful for the study of such properties in four dimensions. In order to further explore its possible use in dimension 4, we study the Ricci flow on locally homogeneous four-manifolds in this paper. We find that unlike in the case of three dimensions ([14], [8]), some of the families of locally homogeneous metrics can not be diagonalized because even if one diagonalizes the initial metric, the flow destroys the diagonalization of the metric at later times. The analysis of the ODE system given by Ricci flow for locally homogeneous manifolds is considerably simplified if the flow preserves the diagonalization. In this paper, we identify some families of initial metrics such that the Ricci flow preserves their diagonalization. For these families, we find that the behavior of the flow is very 345

2 346 James Isenberg, Martin Jackson and Peng Lu close to that seen in dimension three ([8]): either (a) the volume-normalized Ricci flow converges to a metric of constant sectional curvature or constant holomorphic bisectional curvature (CP 2 and CH 2 ); or (b) as t +, the Ricci flow collapses to a lower dimensional flat manifold with the curvature decaying at the rate 1 t ; or (c) the Ricci flow approaches, either in finite time or in infinite time, a direct product of lower dimensional geometries with constant sectional curvature. After describing locally homogeneous geometric structures in dimension 4 in Section 1, we consider in Section 2 the case that the homogeneous space X is a Lie group. We identify families of initial metrics whose diagonalization is preserved by the Ricci flow, and then we discuss the long time behavior of the Ricci flow for those families. In Section 3, we discuss the long time behavior of the Ricci flow for the remaining cases. Since the Ricci flow on closed manifolds preserves the isometry group, for any locally homogeneous closed 4-manifolds, we discuss the Ricci flow on their universal covering spaces. 2. Compact locally homogeneous 4-geometries. We identify a class of four dimensional homogeneous geometries by specifying a simply connected four manifold M, aliegroupg that acts transitively on M, and the minimal isotropy group I of the action. We only consider those (M,G,I) inwhichm is the universal cover of a closed manifold M q. Such a class we call a compact four dimensional homogeneous geometry. Foreach (M,G,I), there is a collection of Riemannian metrics on M for which G is the isometry group. These are the lifts of the locally homogeneous metrics on M q List of compact four dimensional homogeneous geometries. Let H n be the simply-connected hyperbolic n-manifold and S n be the simply-connected round n-sphere. We denote the group of isometries of H n by H(n). We summarize the compact three dimensional homogeneous geometries in the following table.

3 Ricci flow on locally homogeneous closed 4-manifolds 347 Manifold M 3 Lie group G Isotropy group I R 3 R 3 {0} S 3 SU(2) {e} ŜL(2, R) ŜL(2, R) {e} Nil 3 Nil 3 {e} Ŝol 3 Ŝol3 {e} R 3 E(2) {e} S 2 R SO(3) R SO(2) {0} H 2 R H(2) R SO(2) {0} H 3 H(3) SO(3) Here, ŜL(2, R) is the universal cover of the special linear group SL(2, R); its lie algebra sl 2 has a basis X 1,X 2,X 3 such that the Lie bracket is given by [X 1,X 2 ]= X 3, [X 2,X 3 ]=X 1, [X 3,X 1 ]=X 2. Nil 3 is the 3-dimensional Heisenberg group consisting of matrices of the form 1 c 1 c c 3 ; its Lie algebra n 3 has a basis X 1,X 2,X 3 such that the Lie bracket is given by [X 1,X 2 ]=X 3, [X 2,X 3 ]=0, [X 3,X 1 ]=0. Ŝol 3 is the simply-connected solvable Lie group whose Lie algebra sol 3 has abasisx 1,X 2,X 3 satisfying [X 1,X 2 ]=0, [X 2,X 3 ]= X 2, [X 3,X 1 ]= X 1. E(2) is also a solvable Lie group whose Lie algebra L(E 2 ) has a basis X 1,X 2,X 3 satisfying [X 1,X 2 ]=0, [X 2,X 3 ]= X 1, [X 3,X 1 ]= X 2. The Lie algebra su(2) of SU(2) can be described by [X 1,X 2 ]=X 3, [X 2,X 3 ]=X 1, [X 3,X 1 ]=X 2. The compact four dimensional homogeneous geometries have been classified by Ishihara [9]. We list them in the following table (see [18]).

4 348 James Isenberg, Martin Jackson and Peng Lu Manifold M 4 Lie group G Isotropy group I Nil 4 Nil 4 {e} Solm,n 4 Solm,n 4 {e} Sol1 4 Sol1 4 {e} Sol0 4 Sol0 4 {e} ŜL(2, R) R ŜL(2, R) R {e} Nil 3 R Nil 3 R {e} S 3 R SU(2) R {e} R 4 E(2) R {e} R 4 R 4 {0} S 2 S 2 SO(3) SO(3) SO(2) SO(2) S 2 R 2 SO(3) R 2 SO(2) {0} S 2 H 2 SO(3) H(2) SO(2) SO(2) H 2 R 2 H(2) R 2 SO(2) {0} H 2 H 2 H(2) H(2) SO(2) SO(2) CP 2 SU(3) U(2) CH 2 SU(1, 2) U(2) H 3 R H(3) R SO(3) {0} S 4 SO(5) SO(4) H 4 H(4) SO(4) Nil 4, Solm,n 4, Sol4 1 and Sol4 0 are simply connected 4-dimensional Lie groups; we describe their Lie algebras in Section 2.2. Note that Solm,n 4 includes Ŝol 3 R. CH 2 is complex hyperbolic space which has Kähler symmetric space structure (see [12], pp ). Note that there is another locally homogeneous space M = F 4 listed in [18]. This is not a compact homogeneous geometry because it does not have compact quotients. The isometry group G contains a discrete subgroup Γ such that F 4 /Γ has finite volume. One can find a more detailed description of four dimensional locally homogeneous geometries in Part II of [6]. The Ricci flow study for those classes with trivial isotropy group requires substantial new analysis; we group these in a category labelled A. We describe these classes in Section 2.2. Those classes with non-trivial isotropy group are grouped in category (Section 2.3).

5 Ricci flow on locally homogeneous closed 4-manifolds Four dimensional unimodular Lie groups. Recall that a Lie group G is called co-compact if G contains a discrete subgroup Γ such that G/Γ is compact. Each Lie group in (A) is co-compact. A co-compact Lie group has unimodular Lie algebra ([14] Lemma 6.2). Instead of studying Ricci flow on spaces in (A), we broaden the discussion to Ricci flow on 4-dimensional unimodular Lie groups. According to the classification of the 4-dimensional unimodular Lie algebras ([13]), for each such algebra, there is some basis X 1,X 2,X 3,X 4 such that the Lie bracket takes the form indicated below. We adopt the notation 1 in [13]. A1. ClassU1[(1, 1, 1)]. [X 2,X 3 ]=0, [X 3,X 1 ]=0, [X 1,X 2 ]=0, [X 1,X 4 ]=0, [X 2,X 4 ]=0, [X 3,X 4 ]=0. This corresponds to (M,G,I) =(R 4, R 4, {0}) whereg acts on M by translation. A2. ClassU1[1, 1, 1]. [X 2,X 3 ]=0, [X 3,X 1 ]=0, [X 1,X 2 ]=0, [X 1,X 4 ]=X 1, [X 2,X 4 ]=kx 2, [X 3,X 4 ]= (k +1)X 3, where, without loss of generality, we assume k 1 2 since otherwise we can interchange X 2 and X 3. Only the following special cases correspond to compact homogeneous geometries. (A2i) if k = 0, the Lie algebra is isomorphic to the direct sum sol 3 R. This corresponds to (M,G,I) =(Ŝol3 R, Ŝol3 R, {e}). (A2ii) if k = 1, the corresponding geometry is (M,G,I) =(Sol0 4,Sol4 0, {e}); this can be seen by choosing e 1 = X 1,e 2 = X 2,e 3 = X 3,ande 4 = X 4 on p.273 in [18]. (A2iii) if there is a number α>0 such that the exponentials of α, β =. kα and γ =. (k +1)α are roots of λ 3 mλ 2 + nλ 1=0forsomem, n N and m n, then one has (M,G,I) =(Solm,n 4,Sol4 m,n, {e}) for the geometry. This can be seen by choosing e 1 = αx 1,e 2 = αx 2,e 3 = αx 3,ande 4 = αx 4 on p. 274 and p. 270 in [18]. 1 For example, U stands for unimodular and the various integers 1,2,3 refer to certain characteristics of the Lie algebra structure; see [13] for details.

6 350 James Isenberg, Martin Jackson and Peng Lu A3. ClassU1[Z, Z,1]. [X 2,X 3 ]=0, [X 3,X 1 ]=0, [X 1,X 2 ]=0, [X 1,X 4 ]=kx 1 + X 2, [X 2,X 4 ]= X 1 + kx 2, [X 3,X 4 ]= 2kX 3, where k is a real number. If k = 0, this corresponds to the geometry (M,G,I) = (R 4,E(2) R, {e}). Other values of k do not correspond to compact homogeneous geometries. A4. ClassU1[2, 1] with µ =0. [X 2,X 3 ]=0, [X 3,X 1 ]=0, [X 1,X 2 ]=0, [X 1,X 4 ]=X 2, [X 2,X 4 ]=0, [X 3,X 4 ]=0. This Lie algebra is isomorphic to the direct sum n 3 R where n 3 is the Lie algebra of Nil 3. Hence, in this case (M,G,I) =(Nil 3 R,Nil 3 R, {e}). A5. ClassU1[2, 1] with µ =1. [X 2,X 3 ]=0, [X 3,X 1 ]=0, [X 1,X 2 ]=0, [X 1,X 4 ]= 1 2 X 1 + X 2, [X 2,X 4 ]= 1 2 X 2, [X 3,X 4 ]=X 3. This does not correspond to any of the compact homogeneous geometries. A6. ClassU1[3]. [X 2,X 3 ]=0, [X 3,X 1 ]=0, [X 1,X 2 ]=0, [X 1,X 4 ]=X 2, [X 2,X 4 ]=X 3, [X 3,X 4 ]=0. This corresponds to the geometry (M,G,I) =(Nil 4,Nil 4, {e}) whichcanbe seen by choosing e 1 = X 1,e 2 = X 2,e 3 = X 3,ande 4 = X 4 on p. 274 in [18]. A7. ClassU3I0. [X 1,X 4 ]=0, [X 2,X 4 ]=0, [X 3,X 4 ]=0, [X 2,X 3 ]=X 4, [X 3,X 1 ]=X 2, [X 1,X 2 ]= X 3. This corresponds to the geometry (M,G,I) =(Sol 4 1,Sol4 1, {e}) whichcan be seen by choosing e 1 = X 1,e 2 = X 2 + X 3,e 3 = X 2 X 3,e 4 = 2X 4 on p. 272 in [18].

7 Ricci flow on locally homogeneous closed 4-manifolds 351 A8. ClassU3I2. [X 1,X 4 ]=0, [X 2,X 4 ]=0, [X 3,X 4 ]=0, [X 2,X 3 ]= X 4, [X 3,X 1 ]=X 2, [X 1,X 2 ]=X 3. This does not correspond to any of the compact homogeneous geometries. A9. ClassU3S1. [X 1,X 4 ]=0, [X 2,X 4 ]=0, [X 3,X 4 ]=0, [X 2,X 3 ]=X 1, [X 3,X 1 ]=X 2, [X 1,X 2 ]= X 3. This Lie algebra is isomorphic to the direct sum sl 2 R. This corresponds to the geometry (M,G,I) =(ŜL(2, R) R, ŜL(2, R) R, {e}). A10. ClassU3S3. [X 1,X 4 ]=0, [X 2,X 4 ]=0, [X 3,X 4 ]=0, [X 2,X 3 ]=X 1, [X 3,X 1 ]=X 2, [X 1,X 2 ]=X 3. This Lie algebra is isomorphic to the direct sum su(2) R. This corresponds to the geometry (M,G,I) =(S 3 R,SU(2) R, {e}) Compact four dimensional homogeneous geometries with non-trivial isotropy group. Now we list the compact 4-dimensional homogeneous geometries (M 4,G,I) for which dimension of G is bigger than 4. Recall H(n) is the isometry group of the simply-connected hyperbolic n-manifolds H n. 1. (M,G,I) =(H 3 R,H(3) R,SO(3)) 2. (M,G,I) =(S 2 R 2,SO(3) R 2,SO(2) {0}) 3. (M,G,I) =(H 2 R 2,H(2) R 2,SO(2) {0}) 4. (M,G,I) =(S 2 S 2,SO(3) SO(3),SO(2) SO(2)) 5. (M,G,I) =(S 2 H 2,SO(3) H(2),SO(2) SO(2)) 6. (M,G,I) =H 2 H 2,H(2) H(2),SO(2) SO(2)) 7. (M,G,I) =(CP 2,SU(3),U(2)) 8. (M,G,I) =(CH 2,SU(1, 2),U(2)) 9. (M,G,I) =(S 4,SO(5),SO(4)) 10. (M,G,I) =(H 4,H(4),SO(4))

8 352 James Isenberg, Martin Jackson and Peng Lu 3. The Ricci flow on 4-dimensional unimodular Lie groups. Recall that our strategy is to analyze the Ricci flow on a simply connected manifold M that is the universal cover of a closed manifold M q.forafixed class (M,G,I) and a fixed initial homogeneous metric g 0 compatible with the class, let g(t) be the homogeneous solution of the Ricci flow t g(t) = 2Ric(g(t)) g(0) = g 0. On the closed manifold M q, we consider the volume-normalized Ricci flow g N ( t) as the solution of t g N ( t) = 2Ric( g N ( t)) + r N 2 g N g N (0) = g 0 where r N is the scalar curvature of g N. Note that averaging of the scalar curvature is not needed for homogeneous metrics. We also note that the Ricci flow equation for homogeneous metrics reduces to a system of ordinary differential equations. For each class (Ai) listed in Section 2.2, we describe the families of initial metrics which are diagonal and remain diagonal under the Ricci flow and then study their long time behavior. To address the diagonalization issue, we use the following strategy: Fix a homogeneous metric h and let {X i } be any basis of left-invariant vectors on the Lie group G with the bracket structure [X i,x j ]=C k ij X k. For those classes (e.g., A4, A9 and A10) in which the Lie group G is a product group G 1 R with dim(g 1 )=3,wechooseX 1,X 2,X 3 as left invariant vector fields on G 1 and X 4 = u on R. Let {Y i } be any basis orthogonal with respect to h; thatis, h(y i,y j )=λ i δ ij. Let the transformation from {X i } to {Y i } be given by Y i =Λ k i X k. Computing the bracket structure for {Y i },weget [Y i,y j ]=[Λ k i X k, Λ l j X l]=λ k i Λl j Cm kl X m =Λ k i Λl j Cm kl (Λ 1 ) n m Y n. Thus, [Y i,y j ]= C n ij Y n where C n ij =Λk i Λl j (Λ 1 ) n m Cm kl.

9 Ricci flow on locally homogeneous closed 4-manifolds 353 Now, compute the Ricci curvature of h using the orthonormal basis {Ȳi} defined by Ȳi = 1 λi Y i. For this, we use the following Ricci curvature formula for unimodular Lie groups from Corollary 7.33 p. 184 [1], Ric(W, W )= 1 [W, Ȳ i ] 2 1 [ W, [W, 2 2 Ȳi] ], Ȳi + 1 [Ȳi, 2 Ȳj],W 2. i i Finally, check if any positive values of the parameters λ i produce a diagonal Ricci tensor. Only for these values does the metric remain diagonal under the Ricci flow. We follow this strategy and use the same notation in the rest of this section. Remark In our search for families of initial metrics which remain diagonal under the Ricci flow, we have chosen special Y i so that the Lie brackets [Y i,y j ] are simple. Presumably, there are other families of initial metrics and other bases Y i for which the diagonalization is preserved by the Ricci flow. This possibility has not yet been explored. Our calculations do show that there are (M,G,I) and bases Y i such that the property that the initial metric has components (g 0 ) a4 =0,a =1, 2, 3, is preserved under Ricci flow. To study the decay of the curvature tensor, we use the following sectional curvature formula for Lie groups from Theorem 7.30 p. 183 [1]. For the Lie algebra g of G, define the operator U : g g g by 2 U(X, Y ),Z = [Z, X],Y + X, [Z, Y ] for all Z g; (2) then the curvature is given by i<j (1) R(X, Y )X, Y = 3 4 [X, Y ] [X, [X, Y ]],Y 1 2 [Y,[Y,X]],X + U(X, Y ) 2 U(X, X),U(Y,Y ). (3) 3.A1. U1[(1,1,1)]. For (M,G,I) =(R 4, R 4, {0}), G acts on M by translation h(x) =h + x for h G. Any homogeneous metric g 0 on M must be of the form g 0 = λ 1 dx 1 dx 1 ++λ 2 dx 2 dx 2 + λ 3 dx 3 dx 3 + λ 4 dx 4 dx 4 for some constants λ i > 0. The metric g 0 is flat; hence g(t) g 0 for <t<.

10 354 James Isenberg, Martin Jackson and Peng Lu 3.A2. U1[1, 1, 1]. For U1[1, 1, 1], we use Y i =Λ k i X k with Λ= a a 2 a a 4 a 5 a 6 1 to diagonalize the initial metric g 0. Proposition 3.1. For the class U1[1, 1, 1] suppose the initial metric g 0 is diagonal in the basis Y i.then (i) if k 1, 1 2, the Ricci flow solution g(t) remains diagonal in the basis Y i if and only if a 1 = a 2 = a 3 =0; (ii) if k =1, the Ricci flow solution g(t) remains diagonal in the basis Y i if and only if a 2 = a 3 =0; (iii) if k = 1 2, the Ricci flow solution g(t) remains diagonal in the basis Y i if and only if a 1 = a 2 =0; Proof. We compute [Y 2,Y 3 ]=0, [Y 3,Y 1 ]=0, [Y 1,Y 2 ]=0, [Y 1,Y 4 ]=Y 1, [Y 2,Y 4 ]=ky 2 + αy 1, [Y 3,Y 4 ]= (k +1)Y 3 + βy 2 + γy 1. where α =(1 k)a 1, β =(1+2k)a 3, and γ =(k +2)a 2 (1 + 2k)a 1 a 3. Let W = w 1 Ȳ 1 + w 2 Ȳ 2 + w 3 Ȳ 3 + w 4 Ȳ 4. We compute [W, Ȳi] firstandthen use (1) with h = g 0 to compute the coefficients of w i w j in Ric(W, W ). We find that the off-diagonal components of the Ricci tensor in the basis {Ȳi} are given by ( ) Ric(Ȳ1, Ȳ2) βγλ2 +(k 1)αλ 3 λ1 = 2 λ 2 λ 3 λ 4 Ric(Ȳ1, Ȳ3) = (2 + k)γ λ 1 2 λ 3 λ 4 Ric(Ȳ2, Ȳ3) = αγλ 1 +(1+2k)βλ 2 2 λ 2 λ3 λ 4 Ric(Ȳ1, Ȳ4) =Ric(Ȳ2, Ȳ4) =Ric(Ȳ3, Ȳ4) =0

11 Ricci flow on locally homogeneous closed 4-manifolds 355 In order for these off-diagonal components to be zero, we must have α = β = γ = 0. The proposition follows. Now we discuss the long time behavior for the families of the initial metrics in Proposition 3.1 We start with (A2iv) and later show that the other three cases are covered by the same analysis. (A2iv) In this case, k 0, 1, 1 2.Wehave and Y 1 = X 1, Y 2 =X 2, Y 3 = X 3, Y 4 =X 4 + a 4 X 1 + a 5 X 2 + a 6 X 3. [Y 2,Y 3 ]=0, [Y 3,Y 1 ]=0, [Y 1,Y 2 ]=0, [Y 1,Y 4 ]=Y 1, [Y 2,Y 4 ]=ky 2, [Y 3,Y 4 ]= (k +1)Y 3. The bases Y i and X i both satisfy the same Lie bracket relations so either canbeusedinthericciflowanalysis.weusex i. Let θ i be the frame of 1-forms dual to X i. Assume the Ricci flow solution takes the special form g(t) =A(t)(θ 1 ) 2 + (t)(θ 2 ) 2 + C(t)(θ 3 ) 2 + D(t)(θ 4 ) 2 (3.5) with g 0 = λ 1 (θ 1 ) 2 + λ 2 (θ 2 ) 2 + λ 3 (θ 3 ) 2 + λ 4 (θ 4 ) 2. Then, X1 = 1 A X 1,, X 4 = 1 D X 4 is an orthonormal frame with respect to the metric g. LetW = w 1 X1 + w 2 X2 + w 3 X3 + w 4 X4 and then compute [W, X 1 ]= 1 w 4 X1 [W, X 2 ]= k w 4 X2 D D [W, X 3 ]= k +1 w 4 X3 [W, X 4 ]= 1 w 1 X1 + k w 2 X2 k +1 w 3 X3. D D D D We have from (1) with h = g So Ric(W, W )=0 w w w2 3 2(k2 + k +1) D w 2 4. Ric(X 1,X 1 )=Ric(X 2,X 2 )=Ric(X 3,X 3 )=0, Ric(X 4,X 4 )=D Ric( X 4, X 4 )= 2(k 2 + k +1),

12 356 James Isenberg, Martin Jackson and Peng Lu and the Ricci flow is da dt =0, dc dt =0, The solution is given by d dt =0, dd dt =4(k2 + k +1). A(t) =λ 1, (t) =λ 2, C(t) =λ 3, and D(t) =λ 4 +4(k 2 + k +1)t. Hence for the subfamily in Proposition 3.1, the Ricci flow does not move in three directions and expands in the fourth direction at a speed linear in t. Next, we compute the curvature decay of g(t). From (2) we find U(X 1,X 1 )= A D X 4 U(X 2,X 2 )= k D X 4 (k +1)C U(X 3,X 3 )= D X 4 U(X 4,X 4 )=0 U(X 1,X 2 )=0 U(X 1,X 3 )=0 U(X 2,X 3 )=0 U(X 1,X 4 )= 1 2 X 1 U(X 2,X 4 )= k 2 X 2 U(X 3,X 4 )= k +1 2 X 3. From (3) with h = g we find the sectional curvatures K(X 1,X 2 )= k D, K(X 1,X 3 )= k +1 D, K(X k(k +1) 2,X 3 )= D, K(X 1,X 4 )= 1 D, K(X 2,X 4 )= k2 D, K(X (k +1)2 3,X 4 )= D. These curvatures of the solution g(t) decay at the rate 1/t. Recall on closed manifolds the volume-normalized solution g N ( t) relates to g(t) by scaling, where t is a function of t with lim t t =. From (3.5), it is easy to see that the behavior of g N ( t) as t is the same as the behavior of g N (t) ast where g N (t) =. ( ) λ 1 λ 2 λ 3 λ 1/4 4 g(t). A(t)(t)C(t)D(t) In discussing the long time behavior of volume normalized Ricci flow solutions, from this point on, we work with g N (t). For convenience, we call this the volume-normalized solution. Let us pick a point p M q. It is clear that the volume-normalized solution (M q,g N (t),p) collapses to a line in the the pointed Gromov Hausdorff topology.

13 Ricci flow on locally homogeneous closed 4-manifolds 357 (3.A2i) This is a special case of (A2iv) if we allow k = 0, so the analysis in (A2iv) applies. Note that since the Lie algebra is the direct sum sol 3 R, we can get the same conclusions from the analysis in [8](pp ). (3.A2ii) In this case, k =1. Wehave and Y 1 = X 1, Y 2 =X 2 + a 1 X 1, Y 3 = X 3, Y 4 =X 4 + a 4 X 1 + a 5 X 2 + a 6 X 3. [Y 2,Y 3 ]=0, [Y 3,Y 1 ]=0, [Y 1,Y 2 ]=0, [Y 1,Y 4 ]=Y 1, [Y 2,Y 4 ]=Y 2, [Y 3,Y 4 ]= 2Y 3. Note that this is the Lie algebra structure in (A2iv) if we allow k = 1. Hence, the analysis of (A2iv) applies with the same conclusion. (3.A2iii) In this case, k = 1 2.Wehave and Y 1 = X 1, Y 2 =X 2, Y 3 = a 3 X 2 + X 3, Y 4 =X 4 + a 4 X 1 + a 5 X 2 + a 6 X 3. [Y 2,Y 3 ]=0, [Y 3,Y 1 ]=0, [Y 1,Y 2 ]=0, [Y 1,Y 4 ]=Y 1, [Y 2,Y 4 ]= 1 2 Y 2, [Y 3,Y 4 ]= 1 2 Y 3. Note that this is the Lie algebra structure in (A2iv) if we allow k = 1 2. Hence, the analysis of (A2iv) applies with the same conclusion. 3.A3. U1[Z, Z,1]. For U1[Z, Z,1], we use Y i =Λ k i X k with 1 a 2 a 3 0 Λ= 0 1 a a 4 a 5 a 6 1 to diagonalize the initial metric g 0. Proposition 3.2. For the class U1[Z, Z,1], suppose the initial metric g 0 is diagonal in the basis Y i. Then, the Ricci flow solution g(t) remains diagonal in the basis Y i if and only if a 1 = a 2 = a 3 =0.

14 358 James Isenberg, Martin Jackson and Peng Lu Proof. We compute [Y 2,Y 3 ]=0, [Y 3,Y 1 ]=0, [Y 1,Y 2 ]=0, [Y 3,Y 4 ]= 2kY 3 [Y 1,Y 4 ]=(k α)y 1 +(α 2 +1)Y 2 + βy 3, [Y 2,Y 4 ]= Y 1 +(k + α)y 2 + γy 3. where α = a 2, β = a 2 a 3 a 1 a 2 2 a 1 3ka 3, and γ = a 3 3ka 1 a 1 a 2. We compute the off-diagonal components of the Ricci tensor in the basis {Ȳi} using (1) as in Section 2.A2 and get Ric(Ȳ1, Ȳ2) = 2αλ 1 +2α(1 + α 2 )λ 2 + βγλ 3 2 λ 1 λ 2 λ ( 4 ) Ric(Ȳ1, Ȳ3) γλ1 +(α 3k)βλ 2 λ3 = 2 λ 1 λ 2 λ ( 4 (α Ric(Ȳ2, Ȳ3) +3k)γλ1 +(1+α 2 ) )βλ 2 λ3 = 2λ 1 λ2 λ 4 Ric(Ȳ1, Ȳ4) =Ric(Ȳ2, Ȳ4) =Ric(Ȳ3, Ȳ4) =0. In order for these off-diagonal components to be zero, we must have α = β = γ = 0 and the proposition follows. If α = β = γ =0,theY i and X i both satisfy the same Lie bracket relations. As in Section 2.A2, we use X i in carrying out the analysis of the long time behavior of the Ricci flow solution for the family in Proposition 3.2. Proceeding as in Section 2.A2, we find [W, X 1 ]= k w 4 X1 D AD w 4 X 2 [W, X A 2 ]= D w 4 X 1 k w 4 X2, D [W, X 4 ]=( k A w 1 D D w 2) X 1 +( AD w 1 + We have from (1) [W, X 3 ]= 2k D w 4 X3 k w 2 ) X 2 2k w 3 X3. D D Ric(W, W )= A2 2 2AD w2 1 A2 2 2AD w w2 3 (A )2 +12k 2 A w4 2 2AD,

15 Ricci flow on locally homogeneous closed 4-manifolds 359 so the Ricci flow is da dt = 2 A2 D, d dt = A 2 2 AD, dc dt =0, dd dt = (A )2 +12k 2 A. A Clearly, C(t) =λ 3. If λ 1 = λ 2,then A(t) =λ 1, (t) =λ 2, and D(t) =λ 4 +12k 2 t. If λ 1 λ 2,wemayassumeλ 2 >λ 1 without loss of generality by the symmetry of A and in this system. A simple computation gives 1 da A dt + 1 d dt =0, so the product A = λ 1 λ 2 for all t. Another computation gives d + )2 [ A] = (A ( A), dt AD so A is positive and is decreasing in t. From the equation for dd dt,it follows easily that 12k 2 dd dt 12k2 + λ 2, λ 1 and so λ 4 +12k 2 t D(t) λ 4 +(12k 2 + λ 2 )t. λ 1 Since, it follows that Integrating the inequality, we get (A + ) 2 AD 4 D 4 λ 4 +(12k 2 + λ 2 λ 1 )t, 1 d A dt [ A] 4 λ 4 +(12k 2 + λ 2 λ 1 )t. A (λ 2 λ 1 ) (1+( 12k2 + λ ) k )t 2 + λ 2 λ 1. λ 4 λ 1 λ 4

16 360 James Isenberg, Martin Jackson and Peng Lu Hence for the family in Proposition 3.2, When K 0 the long-time behavior of the solution g(t) ast + is (see [8] for K =0) A(t) λ 1 λ 2, (t) λ 1 λ 2, C(t) =λ 3, D(t) linearly. Next, we compute the curvature decay of g(t). From (2), we find U(X 1,X 1 )= ka D X 4 U(X 2,X 2 )= k D X 4 U(X 3,X 3 )= 2kC D X 4 U(X 4,X 4 )=0 U(X 2,X 3 )=0 U(X 1,X 2 )= A 2D X 4 U(X 1,X 3 )=0 U(X 1,X 4 )= k 2 X 1 A 2 X 2 U(X 2,X 4 )= 2A X 1 + k 2 X 2 U(X 3,X 4 )= kx 3. From (3), with h = g, we find the sectional curvatures A K(X 1,X 2 )= + A 2 4k2 4D K(X 2,X 3 )= 2k2 D K(X 2,X 4 )= 3 A + A +2 4k2 D K(X 1,X 3 )= 2k2 D A K(X 1,X 4 )= 3 A +2 4k2 4D K(X 3,X 4 )= 4k2 D. Hence for the family in Proposition 3.2, the curvatures of the solution g(t) decay at the rate 1/t. Let us pick a point p M q. It is clear that the volume-normalized solution (M q,g N (t),p) collapses to a line in the pointed Gromov Hausdorff topology. 3.A4. U1[2, 1]. For U1[2, 1], we use Y i =Λ k i X k with 1 a 2 a 3 0 Λ= a a 4 a 5 a 6 1 to diagonalize the initial metric g 0. Proposition 3.3. For the class U1[2, 1] suppose the initial metric g 0 is diagonal in the basis Y i. Then the Ricci flow solution g(t) remains diagonal in the basis Y i for all t 0.

17 Ricci flow on locally homogeneous closed 4-manifolds 361 Proof. We compute [Y 2,Y 3 ]=0, [Y 3,Y 1 ]=0, [Y 1,Y 2 ]=0, [Y 1,Y 4 ]=Y 2, [Y 2,Y 4 ]=0, [Y 3,Y 4 ]=0. This bracket structure is identical to that of the basis {X i }. We compute the off-diagonal components of the Ricci tensor in the basis {Ȳi} using (1) as in Section 2.A2 and find Ric(Ȳi, Ȳj) = 0 for all i<j.the proposition is proved. Since Y i and X i both satisfy the same Lie bracket relations, as in Section 2.A2, we use X i and can carry out the analysis of the long time behavior of the Ricci flow solution for the family in Proposition 3.3 Proceeding as in Section 2.A2, we find the Ricci tensor Ric(W, W )= 2AD w2 1 + Hence, the Ricci flow is 2AD w w 2 3 2AD w2 4. da dt = D, dc dt =0, d dt = 2 AD, dd dt = A. From d dt ( A D )= d dt (A) = 0, we get A = λ 1 (1 + 3λ 2 λ 1 λ 4 t) 1/3 = λ 2 (1 + 3λ 2 λ 1 λ 4 t) 1/3 C = λ 3 D = λ 4 (1 + 3λ 2 λ 1 λ 4 t) 1/3 Hence, for the family in Proposition 3.3, the long time behavior of the solution g(t) ast + is A(t) +, (t) 0 +, C(t) =λ 3, D(t) +. Next, we compute the curvature decay for g(t). From (2), we find U(X 1,X 2 )= 2D X 4, U(X 2,X 4 ) = 2A X 1 and all other U(X i,x j )=0. From (3) with h = g, we find the sectional curvatures K(X 1,X 2 )=K(X 2,X 4 )= 4AD, K(X 1,X 4 )= 3 4AD,

18 362 James Isenberg, Martin Jackson and Peng Lu and all other K(X i,x j ) = 0. Hence for the family in Proposition 3.3, the curvatures of the solution g(t) decay at the rate 1/t. We now pick a point p M q. It is clear that the volume-normalized solution (M q,g N (t),p) collapses to a plane in the pointed Gromov Hausdorff topology. Note that since the Lie algebra is the direct sum n 3 R, we can get the same conclusions from the analysis in [8](p. 734). 3.A5. U1[2, 1]. For U1[2, 1], we use Y i =Λ k i X k with 1 a 2 a 3 0 Λ= 0 1 a a 4 a 5 a 6 1 to diagonalize the initial metric g 0. Proposition 3.4. For the class U1[2, 1], suppose the initial metric g 0 is diagonal in the basis Y i. Then the Ricci flow solution g(t) remains diagonal in the basis Y i if and only if a 1 = a 3 =0. Proof. We compute [Y 2,Y 3 ]=0, [Y 3,Y 1 ]=0, [Y 1,Y 2 ]=0, [Y 1,Y 4 ]= 1 2 Y 1 + Y 2 + βy 2, [Y 2,Y 4 ]= 1 2 Y 2 + αy 3, [Y 3,Y 4 ]=Y 3. where α = 3 2 a 1 β = 3 2 (a 3 a 1 ). We compute the off-diagonal components of the Ricci tensor in the basis {Ȳi} using (1) as in Section 2.A2 and get Ric(Ȳ1, Ȳ2) = αβλ 3 2 Ric(Ȳ1, λ 1 λ 2 λ Ȳ3) = 3β λ3 4 4 λ 1 λ 4 Ric(Ȳ1, Ȳ4) =0 Ric(Ȳ2, Ȳ3) = ( 3αλ 1 +2βλ 2 ) λ 3 4λ 1 λ2 λ 4 Ric(Ȳ2, Ȳ4) =0 Ric(Ȳ3, Ȳ4) =0. In order for these off-diagonal components to be zero, we must have α = β = 0 and the proposition follows.

19 Ricci flow on locally homogeneous closed 4-manifolds 363 If α = β =0,theY i and X i both satisfy the same Lie algebra bracket relations. As in Section 2.A2, we use X i and carry out the analysis of the long time behavior of the Ricci flow solution for the family in Proposition 3.4. Proceeding as in Section 2.A2, we find [W, X 1 ]= 1 2 D w 4 X 1 [W, X 2 ]= 1 [W, X 4 ]= 1 2 D w 1 X 1 + AD w 4 X 2, 2 D w 4 X 2, [W, X 3 ]= 1 w 4 X3, D ( ) AD w D w 2 X D w 3 X3. We have from (1) Ric(W, W )= 2AD w AD w w ( 3 D + AD )w2 4, so the Ricci flow is da dt = D = dc dt =0, It is clear that From 1 da A dt increasing, so + 1 from which we get AD A, d dt = 2 AD = AD, dd dt =3+ A. C(t) =λ 3. (4) d dt =0,wegetA = λ 1 λ 2. Since d [ A ] dt = 2 D, A/ is 3 dd dt =3+ A 3+λ 2 λ 1, 3t + λ 4 D(t) (3 + λ 2 )t + λ 4. (5) λ 1 To bound A(t) from below and above, we compute da 2 dt from which it follows that =2A D = 2λ 1λ 2 D, 2λ 2 1 λ 2 da2 (3λ 1 + λ 2 )t + λ 1 λ 4 dt 2λ 1λ 2 3t + λ 4.

20 364 James Isenberg, Martin Jackson and Peng Lu Then, by, integrating we obtain ( 2λ 2 λ 1 ln 1+ 3λ ) 1 + λ 2 t +1 A(t) λ 1 3λ 1 + λ 2 λ 1 λ 4 From A = λ 1 λ 2,wethenget λ 2 ( ) 2λ 2 3λ 1 ln 1+ 3 λ 4 t (t) +1 2λ 2 3λ 1 +λ 2 ln ( λ 2 2λ 2 3λ 1 ln 1+ 3λ 1+λ 2 λ 1 λ 4 ( 1+ 3 ) t +1. λ 4 (6) ). (7) t +1 Hence, for the family in Proposition 3.4, the long time behavior of the solution g(t) ast + is A(t) +, (t) 0 +, C(t) =λ 3, D(t) +. Next, we compute the curvature decay of g(t). From (2) we find U(X 1,X 1 )= A 2D X 4 U(X 2,X 2 )= 2D X 4 U(X 3,X 3 )= C D X 4 U(X 4,X 4 )=0 U(X 1,X 2 )= 2D X 4 U(X 1,X 3 )=0 U(X 2,X 3 )=0 U(X 1,X 4 )= 1 4 X 1 U(X 2,X 4 )= 2A X X 2 U(X 3,X 4 )= 1 2 X 3. From (3) with h = g, we find the sectional curvatures K(X 1,X 2 )= 1+ A 4D K(X 1,X 3 )= 1 2D K(X 2,X 3 )= 1 2D K(X 1,X 4 )= 1+3 A K(X 2,X 4 )= 1+ A K(X 3,X 4 )= 1 4D 4D D. Hence, for the family in Proposition 3.4, the curvatures of the solution g(t) decay at the rate 1/t. We now pick a point p M q. It is clear that (M q,g N (t),p) collapses to a line in the pointed Gromov Hausdorff topology. 3.A6. U1[3]. For U1[3], we use Y i =Λ k i X k with 1 a 2 a 3 0 Λ= 0 1 a a 4 a 5 a 6 1 to diagonalize the initial metric g 0.

21 Ricci flow on locally homogeneous closed 4-manifolds 365 Proposition 3.5. For the class U1[3], suppose the initial metric g 0 is diagonal in the basis Y i. Then the Ricci flow solution g(t) remains diagonal in the basis Y i if and only if a 1 = a 2. Proof. We compute [Y 2,Y 3 ]=0, [Y 3,Y 1 ]=0, [Y 1,Y 2 ]=0, [Y 1,Y 4 ]=Y 2 + αy 3, [Y 2,Y 4 ]=Y 3, [Y 3,Y 4 ]=0. where α = a 2 a 1. We compute the off-diagonal components of the Ricci tensor in the basis {Ȳi} as in Section 2.A2 and get Ric(Ȳ1, Ȳ2) = αλ 3 2, Ric(Ȳ2, λ 1 λ 2 λ Ȳ3) = α λ2 λ 3, 4 2λ 1 λ 4 and Ric(Ȳ1, Ȳ3) =Ric(Ȳ1, Ȳ4) =Ric(Ȳ2, Ȳ4) =Ric(Ȳ3, Ȳ4) = 0. In order for these off-diagonal components to be zero, we must have α =0andthe proposition is proved. If α =0,theY i and X i both satisfy the same Lie bracket relations. As in Section 2.A2, we use X i and carry out the analysis of the long time behavior of the Ricci flow solutions for the family in Proposition 3.5. Proceeding as in Section 2.A2, we compute C [W, X 1 ]= AD w 4 X 2 [W, X 2 ]= [W, X 3 ]=0 [W, X 4 ]= We have from (1) Ric(W, W )= 2AD w ( AD so the Ricci flow is da dt = D dc dt = C2 D Note that 1 A 1 C da dt = AD dc dt = C D C D )w2 2 + D w 4 X 3, AD w 1 X 2 + C D w 2 X 3. C 2D w ( AD + d AC 2 = dt AD dd dt = A + C. 1 d dt = C D AD 1 dd D dt = AD + C D. C D )w2 4,

22 366 James Isenberg, Martin Jackson and Peng Lu Hence, AC = λ 1 λ 2 λ 3 and compute Solving these gives E(t) = A CD = λ 1 λ 3 λ 4. Define E. = de dt = 3E2 E 0 3E 0 t +1 df dt = 3F 2. F (t) = F 0 3F 0 t +1 AD and F. = C D,we. where E 0 = λ 2. λ 1 λ 4 and F 0 = λ 3 λ 2 λ 4. Using these in the equations for (1/A)(dA/dt) and(1/c)(dc/dt), we can integrate to get A(t) =λ 1 (3E 0 t +1) 1/3 C(t) =λ 3 (3F 0 t +1) 1/3. (8) Using these with the conserved quantities AC and A CD,weget (t) =λ 2 (3E 0 t +1) 1/3 (3F 0 t +1) 1/3 D(t) =λ 4 (3E 0 t +1) 1/3 (3F 0 t +1) 1/3. (9) Hence, for the family in Proposition 3.5, the long time behavior of the solution g(t) is A(t) + (t) (λ 1 λ 2 λ 3 ) 1/3 C(t) 0 + D(t) +. Next, we compute the curvature decay of g(t). From (2), we find U(X 1,X 1 )=0 U(X 2,X 2 )=0 U(X 3,X 3 )=0 U(X 4,X 4 )=0 U(X 1,X 2 )= 2D X 4 U(X 1,X 3 )=0 U(X 2,X 3 )= C 2D X 4 U(X 1,X 4 )=0 U(X 2,X 4 )= 2A X 1 U(X 3,X 4 )= C 2 X 2. From (3) with h = g, we find the sectional curvatures K(X 1,X 2 )= A 4D K(X 1,X 4 )= 3 A 4D K(X 1,X 3 )=0 K(X 2,X 3 )= 4D A K(X 2,X 4 )= 3 C C K(X 3,X 4 )= 4D 4D. Hence, for the family in Proposition 3.5, the curvatures of the solution g(t) decay at the rate 1/t. Now pick a point p M q, it is easy to see that the volume-normalized solution (M q,g N (t),p) collapses to a plane in the pointed Gromov Hausdorff topology. C

23 Ricci flow on locally homogeneous closed 4-manifolds A7. U3I0. For U1[2, 1], we use Y i =Λ k i X k with 1 a 4 a 5 a 6 Λ= 0 1 a 2 a a to diagonalize the initial metric g 0. Proposition 3.6. For the class U3I0, suppose the initial metric g 0 is diagonal in the basis Y i.let α. = a 2 β. = a 1 a 2 a 3 a 4 γ. = a 1 a 1 a a 2a 3 + a 2 a 4 a 5. Then the Ricci flow solution g(t) remains diagonal in the basis Y i if and only if either (i) α = β = γ =0;or (ii) β = γ =0and λ 2 =(1 α 2 )λ 3. Proof. We compute [Y 1,Y 4 ]=0 [Y 2,Y 4 ]=0 [Y 3,Y 4 ]=0 [Y 2,Y 3 ]=Y 4 [Y 3,Y 1 ]=Y 2 αy 3 + βy 4 [Y 1,Y 2 ]= αy 2 +(α 2 1)Y 3 + γy 4. We compute the off-diagonal components of the Ricci tensor in the basis {Ȳ i } using (1) as in Section 2.A2 and get Ric(Ȳ1, Ȳ2) βλ 4 = 2 λ 1 λ 2 λ 3 Ric(Ȳ1, Ȳ3) γλ 4 = 2 λ 1 λ 3 λ 2 Ric(Ȳ1, Ȳ4) =0 Ric(Ȳ2, Ȳ3) = 2αλ 2 +2α(1 α 2 )λ 3 + βγλ 4 2λ 1 λ2 λ 3 Ric(Ȳ2, Ȳ4) = (βλ 2 αγλ 3 ) λ 4 2λ 1 λ2 λ 3 Ric(Ȳ3, Ȳ4) = ( αβλ 2 +(α 2 1)γλ 3 ) λ 4 2λ 1 λ 2 λ3

24 368 James Isenberg, Martin Jackson and Peng Lu In order for these off-diagonal components to be zero, we must have either (i) or (ii) in the proposition. To finish the proof of (ii), we need to ensure that the condition (t) =(1 α 2 )C(t) holdsforallt>0. We prove this near the end of this subsection. A7i First, we study the family (i) in Proposition 3.6. If α = β = γ =0, the bases Y i and X i both satisfy the same Lie bracket relations. As in Section 2.A2, we use X i and carry out the analysis of the long time behavior of the Ricci flow solutions for the family (i) in Proposition 3.6. Proceeding as in Section 2.A2, we find [W, X 1 ]= AC w 3 X C 2 + A w 2 X 3 [W, X C 2 ]= A w 1 X D 3 C w 3 X 4 [W, X 3 ]= AC w 1 X D 2 + C w 2 X 4 [W, X 4 ]=0. We have from (1) Ric(W, W )= 1 2 ( AC + C A + 2 A )w ( C A + D C AC )w ( AC + D C C A )w D 2 C w2 4, so the Ricci flow equation is da dt = C + C +2 dc dt = A + D C2 A Straightforward calculations give us 1 d C dt [C]=2 D C d dt = C A + D C 2 AC dd dt = D2 C. 1 dd D dt = D C, from which we conclude that CD 2 = λ 2 λ 3 λ 2 dd 4. Now, we can write D4 and solve to get λ 2 λ 3 λ 2 4 dt = ( D(t) =λ λ ) 1/3 4 t. (10) λ 2 λ 3

25 Ricci flow on locally homogeneous closed 4-manifolds 369 Another set of simple calculations give us 1 d ( + C)2 [ C] =AD C dt AC 1 d ( + C)2 [AD] = AD AD dt AC from which we conclude that AD( C) =λ 1 λ 4 (λ 2 λ 3 ). We get ( C) 2 C A2 = A2 D 2 ( C) 2 CD 2 = λ2 1 (λ 2 λ 3 ) 2. =4k 2 λ 2 λ 3. 3 for some k 3 0. With this and the identity we can write Integrating gives us ( + C) 2 C A k 3 tan 1 ( A k 3 ) = ( C)2 C da dt = 4k2 3 A , ( ) =4t + λ 1 k 3 tan 1 λ1. (11) k 3 For large t, wehavea(t) 4t. Using the conserved quantities AD( C) andcd 2, we could solve to get (t) and C(t) explicitly. More importantly, we see that for large t, (t) C(t) 1 D(t) t1/3. (12) Hence, for the family (i) in Proposition 3.6, the long time behavior of the solution g(t) is A(t) + (t) + C(t) + D(t) 0 +. Next, we compute the curvature decay of g(t). From (2), we find U(X 1,X 1 )=0 U(X 2,X 2 )=0 U(X 3,X 3 )=0 U(X 4,X 4 )=0 U(X 1,X 2 )= 2C X 3 U(X 1,X 3 )= C 2 X 2 U(X 2,X 3 )= + C 2A X 1 U(X 1,X 4 )=0 U(X 2,X 4 )= D 2C X 3 U(X 3,X 4 )= D 2 X 2.

26 370 James Isenberg, Martin Jackson and Peng Lu From (3) with h = g, we find the sectional curvatures K(X 1,X 2 )= K(X 1,X 4 )=0 K(X 2,X 4 )= C 3 C 2 4A D 4C C K(X 1,X 3 )= 3 C 2 4A K(X 2,X 3 )= 3D 4C + C + C +2 4A K(X 3,X 4 )= D 4C. Hence, for the family (i) in Proposition 3.6, the curvatures of the solution g(t) decayattherate1/t. For the volume-normalized solution g N (t), the metric components have the following long time behavior: A N (t) +, N (t) andc N (t) approach some positive constant, and D N (t) 0 +. It follows that the volume-normalized solution (M q,g N (t),p) collapses to a stripinthepointedgromov Hausdorfftopologyforp M q. A7ii For the rest of this subsection, we address the family (ii) in Proposition 3.6. Suppose β = γ = 0. The Lie brackets from the proof of Proposition 3.6 take the form [Y 1,Y 4 ]=0 [Y 2,Y 4 ]=0 [Y 3,Y 4 ]=0 [Y 2,Y 3 ]=Y 4 [Y 3,Y 1 ]=Y 2 αy 3 [Y 1,Y 2 ]= αy 2 +(α 2 1)Y 3. Recall we must show the condition (t) =(1 α 2 )C(t) is preserved under Ricci flow. Let ω i be the frame dual to Y i. Assume the Ricci flow solution g takes the form g(t) =A(t)(ω 1 ) 2 + (t)(ω 2 ) 2 + C(t)(ω 3 ) 2 + D(t)(ω 4 ) 2 with g 0 = λ 1 (ω 1 ) 2 + λ 2 (ω 2 ) 2 + λ 3 (ω 3 ) 2 + λ 4 (ω 4 ) 2. Let. Ȳ1 = 1 A Y 1,,. Ȳ4 = 1 D Y 4 and let W = w 1 Ȳ 1 + w 2 Ȳ 2 + w 3 Ȳ 3 + w 4 Ȳ 4. We first compute [W, Ȳi] and then compute the Ricci curvature of g(t) using (1). We find that Ric(W, W )isgivenby Ric(W, W )= 2 +2(1+α 2 )C +(1 α 2 ) 2 C 2 w1 2 2AC + AD + 2 (1 α 2 ) 2 C 2 w2 2 2AC + AD 2 +(1 α 2 ) 2 C 2 w3 2 2AC + D 2C w2 4 + α( +(1 α2 )C) A w 2 w 3. C

27 Ricci flow on locally homogeneous closed 4-manifolds 371 The Ricci flow equation is da dt = 2 +2(1+α 2 )C +(1 α 2 ) 2 C 2 C dc dt = AD + 2 (1 α 2 ) 2 C 2 A Hence, d dt = AD 2 +(1 α 2 ) 2 C 2 AC dd dt = D2 C. d dt ( +(1 α2 )C) = AD ( +(1 α2 )C) 2 ( +(1 α 2 )C ). AC Since +(1 α 2 )C =0attimet = 0, it remains 0 for all time which implies that g(t) is diagonal in the basis Y i. With +(1 α 2 )C = 0, the Ricci flow equations reduce to da dt =4 d dt =(1 α2 ) D dd dt = (1 α2 ) D2 2. A simple calculation gives d dt (2 D 2 ) = 0 which implies D = λ 2 λ 4. Now, we can solve the Ricci flow equations to obtain A = λ 1 +4t = ( λ (1 α2 )λ 2 λ 4 t ) 1/3 C = 1 ( λ 3 1 α (1 α 2 )λ 2 λ 4 t ) 1/3 ( D = λ 2 λ 4 λ (1 α 2 )λ 2 λ 4 t ) 1/3. Hence for the family (ii) in Proposition 3.6, the long time behavior of the Ricci flow g(t) ast is A(t) + (t) + C(t) + D(t) 0 + Finally, we compute the curvature decay of g(t). From (2), we find U(Y 1,Y 1 )=0 U(Y 2,Y 2 )= α A Y 1 U(Y 3,Y 3 )= αc A Y 1 U(Y 4,Y 4 )=0 U(Y 1,Y 2 )= α 2 Y 2 + 2C Y 3 U(Y 1,Y 3 )= (1 α2 )C Y 2 α 2 2 Y 3 U(Y 1,Y 4 )=0 U(Y 2,Y 3 )= +(1 α2 )C Y 1 2A U(Y 2,Y 4 )= D 2C Y 3 U(Y 3,Y 4 )= D 2 Y 2.

28 372 James Isenberg, Martin Jackson and Peng Lu From (3) with h = g, we find the sectional curvatures K(Y 1,Y 2 )= 1 K(Y 1,Y 3 )= 1 A A K(Y 1,Y 4 )=0 K(Y 2,Y 4 )= D 4C K(Y 2,Y 3 )= 3D 4C + 1 A K(Y 3,Y 4 )= D 4C. Hence for the family (ii) in Proposition 3.6, the curvatures of the solution g(t) decayattherate1/t. For the volume-normalized solution g N (t), the metric components have the following long time behavior: A N (t) +, N (t) andc N (t) approach some positive constants, and D N (t) 0 +. It follows that the volume-normalized solution (M q,g N (t),p) collapses to a stripinthepointedgromov Hausdorfftopologyforp M q. For U3I2, we use Y i =Λ k i X k with to diagonalize the initial metric g 0. 3.A8. U3I2. 1 a 4 a 5 a 6 Λ= 0 1 a 2 a a Proposition 3.7. For the class U3I2, suppose the initial metric g 0 is diagonal in the basis Y i. Then, the Ricci flow solution g(t) remains diagonal if and only if a 2 =0, a 1 = a 5 and a 3 = a 4. Proof. We compute [Y 1,Y 4 ]=0 [Y 2,Y 4 ]=0 [Y 3,Y 4 ]=0 [Y 2,Y 3 ]= Y 4 [Y 3,Y 1 ]=Y 2 + αy 3 + βy 4 [Y 1,Y 2 ]=αy 2 +(1+α 2 )Y 3 + γy 4. where α = a 2 β = a 1 a 2 a 3 + a 4 γ = a 1 a 1 a a 2 a 3 a 2 a 4 + a 5.

29 Ricci flow on locally homogeneous closed 4-manifolds 373 We compute the off-diagonal components of the Ricci tensor in the basis {Ȳi} using (1) as in Section 2.A2 and get Ric(Ȳ1, Ȳ2) = βλ 4 2 λ 1 λ 2 λ 3 Ric(Ȳ1, Ȳ3) = γλ 4 2 λ 1 λ 3 λ 2 Ric(Ȳ1, Ȳ4) =0 Ric(Ȳ2, Ȳ3) = 2αλ 2 +2α(1 + α 2 )λ 3 + βγλ 4 2λ 1 λ2 λ 3 Ric(Ȳ2, Ȳ4) = (βλ 2 + αγλ 3 ) λ 4 Ric(Ȳ3, 2λ 1 λ2 λ Ȳ4) = (αβλ 2 +(1+α 2 )γλ 3 ) λ4. 3 2λ 1 λ 2 λ3 In order for these off-diagonal components to be zero, we must have α = β = γ = 0 and the proposition follows. If α = β = γ = 0, the bases Y i and X i both satisfy the same Lie bracket relations. As in Section 2.A2, we use X i and carry out the analysis of the long time behavior of the Ricci flow solutions for the family in Proposition 3.7. Proceeding as in Section 2.A2, we find [W, X 1 ]= AC w 3 X C 2 A w 2 X 3 [W, X C 2 ]= A w 1 X D 3 + C w 3 X 4 [W, X 3 ]= AC w 1 X D 2 C w 2 X 4 We have from (1) [W, X 4 ]=0. Ric(W, W )= 1 2 ( 2 A C A AC )w ( AC C A D C )w ( C A AC D C )w2 3 + D 2C w2 4. The Ricci flow is da dt = C + C 2 dc dt = C2 A + A + D d dt = 2 AC + C A + D C dd dt = D2 C. The equations here are similar to those of the case A7(i), with the only difference being in the equation for A. ecause the equations for, C and D

30 374 James Isenberg, Martin Jackson and Peng Lu are the same, we know that CD 2 = λ 2 λ 3 λ 2 4. It follows that dd dt = D4, λ 2 λ 3 λ 2 4 and hence, ( D(t) =λ λ ) 1/3 4 t. (13) λ 2 λ 3 Calculations similar to those in the case A7(i) show that AD( + C) = λ 1 λ 4 (λ 2 + λ 3 ). So A 2 ( C + C k 4 0, and +2)= (AD(+C))2 CD 2 da dt = k2 4 A = k 2 4 is a constant where Integrating the equation gives us ( ) k 4 2A 2 tanh 1 A =4t + k 5 (14) k 4 where k 5 is a constant. Since A increases for all t and tanh x asymptotes to 1, we see that A(t) k 4 /2ast +. Using the conserved quantity CD 2 and AD( + C), we conclude that for the family in Proposition 3.7 both and C grow at the rate t 1/3,and thelongtimebehaviorofthericciflowg(t) ast + is A(t) k 4 /2 (t) + C(t) + D(t) 0 +. Next, we compute the curvature decay of g(t). From (2), we find U(X 1,X 1 )=0 U(X 2,X 2 )=0 U(X 3,X 3 )=0 U(X 4,X 4 )=0 U(X 1,X 2 )= 2C X 3 U(X 1,X 3 )= C 2 X 2 U(X 2,X 3 )= + C 2A X 1 U(X 1,X 4 )=0 U(X 2,X 4 )= D 2C X 3 U(X 3,X 4 )= D 2 X 2. From (3) with h = g, we find the sectional curvatures C K(X 1,X 2 )= 3 C +2 4A K(X 2,X 3 )= 3D 4C + C + C 2 4A K(X 2,X 4 )= D 4C K(X 1,X 3 )= 3 C + C +2 4A K(X 1,X 4 )=0 K(X 3,X 4 )= D 4C.

31 Ricci flow on locally homogeneous closed 4-manifolds 375 Here, the decay rate is not obvious for all sectional curvatures. Note that C 1. The decay rate of da C 1 follows from the equation of dt. It suffices to show that da dt decays at the rate e ct for some c>0. From (14), we get A = k [ tanh t + 2 A + 2k ] 5 k 4 k 4 k 4 and the decay rate of da dt follows from taking the time derivative of this equation. Thus C 1 decays exponentially. Hence, for the family in Proposition 3.7, the curvatures of the solution g(t) decay at the rate 1/t. For the volume-normalized solution g N (t), the metric components have the following long time behavior: A N (t) 0 +, N (t) andc N (t) approach some positive constants, and D N (t) 0 +. The volume-normalized flow (M q,g N (t),p) collapses to a plane in the pointed Gromov Hausdorff topology for p M q. 3.A9. U3S1. The Lie bracket relations for cases 3.A9 and 3.A10 differ only in [X 1,X 2 ]. To unify some of the calculations for these two cases, we introduce a constant δ and write [X 1,X 2 ]=δx 3 with δ = 1 corresponding to 3.A9 and δ =1 corresponding to 3.A10. For U3S1 andu3s3, we use Y i =Λ k i X k with Λ= a 1 a 2 a 3 1 to diagonalize the initial metric g 0. Proposition 3.8. For the class U3S1, suppose the initial metric g 0 is diagonal in the basis Y i.then (i) if λ 1 λ 2, the Ricci flow solution g(t) remains diagonal if and only if a 1 = a 2 = a 3 =0;and (ii) if λ 1 = λ 2, the Ricci flow solution g(t) remains diagonal if and only if a 1 = a 2 =0. Proof. We compute [Y 1,Y 4 ]= a 3 Y 2 + δa 2 Y 3 [Y 2,Y 4 ]=a 3 Y 1 δa 1 Y 3 [Y 3,Y 4 ]= a 2 Y 1 + a 1 Y 2 [Y 2,Y 3 ]=Y 1, [Y 3,Y 1 ]=Y 2, [Y 1,Y 2 ]=δy 3.

32 376 James Isenberg, Martin Jackson and Peng Lu We compute the off-diagonal components of the Ricci tensor in the basis Ȳi using (1) as in Section 2.A2 and get Ric(Ȳ1, Ȳ2) = (λ2 3 λ 1λ 2 )a 1 a 2 2λ 3 λ 4 λ1 λ 2 Ric(Ȳ1, Ȳ3) = (λ2 2 δλ 1λ 3 )a 1 a 3 2λ 2 λ 4 λ1 λ 3 Ric(Ȳ2, Ȳ3) = (λ2 1 δλ 2λ 3 )a 2 a 3 2λ 1 λ 4 λ2 λ 3 Ric(Ȳ1, Ȳ4) = (λ 2 δλ 3 ) 2 a 1 2λ 2 λ 3 λ1 λ 4 Ric(Ȳ2, Ȳ4) = (λ 1 δλ 3 ) 2 a 2 2λ 1 λ 3 λ2 λ 4 Ric(Ȳ3, Ȳ4) = (λ 1 λ 2 ) 2 a 3 2λ 1 λ 2 λ3 λ 4. The diagonal components are given by Ric(Ȳ1, Ȳ1) = (λ2 1 λ2 3 )λ 2a 2 2 +(λ2 1 λ2 2 )λ 3a ( λ 2 1 (λ 2 δλ 3 ) 2) λ 4 2λ 1 λ 2 λ 3 λ 4 Ric(Ȳ2, Ȳ2) = (λ2 2 λ2 3 )λ 1a 2 1 +(λ2 2 λ2 1 )λ 3a ( λ 2 2 (λ 1 δλ 3 ) 2) λ 4 2λ 1 λ 2 λ 3 λ 4 Ric(Ȳ3, Ȳ3) = (λ2 3 λ2 2 )λ 1a 2 1 +(λ2 3 λ2 1 )λ 2a ( λ 2 3 (λ 1 λ 2 ) 2) λ 4 2λ 1 λ 2 λ 3 λ 4 Ric(Ȳ4, Ȳ4) = (λ 2 δλ 3 ) 2 λ 1 a 2 1 +(λ 1 δλ 3 ) 2 λ 2 a 2 2 +(λ 1 λ 2 ) 2 λ 3 a λ 1 λ 2 λ 3 λ 4 (15) For δ = 1, in order for these off-diagonal components to be zero, we must have either (i) or (ii) in Proposition 3.8. As in case A7ii, to finish the proof of (ii), in Proposition 3.8, we need to ensure that the condition A(t) =(t)holdsforallt>0. We prove this at the end of this subsection. Remark. Note that there are many initial metrics g 0 that cannot be diagonalized by the choice of Λ we use here. For 3.A9 and 3.A10, the Lie group G is a product G 1 R with dim(g 1 ) = 3. After transforming with Λ as given above, one can use a Milnor frame on G 1 (with respect to a chosen initial metric on G 1 ) to further diagonalize, in which case the Lie algebra takes the form [Y 1,Y 4 ]=a 1 Y 1 + a 2 Y 2 + a 3 Y 3 [Y 2,Y 4 ]=b 1 Y 1 + b 2 Y 2 + b 3 Y 3 [Y 3,Y 4 ]=c 1 Y 1 + c 2 Y 2 + c 3 Y 3 [Y 2,Y 3 ]=Y 1 [Y 3,Y 1 ]=Y 2 [Y 1,Y 2 ]=δy 3

33 Ricci flow on locally homogeneous closed 4-manifolds 377 with a 1 + b 2 + c 3 = 0 from the unimodular condition. With these, the off-diagonal components of the Ricci curvature are given by Ric(Ȳ 1, Ȳ 2 )= c 1 c 2 λ 1 λ 2 + b 1 (b 2 a 1 )λ 1 λ 3 + a 2 (a 2 b 2 )λ 2 λ 3 a 3 b 3 λ λ 1 λ2 λ 3 λ 4 Ric(Ȳ1, Ȳ3) = c 1(2a 1 + b 2 )λ 1 λ 2 + b 1 b 3 λ 1 λ 3 a 2 c 2 λ a 3(2a 1 + b 2 )λ 2 λ 3 2 λ 1 λ 2 λ3 λ 4 Ric(Ȳ1, Ȳ4) = (λ 2 δλ 3 )(c 2 λ 2 + b 3 λ 3 ) 2 λ 1 λ 2 λ 3 λ4 Ric(Ȳ2, Ȳ3) = b 1c 1 λ 2 1 c 2(a 1 +2b 2 )λ 1 λ 2 + b 3 (a 1 +2b 2 )λ 1 λ 3 + a 2 a 3 λ 2 λ 3 2 λ 1 λ2 λ3 λ 4 Ric(Ȳ2, Ȳ4) = (λ 1 δλ 3 )(c 1 λ 1 + a 3 λ 3 ) 2 λ 1 λ2 λ 3 λ4 Ric(Ȳ3, Ȳ4) = (λ 1 λ 2 )(b 1 λ 1 + a 2 λ 2 ) 2 λ 1 λ 2 λ3 λ4 One can analyze these expressions to determine conditions under which Ricci flow preserves the diagonalization of an initial metric. The complexity of these expressions leads to many cases that must be analyzed so we have limited our attention to the transformation matrix Λ given above with the results given in Proposition 3.8 for A9 and Proposition 3.9 for A10. 3.A9i First, we study family (i) in Proposition 3.8. For a 1 = a 2 = a 3 = 0, we have Y i = X i.themetricg(t) is a product metric on ŜL(2, R) R g(t) =g SL (t)+λ 4 du 2 where g SL (t) =A(t)(θ 1 ) 2 + (t)(θ 2 ) 2 + C(t)(θ 3 ) 2 is a Ricci flow solution on ŜL(2, R). From (15), we get the Ricci flow equations da dt = ( + C)2 A 2 C d dt = (A + C)2 2 AC dc dt = (A )2 C 2. A The long time behavior of g SL (t) has been analyzed in [8]: the curvatures of g SL (t) decayattherate1/t and the volume-normalized Ricci flow g SL ( t) collapses to a plane. Hence, the volume-normalized solution (M q,g N (t),p) collapses to a plane in the pointed Gromov Hausdorff topology for p M q. 3.A9ii For the rest of this subsection we address family (ii) in Proposition 3.8 where λ 1 = λ 2 and a 1 = a 2 = 0. From (15), we conclude that the Ricci flow equation of g(t) is

34 378 James Isenberg, Martin Jackson and Peng Lu da dt = ( + C)2 A 2 C dc dt = (A )2 C 2 A + A2 + 2 D a2 3 d dt = (A + C)2 2 AC dd (A + )2 = dt A a A2 2 AD a2 3 Recall that we must show that the condition A(t) = (t) is preserved under Ricci flow. To this end, we compute [ d C 2 (A + ) 2 ] (A + )2 (A ) = dt AC AD a2 3 (A ). Since A =0attimet = 0, this implies that A(t) =(t) andg(t) remains diagonal in the basis Y i. With A =, the Ricci flow equations reduce to da dt = C A +2 A simple computation shows d dt da dt 2+ λ 3 λ 1 and dc dt = dd C2 A 2 dt =4a2 3. ( C ) A = 2A 1 ( C A +(C A )2) 0; hence 2 2t + λ 1 A(t) =(t) (2 + λ 3 λ 1 )t + λ 1. (16) From the equation for dc dt and (16), we get C2 these inequalities we find (2t+λ 1 ) 2 dc dt 0. Integrating Finally, 2λ 1 λ 3 2λ 1 + λ 3 C(t) λ 3. (17) D(t) =4a 2 3t + λ 4. (18) Hence, for family (ii) in Proposition 3.8, with a 3 0, the long time behavior of of the Ricci flow g(t) ast + is A(t) + (t) + C(t) constant > 0 D(t) +. Next, we compute the curvature decay of g(t) as in A7ii. From (2), we find (using A(t) =(t)) U(Y 1,Y 3 )= A + C 2A Y 2 U(Y 2,Y 3 )= A + C 2A Y 1 U(Y 1,Y 4 )= 1 2 a 3Y 2 U(Y 2,Y 4 )= 1 2 a 3Y 1,

35 Ricci flow on locally homogeneous closed 4-manifolds 379 and all other U(Y i,y j ) = 0. From (3) with h = g, we find the sectional curvatures K(Y 1,Y 2 )= 4+3C A 4A K(Y 1,Y 3 )=K(Y 2,Y 3 )= C 4A 2 and all other K(Y i,y j ) = 0. Hence for family (ii) in Proposition 3.8(ii), the curvatures of the solution g(t) decay at the rate 1/t. It follows that the volume-normalized solution (M q,g N (t),p) collapses to Euclidean space R 3 in the pointed Gromov Hausdorff topology. 3.A10. U3S3. Using the setup given in A9, we prove the following. Proposition 3.9. For the class U3S1, suppose the initial metric g 0 is diagonal in the basis Y i.then (i) if λ 1,λ 2,λ 3 are all different, the Ricci flow solution g(t) remains diagonal if and only if a 1 = a 2 = a 3 =0; (ii) if λ j = λ k λ i for some permutation {i, j, k} of {1, 2, 3}, thericci flow solution g(t) remains diagonal if and only if a j = a k =0;and (iii) if the initial metric satisfies λ 1 = λ 2 = λ 3, the Ricci flow solution g(t) remains diagonal for any a 1, a 2,anda 3. Proof. Set δ = 1 in the proof of Proposition 3.8. In order for the offdiagonal Ricci components to be zero, we must have either (i) or (ii) or (iii) in Proposition 3.9. As in previous cases, to finish the proof of (ii) in Proposition 3.9, we need to ensure that the condition (t) = C(t) holds for all t>0(usingj =2,k = 3 without loss of generality). Also, to finish the proof of (iii) in Proposition 3.9, we need to ensure that the condition A(t) =(t) =C(t) holdsforallt>0. These are verified below. 3.A10i If a 1 = a 2 = a 3 =0,wehaveY i = X i. The metric is a product metric on S 3 R g(t) =g S 3(t)+λ 4 du 2 where g S 3(t) =A(t)(θ 1 ) 2 + (t)(θ 1 ) 2 + C(t)(θ 1 ) 2 is a Ricci flow solution on S 3. From (15), we get the Ricci flow equations da dt = ( C)2 A 2 C d dt = (A C)2 2 AC dc dt = (A )2 C 2. A (19)

36 380 James Isenberg, Martin Jackson and Peng Lu The volume-normalized flow associated with g S 3(t) has been analyzed in [8] and is found to converge to a round sphere. Hence, the behavior of g(t) is clear. 3.A10ii. For family (ii) in Proposition 3.9, without loss of generality, we may assume that i =1,j =2,andk =3soλ 2 = λ 3 and a 2 = a 3 =0. From (15), we conclude that the Ricci flow equation of g(t) is da dt = ( C)2 A 2 d C dt = (A C)2 2 2 C 2 AC CD a2 1 dc dt = (A )2 C C 2 dd A D a2 C)2 1 = ( dt C a2 1. Recall that we must show that the condition (t) = C(t) is preserved under Ricci flow. This follows from [ d A 2 ( + C) 2 ] ( + C)2 ( C) = dt AC CD a2 1 ( C). With = C, the Ricci flow equations reduce to da dt = A2 2 d dt = A 2 dd dt =0. This is a special case of equation (19) with = C, so the conclusions from 3.A10i hold here. 3.A10iii For family (iii) in Proposition 3.9, λ 1 = λ 2 = λ 3. From (15), we conclude that the Ricci flow equation of g(t) is da dt = C 2 )a 2 (A2 2 +(A2 2 )Ca ( A 2 ( C) 2) D CD d dt = C 2 )Aa 2 (2 1 +(2 A 2 )Ca ( 2 (A C) 2) D ACD dc dt = 2 )Aa 2 (C2 1 +(C2 A 2 )a ( C 2 (A ) 2) D AD dd dt = ( C)2 Aa 2 1 +(A C)2 a 2 2 +(A )2 Ca 2 3. AC Recall we need to show A(t) = (t) = C(t) is preserved under Ricci flow. This follows from d dt (A ) =M 11(A )+M 12 (A C) d dt (A C) =M 21(A )+M 22 (A C)