On the surjectivity of the power maps of a class of solvable groups

Theorem 1.1.

Let G be a group and N a normal subgroup of G. Suppose that N is F-nilpotent with respect to a field F, and that the conjugation action of G on N is F-linear. Let N=N0⊃N1⊃⋯⊃Nr={e} be the central series of N. Let A=G/N, x∈G and a=x⁢N∈A. Let k∈N be coprime to the characteristic of F. Let B={b∈A∣bk=a} and let B* be the subset consisting of all b in B such that for any j=1,…,r any element of Nj-1/Nj which is fixed under the action of a is also fixed under the action of b. Then we have the following:

1. For any b∈B*, n∈N, there exists y∈G such that y⁢N=b and yk=x⁢n.

2. If A is abelian, and x⁢n∈Pk⁢(G) for all n∈N, then B* is nonempty.

Corollary 1.2.

Let G be the group of F-points of a solvable algebraic group G defined over F and let N be the group of F-points of the unipotent radical Ru⁢(G) of G. Let k be coprime to the characteristic of F. Let x be a semisimple element in G. Then x⁢n∈Pk⁢(G) for all n∈N if and only if there exists y∈Z⁢(ZG⁢(x)) such that yk=x.

Corollary 1.3.

Let G be a connected solvable Lie group, N the nilradical of G and H a Cartan subgroup of G. Let h∈H and k∈N. Then h⁢n∈Pk⁢(G) for all n∈N if and only if there exists g∈H such that gk=h and g∈ZH⁢(X) for every X∈L⁢(N) such that h∈ZH⁢(X). In particular, Pk:G→G is surjective if and only if Pk:ZH⁢(X)→ZH⁢(X) is surjective for all X∈L⁢(N).

Corollary 1.4.

Let G be a connected solvable Lie group and N a connected nilpotent closed normal subgroup of G such that G/N is abelian. Let A=G/N and N=N0⊃N1⊃⋯⊃Nr={e} be the central series of N. Let x∈G and a=x⁢N/N. Then xn is exponential in G for all n∈N if and only if there exists a one-parameter subgroup B of A containing a such that for any j=1,…,r any point of Nj-1/Nj which is fixed by the action of a is also fixed by the action of B.

Corollary 1.5.

Let G be a connected Lie group such that Pk:G→G is surjective. Let R be the (solvable) radical of G and S=G/R. Suppose that S has a unipotent one-parameter subgroup U such that ZS⁢(U) does not contain any element whose order divides k. Then Pk:R→R is surjective.

Proposition 2.1.

Let F and V be as above. Let G be a group with (a copy of) V as a normal subgroup such that G/V is cyclic, and the conjugation action of G on V is F-linear. Let σ:G→GL⁢(V) denote the induced action. Let k∈N be coprime to the characteristic of F. Let g∈G and x=gk. Then the following statements hold:

1. If F⁢(σ⁢(x))=(0), then for every v∈V, there exists a unique w∈V such that x⁢v=w⁢x⁢w-1.

2. If F⁢(σ⁢(x))=F⁢(σ⁢(g)), then x⁢v∈Pk⁢(G) for all v∈V.

Proof.

(i) Let I denote the identity transformation. The hypothesis implies that (σ⁢(x)-1-I)∈GL⁢(V). Hence for any v∈V there exists an element w∈V such that

in the group structure of G; hence we have x⁢v=w⁢x⁢w-1. The uniqueness of w follows from the fact that if x⁢v=w1⁢x⁢w1-1=w2⁢x⁢w2-1, then w2-1⁢w1∈F⁢(σ⁢(x)) which is given to be trivial.

(ii) Let W be the largest σ⁢(x) invariant subspace of V such that the restriction of σ⁢(x) to W is unipotent. Then W is also σ⁢(g)-invariant, and the hypothesis together with the remark preceding the proposition (applied to W in place of V) implies that the restriction of σ⁢(g) to W is unipotent. We now proceed by induction on the dimension of W. We note that if W is zero-dimensional then F⁢(σ⁢(x))=0 and in this case the desired assertion follows from part (i). Now consider the general case. Let U=F⁢(σ⁢(x)) which by the hypothesis is also F⁢(σ⁢(g)). Then U is a normal subgroup of G and G/U contains V/U as a normal subgroup with cyclic quotient and the action of its generator on V/U is given by the quotient of the σ⁢(g)-action on V/U. As the restriction of σ⁢(g) to W is unipotent, U is of positive dimension and W/U has dimension less than W. We note also that W/U is the largest σ⁢(g)-invariant subspace of V/U on which the σ⁢(x)-action is unipotent. Hence by the induction hypothesis for any v∈V there exists η∈G such that x⁢v⁢U=ηk⁢U. Thus there exists u′∈U such that x⁢v=ηk⁢u′. Since k is coprime to the characteristic of 𝔽 there exists u∈U such that u′=uk (we use the multiplicative notation since U is now being viewed as a subgroup of G). Also, as σ⁢(g) fixes u, the latter is contained in the center of G. Hence x⁢v=ηk⁢u′=ηk⁢uk=(η⁢u)k. Thus x⁢v∈Pk⁢(G). ∎

Proposition 2.2.

Let V be a finite-dimensional F-vector space. Let G be a group with V as a normal subgroup such that G/V is abelian. Let σ:G→GL⁢(V) be the induced action. Let k∈N, g∈G and x=gk. Let A=G/V, a=x⁢V, b=g⁢V∈A. Suppose that F⁢(a)≠F⁢(b). Let v∈V be such that there exists y∈G satisfying y⁢V=b and yk=x⁢v. Then v∈X⁢(σ⁢(A),V).

Proof.

Using the Jordan canonical form, we see that there exists a unique minimal a-invariant subspace, say U, of V such that the action of a on V/U is trivial while the b-action on V/U has no nonzero fixed point. The condition in the hypothesis implies that U is a proper subspace of V. As A is abelian it follows that U is A-invariant. Moreover, U is contained in X⁢(σ⁢(A),V) as we can find a subspace W∈𝔚 containing it. We show that any v satisfying the condition in the hypothesis is contained in U. Let such a v be given and let y∈G be such that y⁢V=b and yk=x⁢v. Then there exists w∈V such that y=g⁢w and we have

where θ=σ⁢(g)-(k-1)+⋯+σ⁢(g)-1+I, and hence v=θ⁢w. Let V′=V/U. Then any eigenvalue of the factor of σ⁢(g) to V′ is contained in {λ∈ℂ∣λk=1,λ≠1}. Also, (σ⁢(g)-1-I)⁢θ=σ⁢(g)-k-I=σ⁢(x)-1-I factors to the zero transformation on V′. Since σ⁢(g) has no nonzero fixed point on V′, the restriction of (σ⁢(g)-1-I) to V′ is invertible. Hence θ factors to the zero transformation on V′=V/U. Therefore v=θ⁢w∈U, as desired. ∎

Proof of (i).

In proving this part, without loss of generality we may assume that G is the product of 〈g〉 with N, where g∈G is such that g⁢N=b, with b a given element from the subset B*; while the assumption is not crucial it makes the proof more transparent. Let N=N0⊃N1⊃⋯⊃Nr={e} be the central series of N. We proceed by induction on r. For r=1 the desired statement is immediate from Proposition 2.1 (ii). Now consider the general case. Let n∈N be given. Now G/Nr-1 is a product of 〈g⁢Nr-1〉 and N/Nr-1 in G/Nr-1, with g∈G as above, and the condition in the hypothesis of the theorem is satisfied for G/Nr-1 (with N/Nr-1, a1=g1k⁢Nr-1 and b1=g1⁢Nr-1 in place of Na and b, respectively); we note that the corresponding actions on Nj/Nj+1 coincide with those of a and b, respectively. Since G/Nr-1 is an 𝔽-nilpotent group of length less than r, by the induction hypothesis there exists g1∈G such that g1⁢N=b and x⁢n⁢Nr-1=g1k⁢Nr-1. Hence there exists v∈Nr-1 such that x⁢n=g1k⁢v. Now consider the product, say G1, of the subgroups 〈g1〉 and Nr-1 in G; we note that Nr-1 is an 𝔽-vector space, and the condition as in Proposition 2.1 (ii) is satisfied. Hence we get that there exists y∈G1 such that g1k⁢v=yk; thus we have y∈G such that y⁢N=b and x⁢n=yk, and so x⁢n∈Pk⁢(G), which proves (i). ∎

Proof of (ii).

For any subset S of B we denote by ES the set of n in N for which there exists y∈G such that y⁢N∈S and yk=x⁢n. We shall show that ES=N if and only if S∩B* is nonempty; when the condition in (ii) holds, for S=B we have ES=N and this implies assertion (ii) in the theorem. We proceed by induction on r, the length of the central series. For r=1 the assertion follows immediately from Propositions 2.1 and 2.2. Now consider the general case. Let S be a subset of B. If S∩B* is nonempty, then assertion (i) of the theorem, proved above, shows that ES=N. Now suppose that ES=N. Let V=Nr-1, G′=G/V and let B′ be the subset of B consisting of b such that for all indices j=1,…,r-1 any point of Nj-1/Nj which is fixed by a is also fixed by b. Since ES=N, for all n∈N there exists an element y∈G such that y⁢N∈S and yk⁢V=x⁢n⁢V, and hence the induction hypothesis implies that S∩B′ is nonempty. Also, applying the induction hypothesis to the set S\B′, we get that there exists n0∈N for which there does not exist any y∈G such that y⁢N∈S\B′ and yk⁢V=x⁢n0⁢V. Now consider x⁢n0⁢v, with v∈V. Since ES=N, there exists y∈G (depending on v) such that y⁢N∈S and yk=x⁢n0⁢v, and the preceding conclusion shows that in fact y⁢N∈S∩B′. To show that S∩B* is nonempty, we have to show that there exists b∈S∩B′ such that every fixed point of a on V is fixed by b. Suppose that this is not true; thus F⁢(a)≠F⁢(b) for all b∈S∩B′, in the notation as before, with respect to the action on V as above. Since the action of a on V is the same as the (conjugation) action of x⁢n0, Proposition 2.2 implies that any v∈V for which there exists an element y∈G such that y⁢N∈S∩B′ and yk=x⁢n0⁢v is contained in X⁢(σ⁢(A),V), in the notation as before, σ being the conjugation action. This contradicts the conclusion as above, since X⁢(σ⁢(A),V) is a proper subset of V. This shows that S∩B* is nonempty, and completes the proof of the theorem. ∎

Remark 4.1.

Let N=N0⊃N1⊃⋯⊃Nr={e} be the central series of N, with e the identity element. For j=0,…,r-1 let pj:Nj→Nj/Nj+1 be the quotient homomorphism. For j=0,1,…,r-1 let 𝔐𝔧 be the Ad⁢(T)-invariant subspace of the Lie subalgebra of Nj complementary to the Lie subalgebra of Nj+1, and let 𝔓𝔧 be the collection of (algebraic) one-parameter subgroups ρ corresponding to one-dimensional subspaces of 𝔐𝔧. Let 𝔓=⋃j𝔓𝔧. By considering the decomposition of the Lie algebra of N with respect to the action of T it can be seen that, for any t∈T, ZN⁢(t) is generated by the collection of one-parameter subgroups ρ∈𝔓j, j=0,…,r-1, such that any v∈pj⁢(ρ) is fixed under the action of t on Nj/Nj+1. Moreover, any one-dimensional subspace of Nj/Nj+1 which is pointwise fixed under the action of t is of the form pj⁢(ρ) for some ρ∈𝔓j centralised by t.

Proof of Corollary 1.2.

Let T be the subgroup as above. Then every semisimple element of G has a conjugate in T and hence it suffices to prove the assertion in the corollary for x in T. Let x∈T be given and let a=x⁢N∈A=G/N. We note that for any t∈T, ZG⁢(t)=T⁢ZN⁢(t). Suppose there exists y∈Z⁢(ZG⁢(x)) such that yk=x. Thus y has the form sn, s∈T, n∈ZN⁢(x), with sn, and hence n, commuting with all elements of T. Hence x=yk=sk⁢nk, which implies that sk=x and nk=e. Since k is coprime to the characteristic of 𝔽, we get that n=e and hence y=s∈T. Since any ρ∈𝔓j, 0≤j≤r-1, which is centralised by x is also centralised by y, by Remark 4.1 any v∈Nj/Nj+1, 0≤j≤r-1, is fixed under the action of y. Thus for a=x⁢N as above there exists b=y⁢N for which the condition of Theorem 1.1 is satisfied. Hence by the theorem x⁢n∈Pk⁢(G) for all n∈N.

Conversely, suppose that x⁢n∈Pk⁢(G) for all n∈N. Then by Theorem 1.1 there exists b∈G/N such that bk=a and any v∈Nj-1/Nj, j=1,…,r, which is fixed under the action of a is also fixed under the action of b. Let g∈G be such that g⁢N=b. Let g=y⁢n with y∈T and n∈N. Then as bk=a we get yk=x. Also, the actions of b and y on any Nj/Nj+1, j=0,…,r-1 coincide. Thus any v∈Nj-1/Nj, j=1,…,r which is fixed under the action of x is also fixed under the action of y. From Remark 4.1 we get that every one-parameter subgroup ρ∈𝔓j, 0≤j≤r-1, which is centralised by x is centralised by y, and in turn that ZN⁢(x)=ZN⁢(y). Since ZG⁢(x)=T⁢ZN⁢(x), we get that ZG⁢(x)=ZG⁢(y), which shows that y∈Z⁢(ZG⁢(x)). ∎

Corollary 5.1.

Let G be a connected Lie group and let N be a simply connected nilpotent closed normal subgroup of G such that G/N is abelian. Let A=G/N, x∈G and x⁢N=a∈A. Let N=N0⊃N1⊃⋯⊃Nr={e} be the central series of N. Let k∈N. Then x⁢n∈Pk⁢(G) for all n∈N if and only if there exists b∈A with bk=a such that for any j=0,…,r-1 any fixed point of a in Nj/Nj+1 is also fixed by b.

Corollary 5.2.

Let the notation be as in Corollary 5.1. Then x⁢n∈Pk⁢(G) for all n∈N if and only if there exists b∈A such that for any irreducible subquotient (M,M′), where M and M′ are closed connected normal subgroups of G contained in N, if the action of a on M/M′ is trivial, then the action of b on M/M′ is trivial.

Remark 5.3.

Let x,y∈H be such that yk=x. Then for all j=0,…,r-1, every v∈Nj/Nj+1 which is fixed under the action of x is also fixed under the action of y if and only if for every X∈L⁢(N) such that Ad⁢(x)⁢X=X we also have Ad⁢(y)⁢X=X.

Remark 5.4.

The nilradical N has a unique maximal compact subgroup C; moreover C is a connected subgroup contained in the center of G and N/C is simply connected. Consequently, for g∈G, for any k∈ℕ, g∈Pk⁢(G) if and only if g⁢C∈Pk⁢(G/C).

Proof of Corollary 1.3.

In view of Remark 5.4 we may assume N to be simply connected. Thus N is a ℝ-nilpotent group and the G-action on N is ℝ-linear. Let h∈H be given. Suppose that there exists an element g∈H such that gk=h and g∈ZH⁢(X) for all X∈L⁢(N) such that h∈ZH⁢(X). Then Theorem 1.1 together with Remark 5.3 implies that h⁢n∈Pk⁢(G) for all n∈N. Conversely, suppose that h⁢n∈Pk⁢(G) for all n∈N. Let a=h⁢N. By Theorem 1.1 there exists b∈A=G/N such that any v∈Nj/Nj+1, 0≤j≤r-1, which is fixed by a is also fixed by b. Since G=H⁢N and H is a connected nilpotent Lie group, there exists g∈H such that g⁢N=b and gk=h. By Remark 5.3 we also have g∈ZH⁢(X) for all X∈L⁢(N) such that h∈ZH⁢(X), since on each Nj/Nj+1 the action of g is the same as that of b. This proves the first assertion in the corollary.

Now suppose that Pk:G→G is surjective. Let X∈L⁢(N) be given. For any h∈ZH⁢(X) we have h⁢n∈Pk⁢(G) for all n∈N, and hence by the above there exists g∈ZH⁢(X) such that gk=h, which shows that Pk:ZH⁢(X)→ZH⁢(X) is surjective. Conversely, suppose that Pk:ZH⁢(X)→ZH⁢(X) is surjective for all X∈L⁢(N). Since G=H⁢N it suffices to show that h⁢n∈Pk⁢(G) for all h∈H and n∈N. Let h∈H be given, and let Δ′={s∈Δ∣λ⁢(h,s)=1}. For all s∈Δ′ let Xs∈L⁢(N)s be such that Xs≠0, and let X=∑s∈Δ′Xs. Then h∈ZH⁢(X) and hence by the first part there exists g∈ZH⁢(X) such that gk=h. Then g∈H and, since Ad⁢(g)⁢(X)=X, we have Ad⁢(g)⁢Xs=Xs for all Xs, for all s∈Δ′, and hence λ⁢(g,s)=1 for all s∈Δ′. This in turn implies that g∈ZH⁢(Y) for all Y∈L⁢(N) such that h∈ZH⁢(Y). The first part of the corollary proved above now implies that h⁢n∈Pk⁢(G) for all n∈N, as sought to be shown. ∎

Corollary 5.5.

Let G be a connected solvable Lie group and k∈N. For any x∈Pk⁢(G) there exists a Pk-regular element y in G such that yk=x. If Pk is surjective, then Pk:Z⁢(G)→Z⁢(G) is surjective.

Proof.

We note that in view of Remark 5.4 we may assume that N simply connected. First suppose x∈H, a Cartan subgroup. Since x⁢n∈Pk⁢(G) for all n∈N by Corollary 1.3 there exists y∈H such that yk=x and Ad⁢(y)⁢X=X for every X∈L⁢(N) such that Ad⁢(x)⁢X=X. The latter condition implies in particular that y is Pk-regular.

Now consider any x∈G, say x=h⁢u, h∈H and u∈N. Then there exists a Pk-regular element g∈H such that gk=h. It is known that for a Pk-regular element g and u∈N there exists a w∈N such that gk⁢u=(g⁢w)k (see [2, Proposition 3.5]). Thus x=h⁢u=gk⁢u=(g⁢w)k, we get y=g⁢w as a Pk-regular element such that yk=x.

Now suppose that Pk is surjective and let x∈Z⁢(G) be given. By the above assertions there exists a Pk-regular element y such that yk=x. Since Ad⁢(x) is trivial, y being a Pk-regular element with yk=x implies that Ad⁢(y) is trivial, namely y∈Z⁢(G). Hence Pk:Z⁢(G)→Z⁢(G) is surjective. ∎

Corollary 5.6.

Let G be a simply connected solvable Lie group, x∈G and k≥2. Let N be the nilradical of G. Then the following are equivalent:

1. x⁢N⊂Pk⁢(G).

2. There exists g∈G such that gk=x and

   Spec⁢(g)∩{λ∈ℂ∣|λ|=1,λ≠1}=∅.

3. Spec⁢(g)∩{λ∈ℂ∣|λ|=1,λ≠1}=∅ for all g∈G such that gk=x.

Proof.

That statement (i) implies (ii) follows immediately from Corollary 1.3 and the fact that if g=h⁢u, where h∈H, a Cartan subgroup, and u∈N, then Spec⁢(g)=Spec⁢(h); this part does not involve G being simply connected.

To prove the other two assertions, we first note the following. Let A=G/N and a=x⁢N. Since G is simply connected, it follows that A is a vector space. Hence a has a unique kth root b in A. Thus for any g∈G such that gk=x we have g⁢N=b. This firstly shows that (ii) implies (iii), since Spec⁢(g) is determined by b. Finally, suppose (iii) holds and let g∈G be such that g⁢N=b, the unique kth root of a in A. If there exists v∈Nj/Nj+1, where N=N0⊃⋯⊃Nr={e} is the central series of N, which is fixed by the action of a but not by that of b, then Ad⁢(g) would have an eigenvalue λ such that λ≠1 and λk=1. Since this is ruled out, by Theorem 1.1 we get that x⁢N∈Pk⁢(G), thus proving (i). ∎

Lemma 6.1.

Let k∈N. Let U be a one-parameter subgroup of S such that ZS⁢(U) does not contain any element whose order divides k. Let u be a nontrivial element in U and v∈ZS⁢(U) be such that vk=u. Then v∈U.

Proof.

As U is a one-parameter subgroup, there exists w∈U such that wk=u. Since v∈ZS⁢(U), it follows that

the identity element. Thus v⁢w-1∈ZS⁢(u) and its order divides k. Hence by hypothesis v⁢w-1=e, and so v=w∈U. ∎

Proposition 6.2.

Let G be a connected Lie group, let R be the radical of G, and let S=G/R. Let k∈N be such that Pk:G→G is surjective. Let U be a unipotent one-parameter subgroup of S such that ZS⁢(U) does not contain any nontrivial element whose order divides k. Let H be the closed subgroup of G containing R such that H/R=U. Let h∈H be an element not contained in R. Then h∈Pk⁢(H).

Proof.

As Pk is surjective, there exists g∈G such that gk=h. Let v=g⁢R∈S and u=h⁢R∈U. Since v commutes with u and U is a unipotent one-parameter subgroup, it follows that v commutes with all elements of U, viz. v∈ZS⁢(U). As h∉R, u is a nontrivial element of U. Also, since gk=h, we have vk=u, and hence by Lemma 6.1 we get v∈U. Therefore g∈H and in turn h∈Pk⁢(H). ∎

Proof of Corollary 1.5.

Let x∈R be given. Let H be as above. We note that since H/R is isomorphic to ℝ, to prove that x∈Pk⁢(R) it suffices to prove that x∈Pk⁢(H). Also, since N is the nilradical of G, it follows that G/N is reductive, and hence H/N is abelian. Now let {ut} be the one-parameter subgroup of G such that U={ut⁢R}. Let u=u1 and y=u⁢x∈H. Let A=H/N and a=x⁢N, a′=y⁢N∈A. We note that by Proposition 6.2, y⁢n∈Pk⁢(H) for all n∈N. Hence by Corollary 5.2 there exists b′∈A such that b′⁣k=a′ and if the a′-action on an irreducible subquotient (M,M′) of N, with respect to the H-action (see Section 4 for definition) is trivial, then the b′-action is also trivial. Let b=(u1k⁢N)-1⁢b′∈A; then we have bk=a. We note that the action of U on the Lie algebra of N is unipotent and by the irreducibility condition this implies that for any t the action of ut⁢N on M/M′ is trivial. Hence the action of a on M/M′ coincides with the action of a′, and similarly the action of b coincides with that of b′. Thus we see that bk=a and if the action of a on an irreducible subquotient M/M′ is trivial then so is the action of b. Hence by Corollary 5.2x⁢n∈Pk⁢(H) for all n. Thus x∈Pk⁢(H) and as noted above it follows that x∈Pk⁢(R). ∎