Invariant Ricci-flat Metrics of Cohomogeneity One with Wallach Spaces as Principal Orbits
IInvariant Ricci-flat Metrics of Cohomogeneity One with WallachSpaces as Principal Orbits
Hanci ChiMarch 6, 2019
Abstract
We construct a continuous 1-parameter family of smooth complete Ricci-flat metrics of co-homogeneity one on vector bundles over CP , HP and OP with respective principal orbits G/K the Wallach spaces SU (3) /T , Sp (3) / ( Sp (1) Sp (1) Sp (1)) and F / Spin(8). Almost all theRicci-flat metrics constructed have generic holonomy. The only exception is the complete G metric discovered in [7][25]. It lies in the interior of the 1-parameter family on (cid:86) − CP . All theRicci-flat metrics constructed have asymptotically conical limits given by the metric cone overa suitable multiple of the normal Einstein metric on G/K . Contents
A Riemannian manifold (
M, g ) is
Ricci-flat if its Ricci curvature vanishes:Ric( g ) = 0 . (1.1)A Ricci-flat manifold is the Euclidean analogy of a vacuum solution of the Einstein field equations.1 a r X i v : . [ m a t h . DG ] M a r n this article, we study complete noncompact Ricci-flat manifolds of cohomogeneity one. ARiemannian manifold ( M, g ) is of cohomogeneity one if a Lie Group G acts isometrically on M suchthat the principal orbit G/K is of codimension one. The Ricci-flat condition (1.1) is then reducedto a system of ODEs.Many examples of cohomogeneity one Ricci-flat metrics have special holonomy. These includethe first example of an inhomogeneous Einstein metric, which is also a K¨ahler metric. It wasconstructed in [10] on a non-compact open set of C n . A complete Calabi–Yau metric was constructedon T ∗ S independently in [11][22]. The construction was generalized to T ∗ CP n in [11] and thoseRicci-flat metrics are hyper-K¨ahler. Cohomogeneity one K¨ahler–Einstein metrics were constructedon complex line bundles over a product of compact K¨ahler–Einstein manifolds in [2][18]. Completemetrics with G or Spin(7) holonomy can be found in [7][25] [16][17][24].Ricci-flat metrics with generic holonomy, for example, were constructed on various vector bundlesin [2][4][31][13]. It is further shown in [9][8] that for infinitely many dimensions, there exist exampleswhich are homeomorphic but not diffeomorphic. The case where the isotropy representation of theprincipal orbit contains exactly two inequivalent irreducible summands was studied in [4][33]. Inthis article, we consider examples with three inequivalent summands. Specifically, let ( G, H, K ) beone of I. ( SU (3) , S ( U (2) U (1)) , S ( U (1) U (1) U (1))) , II. ( Sp (3) , Sp (2) Sp (1) , Sp (1) Sp (1) Sp (1)) , III. ( F , Spin(9) , Spin(8)) . (1.2)For these triples, we construct Ricci-flat metrics on the corresponding cohomogeneity one vectorbundles M with unit sphere bundle H/K (cid:44) → G/K → G/H.
The singular orbits
G/H ’s are re-spectively CP , HP and OP . The principal orbits G/K ’s are
Wallach spaces . They appearedexplicitly in Wallach classification of even dimensional homogeneous manifolds with positive sec-tional curvature [30]. Throughout this paper, the letters j, k, l will denote three distinct numbersin { , , } whenever more than one of them appear in a formula together. Let d = dim( H/K ) and n = dim( G/K ). As will be shown in Section 2.2, each M is in fact an irreducible (sub)bundle of (cid:86) d − T ∗ ( G/H ).In all three cases, we can rescale the normal metric on
G/K to a metric Q , whose restriction on H/K is the standard metric with constant sectional curvature 1. Take Q as the background metricfor G/K . As will be shown in Section 2.1, the isotropy representation g / k has Z -symmetry amongits three inequivalent irreducible summands. By Schur’s lemma, any G -invariant metric on G/K has the form g G/K = f Q | p ⊕ f Q | p ⊕ f Q | p (1.3)for some f j >
0. Correspondingly, the Ricci endomorphism r of G/K , defined by g G/K ( r ( · ) , · ) =Ric( · , · ), has the form r = r Q | p ⊕ r Q | p ⊕ r Q | p , (1.4)where r j = af j + b (cid:32) f j f k f l − f k f j f l − f l f j f k (cid:33) (1.5)for some constants a and b . Their values were computed in [28]. We have Remark 1.1.
A basic observation on a and b is a − b = d − . This is not surprising since Q isthe sectional curvature 1 metric on S d . Another observation is a − b ≥ , where the equality isachieved in Case I. These observations are frequently used in this article, especially in Section 3.1and Section 3.2. 2able 1Case d n a b I 2 6
32 14
II 4 12 4 III 8 24 9 1 .Note that all three possible f j f k f l ’s appear in (1.5). An important motivation for our choices ofprincipal orbits to consider is to study the complications that arise from the simultaneous presenceof the terms f f f , f f f and f f f . If two of f j ’s are identical, say f ≡ f , the Ricci endomorphismtakes a simpler form, with r = a − bf + b f f and r ≡ r = af − b f f . The Ricci-flat ODE systemfor this special case then reduces to the one for g / k with two inequivalent irreducible summandsconsidered in [4][33]. It is noteworthy that the functional (cid:98) G introduced in [4] does not have anypositive real root for Case I. Nevertheless, the two summands case can be viewed as the subsystemof the ODE system studied in this article. The invariant compact set constructed in Section 3.1 canbe used to prove the existence of complete Ricci-flat metric for this special case. With the condition f ≡ f relaxed, we prove the following theorem. Theorem 1.2.
There exists a continuous 1-parameter family of non-homothetic complete smoothinvariant Ricci-flat metrics on each M . Remark 1.3.
Ricci-flat metrics constructed in Case II and Case III all have generic holonomy. InCase I, the 1-parameter family of smooth Ricci-flat metrics contains in its interior the completesmooth G metric that was first constructed in [7][25]. The other metrics in that family all havegeneric holonomy. Therefore, for M in Case I, the moduli space M G of G metric is not isolatedin M the moduli space of Ricci-flat metric in the C sense. Such a phenomenon cannot occur ona simply connected spin closed manifold, for example, by Theorem 3.1 in [32]. Definition 1.4.
Let (
N, g N ) and ( M, g M ) be Riemannian manifolds of respective dimension n and n + 1. Let t be the geodesic distance from some point on M . Then M has one asymptotically conical(AC) end if there exists a compact subset ˇ M ⊂ M such that M \ ˇ M is diffeomorphic to (1 , ∞ ) × N with g M = dt + t g N + o (1) as t → ∞ . With further analysis on the asymptotic behavior of Ricci-flat metrics in Theorem 1.2, we areable to prove the following:
Theorem 1.5.
Each Ricci-flat metric in Theorem 1.2 has an AC end with limit the metric coneover a suitable multiple of the normal Einstein metric on
G/K . Remark 1.6.
In Case I, the normal Einstein metric on the principal orbit SU (3) /T admits a(strict) nearly K¨ahler structure. Hence the metric cone over G/K is the singular G metric whichwas first constructed in [6]. The other two principal orbits, however, do not admit (strict) nearlyK¨ahler structure [21]. This paper is structured as followings. In Section 2.1 and 2.2, we discuss some details of thegeometry of the cohomogeneity one manifolds M . Based on the work in [23], we reduce (1.1) to asystem of ODEs (2.9) with a conservation law (2.10). A G -invariant Ricci-flat metric around G/H
3s hence represented by an integral curve defined on [0 , (cid:15) ). We derive the condition for smoothextension to
G/H using Lemma 1.1 in [23]. If in addition, the integral curve is defined on [0 , ∞ ),the corresponding Ricci-flat metric is complete.In Section 2.3, we apply the coordinate change introduced in [19][20]. The ODE system istransformed to a polynomial one. Invariant Einstein metrics on G/H and
G/K are transformedto critical points of the new system. We carry out linearizations at these critical points and provethe local existence of invariant Ricci-flat metrics around
G/H . An integral curve defined on [0 , (cid:15) ) istransformed to a new one that is defined on ( −∞ , (cid:15) (cid:48) ) for some (cid:15) (cid:48) ∈ R . Each integral curve representsa Ricci-flat metric on M up to homothety. It is determined by a parameter s that controls theprincipal curvature of G/H at t = 0. To show the completeness of the metric is equivalent toproving that the new integral curve is defined on R .The proof of completeness of the metric is divided into two sections. In Section 3.1, we constructa compact invariant set whose boundary contains critical points that represent the invariant metricon G/H and the normal Einstein metric on
G/K . The construction is almost the same for all threecases with a little difference in Case I. Section 3.2 proves that as long as s is close enough to zero,integral curves of Ricci-flat metrics enter the compact invariant set constructed in Section 3.1 infinite time, hence proving the completeness.In Section 4, we analyze the asymptotic behavior of all the Ricci-flat metrics constructed inSection 3.2. There also exist solutions to the polynomial system that represent singular Ricci-flatmetrics. They are discussed in Section 5. Results in this article are summarized by a plot at theend.With similar techniques introduced in Section 3, we can also show that there exists a 2-parameterfamily of Poincar´e–Einstein metrics on each M . More details will appear in another upcomingarticle. Acknowledgement.
The author is grateful to his PhD supervisor, Prof. McKenzie Wang forhis guidance and encouragement.
In this section, we derive the system of ODEs whose solutions give Ricci-flat metrics of cohomo-geneity one on M .Since M is of cohomogeneity one, there is a G -diffeomorphism between M \ ( G/H ) and (0 , ∞ ) × G/K . We construct a Ricci-flat metric g on M by setting (0 , ∞ ) as a geodesic and assigning a G -invariant metric g G/K to each hypersurface { t } × G/K , i.e., define g = dt + g G/K ( t ) (2.1)on M . By [23], if g G/K ( t ) satisfies ˙ g G/K = 2 g G/K ( L · , · ) , (2.2)˙ L = − tr( L ) L + r, (2.3)tr( ˙ L ) = − tr( L ) , (2.4) d (tr( L )) + δ ∇ L = 0 , (2.5)on (0 , (cid:15) ), where δ ∇ : Ω ( G/K, T ( G/K )) → T ∗ ( G/K ) is the divergence operator composed with themusical isomorphism, then g is a Ricci-flat metric on (0 , (cid:15) ) × G/K .4ote that (2.2) provides a formula for computing L ( t ) the shape operator of hypersurface { t } × G/K for each t ∈ (0 , (cid:15) ). By [1] and [23], Equation (2.5) automatically holds for a C metric satisfying(2.2) and (2.3) if there exists a singular orbit of dimension smaller than dim( G/K ). Canceling theterm tr( ˙ L ) using (2.3) and (2.4) yields the conservation law R − (tr( L )) + tr( L ) = 0 . (2.6)We shall focus on deriving specific formulas for (2.2),(2.3) and (2.6) on M . It requires a closerlook at isotropy representations of G/K and
G/H . We fix notations first. Each irreducible complexrepresentation is characterized by inner products between the dominant weight and simple roots onnodes of the corresponding Dynkin diagram. We use [ a ] for class A = B = C ; [ a, b ] for C = B with the shorter root on the right end; [ a, b, c, d ] for B with the shorter root on the right end.Furthermore, let t be the Lie algebra of S ( U (1) U (1) U (1)). Choose Q -orthogonal decomposition t = t ⊕ t , where t = span R i − i , t = span R i i − i . Let θ aj denote the complexified irreducible representation of circle generated by t j with weight a . We use Λ and ∆ ± to respectively denote the complexified standard representation and spinrepresentations of Spin(8). We use I to denote the trivial representation. Proposition 2.1.
The formula of g G/K is given by (1.3).
Proof.
With (
G, H, K ) listed in (1.2), we have the following Q -orthogonal decomposition for g : g = h ⊕ q as a representation of Ad ( G ) | H = ( k ⊕ p ) ⊕ ( p ⊕ p ) as a representation of Ad ( G ) | K . (2.7)Irreducible K -modules p j ’s are all of dimension d but they are inequivalent to each other. Specifi-cally, we have Table 2. By Schur’s lemma, a G -invariant metric on G/K has the form of (1.3).Table 2Case p ⊗ C p ⊗ C p ⊗ C I θ ⊗ I θ ⊗ θ θ − ⊗ θ II [1] ⊗ [1] ⊗ I [1] ⊗ I ⊗ [1] I ⊗ [1] ⊗ [1]III Λ ∆ +8 ∆ − Proposition 2.2.
The formula of Ricci endomorphism on (
G/K, g
G/K ) is given by (1.4) and (1.5)with constants a and b listed in Table 1. Proof.
Since the Ricci endomorphism is also G -invariant, it has the form of (1.4). To compute itsformula, use (7.39) in [3] to derive the scalar curvature on G/K and then apply variation. For eachcase, since [ p j , p j ] ⊂ k and [ p j , p k ] ⊂ p l , each r j in (1.4) has the form of (1.5).Take M as an associated vector bundle to principal H -bundle G → G/H of cohomogeneityone. As the orbit space is of dimension one, the action of H on the unit sphere of R d +1 must betransitive. Then the group K is taken as an isotropy group of a fixed nonzero element in R d +1 , say5 = (1 , , . . . , H/K = S d . Hence G/K is indeed a unit sphere bundle over
G/H .In this setting, g G/K ( t ) is an S ( p ⊕ p ⊕ p ) K -valued function with each f j in (1.3) as a positivefunction. Correspondingly, the Ricci endomorphism r in (1.4) is also an S ( p ⊕ p ⊕ p ) K -valuedfunction. Proposition 2.3.
For (
G, H, K ) listed in (1.2), Ricci-flat conditions (2.2) (2.3) and (2.6) respec-tively become L = ˙ f f Q | p ⊕ ˙ f f Q | p ⊕ ˙ f f Q | p , (2.8)¨ f j f j − (cid:32) ˙ f j f j (cid:33) = − (cid:32) d ˙ f f + d ˙ f f + d ˙ f f (cid:33) ˙ f j f j + r j , j = 1 , , − d (cid:88) j =1 (cid:32) ˙ f j f j (cid:33) = − (cid:88) j =1 d ˙ f j f j + R. (2.10) Proof.
The proof is complete by computation results in Proposition 2.2 and Proposition 2.3.In summary, constructing a smooth complete cohomogeneity one Ricci-flat metric on M isessentially equivalent to solving g G/K ( t ) that satisfies (2.8), (2.9) and (2.10). The fundamentaltheorem of ODE guarantees the existence of solution on neighborhood around { t } × G/K for any t ∈ (0 , ∞ ). In order to have a smooth complete Ricci-flat metric on M , we need to show that1. (Smooth extension) the solution exists on a tubular neighborhood around G/H and extendssmoothly to the singular orbit;2. (Completeness) the solution exists on (0 , ∞ ) × G/K .We discuss the smooth extension in Section 2.2 and Section 2.3. The proof for completeness isin Section 3.
It is not difficult to guarantee the smoothness of g G/K ( t ) at t = 0 as a S ( p ⊕ p ⊕ p )-valued function.However, the smooth function does not guarantee the smooth extension of g = dt + g G/K ( t ) asa metric on G/H as t →
0. By Lemma 1.1 in [23], the question boils down to studying the slicerepresentation χ = R d +1 of M and the isotropy representation q of G/H . We rephrase the lemmabelow.
Lemma 2.4. [23] Let g ( t ) : [0 , ∞ ) → S ( χ ⊕ q ) K be a smooth curve with Taylor expansion at t = 0 as (cid:80) ∞ l =0 g l t l . Let W l = Hom( S l ( χ ) , S ( χ ⊕ q )) H be the space of H -equivariant homogeneouspolynomials of degree l . Let ι : W l → S ( χ ⊕ q ) denote the evaluation map at v = (1 , , . . . , .Then the map g ( t ) has a smooth extension to G/H as a symmetric tensor if and only if g l ∈ ι ( W l ) for all l . To compute W l , we need to identify χ and q first. Since H acts transitively on H/K , theslice representation χ = R d +1 of M is irreducible and hence can be identified. Recall that q is anirreducible H -module in decomposition (2.7). Hence we have Table 3.6able 3Case χ ⊗ C as an H -module χ ⊗ C as a K -module q ⊗ C as an H -moduleI [2] ⊗ I R ⊕ ( θ ⊗ I ) ([1] ⊗ θ ) ⊕ ([1] ⊗ θ − )II [1 , ⊗ I R ⊕ ([1] ⊗ [1] ⊗ I ) [0 , ⊗ [1]III [1 , , , R ⊕ Λ [0 , , ,
1] .
Remark 2.5.
Recall the background metric Q on G/K is chosen that Q | p is the standard metricon S d . Therefore, the Euclidean inner product (cid:104)· , ·(cid:105) on χ can be written in “polar coordinate” as dt + t Q | p . As shown in the first column of Table 3, the action of H is essentially the standardrepresentation of Spin( d + 1) on χ and it preserves (cid:104)· , ·(cid:105) . In the following discussion, we take (cid:104)· , ·(cid:105) ⊕ Q | p ⊕ Q | p as the background metric of T p M = χ ⊕ T p ( G/H ) for p = [ H ] ∈ G/H .Compare the second column of Table 3 to the first column of Table 2. It is clear that χ = R ⊕ p as a K -module. Since χ and q are inequivalent H -modules, we have S ( χ ⊕ q ) K = S ( χ ) K ⊕ S ( q ) K . (2.11)Hence we have decomposition W l = W + l ⊕ W − l where W + l and W − l are respectively valued in S ( χ )and S ( q ). We are ready to compute each W ± l . Proposition 2.6.
For each M , we have W + l ∼ = R l = 00 l ≡ R l ≡ , l ≥ , W − l ∼ = R Proof.
From Table 3, we can derive the decomposition of complexified symmetric products S l ( χ ) ⊗ C and S l ( q ) ⊗ C as H -modules, as shown in Table 4 below. The proof is complete.Table 4Case S m − ( χ ) ⊗ C S m ( χ ) ⊗ C S ( q ) ⊗ C I (cid:76) mi =1 ([4 i − ⊗ I ) (cid:76) mi =0 ([4 i ] ⊗ I ) ([2] ⊗ θ ) ⊕ ([2] ⊗ θ − ) ⊕ ([2] ⊗ I ) ⊕ I II (cid:76) mi =1 ([2 i − , ⊗ I ) (cid:76) mi =0 ([2 i, ⊗ I ) ([0 , ⊗ [2]) ⊕ ([1 , ⊗ I ) ⊕ I III (cid:76) mi =1 [2 i − , , , (cid:76) mi =0 [2 i, , ,
0] [0 , , , ⊕ [1 , , , ⊕ I In order to apply Lemma 2.4, we need to find generators of each W ± l in Proposition 2.6. Notethat W ± l can be viewed as subspaces of W ± l +2 by multiplying each element with (cid:80) di =0 x i . Hence weonly need to find generators of W ± , W +2 and W − . It is clear that W +0 is spanned by I d +1 ∈ S ( χ )and W − is spanned by I d ∈ S ( q ). It is also clear that W +2 is generated by the identity map and( (cid:80) di =0 x i ) I d +1 . Note that the identity map in the form of a homogeneous polynomial is a symmetricmatrix Π with Π ij = x i x j for i, j ∈ { , , . . . , d } .The computation for W − is a bit more complicated. We follow Chapter 14 in [26] and consider χ = R ⊕ F with F as one of C , H and O for Case I, II and III, respectively. Proposition 2.7. W − is generated by the R -linear mapΦ : χ → S ( q )( x , x ) (cid:55)→ (cid:20) x I d L x L ¯ x − x I d , (cid:21) L x is the real matrix representation of left multiplication of x ∈ F , as shown in Table 5 below.Table 5Case I II III L x (cid:20) x − x x x (cid:21) x − x − x − x x x − x x x x x − x x − x x x x − x − x − x − x − x − x − x x x − x x − x x x − x x x x − x − x − x x x x − x x x − x x − x x x x x x x − x − x − x x − x x − x x x x − x x − x − x x x − x x x x x − x − x x x − x x Proof.
Consider i Φ( χ ) a subspace of C ⊗ R S ( q ). Since ( i Φ( x , x )) = − ( x + (cid:107) x (cid:107) ) I d +1 , it isclear that the matrix multiplication of i Φ( χ ) generates a Clifford algebra and hence Spin( d + 1).Specifically, the group is generated by elements Ξ( y , y ) := Φ( − , )Φ( y , y ) with y + (cid:107) y (cid:107) = 1.Since each F is an alternative algebra that satisfies Moufang identity, computations show Ad (Ξ( y , y ))(Φ( x , x )) = Ξ( y , y )(Φ( x , x ))Ξ( y , y ) − = Φ( z , z ) , (2.12)where z = ( y − (cid:107) y (cid:107) ) x + 2 y (cid:104) y , x (cid:105) and z = y x − x y y − ( y ¯ x ) y . Hence Φ( χ ) is an Ad Spin( d +1) -invariant subspace in S ( q ). Moreover, since( Ad (Ξ( y , y ))(Φ( x , x ))) = (Φ( x , x ))) = ( x + (cid:107) x (cid:107) ) I d +1 , The adjoint action on Φ( χ ) induces the standard representation Λ d +1 on R d +1 . Therefore,Φ : ( χ, Λ d +1 ) → (cid:0) Φ( χ ) , Ad Spin( d +1) (cid:1) is H -equivariant and generates W − .With the generators known, we are ready to prove the following proposition. Proposition 2.8.
The necessary and sufficient conditions for a metric g = dt + g G/K ( t ) on M toextend to a smooth metric in a tubular neighborhood of the singular orbit G/H arelim t → ( f , f , f , ˙ f , ˙ f , ˙ f ) = (0 , h , h , , − h , h ) (2.13)for some h > h ∈ R . Proof.
The metric g in LHS of (2.1) can be identified with a map g ( t ) : [0 , (cid:15) ) → S ( χ ) K ⊕ S ( q ) K (2.14)with Taylor expansion g ( t ) = ∞ (cid:88) l =0 g l t l . (2.15)8rite g ( t ) = D ( t ) ⊕ J ( t ), where D ( t ) : [0 , ∞ ) → S ( χ ) and J ( t ) : [0 , ∞ ) → S ( q ). The Taylorexpansion (2.15) can be rewritten as D ( t ) = D + D t + D t + . . .J ( t ) = J + J t + J t + . . . (2.16)Since W +2 /W +0 ∼ = R , in principle there is a free variable for the second derivative of a smooth D ( t ). However, with the geometric setting that t is a unit speed geodesic, the choice of D is in factdetermined by D . Hence we take D = I d +1 and D must be a multiple of (cid:16) ( (cid:80) di =0 x i ) I d +1 − Π (cid:17) ( v ) = (cid:20) I d (cid:21) with the multiplier determined by the choice of D . Since H/K is and irreducible sphere,it is expected that there is no indeterminacy from D ( t ). By Lemma 2.4, the smooth condition for D ( t ) with respect to background metric (cid:104)· , ·(cid:105) is D ( t ) = I d +1 + O ( t ) . This is consistent with Lemma9.114 in [3].As g degenerates to an invariant metric on G/H and the isotropy representation of
G/H isirreducible, J is a positive multiple of I d . The evaluation of Φ at v in Proposition 2.7 is (cid:20) I d − I d (cid:21) . Hence by Lemma 2.4, the smoothness condition for J ( t ) is J ( t ) = (cid:20) f ( t ) I d f ( t ) I d (cid:21) = c I d + c (cid:20) I d − I d (cid:21) t + O ( t )for some c > c ∈ R .Recall 2.5, note that (cid:104)· , ·(cid:105) = dt + t Q | p . Switch the background metric to dt + Q , we concludethat the smoothness condition for g is f ( t ) = t + O ( t ) f ( t ) = c + c t + O ( t ) f ( t ) = c − c t + O ( t )Then the proof is complete. Remark 2.9.
The Ricci-flat ODE system (2.9) and (2.10) is invariant under the homotheticchange κ ( dt + g G/K ) with ds = κdt . The smooth initial condition 2.13 is transformed to(0 , κh , κh , , h , − h ) . Hence if we abuse the notation. Multiplying h by κ > f j (0)unchanged give the smooth initial condition for metrics in the same homothetic family. Therefore,in the original coordinate, h is the free variable that gives non-homothetic metrics. As shown in(2.27), only h matters in producing different curves in the polynomial system.Combine the analysis in Proposition 2.8 with the main result in [23], we conclude that thereexists a 1-parameter family of Ricci-flat metric on a neighborhood around G/H in M . We derivethe same result in Section 2.3 using a new coordinate. Remark 2.10.
Note that we always have lim t → f f + ˙ f f = 0, i.e., the mean curvature of G/H vanishesat t = 0. This is consistent with Corollary 1.1 in [27]. The last two components of (2.13) showsthat the smooth extension does not require G/H to be totally geodesic. If h in (2.13) vanishes,then we recover cases in [4][33] with f ≡ f . Remark 2.11.
It is worth pointing out that Equations (2.9) and (2.10) are symmetric among f , f and f . Therefore, initial condition (2.13) has two other counterparts where f or f collapsesinitially depending how H is embedded in G . Without loss of generality, we will consider initialcondition (2.13) in this article. 9e end this section by identifying each vector bundle M as a (sub)bundle of ASD d -form oflowest rank. Table 6 lists out H -decomposition of (cid:86) d q ⊗ C and dimension of each irreduciblesummand. The subspace (cid:86) d − q ⊗ C consists of summands in brace brackets. Decomposition belowis mostly computed via software LiE , with reference in [5][29][12].Table 6Case
H H -decomposition of (cid:86) d q ⊗ C and Dimension of each SummandI S ( U (2) U (1)) (cid:86) (cid:0) ([1] ⊗ θ ) ⊕ ([1] ⊗ θ ) (cid:1) = ( I ⊗ θ ) ⊕ ( I ⊗ θ − ) ⊕ I ⊕ { [2] ⊗ I } = + + + { } II Sp (2) Sp (1) (cid:86) [01] ⊗ [1] = [01] ⊗ [2] ⊕ [02] ⊗ I ⊕ I ⊗ [4] ⊕ I ⊕ { [02] ⊗ [2] ⊕ [01] ⊗ I } = + + + + { + } III Spin(9) (cid:86) [0001] = [2010] ⊕ [0020] ⊕ [1002] ⊕ [0200] ⊕ [4000] ⊕ [0010] ⊕ [2000] ⊕ I ⊕{ [2002] ⊕ [0110] ⊕ [1010] ⊕ [3000] ⊕ [0002] ⊕ [1000] } = + + + + + + + + { + + + + + } For Case I, it is known that the trivial representation generates the invariant K¨ahler form on CP . The bundle that we study in this paper is the associated bundle with respect to representation[2] ⊗ I , which is the bundle of ASD 2-form (cid:86) − T ∗ CP that admits a complete smooth G metric[7][25].For Case II, the trivial representation generates a canonical 4-form for Quaternionic K¨ahlermanifolds, as described in [29]. Explicitly, given a Quaternionic K¨ahler manifold with a triple ofcomplex structures ( I, J, K ) and corresponding symplectic forms ( ω I , ω J , ω K ), the canonical 4-formis defined as Ω = ω I ∧ ω I + ω J ∧ ω J + ω K ∧ ω K . By Table 3, M is an associate bundle with respect torepresentation [01] ⊗ I in (cid:86) − q ⊗ C . Therefore, M is indeed an irreducible subbundle of (cid:86) − T ∗ HP .For Case III, the trivial representation generates the canonical 8-form, whose existence is provedin [5]. Explicit formula for the canonical 8-form can be found in [12]. The 9-dimensional represen-tation [1000] is the (twisted) adjoint representation of Spin(9) on R . Similar to Case II, the bundlethat we consider in this paper is an irreducible subbundle of (cid:86) − T ∗ OP .In conclusion, the name “(sub)bundle of ASD d -form of lowest rank” for M is justified. We apply the coordinate change introduced in [19][20] to the Ricci-flat system in this section.The original ODE system is transformed to a polynomial one. As described in Remark 2.16, somecritical points of the new system carry geometric data. Linearizations at these critical points provideguidance on how integral curves potentially behave, which help us to construct a compact invariantset in Section 3 to prove the completeness. 10s predicted by the result in the previous section (Remark 2.9), analysis on the new systemshows that there exists a 1-parameter family of integral curves with each represents a homotheticclass of Ricci-flat metrics on a neighborhood around
G/H .Consider dη = tr( L ) dt. (2.17)Define X j := ˙ f j f j tr( L ) , Z j := f j f k f l tr( L ) . (2.18)And define R j := r j (tr( L )) = aZ k Z l + b (cid:0) Z j − Z k − Z l (cid:1) , G := (cid:88) j =1 dX j , H := (cid:88) j =1 dX j . Use (cid:48) to denote derivative with respect to η . In the new coordinates given by (2.17) and (2.18), thesystem (2.9) is transformed to X X X Z Z Z (cid:48) = V ( X , X , X , Z , Z , Z ) := X ( G −
1) + R X ( G −
1) + R X ( G −
1) + R Z (cid:0) G − H d + 2 X (cid:1) Z (cid:0) G − H d + 2 X (cid:1) Z (cid:0) G − H d + 2 X (cid:1) , (2.19)and the conservation law (2.10) becomes C : G − d (cid:88) j R j = 0 . (2.20)As (cid:16) L ) (cid:17) (cid:48) = G tr( L ) , the original variables can be recovered by t = (cid:90) ηη exp (cid:18)(cid:90) ˜ η ˜ η G d ˜˜ η + ˜ t (cid:19) d ˜ η + t , f j = exp (cid:16)(cid:82) ηη G d ˜ η + t (cid:17) √ Z k Z l . (2.21) Remark 2.12.
The new variables X j ’s record the relative size of each principal curvature of G/K .Variables Z j ’s carry the data of relative size of each f j ’s. Note that Z j Z k = f j f k .In the original coordinates, a smooth solution to (2.9) is an integral curve with variable t ∈ [0 , (cid:15) ).Since by (2.17), lim t → η = lim t → ln (cid:0) f d f d f d (cid:1) + ˆ η = −∞ , the original solution is transformed to an integralcurve with variable η ∈ ( −∞ , (cid:15) (cid:48) ) for some (cid:15) (cid:48) ∈ R . Note that the graph of the integral curve doesnot change when homothetic change is applied to the original variable. Hence each integral curveto the new system represent a solution in the original coordinate up to homothety. Remark 2.13.
It is clear that the symmetry mentioned in Remark 2.11 remains among pairs( X j , Z j )’s in the new system (2.19) with (2.20). In addition, by the observation on Z j ’s derivative.It is clear that they do not change sign along the integral curve. Without loss of generality, we focuson the region where these three variables are positive. This observation provides basic estimatesneeded in our construction of compact invariant set (the set P introduced in (3.1)).11 emark 2.14. It is clear that
H ≡ X j . In fact, since H (cid:48) = ( H− G − C , the set C ∩ {H ≡ } is flow-invariant. Furthermore, C ∩ {H ≡ } is diffeomorphic to a levelset dX + dX + d (cid:18) d − X − X (cid:19) − d (cid:88) j R j = 0in R . Therefore, C ∩ {H ≡ } is a 4-dimensional smooth manifold by the inverse function theorem.System (2.19) can be restricted to a 4-dimensional subsystem on C ∩ {H ≡ } . Proposition 2.15.
The complete list of critical points of system (2.19) in C ∩ {H ≡ } is thefollowing:I. the set (cid:110) ( x , x , x , , , | (cid:80) j x j = d , (cid:80) j x j = d (cid:111) ;II. (cid:18) − d , d , d , ± d (cid:113) − db , , (cid:19) and its counterparts with pairs ( X j , Z j )’s permuted. This criticalpoint occurs only for Case I;III. (cid:0) d , , , , ± d , ± d (cid:1) and its counterparts with pairs ( X j , Z j )’s permuted;IV. (cid:16) n , n , n , ± bd − n (cid:113) ( n − d − b ( a +2 b ) , ± bd − n (cid:113) ( n − d − b ( a +2 b ) , ± n (cid:113) ( n − d − b ( a +2 b ) (cid:17) and its counterparts withpairs ( X j , Z j )’s permuted;V. (cid:16) n , n , n , ± n (cid:113) n − a − b , ± n (cid:113) n − a − b , ± n (cid:113) n − a − b (cid:17) . Proof.
The proof is processed by direct computations.By Remark 2.13, we focus on critical points with non-negative Z j ’s. Remark 2.16.
Some critical points in Proposition 2.15 have further geometric significance. • p := (cid:0) d , , , , d , d (cid:1) This critical point is the initial condition (2.13) under the new coordinate (2.17)-(2.18), i.e.,(2.13) becomes lim η →−∞ ( X , X , X , Z , Z , Z ) = p . Hence we study integral curves emanatingfrom p . In order to prove the completeness, we construct a compact invariant set in Section3 that contains p in its boundary and traps the integral curve initially.By Remark 2.11 and Remark 2.13, its two other counterparts p (cid:48) = (cid:0) , d , , d , , d (cid:1) and p (cid:48)(cid:48) = (cid:0) , , d , d , d , (cid:1) also have the similar geometric meaning depending on how H is embedded in G . • p := (cid:16) n , n , n , n (cid:113) n − a − b , n (cid:113) n − a − b , n (cid:113) n − a − b (cid:17) This critical point is symmetric among all ( X j , Z j )’s. Note that f j f k ( p ) = Z j Z k ( p ) = 1, all f j ’sare equal at this point. We prove in Section 4 that p represents an AC end for the completeRicci-flat metric represented by the integral curve emanating from p . The conical limit is ametric cone over a suitable multiple of the normal Einstein metric on G/K . • p := (cid:16) n , n , n , bd − n (cid:113) ( n − d − b ( a +2 b ) , bd − n (cid:113) ( n − d − b ( a +2 b ) , n (cid:113) ( n − d − b ( a +2 b ) (cid:17) Since r j ( p ) are all equal, this point represent an invariant Einstein metric on G/K otherthan the one represented by p . In the following text, we call the metric the “alternative12instein metric”. For Case I, it is a K¨ahler–Einstein metric. It has two other counterpartswith permuted Z j ’s.Although we do not find any integral curve with its limit as p , we show in Section 5 that thereexists an integral curve emanating from p and tends to p , representing a singular Ricci-flatmetric with a conical singularity and an AC end.The linearization L of vector field V in (2.19) is G − dX dX X dX X bZ aZ − bZ aZ − bZ dX X G − dX dX X aZ − bZ bZ aZ − bZ dX X dX X G − dX aZ − bZ aZ − bZ bZ (2 dX + 1) Z (2 dX − Z (2 dX − Z G − H d + 2 X dX − Z (2 dX + 1) Z (2 dX − Z G − H d + 2 X dX − Z (2 dX − Z (2 dX + 1) Z G − H d + 2 X (2.22)With (2.22) we can compute the dimension of the unstable subspace at p . As we are consideringsystem (2.19) on C ∩ {H ≡ } , we require each unstable eigenvector to be tangent to C ∩ {H ≡ } .The normal vector field to the hypersurfaces C and {H ≡ } are respectively N C = dX dX dX adZ + adZ − bdZ adZ + adZ − bdZ adZ + adZ − bdZ , N {H≡ } = . (2.23) Lemma 2.17.
The unstable subspace of system (2.19) at p , restricted on C ∩ {H ≡ } , is ofdimension 2.Proof. Hence the linearization at p is L ( p ) = d − a − bd a − bd d − ad bd − bd d − ad − bd bd d d d − d d − d d . (2.24)Eigenvalues and corresponding eigenvectors of (2.24) are λ = 1 d , λ = λ = 2 d , λ = λ = 1 d − , λ = − .v = − − , v = , v = ad +1 ad +1 , v = − d , v = , v = b − b − . (2.25)With Remark 1.1, Remark 2.14 and (2.23), it is clear that T p ( C ∩ {H ≡ } ) = span { v , ( d + 1) v − av , v + ( d − v , v } . By (2.25), an unstable subspace at p is spanned by v and ( d + 1) v − av .13olutions of the linearized equations at p have the form X X X Z Z Z = p + s e ηd (( d + 1) v − av ) + s e ηd v = d d d + s e ηd − aaad + 1 − a − a + s e ηd − − , (2.26)for some s > s ∈ R . Recall Remark 2.13. In order to let Z be positive initially, theassumption s > p and [ s : s ] in RP . We fix s > s >
0, there is no ambiguity to use γ s to denote an integral curve to system (2.19) on (2.20) with γ s ∼ p + s e ηd (( d + 1) v − av ) + s e ηd v near p .Analysis above shows that there exists a 1-parameters family of short-time existing integralcurves of system (2.19) on (2.20). Since each curve corresponds to a homothetic class of Ricci-flatmetrics defined on a neighborhood around singular orbit G/H , there exists a 1-parameters familyof non-homothetic Ricci-flat metrics defined on a neighborhood around
G/H . Recall Remark 2.9,the result is consistent with the main theorem in [23].
Remark 2.18.
By the unstable version of Theorem 4.5 in [15], from (2.13) we know that2 h √ d = lim t → (cid:16) ˙ f f − ˙ f f (cid:17) √ f f (cid:112) tr( L ) f = lim η →∞ X − X √ Z = 2 s (cid:112) ( d + 1) s . (2.27)Hence the parameter s vanishes if and only if h does. The solution with s = 0 corresponds to thesubsystem of (2.19) where ( X , Z ) ≡ ( X , Z ) is imposed, which corresponds to the subsystem ofthe original system (2.9) where f ≡ f is imposed. The reduced system is essentially the same asthe one for the case where the isotropy representation has two inequivalent irreducible summands.For Case I, γ represents the smooth complete G metric in [7][25]. For Case II and Case III,Ricci-flat metrics with s = 0 are proved to be complete in [4][33].Our construction does not assume the vanishing of s . By the symmetry of the ODE system,we mainly focus on the situation where s ≥ γ s is defined on R , then by Lemma 5.1 in [9], functions f j ( t )’s aredefined on [0 , ∞ ). Therefore, Theorem 1.2 is proved once γ s is shown to be defined on R . With smooth extension of metrics represented by γ s proved, the next step is to show that γ s isdefined on R so that the Ricci-flat metric it represents is complete. Our construction is divided intotwo parts. The first part is to find an appropriate compact invariant set ˆ S with p sitting on itsboundary. Although p is in the boundary of ˆ S , integral curves are not trapped in the set initiallyunless s = 0. In the second step, we construct another compact set that serves as an entrancezone . It traps γ s initially as long as s is close enough to zero. Moreover, integral curves trappedin this set cannot escape through some part of its boundary and they are forced to enter ˆ S . Hencesuch a γ s must be defined on R . 14 .1 Compact Invariant Set We describe the first step in this section. There is a subtle difference between the compact invariantset for Case I and ones for Case II and III. We first construct the set for Case II and III since it issimpler.Let ρ = (cid:113) a +2 b . It is clear that ρ ≥ P = { Z , Z , Z ≥ } ˜ S = (cid:92) j =1 { Z − Z j ≥ , X − X j + ρ ( Z − Z j ) ≥ , X ≥ } . (3.1)And define S = C ∩ {H ≡ } ∩ P ∩ ˜ S . (3.2)Before doing further analysis on S , we give some explanations as to why it is constructed inthis way. Note that the positivity of Z j ’s are immediate by Remark 2.13. The first inequality in ˜ S is to require Z to be the largest variable among Z j ’s. Equivalently, it requires f to be the largestamong f j ’s in the original coordinate. This condition is indicated by the subscript of ˜ S and S . Adirect consequence of this assumption is that we can assume X ≥ γ s as shown in (3.9).It is easy to check that p ∈ S hence the set is nonempty. Each inequality in (3.2) defines aclosed subset in R whose boundary is defined by the equality. Therefore, a point x ∈ ∂S if thereexists at least one defining inequality in (3.1) reaches equality at x . For Case II and III, functions X , Z , Z − Z , X − X + ρ ( Z − Z ) (3.3)among those in (3.1) vanish at p . The point is hence in ∂S . Substitute (2.26) to functions in(3.3). It is clear that γ s is trapped in S initially if s ≥
0. By Remark 2.18, we know that γ istrapped in ∂S with ( X , Z ) ≡ ( X , Z ). Proposition 3.1.
In the set S ∩ { bZ − a ( Z + Z ) ≤ } , we have estimate Z + Z ≤ (cid:115) n − n ( a − b ) . (3.4) Proof.
By the conservation law (2.20), it follows that0 ≥ n − da ( Z Z + Z Z + Z Z ) − db ( Z + Z + Z ) . (3.5)Note that the RHS of (3.5) is symmetric between Z and Z . It is convenient to find the maximumof Z + Z on S ∩ { Z ≥ Z } first. By the symmetry between Z and Z in (3.5), such a maximumis the maximum of Z + Z in S . With the assumption Z ≥ Z , we write Z = νZ for some ν ∈ [0 , ν . Then (3.5) becomes0 ≥ n − da ( Z Z + νZ Z + νZ ) − db ( ν Z + Z + Z )= 1 n − d ( − bZ + a (1 + ν ) Z Z + ( aν − b (1 + ν )) Z ) . (3.6)Define F ( Z ) = − bZ + a (1+ ν ) Z Z +( aν − b (1+ ν )) Z . Consider the set S ∩{ bZ − a ( Z + Z ) ≤ } ∩ { Z = νZ } , we have Z ≤ Z ≤ a b (1 + ν ) Z . ν and Z , the minimum of F in S ∩ { bZ − a ( Z + Z ) ≤ } ∩ { Z = νZ } isreached at Z = Z . Therefore, computation (3.6) continues as0 ≥ n − d (cid:0) − b + a (1 + ν ) + ( νa − b (1 + ν )) (cid:1) Z = 1 n − d (cid:0) − bν + 2 aν + a − b (cid:1) Z . (3.7)The coefficient of Z in (3.7) can be easily checked to be positive. It follows that( Z + Z ) = (1 + ν ) Z ≤ (cid:18) − n (cid:19) (1 + ν ) d ( − bν + 2 aν + a − b )= (cid:18) − n (cid:19) d (cid:16) − b + 2( a + b ) ν − ( a + 3 b ) ν ) (cid:17) . (3.8)Consider function h (cid:16) ν (cid:17) = − ( a + 3 b ) ν ) + 2( a + b ) ν − b . Since by Remark 1.1, we have ≤ a + ba +3 b ≤ , the minimum of h is either h (cid:0) (cid:1) or h (1). Computation shows h (cid:0) (cid:1) < h (1). Weconclude that ( Z + Z ) ≤ (cid:0) − n (cid:1) d h ( ) = 4 n − n ( a − b ) . Hence the proof is complete. Note that theequality in (3.4) is reached by p . Proposition 3.2.
For Case II and III, integral curves γ s to system (2.19) on C ∩ {H ≡ } emanating from p with s ≥ S . Proof.
Two perspectives can be taken in the following computations that frequently appear throughout this article. First is to view algebraic expressions in (3.1) as functions along γ s and they allvanish at p . Integral curves emanating from p being trapped in S initially is equivalent to thesedefining functions being positive near p . To show that γ s does not escape S is to show thenon-negativity of these functions along the integral curves. Suppose one of these functions vanishesat some point along the integral curves for the first time. We want to show that its derivative atthat point is non-negative.The second perspective is to consider ∂S as a union of subsets of a collection of linear andquadratic varieties. Require the restriction of the vector field V in (2.19) on each of these subsetsto point inward S . If such a requirement is met, then it is impossible for the integral curves toescape if they are initially in S . Both perspectives lead to the same computation of inner productbetween V and the gradient of each defining function in (3.1). Then require the inner product tobe non-negative if the gradient points inward S . It might not be true that the inner product isnon-negative on each variety globally. But all we need is the non-negativity on its subsets that ∂S consists of.By definition of S , we automatically have R = aZ Z + b ( Z − Z − Z ) = (cid:26) Z ( aZ − bZ ) + b ( Z − Z ) ≥ Z ≥ Z Z ( aZ − bZ ) + b ( Z − Z ) ≥ Z ≥ Z . (3.9)On X = 0, we have (cid:104)∇ ( X ) , V (cid:105)| X =0 = R ≥ X is non-negative along every γ s that is trapped in S initially.Next we need to show that the integral curves cannot escape from the part of ∂ ˜ S that is in16 S . For distinct j, k ∈ { , } , it follows that (cid:104)∇ ( Z − Z j ) , V (cid:105)| Z − Z j =0 = Z (cid:18) G − d + 2 X (cid:19) − Z j (cid:18) G − d + 2 X j (cid:19) = 2 Z ( X − X j ) since Z − Z j = 0 ≥ ρZ ( Z j − Z ) by definition of S = 0 since Z − Z j = 0 . Although it is not clear if X − X j ≥ γ s , we impose a weaker condition, which is thesecond inequality in ˜ S . What it means is to allow Z − Z j to decrease, yet the rate of its decreasingcannot be too steep so that Z − Z j increases before it could decrease to zero. Fortunately, theweaker condition does hold along the integral curves. (cid:104)∇ ( X − X j + ρ ( Z − Z j )) , V (cid:105)| X − X j + ρ ( Z − Z j )=0 = ( X − X j + ρ ( Z − Z j )) ( G −
1) + R − R j + ρZ (cid:18) − d + 2 X (cid:19) − ρZ j (cid:18) − d + 2 X j (cid:19) = ( Z − Z j ) (cid:18) b ( Z + Z j ) − aZ k + ρ (cid:18) − d (cid:19) + 2 ρX − ρ Z j (cid:19) since X j = X + ρ ( Z − Z j ) ≥ ( Z − Z j ) (cid:18) bZ − a ( Z j + Z k ) + ρ (cid:18) − d (cid:19)(cid:19) since X ≥ S = ( Z − Z j ) (cid:18) bZ − a ( Z + Z ) + ρ (cid:18) − d (cid:19)(cid:19) . (3.10)If 2 bZ − a ( Z + Z ) ≥
0, then the last line of computation above is obviously non-negative. If2 bZ − a ( Z + Z ) ≤
0, then (3.10) continues as ≥ ( Z − Z j ) (cid:18) ( b − a )( Z + Z ) + ρ (cid:18) − d (cid:19)(cid:19) (3.11)since Z ≥ Z + Z in S . Apply Proposition 3.1, we know that (3.11) is non-negative if ρ ( d − d ( a − b ) ≥ (cid:115) n − n ( a − b ) . (3.12)Straightforward computations show thatCase ρ ρ ( d − d ( a − b ) 2 (cid:113) n − n ( a − b ) I 1
25 23 II (cid:113)
52 3 √ ≈ . √ ≈ . (cid:113)
112 7 √ ≈ . √ ≈ . . Inequality (3.12) holds only for Case II and III. Hence for Case II and III, integral curves γ s emanating from p does not escape S if s ≥ S , inequality (3.10) has room to be improved as we droppeda non-negative term 2 ρX in the computation. It turns out (3.10) can be proved to be non-negativefor Case I with an additional inequality, as demonstrated in Proposition 3.4.17e move on to Case I. Recall that the construction in Proposition 3.2 is not successful justbecause inequality (3.12) does not hold in this case. To fix this issue, an additional inequality isneeded. Define F j := X k + X l − Z j . (3.13)Computations show (cid:104)∇ F j , V (cid:105) = F j ( G −
1) + 3 Z j (cid:18) F j − F k − F l (cid:19) . Remark 3.3.
The condition F ≡ F ≡ F ≡ G condition on cohomogeneity onemanifold with principal orbit SU (3) /T . Hence ∩ j =1 { F j ≡ } is flow-invariant and it contains theintegral curve γ that represents the complete smooth G metric on M , which is firstly discoveredin [7][25].In the following text, we still use ˜ S and S to denote invariant sets constructed. If necessary,we use the phrase such as “ S for Case I” to refer to the case in particular. Define˜ S = (cid:92) j =1 { Z − Z j ≥ , F j − F ≥ , X ≥ } ∩ { F + 3 F − F ≥ } . (3.14)And define S = C ∩ {H ≡ } ∩ P ∩ ˜ S . (3.15)Note that F j − F ≥ S in (3.1) with ρ = 1.It is easy to check that p ∈ S hence S is nonempty. Since functions X , Z , Z − Z , F j − F and 3 F + 3 F − F vanish at p among those in (3.14), the point is in ∂S . With the same argumentas the one for Case II and III, we know that γ s is trapped in S initially if s ≥ Proposition 3.4.
Integral curves γ s to system (2.19) on C ∩ {H ≡ } emanating from p with s ≥ S . Proof.
The idea of proving Proposition 3.4 is the same as the one of Proposition 3.2. Besides,almost all computations for Proposition 3.2 still hold except the one for F j − F ≥ (cid:104)∇ ( F j − F ) , V (cid:105)| F j − F =0 = ( F j − F ) ( G −
1) + 3 Z j (cid:18) F j − F k − F (cid:19) − Z (cid:18) F − F j − F k (cid:19) = 3 Z j (cid:18) F j − F k − F (cid:19) − Z (cid:18) F − F j − F k (cid:19) since F j = F = 3 Z j (cid:18) F j − F k − F j (cid:19) − Z (cid:18) F j − F j − F k (cid:19) since F j = F = F k
32 ( Z − Z j ) + F j ( Z − Z j )= 12 ( Z − Z j )(3 F j + 3 F k − F ) since F j = F ≥ . (cid:104)∇ ( F j − F ) , V (cid:105)| F j − F =0 hence becomes sharper. Finally, we need to show thatthe additional inequality holds along the integral curves. Indeed, since (cid:104)∇ (3 F + 3 F − F ) , V (cid:105)| F +3 F − F =0 = (3 F + 3 F − F ) ( G − Z F − F − F ) + 3 Z F − F − F ) − Z (cid:18) F − F − F (cid:19) = 3 Z F − F − F ) + 3 Z F − F − F ) since 3 F + 3 F − F = 0= 3 Z F − F ) + 3 Z F − F ) since 3 F + 3 F − F = 0 ≥ S for i = 1 , F + 3 F − F remains non-negative along the integral curves. Therefore, integral curves γ s donot escape S in Case I if s ≥ Remark 3.5.
One may want to integrate the additional inequality in S for Case I to the othertwo cases so that all cases can be discussed by a single construction. Specifically, one can define F j := X k + X l − ρZ j . Then the additional inequality analogous to 3 F + 3 F − F ≥ aF + aF − bF ≥ . But (cid:104)∇ ( aF + aF − bF ) , V (cid:105)| aF + aF − bF =0 = aZ k ( a + 2 b )( F − F ) + aZ k ( a + 2 b )( F − F ) + ζk ( aZ + aZ − bZ ) , where ζ = (3 − d ) a − (2+2 d ) b d ≤
0. It only vanishes in Case I. Hence whether aF + aF − bF isnon-negative along the integral curves in S is not clear. The analogous F j defined for Case II andCase III may not have too much meaning after all because there is no special holonomy for odddimension other than 7.We are ready to construct the compact invariant set mentioned at the beginning of this section.Define ˆ S = S ∩ { Z + Z − Z ≥ } ∩ { Z ( X − X ) + Z ( X − X ) ≥ } for all three cases. We have the following lemma. Lemma 3.6. ˆ S is a compact invariant set.Proof. Because Z + Z − Z ≥ S , we can apply Proposition 3.1 so that Z + Z is boundedabove. Then all Z j ’s are bounded in ˆ S . By conservation law (2.20), we immediately conclude thatall variables are bounded. The compactness of ˆ S is hence proved.To check that ˆ S is flow invariant, consider the hyperplane Z + Z − Z = 0. It follows that (cid:104)∇ ( Z + Z − Z ) , V (cid:105)| Z + Z − Z =0 = ( Z + Z − Z ) (cid:18) G − d (cid:19) + 2 Z X + 2 Z X − Z X = 2 Z ( X − X ) + 2 Z ( X − X ) since Z + Z − Z = 0 ≥ S .
19n hypersurface Z ( X − X ) + Z ( X − X ) = 0, we have (cid:104)∇ ( Z ( X − X ) + Z ( X − X )) , V (cid:105)| Z ( X − X )+ Z ( X − X )=0 = (cid:28) ∇ (cid:18) Z (cid:18) Z Z ( X − X ) + Z Z ( X − X ) (cid:19)(cid:19) , V (cid:29)(cid:12)(cid:12)(cid:12)(cid:12) Z ( X − X )+ Z ( X − X )=0 = Z (cid:18) G − d + 2 X (cid:19) (cid:18) Z Z ( X − X ) + Z Z ( X − X ) (cid:19) + Z (cid:18) Z Z ( X − X ) + 2 Z Z ( X − X ) (cid:19) + Z (( X − X ) ( G −
1) + R − R ) + Z (( X − X ) ( G −
1) + R − R )= 2 Z ( X − X ) + 2 Z ( X − X ) + Z ( R − R ) + Z ( R − R )since Z ( X − X ) + Z ( X − X ) = 0 ≥ Z ( R − R ) + Z ( R − R )= Z ( Z − Z )( aZ − b ( Z + Z )) + Z ( Z − Z )( aZ − b ( Z + Z )) (3.16)For distinct j, k ∈ { , } , take A j = Z j ( Z − Z j ) and B j = aZ k − b ( Z + Z j ). Apply identity A B + A B = 12 (( A + A )( B + B ) + ( A − A )( B − B )) . Then the computation (3.16) continues as= 12 ( Z ( Z − Z ) + Z ( Z − Z ))(( a − b )( Z + Z ) − bZ ) + 12 ( Z − Z ) ( Z + Z − Z )( a + 2 b ) ≥
12 ( Z ( Z − Z ) + Z ( Z − Z ))( a − b ) Z since Z + Z ≥ Z ≥ . (3.17)Therefore, ˆ S is flow-invariant. Remark 3.7.
By the symmetry between ( X , Z ) and ( X , Z ), constructions of S and ˆ S abovecan be carried over to defining S and ˆ S . With the same arguments, it can be shown that γ s doesnot escape S whenever s ≤ S is a compact invariant set. Remark 3.8.
It is clear that p ∈ ∂ ˆ S . One can check that γ is trapped in ˆ S initially. Hence thelong time existence for γ is proved. By Remark 2.18, it is trapped in ˆ S ∩ { X ≡ X , Z ≡ Z } .Hence ˆ S can be used to prove the long time existence for the special case where ( X , Z ) ≡ ( X , Z )is imposed. In fact, the compact invariant set for cohomogeneity one manifolds of two summandscan be constructed by a little modification on ˆ S ∩ { X ≡ X , Z ≡ Z } , reproducing the sameresult in [4][33]. For Case I in particular, γ represents the complete G metric discovered in in[7][25]. Remark 3.9.
Not only R is non-negative in ˆ S . This is in fact the case for all R j ’s. For distinct j, k ∈ { , } , we have R j = aZ Z k + b ( Z j − Z k − Z ) ≥ aZ Z k + b ( Z j + Z k − ( Z j + Z k ) ) by definition of ˆ S = aZ Z k − bZ j Z k ≥ ( a − b ) Z j Z k by definition of ˆ S ≥ . γ is that hypersur-face has positive Ricci tensor for all t ∈ (0 , ∞ ). As discussed in Remark 3.23, Ricci-flat metricsrepresented by γ s with s (cid:54) = 0 does not hold such a property.Although γ s is trapped in S if s ≥
0, functions Z + Z − Z and Z ( X − X ) + Z ( X − X )are negative initially if s >
0. Hence γ s is not trapped in ˆ S initially if s >
0. To include thecase where s >
0, we need to enlarge ˆ S a little bit so that it initially traps all γ s with s closeenough to zero. That leads us to the second step of our construction. In this section, we assume s > S . We construct an entrance zone thatforces γ s to enter ˆ S eventually. Our goal is to show that for all small enough s > γ s willenter ˆ S in a compact set. As shown in computation (3.16), it is more convenient to computewith variables ω = Z Z and ω = Z Z , whose respective derivatives are ω (cid:48) = 2 ω ( X − X ) and ω (cid:48) = 2 ω ( X − X ) . By the definition of S , we have Z ≥ Z , Z . Therefore ω , ω ∈ [0 , ω ω -plane as shown Figure 1. Whatever γ s looks like, we can always project its Z and Z coordinate to ω ω -plane. And we want to provethe projection is bounded away from (0 ,
0) and hopefully going through the line l : ω + ω − , which is the projection of hyperplane Z + Z − Z = 0 . Note that any homogeneous variety in Z j ’sof degree D can be projected to an algebraic curve on ω ω -plane by dividing by Z D . Before theconstruction, we establish the following basic fact.Figure 1: Projection to ω ω -plane Proposition 3.10. Z Z is strictly increasing along γ s as long as Z > Z . Proof.
Initially we have ( X − X )( p ) = d . If Z > Z , we have( X − X ) (cid:48) (cid:12)(cid:12) X − X =0 = ( X − X )( G −
1) + R − R = ( Z − Z )( aZ − bZ − bZ ) ≥ ( Z − Z )( a − b ) Z since Z ≥ Z > Z > . (3.18)21ence X − X > γ s when Z > Z . But then (cid:18) Z Z (cid:19) (cid:48) = 2 Z Z ( X − X ) > Z > Z . Therefore Z Z is strictly increasing along γ s as long as Z > Z .Substitute solution (2.26) of linearized equation to R − R and R − R . It is clear that theyare positive initially. Hence at the beginning, the integral curve is trapped in U = S ∩ { Z + Z − Z ≤ , R − R ≥ , R − R ≥ } , (3.20)whose projection on ω ω -plane for all three cases is illustrated in Figure 2. (a) Case I (b) Case II(c) Case III Figure 2: Projection of U (enclosed by bold line segments) on ω ω -plane for all three casesBy Proposition 3.10, we know that in principal, the projection of γ s on ω ω -plane can getarbitrarily closed to ω − ω = 0, represented the dashed lines in Figure 2. Therefore, an integralcurve that is initially trapped in U has to escape. The question is whether it will escape U through Z + Z − Z = 0, represented by the red line segment. It turns out that a subset of U can beconstructed in a way that it contains a part of Z + Z − Z = 0 and γ s has to escape that subsetthrough Z + Z − Z = 0. Specifically, the construction is based on the following three ideas.1. Since Z + Z − Z ≤ γ s , the main task is to bound Z from above.For computation conveniences, we prefer to bound Z from above by some homogeneous algebraic22arieties in Z j ’s. In other words, defining inequalities of the entrance zone should include Z + Z − Z ≤ B ( Z , Z , Z ) ≥ B in Z j ’s.2. In order to show that γ s does not escape through B = 0, we need to show that (cid:104)∇ ( B ) , V (cid:105)| B =0 is non-negative along γ s in the entrance zone. This idea is discussed in the proof of Proposition3.2. It might be difficult to determine the sign of (cid:104)∇ ( B ) , V (cid:105)| B =0 even we are allowed to mod out B = 0 in the computation result. But notice that B (cid:48) := (cid:104)∇ ( P ) , V (cid:105)| B =0 = 0 vanishes at p , andinequality B (cid:48) ≥ B (cid:48) ≥
0, the trade-off is to show that (cid:104)∇ ( B (cid:48) ) , V (cid:105)| B (cid:48) =0 ≥ γ s inthe entrance zone. The homogeneous polynomial B that we find consists of two parameters. Theyallow us to tune the entrance zone to satisfy some technical inequalities. Once these inequalitiesare satisfied, we can show that (cid:104)∇ ( B (cid:48) ) , V (cid:105)| B (cid:48) =0 ≥ γ s is forced to escapethrough Z + Z − Z = 0.We first proceed the construction by having those technical inequalities in part 3 ready. Inthis process, the first parameter for B is introduced and how they interact with these technicalinequalities are explained. Then we reveal the definition for B and its last parameter. Proposition 3.11. In S , X + X > γ s always. Proof.
It is clear that X + X is positive initially along the curves. Since (cid:104)∇ ( X + X ) , V (cid:105)| X + X =0 = ( X + X )( G −
1) + R + R = R + R since X + X = 0 ≥ Z ( a ( Z + Z ) − bZ ) > Z ≥ Z and a − b = d − > , (3.21) X + X stays positive along γ s . Proposition 3.12.
For any fixed δ ≥ X − (1 + δ ) X > γ s and stay positive inthe region where R − (1 + δ ) R ≥ Proof.
Substitute solution (2.26) of linearized equation to X − (1 + δ ) X . We have(2 + δ ) s e ηd − aδs e ηd ∼ (2 + δ ) s e ηd > p . Since (cid:104)∇ ( X − (1 + δX )) , V (cid:105)| X − (1+ δ ) X =0 = ( X − (1 + δ ) X ) ( G −
1) + R − (1 + δ ) R = R − (1 + δ ) R since X − (1 + δ ) X = 0 , (3.22)the proof is complete.Define U δ = U ∩ {R − (1 + δ ) R ≥ } . (3.23)It is easy to check that U δ is a subset of U and γ s is initially trapped in U δ if s >
0. Therefore, X − (1 + δ ) X > γ s is in U δ by Proposition 3.12.The fixed value of δ needs to be picked in a certain range for the following two technical reasons.Firstly, we want inequality X − (1 + δ ) X > γ s enters ˆ S . Hence byproposition 3.12, we need to pick δ that make R − (1+ δ ) R ≥ Z + Z − Z = 0.Secondly, because U ⊂ S ∩ { Z − Z > } and the behavior of γ s is better known in U , we want γ s passes though the part of Z + Z − Z = 0 that Z − Z ≥ roposition 3.13. If δ ∈ (cid:16) b − a d − , bd − (cid:17) , then {R − (1 + δ ) R ≥ } contains a subset of { Z + Z − Z = 0 } ∩ { Z − Z > } in U . Proof. If δ ∈ (cid:16) b − a d − , bd − (cid:17) , then we have b − ( d − δ ( d − δ ) ∈ (0 , b − ( d − δ ( d − δ ) ≥ Z Z , then( R − (1 + δ ) R ) | Z + Z − Z =0 = ( Z − Z )(2 b ( Z + Z ) − aZ ) − δ ( aZ Z + b ( Z − Z − Z )) since Z + Z − Z = 0= Z (2 b (2 Z + Z ) − aZ ) − δ ( aZ Z + aZ − bZ − bZ Z ) since Z + Z − Z = 0= Z ( − ( d − δ ) Z + (4 b − ( d − δ ) Z ) Remark 1.1 ≥ . (3.24)The proof is complete. Remark 3.14.
Perhaps a better way to illustrate Proposition 3.13 is to consider the projection onthe ω ω -plane. For R − (1 + δ ) R = 0, we obtain an algebraic curve l : (1 − ω )(2 b (1 + ω ) − aω ) − δ ( aω + b ( ω − ω − . Straightforward computation shows that l intersect with ω + ω = 1 at points (0 ,
1) andFigure 3: δ = 0 . (cid:16) b − δ ( d − a +2 b , ( d − δ ) a +2 b (cid:17) . If δ ∈ (cid:16) b − ad − , bd − (cid:17) , then the second intersection point (cid:16) b − δ ( d − a +2 b , ( d − δ ) a +2 b (cid:17) is in the region where ω − ω >
0. Hence U δ , denoted by the darker area in Figure 3, can includea segment of l in U , represented by the bold segment, that is away from ω − ω = 0. Remark 3.15.
Note that Case I is the only case where the admissible δ must be positive.The entrance zone we construct is a subset of U δ . We impose δ ∈ (cid:16) , bd − (cid:17) . As shown in thefollowing technical proposition, δ > B is also introduced. Proposition 3.16. In U δ , we can find a p large enough such that(( X − X ) + ( p − X − X ))( X − X + ( p + 1)( X − X )) ≥ − G d ( d −
1) (3.25)along γ s in U δ . 24 roof. Since X = d − X − X , we can write inequality (3.25) with respect to ˜ X = X + X and˜ Y = X − X . Straight forward computation shows that inequality (3.25) is equivalent to (cid:18)(cid:18) p − (cid:19) (cid:18) p + 32 (cid:19) + 12( d − (cid:19) ˜ Y − (cid:18) p + 32 (cid:19) ˜ X ˜ Y + (cid:18)
94 + 32( d − (cid:19) ˜ X + 2 p + 1 d ˜ Y − d − d ( d −
1) ˜ X ≥ . (3.26)Note that ˜ X and ˜ Y are positive along γ s in U δ by Proposition 3.11 and 3.12. Moreover, in U δ , wehave X − (1 + δ ) X > γ s by Proposition 3.12. Rewrite this condition in terms of ˜ X and˜ Y so we have (2 + δ ) ˜ Y − δ ˜ X > γ s in U δ . Hence the LHS of (3.26) is larger than (cid:18)(cid:18)(cid:18) p − (cid:19) (cid:18) p + 32 (cid:19) + 12( d − (cid:19) δ δ − (cid:18) p + 32 (cid:19)(cid:19) ˜ X ˜ Y + (cid:18)
94 + 32( d − (cid:19) ˜ X + (cid:18) p + 1 d δ δ − d − d ( d − (cid:19) ˜ X . (3.27)Since δ ∈ (cid:16) , bd − (cid:17) is fixed, we can choose p large enough so that (cid:18)(cid:18) p − (cid:19) (cid:18) p + 32 (cid:19) + 12( d − (cid:19) δ δ ≥ p + 322 p + 1 d δ δ ≥ d − d ( d −
1) (3.28)are satisfied. Then inequality (3.26) is satisfied.Now we are ready to reveal the definition for B and its last parameter. Define B p,k ( Z , Z , Z ) := kZ Z p +13 − Z p ( Z − Z ) . For a fixed δ ∈ (cid:16) , bd − (cid:17) , choose a p that satisfies inequalities (3.28). Then define U ( δ,p,k ) = S ∩ { Z + Z − Z ≤ } ∩ { B p,k ≥ }∩ { ( Z − Z )( X − X ) + ( p ( Z − Z ) − Z )( X − X ) ≥ } , (3.29)More requirements on the choice of p and k are added later. Before that, we prove the following. Proposition 3.17.
For any fixed k > γ s is initially trapped in U ( δ,p,k ) as long as s ∈ (cid:18) , (cid:113) ks ( d +1)16 d (cid:19) . Proof.
With discussion in Section 3.1, we know that γ s is initially in S if s >
0. Since all theother inequalities in (3.29) reach equality at p , we need to substitute solution (2.26) of linearizedequation in each one of them. For Z + Z − Z , we have( d + 1) s e ηd − s e ηd ∼ − s e ηd < s <
0. 25ubstitute solution (2.26) of linearized equation to kZ Z p +13 − Z p ( Z − Z ) , we have k ( d + 1) s e ηd (cid:18) d − as e ηd + 2 s e ηd (cid:19) p +1 − (cid:18) d − as e ηd − s e ηd (cid:19) p s e ηd ∼ (cid:18) d (cid:19) p (cid:18) ks ( d + 1) d − s (cid:19) e ηd . (3.31)Hence kZ Z p +13 − Z p ( Z − Z ) > γ s when s < ks ( d +1)16 d . Finally, for ( Z − Z )( X − X ) + ( p ( Z − Z ) − Z )( X − X ), we have4 s e ηd (cid:18) d − as e ηd − s e ηd (cid:19) + (cid:18) ps e ηd − (cid:18) d − as e ηd − s e ηd (cid:19)(cid:19) s e ηd ∼ (4 + 8 p ) s e ηd > . (3.32)Hence γ s is indeed trapped in U ( δ,p,k ) initially when s ∈ (cid:18) , (cid:113) ks ( d +1)16 d (cid:19) .We now specify our choice for p and k . Projected to the ω ω -plane, the first two inequalitiesin (3.29) is equivalent to ω p (1 − ω ) k ≤ ω ≤ − ω . Write l as a function C ( ω ) = 1 − ω . Define l : C ( ω ) = ω p (1 − ω ) k . It is clear that C − C = 0at ω = 1. Our goal is to choose p and k so that C − C vanishes again at some ω ∗ <
1. Then wedefine ˆ U ( δ,p,k ) to be the compact subset of U ( δ,p,k ) where ω ∈ [ ω ∗ ,
1] and C > C for ω ∈ ( ω ∗ , p and k are chosen to guaranteethat ω ∗ is not too small so that ˆ U ( δ,p,k ) ⊂ U δ . Specifically, we have the following proposition.
Proposition 3.18.
Let p ≥ pp + 1 ≥ ( d − δ ) a + 2 b . (3.33)Let k > k < (cid:18) pp + 1 (cid:19) p p + 1 . (3.34)Then there exists some ω ∗ ∈ (cid:16) pp +1 , (cid:17) such thatˆ U ( δ,p,k ) := U ( δ,p,k ) ∩ { Z − ω ∗ Z ≥ } (3.35)is a compact subset of U δ . Proof.
Although the proposition is true as long as p > , the technical condition p ≥ p exists. Because δ is a fixednumber in (0 , bd − ), we have ( d − δ ) a +2 b < d − ba +2 b = 1 . Hence we can choose p large enough on topof inequalities (3.28) to satisfies inequalities (3.33).Consider the function C = C − C = 1 − ω − ω p (1 − ω ) k = 1 − ω k ( k − ω p (1 − ω )) . (3.36)26t is clear that C vanishes at ω = 1 and C > C = k − ω p (1 − ω ). Since d ˜ C dω = ω p − ( ω − p (1 − ω )) , we have ω (cid:16) , pp +1 (cid:17) pp +1 (cid:16) pp +1 , (cid:17) d ˜ C dω < > C k Decrease Local Minimum Increase k .
Therefore, for an arbitrary p , inequality (3.34) is satisfied if and only if ˜ C (cid:16) pp +1 (cid:17) <
0. Then thereexists some ω ∗ ∈ (cid:16) pp +1 , (cid:17) such that ˜ C ( ω ∗ ) = 0 and ˜ C ( ω ) > ω ∗ , ω ≤
1, that meansfor such an ω ∗ , we must have C ( ω ∗ ) = 0 and C ( ω ) > ω ∗ , ω -coordinate of the intersection point between l and l is ( d − δ ) a +2 b . By (3.33) and Remark1.1, the root ω ∗ discussed above satisfies ω ∗ > pp + 1 ≥ ( d − δ ) a + 2 b > a − ba + 2 b (3.37)We are ready to prove that ˆ U ( δ,p,k ) ⊂ U δ . In other words, with our choice of p and k above,inequalities in the definition (3.29) of U ( δ,p,k ) and (3.35) of ˆ U ( δ,p,k ) imply all inequalities in definition(3.20) of U and (3.23) of U δ .Firstly, we need to show ˆ U ( δ,p,k ) ⊂ U . With − Z ≥ Z − Z and Z ≥ ω ∗ Z satisfied in S , wehave R − R = ( Z − Z )( aZ − bZ − bZ ) ≥ ( Z − Z )(( a + 2 b ) Z ω ∗ − bZ ) ≥ ( Z − Z )(( a + 2 b ) Z ω ∗ − ( a − b ) Z ) Remark 1.1 ≥ S (3.38)and R − R = ( Z − Z )(2 bZ + 2 bZ − aZ ) ≥ ( Z − Z )(( a + 2 b ) ω ∗ Z − ( a − b ) Z ) ≥ S . (3.39)Hence ˆ U ( δ,p,k ) ⊂ U .In ˆ U ( δ,p,k ) , we have R − (1 + δ ) R = ( Z − Z )(2 b ( Z + Z ) − aZ ) − δ ( aZ Z + b ( Z − Z − Z ))= 2 b ( Z − Z ) − aZ ( Z − Z ) − δaZ Z − δbZ + δbZ + δbZ = (2 + δ ) bZ − (1 + δ ) aZ Z + δbZ − ( δb + 2 b ) Z + aZ Z . (3.40)Treat the result of the computation above as a function of Z . It is a parabola centered at (1+ δ ) a (2+ δ )2 b Z . By (3.37), it is clear that − ω ∗ > a +2 b b . Since δ ∈ (cid:16) , ba − b (cid:17) , it is straightforward to deduce that a +2 b b > (1+ δ ) a (2+ δ )2 b . From Z ≥ ω ∗ Z ≥ ω ∗ ( Z + Z ) we also deduce Z ≥ ω ∗ − ω ∗ Z . (3.41)27herefore, we know that Z + Z ≥ − ω ∗ Z ≥ a +2 b b Z ≥ (1+ δ ) a (2+ δ )2 b Z in ˆ U ( δ,p,k ) . Hence R − (1 + δ ) R ≥ ( R − (1 + δ ) R ) | Z = Z + Z ≥ . (3.42)Finally, we need to show that ˆ U ( δ,p,k ) is compact. Since Z − ω ∗ Z ≥
0, we automatically have Z + Z ≥ ω ∗ Z in ˆ U ( δ,p,k ) . By (3.37), we can deduce ω ∗ > a − ba +2 b > ba in ˆ U ( δ,p,k ) , where the lastinequality is from Remark 1.1. Hence a ( Z + Z ) − bZ ≥ U ( δ,p,k ) . Proposition 3.1 can beapplied and all Z j ’s are bounded above. By the conservation law (2.20), we know that all variablesare bounded. Hence ˆ U ( δ,p,k ) is compact. The proof is complete.Figure 4: δ = 0 . p = 12, k = (cid:0) (cid:1) for Case IWe are ready show that ˆ U ( δ,p,k ) is the entrance zone. Lemma 3.19.
For s ∈ (cid:18) , (cid:113) k ( d +1) s d (cid:19) and suitable choice of δ, p and k as described above, theintegral curve γ s escapes ˆ U ( δ,p,k ) through Z + Z − Z = 0 .Proof. Suppose γ s does not escape through Z + Z − Z = 0, then it can only escape througheither kZ Z p +13 − Z p ( Z − Z ) = 0 or ( Z − Z )( X − X ) + ( p ( Z − Z ) − Z )( X − X ) = 0. Weprove that these situations are impossible. 28ince (cid:104)∇ ( kZ Z p +13 − Z p ( Z − Z ) ) , V (cid:105) (cid:12)(cid:12)(cid:12) kZ Z p +13 − Z p ( Z − Z ) =0 = (cid:104)∇ ( Z p +23 ( kω − ω p (1 − ω ) )) , V (cid:105) (cid:12)(cid:12)(cid:12) kZ Z p +13 − Z p ( Z − Z ) =0 = ( p + 2) Z p +23 (cid:18) G − d + 2 X (cid:19) ( kω − ω p (1 − ω ) )+ Z p +23 (2 kω ( X − X ) − pω p ( X − X )(1 − ω ) + 4 ω p (1 − ω ) ω ( X − X ))= Z p +23 (2 kω ( X − X ) − pω p ( X − X )(1 − ω ) + 4 ω p (1 − ω ) ω ( X − X ))since kZ Z p +13 − Z p ( Z − Z ) = 0= Z p +23 (2 ω p (1 − ω ) ( X − X ) − pω p ( X − X )(1 − ω ) + 4 ω p (1 − ω ) ω ( X − X ))since kZ Z p +13 − Z p ( Z − Z ) = 0= 2 Z p ( Z − Z )(( Z − Z )( X − X ) + ( p ( Z − Z ) − Z )( X − X )) ≥ U ( δ,p,k ) , (3.43)it is impossible for γ s to escape ˆ U ( δ,p,k ) through kZ Z p +13 − Z p ( Z − Z ) = 0.For the other defining inequality, we have (cid:104)∇ (( Z − Z )( X − X ) + ( p ( Z − Z ) − Z )( X − X )) , V (cid:105)| ( Z − Z )( X − X )+( p ( Z − Z ) − Z )( X − X )=0 = (cid:104)∇ ( Z ((1 − ω )( X − X ) + ( p (1 − ω ) − ω )( X − X ))) , V (cid:105)| ( Z − Z )( X − X )+( p ( Z − Z ) − Z )( X − X )=0 = Z (cid:18) G − d + 2 X (cid:19) ((1 − ω )( X − X ) + ( p (1 − ω ) − ω )( X − X )))+ Z ((1 − ω )( X − X ) + ( p (1 − ω ) − ω )( X − X )))( G − Z ((1 − ω )( R − R ) + ( p (1 − ω ) − ω )( R − R ))+ Z (2 ω ( X − X )( X − X ) + 2( p + 2) ω ( X − X ) )= Z ((1 − ω )( R − R ) + ( p (1 − ω ) − ω )( R − R ))+ Z (2 ω ( X − X )( X − X ) + 2( p + 2) ω ( X − X ) )since ( Z − Z )( X − X ) + ( p ( Z − Z ) − Z )( X − X )) = 0= ( Z − Z )( R − R ) + ( p ( Z − Z ) − Z )( R − R )+ 2 Z ( X − X )( X − X ) + 2( p + 2) Z ( X − X ) = ( Z − Z )( R − R + p ( R − R )) − Z ( Z − Z )(2 bZ + 2 bZ − aZ )+ 2 Z ( X − X )(( X − X ) + ( p + 2)( X − X ))= ( Z − Z )( R − R + p ( R − R ) + 2 Z ( aZ − bZ − bZ ))+ ( Z − Z )(( X − X ) + ( p − X − X ))( X − X + ( p + 1)( X − X ))since 2 Z ( X − X ) = ( Z − Z )( X − X ) + p ( Z − Z )( X − X ) . (3.44)Because ˆ U ( δ,p,k ) ⊂ U δ , we can apply Proposition 3.16 to the last line of (3.44) and continue the29omputation as ≥ ( Z − Z ) (cid:18) R − R + p ( R − R ) + 2 Z ( aZ − bZ − bZ ) + 1 − G d ( d − (cid:19) = ( Z − Z ) (cid:18) R − R + p ( R − R ) + 2 Z ( aZ − bZ − bZ ) + 1 d − R + R + R ) (cid:19) by (2.20)= (cid:18) − d − (cid:19) bZ + (cid:16) aZ (( p + 1) ω − p ) + ad ( Z + Z ) (cid:17) Z + Z (cid:18) − (cid:18) bp + 4 b + bd − (cid:19) ω + (cid:18) ad − a − b (cid:19) ω + (cid:18) pb − b − bd − (cid:19)(cid:19) . (3.45)The first term of the computation result above is obviously positive. The second term is positivebecause ω ≥ ω ∗ > pp +1 in ˆ U ( δ,p,k ) . The positivity of the last term depends on the one of parabola π ( ω ) = − (cid:18) bp + 4 b + bd − (cid:19) ω + (cid:18) ad − a − b (cid:19) ω + (cid:18) pb − b − bd − (cid:19) . Since we impose p ≥
2, it is clear that π (0) is positive. As the coefficient of the first term isnegative, we know that π has two roots with different signs. It is easy to verify that π (1) = 0.Then we conclude that π is non-negative for all ω ∈ [0 , Z = 0 and Z = Z .Notice that there is no need to check the possibility that γ s may escape through Z − ω ∗ Z = 0.Because when the equality of Z − ω ∗ Z ≥ γ s ( η ∗ ), it implies that thefunction C in (3.36) vanishes at that point. Specifically, we have 1 − ω = ω p (1 − ω ) k at that point.But then Z Z p +13 ≤ Z p +13 ( Z − Z ) = Z p ( Z − Z ) k ≤ Z Z p +13 , which implies kZ Z p +13 − Z p ( Z − Z ) = 0 at that point and this case is included in the computationat the beginning of the proof. Proposition 3.20.
The only critical points in ˆ U ( δ,p,k ) are p and those of Type I. Proof.
By proposition 2.15, it is clear that p and critical points of Type I are in ˆ U ( δ,p,k ) . We firsteliminate critical points with negative Z j entry. Since ˆ U ( δ,p,k ) ⊂ S , we can eliminate critical pointswith Z smaller than the other two Z j ’s. Because X ≥ S , there is no critical points of TypeII. Since Z ≥ pZ ≥ Z in ˆ U ( δ,p,k ) by (3.37) and (3.41), there is no critical points other than p andthose of Type I in ˆ U ( δ,p,k ) . Proposition 3.21.
The function Z Z Z stays positive and increases along γ s . Proof.
Since
H ≡
1, it is clear that
G ≥ n . Hence( Z Z Z ) (cid:48) = Z Z Z (cid:18) G − d (cid:19) ≥ . (3.46)Since Z Z Z is initially positive along γ s , the proof is complete.We are ready to prove the completeness of Ricci-flat metrics represented by γ s with s closeenough to zero. 30 emma 3.22. There exists a k > such that an unstable integral curve γ s to (2.19) on C ∩{H ≡ } emanating from p is defined on R if s ∈ (cid:18) − (cid:113) k ( d +1) s d , (cid:113) k ( d +1) s d (cid:19) .Proof. If s >
0, the curve γ s is initially trapped in ˆ U ( δ,p,k ) as long as s ∈ (cid:18) , (cid:113) k ( d +1) s d (cid:19) .The function Z + Z − Z vanishes at p and it is negative along γ s in ˆ U ( δ,p,k ) . By Lemma3.19, the function Z + Z − Z must vanish at γ s ( η ∗ ) for some η ∗ ∈ R . Then we must have( Z + Z − Z ) (cid:48) ( γ s ( η ∗ )) ≥
0. But( Z + Z − Z ) (cid:48) ( γ s ( η ∗ )) = (cid:104)∇ ( Z + Z − Z ) , V (cid:105) ( γ s ( η ∗ ))= (cid:0) (cid:104)∇ ( Z + Z − Z ) , V (cid:105)| Z + Z − Z =0 (cid:1) ( γ s ( η ∗ ))= ( Z ( X − X ) + Z ( X − X )) ( γ s ( η ∗ )) . (3.47)Hence γ s ( η ∗ ) is in ∂ ˆ S . By Proposition 3.18, we know that ˆ U ( δ,p,k ) is in U , where R − R ≥ R − R ≥ X > X > X along γ s in ˆ U ( δ,p,k ) . Hence the intersection point γ s ( η ∗ ) is not p . By Proposition 3.21, we know that γ s ( η ∗ ) cannot be a critical point of Type I. By Proposition 3.20, we know that γ s ( η ∗ ) is not acritical point. Then by Lemma 3.6, γ s continue to flows inward ˆ S from γ s ( η ∗ ) and never escape.Therefore, such a γ s is defined on R .By symmetry, similar result can be obtained for s ∈ (cid:18) − (cid:113) k ( d +1) s d , (cid:19) . If s = 0, then we areback to the special case by Remark 3.8.By the discussion at the end of Section 2.3, Lemma 3.22 proves the first half of Theorem 1.2. Remark 3.23.
For γ s with s ∈ (cid:18) , (cid:113) k ( d +1) s d (cid:19) , it can be shown that R is negative initially bysubstituting (2.26). Hence the Ricci-flat metrics represented does not have the property introducedin Remark 3.9. By straightforward computation, however, it processes a weaker condition that thescalar curvature of each hypersurface remain positive. In this section, we study the asymptotic behavior of complete Ricci-flat metrics constructed above.Each integral curve γ s mentioned below satisfies the condition in Lemma 3.22, i.e., each γ s istrapped in ˆ U ( δ,p,k ) initially and then enter ˆ S in finite time. Lemma 4.1.
Let γ s be a long time existing integral curve that intersects with ˆ S at a non-criticalpoint γ s ( η ∗ ) . Then function ω + ω > along γ s ( η ) for η ∈ ( η ∗ , ∞ ) .Proof. Note that ( ω + ω )( γ s ( η ∗ )) = 1. By Lemma 3.6, we know that γ s ( η ) ∈ ˆ S for η ≥ η ∗ . Wehave ( ω + ω ) (cid:48) ( γ s ( η ∗ )) = (2 ω ( X − X ) + 2 ω ( X − X ))( γ s ( η ∗ )) ≥ S . (4.1)Suppose ( ω + ω ) (cid:48) ( γ s ( η ∗ )) = 0. Recall in the proofs of Lemma 3.22, we know that X > X > X at γ s ( η ∗ ). By (3.16) and (3.17), we have( ω + ω ) (cid:48)(cid:48) ( γ s ( η ∗ )) ≥ (cid:0) ω ( X − X ) + 4 ω ( X − X ) (cid:1) ( γ s ( η ∗ )) > . (4.2)31uppose there exists η ∈ ( η ∗ , ∞ ) that ( ω + ω )( γ s ( η )) = 1. We know from the computationabove that there exists η ∈ ( η ∗ , η ) such that ( ω + ω )( γ s ( η )) >
1. By mean value theorem,there exists η ∈ [ η , η ] such that ( ω + ω ) (cid:48) ( γ s ( η )) = (2 ω ( X − X )+2 ω ( X − X ))( γ s ( η )) < , a contradiction to the definition of ˆ S . Lemma 4.2.
The variable X is smaller than n along integral curves γ s .Proof. Since
H ≡ X ≤ n is equivalent to X + X − X ≥
0. The function X + X − X ispositive at p . Suppose the function vanishes along γ s at some point in ˆ U ( δ,p,k ) , then we have( X + X − X ) (cid:48) (cid:12)(cid:12) X + X − X =0 = ( X + X − X )( G −
1) + R + R − R = R + R − R since X + X − X = 0= a ( Z Z + Z Z − Z Z ) − b (2 Z − Z − Z ) . (4.3)Consider the computation result above as a function J ( Z ) = − bZ + a ( Z + Z ) Z + 2 bZ + 2 bZ − aZ Z . Since Z + Z ≤ Z ≤ Z ω ∗ in ˆ U ( δ,p,k ) , the positivity of J is implied by those of J ( Z + Z ) and J (cid:16) Z ω ∗ (cid:17) . With the choice p ≥
2, inequality (3.37) implies ω ∗ > pp +1 ≥ ≥ ba . Hence it is sufficientto prove a stronger condition: the positivity of J ( Z + Z ) and J (cid:0) a b Z (cid:1) . We have J ( Z + Z ) = ( a − b )( Z + Z ) − bZ Z ≥ b ( Z − Z ) Remark 1.1 ≥ . (4.4)And we have J (cid:16) a b Z (cid:17) = (cid:18) a b − a (cid:19) Z Z + 2 b ( Z + Z ) ≥ (cid:18) a b − a (cid:19) Z Z + 4 bZ Z ≥ . (4.5)All Z j ’s are positive along γ s . Hence by (4.4) and (4.5), computation (4.3) can vanish only if Z = Z = Z . But with p ≥ Z ≥ ω ∗ Z ≥ Z ≥ Z in ˆ U ( δ,p,k ) . Hence J canonly vanish at the origin of Z -space, which is impossible for γ s to reach by (3.46). Therefore, X + X − X never vanishes along γ s at least till γ s intersect with ∂ ˆ S at some γ s ( η ∗ ). γ s is in ˆ S for η ∈ [ η ∗ , ∞ ). The function X + X − X is positive at γ s ( η ∗ ). Suppose thefunction vanishes at some point along γ s in ˆ S , then( X + X − X ) (cid:48) (cid:12)(cid:12) X + X − X =0 = R + R − R = a ( Z Z + Z Z − Z Z ) − b (2 Z − Z − Z ) ≥ a ( Z Z + Z Z − Z Z ) − a Z − Z − Z ) Remark 1.1 ≥ a Z + Z − Z )(2 Z − Z − Z ) ≥ S . (4.6)32y Proposition 3.21, there is no need to consider the case where each Z j vanishes. For Case I-III,suppose computation above vanishes at some point on γ s . Then one possibility is that Z = Z = Z at that point. But then X − X j + ρ ( Z − Z j ) = X − X j = n − X j ≥ S . Then we must have X j = n for each j . Hence the point must be the critical point p , a contradiction. For Case I in particular, there is an extra possibility where Z = Z = Z atthat point. It is ruled out by Lemma 4.1. Hence X < n along γ s all the way.We can now describe the asymptotic limit of γ s . Lemma 4.3.
The integral curve γ s converges to p .Proof. Since γ s does not hit any critical point in ˆ U ( δ,p,k ) by Lemma 3.22, we can focus on thebehavior of the integral curve in the set ˆ S . By Proposition 3.21, we know that Z Z Z convergesto some positive number along γ s . There exists a sequence { η m } such that lim m →∞ η m = ∞ andlim m →∞ G = n . Hence lim m →∞ X j ( η m ) = n for each j . But then0 = lim m →∞ ( X + X − X ) (cid:48) ( η m )= lim m →∞ (( X + X − X )( G −
1) + R + R − R ) ( η m )= lim m →∞ ( R + R − R ) ( η m ) ≥ . (4.7)Therefore, either lim m →∞ Z ( ω + ω − η m ) = 0 or lim m →∞ (2 Z − Z − Z )( η m ) = 0. It is clear thatlim m →∞ Z ( η m ) (cid:54) = 0 as Z ≥ Z , Z in S and Z Z Z increases along γ s . By Lemma 4.1, we knowthat lim m →∞ ( ω + ω − η m ) (cid:54) = 0. Hence lim m →∞ (2 Z − Z − Z )( η m ) = 0. Since Z ≥ Z , Z in S ,we conclude that lim m →∞ ( Z − Z )( η m ) = lim m →∞ ( Z − Z )( η m ) = 0. With (2.20), we conclude thatlim m →∞ γ s ( η m ) = p . Hence p is in the ω -limit set of γ s .Consider p = (cid:0) n , n , n , α, α, α (cid:1) , where α = n (cid:113) n − a − b . By (2.22), the linearization at p is L ( p ) = n − n n bα ( a − b ) α ( a − b ) α n n − n ( a − b ) α bα ( a − b ) α n n n − a − b ) α ( a − b ) α bα α − α − α − α α − α − α − α α . (4.8)Its eigenvalues and corresponding eigenvectors are λ = 1 n − , λ = λ = β , λ = λ = β , λ = 2 n .v = n − n − n − − nα − nα − nα , v = − β αβ α − , v = − β α β α − , v = − β αβ α − , v = − β α β α − , v = nαnαnα , β = − n − (cid:112) ( n − − n α ( a − b )2 n < , β = − n − − (cid:112) ( n − − n α ( a − b )2 n < . Evaluate (2.23) at p , it is clear that T p ( C ∩ {H ≡ } ) = span { v , v , v , v } . Critical point p is asink. Hence lim η →∞ γ s = p . Lemma 4.4.
Ricci-flat metrics represented by γ s are AC.Proof. For each j , we have lim t →∞ ˙ f j = lim η →∞ X j √ Z k Z l = (cid:114) a − bn − . (4.9)Therefore by Definition 1.4, the Ricci-flat metric represented by γ s has conical asymptotic limit dt + t a − bn − Q. Lemma 4.3 and Lemma 4.4 imply Theorem 1.5.
This section is dedicated to singular Ricci-flat metrics. Note that critical points p and p canbe viewed as integral curves defined on R . They correspond to singular Ricci-flat metrics g = dt + t a − bn − Q . This is consistent with the fact that the Euclidean metric cone over a proper scaledhomogeneous Einstein manifold is Ricci-flat. For Case I in particular, the normal Einstein metricon G/K is strict nearly K¨ahler. Hence the metric cone represented by p is the singular G metricdiscovered in [6]. Note that functions F j ’s in (3.13) do note vanish at p . Therefore, the Euclideanmetric cone over the K¨ahler–Einstein metric has generic holonomy.There are also singular Ricci-flat metrics represented by nontrivial integral curves. Recall Re-mark 3.3, The cohomogeneity one G condition is given by F j ≡ j . Eliminate X j ’s in theconservation law C shows that (cid:78) = C ∩ {H ≡ } ∩ P ∩ { F ≡ F ≡ F ≡ } = { Z + Z + Z − ≡ } ∩ P ∩ { F ≡ F ≡ F ≡ } is an invariant 2-dimensional plane with boundary. Its projection in Z -space is plotted in Figure5. Black squares are critical poitns of Type II. Linearization at these points shows that they aresources. Furthermore, for any ξ ∈ R , (cid:78) ∩ { Z ( Z − Z ) − ξZ ( Z − Z ) ≡ } is a pair of integralcurves that connects three critical points. If ξ (cid:54) = 0 ,
1, then these two integral curves connect p withtwo distinct critical points of Type II. These integral curves represent singular cohomogeneity one G metrics on (0 , ∞ ) × G/K that do not have smooth extension to
G/H [16][14]. They all share thesame AC limit as the metric cone over
G/K equipped with the normal Einstein metric.When ξ = 0 ,
1, then one of the integral curve connects a critical point of Type II with p andthe other one connects a critical point of Type III with p . In particular, if ξ = 1, then we recover γ that represents the G metric, connecting p and p .There are singular metrics with generic holonomy. We construct a new compact invariant setwhose boundary includes p and p . Considerˇ S = S ∩ { X ≡ X , Z ≡ Z } ∩ { X + X − X ≥ } ∩ (cid:8) ( d − Z Z − b Z ≥ (cid:9) . (cid:78) with 0 < ξ ≤ Proposition 5.1. ˇ S is a compact invariant set. Proof.
It is easy to show that { X ≡ X , Z ≡ Z } is flow invariant. In fact, even if we defineˇ S without { X ≡ X , Z ≡ Z } , the set is still compact and invariant. However, considering thesubsystem does make the computation easier.In ˇ S , we have 4 b Z ≤ ( d − Z Z < a Z Z ≤ a ( Z + Z ) . (5.1)Hence we can apply Proposition 3.1 and conclude that inequality (3.4) holds in ˇ S . As Z is boundedabove by d − b √ Z Z in ˇ S , the compactness follows by (2.20).To show that ˇ S is invariant, consider (cid:104)∇ ( X + X − X ) , V (cid:105)| X + X − X =0 = ( X + X − X )( G −
1) + R + R − R = 2 R − R since Z ≡ Z in ˇ S and X + X − X = 0= 2( Z − Z )(( d − (cid:112) Z Z − bZ ) since Z ≡ Z in ˇ S ≥ S . (cid:104)∇ (( d − Z Z − b Z ) , V (cid:105) (cid:12)(cid:12) ( d − Z Z − b Z =0 = ∇ (cid:18) Z (cid:18) ( d − Z Z Z − b (cid:19)(cid:19)(cid:12)(cid:12)(cid:12)(cid:12) ( d − Z Z − b Z =0 = (cid:0) ( d − Z Z − b Z (cid:1) (cid:18) G − d + 2 X (cid:19) + 2( d − Z Z ( X + X − X )= 2( d − Z Z ( X + X − X ) since ( d − Z Z − b Z = 0 ≥ S . Hence ˇ S is a compact invariant set. Lemma 5.2.
There exists an integral curve Γ defined on R emanating from p in ˇ S .Proof. Consider p = (cid:16) n , n , n , n (cid:113) ( n − b ( d − a +2 b ) , n (cid:113) ( n − b ( d − a +2 b ) , n (cid:113) ( n − d − b ( a +2 b ) (cid:17) . For simplicity, de-note Z ∗ = n (cid:113) ( n − b ( d − a +2 b ) . The linearization at p is L ( p ) = d − d d bZ ∗ (cid:16) a ( d − b − b (cid:17) Z ∗ bZ ∗ d d − d (cid:16) a ( d − b − b (cid:17) Z ∗ bZ ∗ bZ ∗ d d d − d − Z ∗ ( d − Z ∗ ( d − Z ∗ Z ∗ − Z ∗ − Z ∗ − Z ∗ Z ∗ − Z ∗ − d − b Z ∗ − d − b Z ∗ − d − b Z ∗ (5.2)Straightforward computation shows that for all cases, L ( p ) is a hyperbolic critical point that hasonly one unstable eigenvalues with the corresponding eigenvector asˇ λ = 12 n (cid:16)(cid:112) ( n − + 96 n ( d − a − b ) Z ∗ − ( n − (cid:17) , ˇ v = b ˇ λb ˇ λ − b ˇ λ bZ ∗ bZ ∗ − d − Z ∗ Evaluate (2.23) at p , it is clear that ˇ v are tangent to C ∩ {H ≡ } . Fix ˇ s >
0, there exists aunique trajectory Γ emanating from p with Γ ∼ p + ˇ s e ˇ λη ˇ v. It is easy to check that p ∈ ∂ ˇ S with only X + X − X and ( d − Z Z − b Z vanished at p . By straightforward computation, we know that Γ is trapped in ˇ S initially. The integral curveis hence defined on R . Functions f j ’s that correspond to solutions Γ are defined on [0 , ∞ ). Lemma 5.3.
The integral curve Γ converges to p .Proof. Since ˇ S is a compact invariant set with X ≤ n . Arguments in Lemma 4.3 and 4.4 carryover. Hence for Γ, we have lim η →∞ Γ = p . 36or each j , we have lim t → ˙ f j = lim η →−∞ X j √ Z k Z = ˇ λ Z ∗ j, k ∈ { , } lim t → ˙ f = lim η →−∞ X √ Z Z = b ˇ λ ( d − Z ∗ . (5.3)Hence f = f ∼ ˇ λ Z ∗ t and f ∼ b ˇ λ ( d − Z ∗ t as t →
0. Since lim t →∞ ˙ f (cid:54) = lim t →∞ ˙ f , Γ represents a singularmetric whose end at t → Theorem 5.4.
Up to homothety, there exists a unique singular Ricci-flat metric on (0 , ∞ ) × G/K that at the end with t → , it admits conical singularity as the metric cone over G/K with alternativeEinstein metric. It has an AC limit at the end with t → ∞ as the metric cone over G/K with normalEinstein metric.
Results of this article can be summarized by the plot in the following page. It shows theprojection of integral curves to (2.19) on the Z -space for Case I. It is computed by MATLAB usingthe 4th order Runge–Kutta method.Integral Curves Metric Type γ Smooth metric with vanished principal curvatures on
G/H ;Nonsmooth metric with non-vanishing mean curvatures on
G/Hγ s , s (cid:54) = 0 Smooth metrics with non-zero principal curvatures on G/H
Nonsmooth metrics with non-vanishing mean curvatures on
G/H
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