International Journal of Mathematics Vol. 18, No. 7 (2007) 783–795 c World Scientific Publishing Company CRITERION OF PROPER ACTIONS FOR 3-STEP NILPOTENT LIE GROUPS TARO YOSHINO Research Institute for Mathematical Sciences Kyoto University, Sakyo-ku Kyoto, 606-8502, Japan [email protected]Received 31 December 2004 Revised 13 June 2006 For a nilpotent Lie group G and its closed subgroup L, Lipsman [13] conjectured that the L-action on some homogeneous space of G is proper in the sense of Palais if and only if the action is free. Nasrin [14] proved this conjecture assuming that G is a 2-step nilpotent Lie group. However, Lipsman’s conjecture fails for the 4-step nilpotent case. This paper gives an affirmative solution to Lipsman’s conjecture for the 3-step nilpotent case. Keywords : Lipsman’s conjecture; proper action; nilpotent Lie group. Mathematics Subject Classification 2000: 22E30, 22E25 1. Introduction As a “nice class” of actions of (non-compact) Lie groups, Palais [15] introduced the notion of “proper actions” (Definition 2.1). It can be regarded as a generalization of properly discontinuous actions. In fact: (1) If a Lie group L is discrete, the L-action is proper if and only if the L-action is properly discontinuous. More generally, if Γ is a discrete subgroup of a Lie group L, then properness of the L-action implies proper discontinuity of the Γ-action. (2) Suppose that a Lie group L acts properly on a manifold M . Then, the quotient space L\M becomes Hausdorff. If this action is furthermore (fixed point) free, the quotient space carries naturally a manifold structure (see Appendix A). Although the definition of proper action itself is simple, it is not easy in general to affirm whether a given action is proper or not. Recently, some progress has been made in the setting that M is a homogeneous space of a Lie group G, and L is its closed subgroup (see, for example, [1, 2, 7–10, 13, 14, 16]; see also a survey paper [12]). The aim of this paper is to prove a criterion of properness in the case where G is a 3-step nilpotent Lie group. Namely, we prove the following theorem. 783 Int. J. Math. 2007.18:783-795. Downloaded from www.worldscientific.com by HENRICH-HEINE-UNIVERSITAET on 12/29/13. For personal use only.
For a nilpotent Lie group G and its closed subgroup L, Lipsman [13] conjectured that
the L-action on some homogeneous space of G is proper in the sense of Palais if andonly if the action is free. Nasrin [14] proved this conjecture assuming that G is a 2-stepnilpotent Lie group. However, Lipsman’s conjecture fails for the 4-step nilpotent case.This paper gives an affirmative solution to Lipsman’s conjecture for the 3-step nilpotentcase.
As a “nice class” of actions of (non-compact) Lie groups, Palais [15] introduced thenotion of “proper actions” (Definition 2.1). It can be regarded as a generalizationof properly discontinuous actions. In fact:
(1) If a Lie group L is discrete, the L-action is proper if and only if the L-actionis properly discontinuous. More generally, if Γ is a discrete subgroup of a Lie groupL, then properness of the L-action implies proper discontinuity of the Γ-action.
(2) Suppose that a Lie group L acts properly on a manifold M . Then, thequotient space L\M becomes Hausdorff. If this action is furthermore (fixed point)free, the quotient space carries naturally a manifold structure (see Appendix A).
Although the definition of proper action itself is simple, it is not easy in generalto affirm whether a given action is proper or not. Recently, some progress has beenmade in the setting that M is a homogeneous space of a Lie group G, and L is itsclosed subgroup (see, for example, [1, 2, 7–10, 13, 14, 16]; see also a survey paper[12]). The aim of this paper is to prove a criterion of properness in the case whereG is a 3-step nilpotent Lie group. Namely, we prove the following theorem.
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Theorem 1.1. Suppose L is a closed subgroup in a connected and simply connected3-step nilpotent Lie group G. Then the action of L on a homogeneous space of G
is proper if and only if the action is free.
Theorem 1.1 gives also a criterion of proper discontinuity of an action of adiscrete group.
Corollary 1.2. Suppose G is a connected and simply connected 3-step nilpotentLie group, and Γ is its discrete subgroup whose Zariski closure is denoted by L. LetX be a homogeneous space of G. Then the Γ-action on X is properly discontinuousif and only if the L-action on X is free.
Here, note that for a subgroup L and its cocompact discrete subgroup Γ,
the Γ-action is properly discontinuous ⇔ the L-action is proper.
Thus, Corollary 1.2 follows from the next theorem, which claims cocompactness ofZariski closure.
Theorem 1.3. Suppose that G is a connected and simply connected nilpotent Liegroup. For any subgroup Γ in G, its Zariski closure contains Γ cocompactly.
Theorem 1.3 is essentialy due to Malcev in the case where Γ is discrete (see[5, Theorem 5.4.3]). An analogous result does not hold in the setting that G is areductive Lie group. By this theorem, we can easily conclude a “nilpotent version”of the following conjecture. We call the L-action on G/H cocompact, if the doublecoset space L\G/H is compact.
Conjecture 1.4 (Kobayashi [12, Conjecture 4.3]). Let (G, H) be a pair ofreductive Lie groups. If a discrete subgroup Γ in G acts on G/H properly discon-tinuously and cocompactly, then there exists a reductive subgroup L acting on G/Hproperly and cocompactly.
Namely, as a “nilpotent version” of Conjecture 1.4, we have:
Theorem 1.5. Let G be a connected and simply connected Lie group, and L beits closed subgroup. If a discrete subgroup Γ in G acts on G/H properly discontin-uously and cocompactly, then the Zariski closure of Γ acts on G/H properly andcocompactly.
After proving Theorem 1.1, the author was informed that Baklouti–Khlif [1]also proved Theorem 1.1 independently.
2. Background
First, we recall the definition of properness in two ways.
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Criterion of Proper Actions for 3-Step Nilpotent Lie Groups 785
(1) Suppose a Lie group L acts on a manifold M . The action of L on M is calledproper if the map
Φ : L × M → M × M, (l, x) �→ (lx, x)
is a proper map. In other words, Φ−1(S) is compact for any compact set S inM × M .
(2) Suppose L and H are closed subgroups in a Lie group G. The triple (L, G, H)of Lie groups is called proper if L ∩ SHS−1 is compact for any compact set S
in G.
Remark 2.2. The triple (L, G, H) is proper ⇔ the action L on G/H is proper.
The first breakthrough about criterions of properness has been done byKobayashi in 1989.
Theorem 2.3 (Kobayashi [7], see also [8]). Let G be a reductive Lie group.Then
(L, G, H) is (CI) ⇔ (L, G, H) is proper,
for any reductive subgroups L and H in G.
Here, (CI) is defined as follows:
Definition 2.4 (Kobayashi [7]). (1) Suppose a Lie group L acts on a manifoldM . The action of L on M is called (CI) if the isotropy group Lx := {l ∈ L | lx = x}is compact for any x ∈ M .
(2) Suppose L and H are closed subgroups in a Lie group G. The triple (L, G, H)is called (CI) if L ∩ gHg−1 is compact for any g ∈ G.
Parallel to Remark 2.2, we have:
Remark 2.5. The triple (L, G, H) is (CI) ⇔ the action L on G/H is (CI).
Remark 2.6. Properness always implies (CI).
Now, let us see the case where G is nilpotent, connected and simply connected.In this case, note that (CI) is equivalent to free (Lemma 3.6). We recall Lipsman’sconjecture.
Conjecture 2.7 (Lipsman’s Conjecture; [13, Conjecture 4.1]). (a) Let H =N(n) be the nilpotent Lie group consisting of n × n upper triangular unipotentmatrices, V the abelian Lie group R
n, and G the semidirect product Lie groupH � V . Then,
(L, G, H) is (CI) ⇔ (L, G, H) is proper
for any connected closed subgroup L in G.
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(b) More generally, let G be a connected and simply connected nilpotent Lie group.Then,
(L, G, H) is (CI) ⇔ (L, G, H) is proper
for any connected closed subgroups L and H in G.
In (a), we have G/H � Rn, so we call (a) Lipsman’s conjecture for R
n. Lipsman’sconjecture for R
n has been proved to be true, if n ≤ 4 (n = 2 by Kobayashi [8], n = 3by Lipsman [13] and n = 4 by [16] respectively). However, Lipsman’s conjecture isfalse for R
5 [17].On the other hand, Nasrin [14] proved (b) in the setting that G is a 2-step
nilpotent Lie group. So let us call (b) with assumption that G is n-step nilpotent,Lipsman’s conjecture for n-step nilpotent. However, it is known that Lipsman’sconjecture for 4-step nilpotent is false [17].
G is n-step nilpotentn = 1 truen = 2 truen = 3n ≥ 4 false
A natural question is whether Lipsman’s conjecture holds for 3-step nilpotent ornot. Theorem 1.1 answers it affirmatively.
3. Preliminaries
Let G be a Lie group and g be its Lie algebra. we shall denote
g0 := g,
gn := [g, gn−1] (n ≥ 1).
Lie algebra g is called n-step nilpotent if gn = {0} and gn−1 = {0}. In this paper,we shall often assume that g is a 3-step nilpotent Lie algebra.
Example 3.1. An abelian Lie algebra is 1-step nilpotent. A Heisenberg algebrais 2-step nilpotent. The Lie algebra consisting of strictly upper triangular 4 × 4matrices is 3-step nilpotent.
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Let p be the canonical projection
p : g → g/g1. (3.1)
For x ∈ g and a subset X ⊂ g, if p(x) ∈ p(X), we shall write
x ∈ X mod g1. (3.2)
In this paper, the following easy observations will be frequently used in theproof.
Observation 3.2. If g is a 3-step nilpotent Lie algebra, then
A ≡ A′, B ≡ B′, C ≡ C′ mod g1
implies
[A, [B, C]] = [A′, [B′, C′]].
Observation 3.3. If g is a 3-step nilpotent Lie algebra, then
A, B, C ∈ l mod g1 ⇒ [A, [B, C]] ∈ l
for any subalgebra l in g.
If G is a connected and simply connected nilpotent Lie group, the exponentialmap
exp: g → G
is diffeomorphic, so its inverse map is denoted by log : G → g.In view of Definitions 2.1(2) and 2.4(2), we introduce the following
Definition 3.4. Suppose L and H are closed subgroups in a Lie group G. Thetriple (L, G, H) is called free if L ∩ gHg−1 = {e} for any g ∈ G.
Parallel to Remarks 2.2 and 2.5, we have:
Remark 3.5. The triple (L, G, H) is free ⇔ the action L on G/H is free.
The next lemma follows from [14, Theorem 2.11 and Lemma 3.6].
Lemma 3.6. Let G be a connected and simply connected nilpotent Lie group. Sup-pose that l and h are Lie algebras of connected closed subgroups L and H in G,
respectively. Then the following three conditions are equivalent.
(i) The triplet (L, G, H) is (CI).(ii) The triplet (L, G, H) is free.(iii) l ∩ (Ad(S)h + s) is compact for any compact sets S ⊂ G and s ⊂ g.
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4. Lemmas
In this section, we prove some lemmas for the proof of main theorem.
Lemma 4.1. If g is a 3-step nilpotent Lie algebra, then
X = log(ere−s) + Ad(e12 (r+s))Y +
124
[X + Y, [Y, X ]] (4.1)
for any X, Y, r, s ∈ g satisfying eX = ereY e−s.
Proof. The assumption eX = ereY e−s implies
X ≡ r + Y − s mod g1,
so we have
[X + Y, [Y, X ]] = [2Y + r − s, [Y, r − s]]
by Observation 3.2. With Campbell–Hausdorff formula, we calculate (LHE)−(RHE)of (4.1):
In the last one, all terms are rewritable in the form [α, [β, Y ]] = ad(α)ad(β)Y.We continue the above calculation with omitting Y and denoting ad(α)ad(β) as αβ.
Criterion of Proper Actions for 3-Step Nilpotent Lie Groups 789
Lemma 4.2. Let l and h be subalgebras in g. Suppose that sequences {Xi} ⊂ l and{Yi} ⊂ h satisfy
Xi − Yi ∈ S mod g1 (for any i ∈ N), (4.2)
for a compact set S ⊂ g. Then, there exist bounded sequences {xi}, {yi} ⊂ g suchthat
xi ∈ l mod g1, (4.3)
yi ∈ h mod g1, (4.4)
Xi − xi ≡ Yi − yi mod g1. (4.5)
Proof. We may and do assume S ⊂ l + h + g1 by replacing S by S ∩ (l + h + g1)if necessary, because this replacement does not change the condition (4.2). Let p
be the canonical projection as in (3.1). Since the natural linear map p(l) × p(h) →p(l + h), (p(X), p(Y )) �→ p(X + Y ) is well-defined and surjective, there exists asplitting
Tl × Th : p(l + h) → p(l) × p(h),
namely,
Tl : p(l + h) → p(l)
Th : p(l + h) → p(h)
are linear maps with the following property: x = Tl(x)+Th(x) for any x ∈ p(l+ h).We define a compact subset T in g/g1 by
T := Tl(p(S)) ∪ (−Th(p(S))).
We now set
xi := Tl(p(Xi − Yi)), yi := −Th(p(Xi − Yi)).
Then {xi} and {yi} are contained in T , and therefore form bounded sequences. Letg′ be a complementary subspace of g1 in g. We shall regard xi and yi as elementsof g′ through the isomorphism g′ � g/g1. Then (4.3)–(4.5) hold. Hence, we haveproved Lemma 4.2.
Lemma 4.3. If g is a 3-step nilpotent Lie algebra, then
X − x ≡ Y − y mod g1
implies
[X + Y, [Y, X ]] = φ(X, x) − φ(Y, y) − φ(y, x) + 3[[x, y], Y ]
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for every X, Y, x and y in g, with
φ(α, β) := [2α − β, [α, β]].
Proof. We calculate [X + Y, [Y, X ]] as in Lemma 4.1. By Observation 3.2, we canregard “equal mod g1” as “equal”.
(X + Y )Y = (2X − x + y)Y
= (2X − x)Y + yY
= (2X − x)(X − x + y) + yY
= (2X − x)(X − x) + (2X − x)y + yY
= (2X − x)(X − x) + (2Y − 2y + x)y + yY
= (2X − x)(X − x) + 2Y y − (2y − x)y + yY.
On the other hand
(X + Y )Y = (2Y − y + x)Y
= (2Y − y)Y + xY.
So we have
[X + Y, [Y, X ]] = [X + Y, [Y, x]] − [X + Y, [Y, y]]
In this section we shall prove main theorem (Theorem 1.1). The implication“proper ⇒ free” is clear by Remark 2.6 and Lemma 3.6. So we shall prove that ifthe action is not proper, then it is not free.
Assume (L, G, H) is not proper (see Definition 2.1), then we can take a compactset S ⊂ G such that L ∩ SHS−1 is non-compact. Then we can take sequences
{Xi} ⊂ l, {Yi} ⊂ h, {ri}, {si} ⊂ log(S)
such that
eXi = erieYie−si ∈ L ∩ SHS−1, ‖Xi‖ → ∞ (i → ∞).
Using Lemma 4.1, we have
Xi = log(erie−si) + Ad(e12 (ri+si))Yi +
124
[Xi + Yi, [Yi, Xi]]. (5.1)
Then we shall prove the following (a) and (b).
(a) The last term 124 [Xi +Yi, [Yi, Xi]] is decomposed as Ai +Bi +[ci, Yi]+di where
Ai ∈ g2 ∩ l, Bi ∈ g2 ∩ h, ci ∈ g1, di ∈ g
and {ci}, {di} are bounded sequences.
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(b) The sequence {Xi − Ai} is not bounded.
If (a) and (b) are proved, the proof of main theorem is completed. In fact,Eq. (5.1) becomes
Xi − Ai = Ad(e
12 (ri+si)
)Yi + Bi + [ci, Yi] + log(erie−si) + di (5.2)
= Ad(e
12 (ri+si)+ci
)(Yi + Bi) + log(erie−si) + di (5.3)
by (a). Since {ri}, {si}, {ci}, {di} are bounded sequences, (5.3) is included inl ∩ (Ad(S)h + s) with adequate compact sets S ⊂ G and s ⊂ g. Then, (b) impliesl∩ (Ad(S)h+ s) is not relatively compact. Thus (L, G, H) is not free by Lemma 3.6.Therefore Theorem 1.1 is proved. Now let us prove (a).
Since we have
Xi − Yi ≡ ri − si mod g1,
we can take bounded sequences {xi}, {yi} ⊂ g satisfying (4.3)–(4.5) by Lemma 4.2.Using Lemma 4.3, we obtain the decomposition:
i := Xi − Ai is bounded, then Xi should be alsobounded mod g1. Since φ(Xi, xi) depends on Xi only “mod g1”, φ(Xi, xi) mustbe bounded. Then Ai = 1
24φ(Xi, xi) is also bounded, and ‖Xi‖ → ∞ implies‖X ′
i‖ → ∞. This contradict to the assumption that X ′i is bounded. Thus (b) has
been proved.The proof of Theorem 1.1 has been completed.
6. Proof of Theorem 1.3
In this section, we shall prove Theorem 1.3. More precisely, we shall prove thatthere exists R ∈ Z>0 satisfying
(A) R(X + Y ) ⊂ log Γ,
(B) R[X, Y ] ⊂ log Γ
for any X and Y in log Γ. This implies that log Γ contains a lattice of
l := R-Span log Γ.
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In fact, we can take {Xi} ⊂ log Γ as a basis of l. Then
Rn−1m∑
i=1
ciXi ∈ log Γ
for any ci ∈ Z by (A). Moreover (B) implies l is a subalgebra of g. Thus theanalytic subgroup L of l contains Γ cocompactly, and L is the Zariski closure of Γ.So Theorem 1.3 is proved.
To prove (A) and (B), we introduce some notations.
[log Γ]1 := log Γ,
[ log Γ]n := {[X, Y ] | X ∈ log Γ, Y ∈ [log Γ]n−1}, (n = 2, 3, . . .)
[ log Γ]n∗ := [log Γ]n ∩ log Γ.
Now we write the following two conditions.
Cn-(a): There exists Rn ∈ Z>0 such that Rn(log Γ + [log Γ]n∗ ) ⊂ log Γ.
Cn-(b): There exists Sn ∈ Z>0 such that Sn([log Γ]n) ⊂ log Γ.
It is clear that (for n = 1) C1-(a) implies (A) and (for n = 2) C2-(b) implies(B). Simply, we denote Cn for Cn-(a) and Cn-(b).
Since G is nilpotent, for sufficiently large n, we have [log Γ]n = {0} , and thisimplies Cn. Thus, what we need is to prove that Cl (for any l > n) implies Cn.
Assume Cl (for any l > n). First, we shall prove Cn-(a). Let X ∈ log Γ andY ∈ [log Γ]n∗ , then
log(exp(X) exp(Y )) ∈ log Γ. (5.4)
By Campbell–Hausdorff formula, we can expand (5.4) as follows.
log(exp(X) exp(Y )) = X + Y +p∑
i=1
1ci
ti. (5.5)
Here, ci ∈ Z and ti ∈ [log Γ]li (li > n). Note that p, ci and li do not depend on X
and Y . So independently, we can define an integer c as follows.
c := LCM(c1Sl1 , . . . , cpSlp).
Here, LCM means “Least Common Multiple”. Then Cli -(b) implies
c
citi ∈ [log Γ]li∗ .
Vanishing the third term of (5.5) by Cli -(a), we have(c
p∏i=1
Rli
)(X + Y ) ∈ log Γ.
Let Rn := c∏p
i=1 Rli , then Cn-(a) has been proved.
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Next, we shall prove Cn-(b). Let X ∈ [log Γ]n, then there exist X1, . . . , Xn ∈log Γ such that X = [X1, . . . , [Xn−1, Xn] . . .]. We define γi ∈ Γ as follows.
γn := exp(Xn),
γn−1 := exp(Xn−1)γn exp(−Xn−1)γ−1n ,
...
γ1 := exp(X1)γ2 exp(−X1)γ−12 .
By Campbell–Hausdorff formula, we can expand log γ1 as follows.
log γ1 = [X1, . . . , [Xn−1, Xn] . . .] +p∑
i=1
1ci
ti. (5.6)
Here, ci ∈ Z and ti ∈ [log Γ]li (li > n). Note that p, ci and li do not depend on X .So independently, we can define an integer c as follows.
c := LCM(c1Sl1 , . . . , cpSlp).
Then Cli-(b) impliesc
citi ∈ [log Γ]li∗ .
Vanishing the second term of (5.6) by Cli -(a), we have(c
p∏i=1
Rli
)X ∈ log Γ.
Let Sn := c∏p
i=1 Rli , then Cn-(b) has been proved.The proof of Theorem 1.3 has been completed.
Appendix A
In this section, we give a proof of the following:
Proposition A.1. If a Lie group G acts on a manifold M properly and freely bydiffeomorphisms, the quotient space G\M naturally has a manifold structure suchthat the canonical projection
π : M → G\Mis a differential map.
This result is not new though it has not, to the author’s knowledge, appeared ina published form. For instance, Kobayashi gave this result in a lecture at the univer-sity of Tokyo in 1994. For the sake of completeness, we show that Proposition A.1is based on the following:
Fact A.2 ([3, 5.9.5], [6, Sec. 4, Theoreme 5]). Let R ⊂ M × M be an equiv-alence relation on a Cr-manifold M . The quotient space M/R has a Cr-manifold
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structure such that the projection p : M → M/R is a submersion, if and only if R
satisfies the following two conditions:
(1) R is a closed submanifold of M × M .(2) The second projection pr 2 : R → M is a submersion.
The manifold structure of M/R such that p is a Cr-map is then unique.
In our setting of Proposition A.1, we define the equivalence relation R ⊂ M×M
by the image of the following map:
Φ : G × M → M × M, (g, p) �→ (gp, p).
Let us show that the conditions (1) and (2) in Fact A.2 are fulfilled for R =Φ(G × M). To see this, it is enough to verify the following conditions (a) and (b).
(a) Φ is a closed map.(b) The differential dΦ is injective.
Suppose first that (a) and (b) are satisfied. Then, the image Φ(G × M) is closedby (a). Since the G-action is free, Φ is injective. Thus Φ : G × M → Φ(G × M)is homeomorphic by (a). Therefore (b) implies that Φ is an embedding, so R =Φ(G × M) is a closed submanifold of M × M . We have shown (1). Then, we haveG×M is diffeomorphic to R. Therefore, it is clear that the natural map G×M → M
is a submersion. Thus, we have shown (2).Finally, let us verify the condition (a) and (b). Since the G-action on M is
proper, the map Φ is a proper map. Thus, it is a closed map by the following:
Fact A.3 [4, Chap. I, Sec. 10, Proposition 1] Every proper mapping is closed.
We proved (a). Let g be the Lie algebra of G. Then X ∈ g defines a vector fieldon M , say X ∈ X(M) by
Xp :=d
dtetXp
∣∣∣∣t=0
.
Note that this vector field is invariant under the action of exp(RX) ⊂ G, andtherefore freeness of the action implies Xp = 0 for any p ∈ M and any nonzeroX ∈ g. Here the differential dΦ : TgG ⊕ TpM → TgpM ⊕ TpM is given by
dΦ(g,p) : (X, Y ) �→ (Xgp, Y ).
Thus, this map is injective. We proved (b), and so completed the proof ofProposition A.1.
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