In this section, we investigate the nilradical of a connected non-compact Lie group G, provided with an ad-invariant symmetric bilinear formκon its Lie algebragthat fulfills condition () as described in Theorem 1.
Lemma 3.4. Letp⊆gbe an abelian subalgebra containing an element X ∈pgenerating a non-precompact one-parameter group. Then the set of elements in pgenerating a non-precompact one-parameter group is dense in p. κ|p×p is positive semidefinite and the dimension of its kernel is at most one.
Proof. The closure of the subgroup generated by pis a connected abelian subgroup and by Lemma 1.9 isomorphic toTm×Rm. m = 0 sinceXgenerates a non-precompact one-parameter group. Let p∗ denote the Lie algebra of the closure andt⊂p∗ the subalgebra corresponding to the torus factor. All elements of p∗ that are not in t generate a non-precompact one-parameter group. By definition, p is not contained in t, hence, the set of elements in p generating a non-precompact one-parameter group is dense in p. The remainder of the assertion now follows from condition ().
3.2. Nilradical 36
Lemma 3.5. Let N be a connected non-compact nilpotent Lie group and n its Lie algebra. Then the set of X ∈ n generating a non-precompact one-parameter group is dense in n.
Proof. If N is abelian, N is isomorphic to Tm ×Rm by Lemma 1.9. m = 0 because N is not compact, it follows that the set of elements inn generating a non-precompact one-parameter group is dense in n.
Now assume that N is not abelian and the statement of the lemma is false. Using Lemma 3.3 and the nilpotency ofn, it follows that adX is trivial for all X of some open ball in n. Thus, z(n) contains elements of an open ball in n and therefore has the same dimension asn. Hence, z(n) =n, that is, N is abelian, contradiction.
The proof of the last lemma also shows the following:
Corollary 3.6. A connected compact nilpotent Lie group is abelian.
In what follows, denote by N the nilradical of G, that is, the largest (with respect to inclusion) connected normal Lie subgroup of G, which is nilpotent. By maximality, N is closed inG. Its Lie algebra n is the largest (with respect to inclusion) nilpotent ideal of g.
According to Corollary 3.6,N is abelian if it is compact. In the remainder of this section, we assume thatN is not compact.
It follows from Lemma 3.5 and condition (), that the restriction ofκton×n is positive semidefinite and its kernel i is an ideal of dimension at most one.
Proposition 3.7. Let κ be any ad-invariant positive semidefinite symmetric bilinear form on n with kernel i having dimension at most one.
Then n =a⊕h is the κ-orthogonal direct sum of an abelian algebra a (which might be trivial) and an subalgebra h being either one-dimensional or isomorphic to a Heisenberg algebra hed of dimension 2d+ 1. h is one-dimensional if and only if n is abelian.
The adjoint representation of G on n/i has precompact image. If κ is not positive definite, then i=RZ, where RZ is the center of h.
Proof. n/iis nilpotent and possesses an ad-invariant positive definite symmetric bilinear form. By Propositions 1.4 and 1.5, n/i is abelian, so [n,n]⊆i. If i={0}, n is abelian.
In the abelian case, we choose RZ to be the kernel ofκ if it has dimension one and ifκ is positive semidefinite, we choose an arbitrary non-zero Z ∈n. Then n=a⊕h, where h =RZ and a is a vector space complement to h, which is orthogonal to Z in the case that κ is positive definite.
Supposei=RZ is one-dimensional. ThenZ ∈z(n), sincenis nilpotent and [n,n]⊆i. It follows that in any case, n is at most two-step nilpotent (this actually shows the second part of Theorem B in [Zim86]). z(n) is abelian, so z(n) = a⊕RZ with some abelian summand a. If n=z(n), n is abelian and we are done. So suppose n is not abelian.
Let h := z(n)⊥ (κ-orthogonal complement in n). Because κ is ad-invariant, h is an subalgebra. Moreover, n = a⊕ h is an orthogonal and direct sum by construction.
Because n is not abelian, dimh≥2 and so
[h,h] = [n,n] =RZ as well asz(h) =z(n)∩h=RZ.
If V is a complementary vector space of RZ in h, then ω : V × V → R defined by [X, Y] =ω(X, Y)Z is an alternating bilinear form, which is non-degenerate (an element of the kernel has to lie in the center ofh). Thus,h is isomorphic to a Heisenberg algebra.
Finally, the image of the adjoint representation of G on n/i is precompact, since it preserves a positive definite scalar product and therefore acts by orthogonal transforma-tions.
Remark. Note that the abelian summand a is in general not canonically defined, but a⊕RZ =z(n) and h =z(n)⊥ are.
Proposition 3.8. Suppose n is not abelian.
(i) The center i=RZ of h∼=hed is central in g.
(ii) All non-central X ∈h ⊆n generate non-precompact one-parameter groups.
(iii) If Y ∈g,Y /∈RZ, commutes with a non-central elementX of n, thenκ(Y, Y)>0.
Proof. (i) h∼=hed by Proposition 3.7. We will show that the adjoint action ofG oni is trivial. Let X, Y ∈ h such that [X, Y] = Z. By Proposition 3.7, the adjoint action of
3.2. Nilradical 38
G on n/i has precompact image. Therefore, we may choose a compact set K ⊂ n such that Adf(X)∈K+i and Adf(Y)∈K+i for all f ∈G. It follows that
Adf(Z) = Adf([X, Y]) = [Adf(X),Adf(Y)]∈[K+i, K+i]⊆[K, K].
It follows that the adjoint action of G on i has precompact image. But i is one-dimensional and Gis connected, so Ad(G) acts trivially on i.
(ii) This follows from Lemma 3.3 and the fact that adX is nilpotent, but not trivial.
(iii) p := span{X, Y, Z} is an abelian subalgebra of dimension at least 2, and by (ii), X generates a non-precompact one-parameter group. By Lemma 3.4, the restriction of κ to p×p is positive semidefinite and its kernel has dimension at most one. But the kernel is already given byRZ, hence,κ(Y, Y)>0.
Proposition 3.9. Let l ⊂ g be a semisimple subalgebra. Then g = l+n is in fact a κ-orthogonal direct sum.
Proof. l∩n is a nilpotent ideal in l. Sincel is semisimple, l∩n={0}.
As above, let i be the kernel of the restriction of κ to n×n. Due to Lemma 3.2, i is an ideal inn.
LetX ∈l, Y ∈n, Z ∈i. Then
κ([Z, X], Y) = κ(Z,[X, Y]) = 0 since [l,n]⊆n. Thus, [l,i]⊆i, that is, i is an ideal in g.
The setm of Y ∈l such that [Y,i] ={0} is an ideal of l: Indeed, let X, Y ∈l such that [Y,i] ={0}. Since [l,i]⊆i,
0 = [X,[Y, Z]] + [Y,[Z, X]] + [Z,[X, Y]] = [Z,[X, Y]]
for Z ∈i by the Jacobi identity.
mhas codimension at most one, becauseihas dimension at most one. Butlis semisimple, thus, it contains no ideals of codimension exactly one. Hence, l centralizesi.
Since lis semisimple, l= [l,l] and by Lemma 3.1, l and thereforeg are κ-orthogonal to i. So we may pass to the quotient g/i=l+n/i or equivalently, we can assume without
loss of generality that κ is positive definite on n. Note that because of Proposition 3.7, n is abelian in this case.
We want to show [l,n] = {0}. By Proposition 3.7, the adjoint action of G on n has precompact image. Thus, we may assume that l is compact. Treating each simple summand of l separately, we also can assume l to be simple.
In the following, we construct an ad-invariant positive definite scalar product on g. Let k be the Killing form of g. If X ∈ n, Y ∈ g, then ad2X(Y) = 0 because n is an abelian ideal. Thus, k restricted to g×n is identically zero. Let X ∈l, Y ∈n. If W ∈n,
adX(adY(W)) = 0, and if W ∈l,
adX(adY(W))∈n.
Thus, k(X, Y) = Tr(adX ◦adY) = 0.
By Lemma 3.2 (ii), k restricted to l×l is a multiple of the negative definite Killing form of l. It is a non-zero multiple, because otherwise k would vanish on the whole of g ×g and g would be solvable (cf. [OV94], Chapter 1, corollary to Theorem 2.1), contradiction.
−k is a positive definite symmetric bilinear form onl, so we can find a (−k)-orthonormal basis which isκ-orthogonal (principal axis transformation). Thus, there is at0 ∈R, such that κ−tk is positive definite on l for all t≥t0.
Obviously, the restriction of κ−tk ton×n is equal to the restriction ofκ and therefore positive definite. We want to find at ≥t0 such that κ−tk is positive definite ong×g. It suffices to obtain
(κ−tk)(λX+Y, λX +Y)>0
for all non-zero λ ∈ R, X ∈ l, Y ∈n. By linearity, we may suppose k(X, X) = −1 and κ(Y, Y) = 1. Then
(κ−tk)(λX+Y, λX +Y)
=λ2κ(X, X) +tλ2+ 2λκ(X, Y) + 1
≥λ2(t+ min
X∈l,k(X,X)=−1κ(X, X))−2λ max
X∈l,k(X,X)=−1 max
Y∈n,κ(Y,Y)=1|κ(X, Y)|+ 1.