MATHEMATICS

V. I. USHAKOV

ON A CLASS OF TOPOLOGICAL GROUPS

(Presented by Academician P. S. Aleksandrov on 22 XII 1961)

In abstract group theory, the so-called finiteness conditions are very popular. In particular, many authors (see, for example, \((^{3,4})\)) have studied groups with finite classes of conjugate elements (\(FC\)-groups). In the present note we consider topological \(\overline{FC}\)-groups (groups with bicompact classes of conjugate elements) and \(\overline{\overline{FC}}\)-groups, in which the closures of all classes of conjugate elements are bicompact. The necessity of considering \(\overline{\overline{FC}}\)-groups alongside \(\overline{FC}\)-groups follows from the fact that in a topological group the classes of conjugate elements are not always closed. All bicompact and all commutative groups belong to the \(FC\)-groups. The class of \(\overline{FC}\)-groups contains locally normal groups, i.e., groups in which every element is contained in a bicompact normal divisor. However, locally normal groups need not be \(FC\)-groups.

We begin with the consideration of connected simply connected \(\overline{FC}\)-groups of Lie.

Lemma. Let \(V\) be an \(n\)-dimensional complex space, \(\Delta \subset E(V)\) an irreducible set of linear transformations of this space; \(A \in E(V)\) a linear transformation of the space \(V\) with eigenvalues \(\lambda_1,\ldots,\lambda_n\). If for every \(B \in \Delta\) there exists a bicompact set \(K_B \subset E(V)\) such that \(e^{kA} B e^{-kA} \in K_B\) for every integer \(k\), then \(\operatorname{Re}\lambda_1=\cdots=\operatorname{Re}\lambda_n\).

Proof. Let \(e_1,\ldots,e_n\) be a basis in which the matrix of the transformation \(A\) has Jordan normal form, \(A'e_i=\lambda_i e_i\), \(i=1,\ldots,n\), \(\widetilde A=A-A'\). Then \(\widetilde A^m=0\) for some \(m\). Since \(A'\) and \(\widetilde A\) commute, we have

\[ e^{kA}Be^{-kA}=e^{kA'}B_k e^{-kA'},\qquad B_k=\sum_{0\le i+j<2m}\frac{(-1)^j k^{i+j}}{i!j!}\,\widetilde A^{\,i}B\widetilde A^{\,j}. \tag{1} \]

Suppose it has already been proved that \(\operatorname{Re}\lambda_1=\cdots=\operatorname{Re}\lambda_r\), \(n>r\ge 1\), and let \(V_r\) be the linear span of the vectors \(e_1,\ldots,e_r\). In view of the irreducibility of \(\Delta\), one may assume that

\[ B(e_s)=\sum_{i=1}^{n} b_i e_i,\qquad b_{r+1}\ne 0 \tag{2} \]

for some \(B\in\Delta\) and \(1\le s\le r\). Further, by virtue of (1) and (2),

\[ e^{kA}Be^{-kA}(e_s)=\sum_{i=1}^{n} b_i(k)e^{(\lambda_i-\lambda_s)k}e_i\in K_B(e_s),\qquad b_i(0)=b_i, \tag{3} \]

where \(b_i(k)\), \(i=1,\ldots,n\), is a polynomial in \(k\) of degree \(\le 2m\). Since the set \(K_B(e_s)\) is bicompact, all coefficients in (3) are bounded, i.e.,

\[ |b_i(k)|e^{\operatorname{Re}(\lambda_i-\lambda_s)k}<c,\qquad 1\le i\le n, \]

for every integer \(k\). Since \(b_{r+1}(k)\) is not identically zero \((b_{r+1}(0)=b_{r+1}\ne 0)\), \(\operatorname{Re}\lambda_{r+1}=\operatorname{Re}\lambda_s\) ( \(b_{r+1}(k)\) will in this case be a constant different from zero).

Theorem 1. A connected semisimple \(\overline{FC}\)-Lie group \(G\) is bicompact.

Proof. Let \(V\) be the Lie algebra of the Lie group \(G\), \(Z\) the center of \(G\), and \(\widetilde G(V)\) the connected component of the group of automorphisms of the algebra \(V\). Then the groups \(G/Z\) and \(\widetilde G(V)\) are isomorphic. Further, \(V\) decomposes into a direct sum of simple algebras \(V_1,\ldots,V_n\). Since \(\widetilde G(V)\) is the direct product of the groups \(\widetilde G(V_i)\), \(i=1,\ldots,n\), it is enough to prove that if \(V\) is a simple Lie algebra and \(\widetilde G(V)\) is an \(\overline{FC}\)-group, then \(V\) is compact.

It is known that if \(V\) is a simple real Lie algebra, then either its complex extension \([V]\) \((^2),\ \S 58,\ C\) is a simple algebra, or \(V\) is obtained from some simple complex algebra \(W\) by depriving the latter of complex operators.

In the first case \(\widetilde G(V)\) can be regarded as an irreducible set of linear transformations of the space \([V]\), for \(V\) and \([V]\) have one and the same basis. Since the Lie algebra of the group \(\widetilde G(V)\) is the adjoint algebra \(P\) of the algebra \(V\) \((^2),\ \S 54,\ E\), by the lemma the real parts of all eigenvalues of any endomorphism \(p_v\) from \(P\) \((v\in V)\) are equal to zero. Then the eigenvalues of the transformation \(p_v^2\) are nonpositive. Hence, and from the simplicity of \(V\), it follows that the scalar square \(\operatorname{Sp} p_v^2\) is positive definite, i.e. \(V\) is compact.

In the second case it is easy to show that \(\widetilde G(W)=\widetilde G(V)\) and \(\widetilde G(W)\) is irreducible in \(W\). Let \(s\) be a regular element of the compact form \(W'\) of the algebra \(W\), and let \(A=ip_s\). Since the eigenvalues of \(A\) are real \((^2),\ \S 62\, A\), by the lemma they are all equal to zero. Hence follows the coincidence of \(W'\) with its regular subalgebra, which is commutative, and this contradicts the simplicity of \(W\). Thus the second case is in fact impossible.

Definition. An element \(g\) of a topological group \(G\) is called bicompact if the closure of the cyclic subgroup generated by \(g\) is bicompact. If all elements of the group are bicompact, then the group is called periodic; if \(G\) has no bicompact elements except the identity, then the group is called pure.

Theorem 2. A pure locally bicompact normal divisor \(H\) of an arbitrary \(\overline{FC}\)-group \(G\) is central.

Proof. Let \(K\) be the connected component of \(H\), and \(R\) its solvable radical. In \(R\) there is a series of subgroups invariant in \(G\), all factors of which are vector groups or tori. The factor group \(K/R\) is bicompact by Theorem 1. From Lemma 3.8 of \((^6)\) it follows that \(R=K\) is a vector group. Since \(H/K\) is a discrete abelian torsion-free group (see \((^5)\)), \(H\) is solvable. If \(H\) is commutative, then for any \(h\in H\) and \(g\in G\)

\[ h^kgh^{-k}g^{-1}=(hgh^{-1}g^{-1})^k, \]

i.e. the subgroup \(\{hgh^{-1}g^{-1}\}\) of the group \(H\), lying in \(S_g\cdot g^{-1}\), is bicompact (\(S_g\) is the class of elements of the group \(G\) conjugate to \(g\)). From the purity of \(H\) it follows that \(hg=gh\). The centrality of an arbitrary pure solvable normal divisor \(H\) is proved by induction on the class of solvability of the group \(H\).

With the aid of Theorem 2 and Lemma 4 of \((^7)\), the nilpotency of a connected solvable \(\overline{FC}\)-group is established. And then, according to the results of V. M. Glushkov \(((^8),\) Theorem 5.1), Lemma 3.8 of \((^6)\), and Theorem 1, we obtain:

Theorem 3. A connected locally bicompact \(\overline{FC}\)-group is an extension of a bicompact group by means of a vector group.

Theorem 4. In a locally bicompact group, every bicompact invariant set consisting of bicompact elements generates, in the topological sense, a bicompact normal divisor.

The proof of this theorem, which generalizes the well-known lemma of Dixmier (see (¹), p. 339), uses Theorem 3 and the lemma from (⁵).

A trivial consequence of Theorem 4 is

Theorem 5. A locally bicompact group is a periodic \(\overline{FC}\)-group if and only if it is locally normal.

The main result of the present paper is the following.

Theorem 6. The set of all bicompact elements of a locally bicompact \(\overline{FC}\)-group \(G\) forms a closed characteristic subgroup \(P(G)\) (the periodic part of \(G\)), the quotient group by which is a torsion-free Abelian group.

Proof. The fact that \(P(G)=P\) is a subgroup of the group \(G\) in the algebraic sense follows from Theorem 4.

We shall prove the closedness of \(P\). Let \(K\) be the connected component of the group \(G\), and let \(P'/K\) be the periodic part of the quotient group \(G/K\). According to (⁵), \(P'\) is an open subgroup of the group \(G\). It is not difficult to prove that \(P'=PK\). Since \(\overline P\) is a locally bicompact group, it contains a neighborhood of the identity \(U\) with bicompact closure. Consider the subgroup
\(A=\{\overline U\}\) of the group \(\overline P\). It is open in \(\overline P\). Using the existence in \(A\) of the bicompact set of generators \(\overline U\) and the everywhere dense set of bicompact elements \(A\cap P\), one can show that \(A\) is bicompact and, consequently, is contained in \(P\). Hence it follows that \(P=\overline P\).

We shall prove the torsion-freeness of the quotient group \(G/P\). The commutativity of \(G/P\) will then follow from Theorem 2. Suppose that there is a bicompact element in \(G/P\). Then it is contained in \(P'\), since, by (⁵), \(G/P'\) is a torsion-free group. Therefore it remains to prove the torsion-freeness of the group \(PK/P\). Consider the quotient group
\[ KP/P(K)=P'/P(K)\cong \widetilde P', \]
where \(P(K)\) is the periodic part of \(K\) (see Theorem 3). Since
\(\widetilde P'=\widetilde P\cdot\widetilde K\) and \(\widetilde P\cap\widetilde K=1\), the mapping
\[ \varphi:\quad \widetilde K\ni \widetilde k=kP(K)\to \widetilde k\widetilde P\in \widetilde P'/\widetilde P \]
is a continuous isomorphism (in the algebraic sense) of the group \(\widetilde K\) onto the whole group \(\widetilde P'/\widetilde P\). By the connectedness and local bicompactness of \(\widetilde K\), \(\varphi\) is a homeomorphism ((²), Theorem 12). Consequently,
\(\widetilde P'/\widetilde P\cong P'/P\) is a torsion-free group.

In the case of discrete groups this theorem becomes the well-known theorem of B. H. Neumann (⁴).

Definition. A topological group \(G\) is called bicompactly generated if it possesses a bicompact system of elements algebraically generating the whole group.

Theorem 7. A locally bicompact group \(G\) is a bicompactly generated \(\overline{FC}\)-group if and only if it is an extension of a bicompact group by means of the direct product of a vector group and a discrete Abelian group without torsion with a finite number of generators.

Proof. It is clear that in the proof only the bicompactness of the periodic part \(P\) of the bicompactly generated \(\overline{FC}\)-group \(G\) is needed. If \(G\) is totally disconnected, then it contains an open bicompact subgroup \(H\), and
\(G=\{H,x_1,\ldots,x_n\}\), where \(x_1,\ldots,x_n\) belong to a bicompact set of generators of the group \(G\). Since \(P\) contains the commutator subgroup \(G'\) of the group \(G\), it is enough to prove the bicompact generation of \(G'\), whence the bicompactness of the closure of the commutator subgroup, and hence the bicompactness of \(P\), will follow. A bicompact set of generators of the group

\(G'\) consists of elements of the following form: 1) \([h,h']\), where \(h,h' \in H\); 2) \([s_i^\varepsilon,h]\), where \(h \in H\), \(s_i \in \overline{S}_{x_i}\), \(i=1,\ldots,n\); \(\varepsilon=\pm 1\); 3) \([s_i^\varepsilon,s_j^{\varepsilon'}]\), \(s_i \in \overline{S}_{x_i}\), \(s_j \in \overline{S}_{x_j}\); \(i,j=1,\ldots,n\); \(\varepsilon,\varepsilon'=\pm 1\) (here \([x,y]\) denotes the commutator of the elements \(x\) and \(y\)).

If \(G\) is an arbitrary locally bicompact group, then consider its connected component \(K\). The factor group \(KP/K\) is bicompact by what has been proved. From the bicompact generatedness of the group \(PK\) it follows that \(P/K \cap P \cong KP/K\) is a bicompact group (\((^2)\), § 20, \(G\)). The bicompactness of \(P\) now follows from the bicompactness of \(K \cap P\).

Corollary 1. A locally bicompact, bicompactly generated, locally normal group is bicompact.

Corollary 2. A locally bicompact, bicompactly generated group is an \(\overline{FC}\)-group if and only if the closure of its commutant is bicompact.

In conclusion I take this opportunity to express my deep gratitude to L. Ivanovskii for valuable advice.

REFERENCES

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