Of basic importantance in the study of nilmanifolds is equidistribution: one needs to know when the sequence $g^n x$ equidistributes and when it is trapped inside a subnilmanifold. It turns out that this problem was already studied by Leon Green in the 60s. To describe the theorem note first that the abelianisation map $G\to G/G_2$ induces a map from $G/\Gamma$ to a torus $G/(G_2\Gamma)$ which respects the action of $G$, and recall that equidistribution on tori is well understood by Weyl's criterion. Leon Green's beautiful theorem then states that $g^n x$ equidistributes in the nilmanifold if and only if its image in the torus $G/(G_2\Gamma)$ equidistributes.

Today at our miniseminar Aled Walker showed us Parry's nice proof of this theorem, which is more elementary than Green's original proof. During the talk there was some discussion about the importantance of various hypotheses such as 'simply connected' and 'Lie'. It turns out that the proof works rather generally for connected locally compact nilpotent groups, so I thought I would record the proof here with minimal hypotheses. The meat of the argument is exactly as in Aled's talk and, presumably, Parry's paper.

Let $G$ be an arbitrary locally compact connected nilpotent group, say with lower central series$$G=G_1\geq G_2 \geq\cdots\geq G_s\geq G_{s+1}=1,$$and let $\Gamma\leq G$ be a closed cocompact subgroup. Under these conditions the Haar measure $\mu_G$ of $G$ induces a $G$-invariant probability measure $\mu_{G/\Gamma}$ on $G/\Gamma$. We say that $x_n\in G/\Gamma$ is

*equidistributed*if for every $f\in C(G/\Gamma)$ we have$$\frac{1}{N}\sum_{n=0}^{N-1} f(x_n) \to \int f \,d\mu_{G/\Gamma}.$$We fix our attention on the sequence$$x_n = g^n x$$for some $g\in G$ and $x\in G/\Gamma$. As before we have an abelianisation map$$\pi: G/\Gamma\to G/G_2\Gamma$$from the $G$-space $G/\Gamma$ to the compact abelian group $G/G_2\Gamma$. We define equidistribution on $G/G_2\Gamma$ similarly. The theorem is then the following.

Theorem: (Leon Green) For $g\in G$ and $x\in G/\Gamma$ the following are equivalent.

1. The sequence $g^n x$ is equidistributed in $G/\Gamma$.

2. The sequence $\pi(g^n x)$ is equidistributed in $G/G_2\Gamma$.

3. The orbit of $\pi(g)$ is dense in $G/G_2\Gamma$.

4. $\chi(\pi(g))\neq 0$ for every nontrivial character $\chi:G/G_2\Gamma\to\mathbf{R}/\mathbf{Z}$.

Item 1 above trivially implies every other item. The implication 4$\implies$3 (a generalised Kronecker theorem) follows by pulling back any nontrivial character of $(G/G_2\Gamma)/\overline{\langle\pi(g)\rangle}$. The implication 3$\implies$2 (a generalised Weyl theorem) follows from the observation that every weak* limit point of the sequence of measures$$ \frac{1}{N}\sum_{n=0}^{N-1} \delta_{\pi(g^n x)}$$must be shift-invariant and thus equal to the Haar measure. So the interesting content of the theorem is 2$\implies$1.

A word about the relation to ergodicity: By the ergodic theorem the left shift $\tau_g:x\mapsto gx$ is ergodic if and only if for almost every $x$ the sequence $g^n x$ equidistributes; on the other hand $\tau_g$ is

Now let $\mu'$ be any $\tau_g$-invariant ergodic measure. By induction we may assume that $\tau_g:G/G_s\Gamma\to G/G_s\Gamma$ is uniquely ergodic, so we must have $p_*\mu' = p_*\mu_{G/\Gamma}$, so$$\mu'(A) = p_*\mu'(p(A)) = p_*\mu_{G/\Gamma}(p(A)) = \mu_{G/\Gamma}(A) = 1.$$But by the ergodic theorem the set of $\mu'$-generic points must also have $\mu'$-measure $1$, so there must be some point which is both $\mu_{G/\Gamma}$- and $\mu'$-generic, and this implies that $\mu'=\mu_{G/\Gamma}$.$\square$

We need one more preliminary lemma about topological groups before we really get started on the proof.

Thus if $G$ is connected then every term $G_1,G_2,G_3,\dots$ in the lower central series of $G$ is connected.

Proof of Leon Green's theorem: As noted it suffices to prove that $\tau_g$ acts ergodically on $G/\Gamma$ whenever it acts ergodically on $G/G_2\Gamma$. By induction we may assume that $\tau_g$ acts ergodically on $G/G_s\Gamma$. So suppose that $f\in L^2(G/\Gamma)$ is $\tau_g$-invariant. By decomposing $L^2(G/\Gamma)$ as a $\overline{G_s\Gamma}/\Gamma$-space we may assume that $f$ obeys$$f(cx)=\gamma(c)f(x)\quad(c\in G_s, x\in G/\Gamma)$$for some character $\gamma:G_s\Gamma/\Gamma\to S^1$. In particular $|f|$ is both $G_s$-invariant and $\tau_g$-invariant, so it factors through a $\tau_g$-invariant function $G/G_s\Gamma\to\mathbf{R}$, so it must be constant, say $1$. Moreover for every $b\in G_{s-1}$ the function$$\Delta_bf(x) = f(bx)\overline{f(x)}$$is $G_s$-invariant, and also a $\tau_g$ eigenvector:$$\Delta_bf(gx) = \gamma([b,g])\Delta_bf(x).$$By integrating this equation we find that either $\gamma([b,g])=1$, so $\Delta_bf$ is constant, or $\int\Delta_bf \,d\mu_{G/\Gamma}= 0$, so either way we have$$\int \Delta_bf\,d\mu_{G/\Gamma}\in \{0\}\cup S^1.$$But since $\int\Delta_bf\,d\mu_{G/\Gamma}$ is a continuous function of $b$ and equal to $1$ when $b=1$ we must have $\gamma([b,g])=1$ for all sufficiently small $b$, and thus for all $b$ by connectedness of $G_{s-1}$ and the identity$$[b_1b_2,g]=[b_1,g][b_2,g].$$Thus setting $\gamma(b)=\Delta_bf$ extends $\gamma$ to a homomorphism $G_{s-1}\to S^1$. In fact we can extend $\gamma$ still further to a function $G\to D_1$, where $D_1$ is the unit disc in $\mathbf{C}$, by setting$$\gamma(a) = \int \Delta_af\,d\mu_{G/\Gamma}.$$Now if $a\in G$ and $b\in G_{s-1}$ then$$\gamma(ba) = \int f(bax) \overline{f(x)}\,d\mu_{G/\Gamma} = \int \gamma(b)f(ax)\overline{f(x)}\,d\mu_{G/\Gamma}=\gamma(b)\gamma(a),$$and$$\gamma(ab) = \int f(abx)\overline{f(x)}\,d\mu_{G/\Gamma} = \int f(ax) \overline{f(b^{-1}x)}\,d\mu_{G/\Gamma} = \int f(ax) \overline{\gamma(b^{-1})}\overline{f(x)}\,d\mu_{G/\Gamma} = \gamma(b)\gamma(a),$$so$$\gamma(b)\gamma(a)=\gamma(ab)=\gamma(ba[a,b]) = \gamma(ba)\gamma([a,b])=\gamma(b)\gamma(a)\gamma([a,b]).$$Since $|\gamma(b)|=1$ we can cancel $\gamma(b)$, so$$\gamma(a)(\gamma([a,b])-1) = 0.$$Finally observe that $\gamma(a)$ is a continuous function of $a$, and $\gamma(1)=1$, so we must have $\gamma([a,b])=1$ for all sufficiently small $a$, and thus by connectedness of $G$ and the identity$$[a_1a_2,b]=[a_1,b][a_2,b]$$we must have $\gamma([a,b])=1$ identically. But this implies that $\gamma$ vanishes on all $s$-term commutators and thus on all of $G_s$, so in fact $f$ factors through $G/G_s\Gamma$, so it must be constant.$\square$

A remark is in order about the possibility that some of the groups $G_i$ and $G_i\Gamma$ are not closed. This should not matter. One could either read the above proof as it is written, noting carefully that I never said groups should be Hausdorff, or, what's similar, instead modify it so that whenever you quotient by a group $H$ you instead quotient by the group $\overline{H}$.

Embarrassingly, it's difficult to come up with a non-Lie group to which this generalised Leon Green's theorem applies. It seems that many natural candidates have the property that $G$ is not connected but $G/\Gamma$ is: for example consider$$\left(\begin{array}{ccc}1&\mathbf{R}\times\mathbf{Q}_2&\mathbf{R}\times\mathbf{Q}_2\\0&1&\mathbf{R}\times\mathbf{Q}_2\\0&0&1\end{array}\right)/\left(\begin{array}{ccc}1&\mathbf{Z}[1/2]&\mathbf{Z}[1/2]\\0&1&\mathbf{Z}[1/2]\\0&0&1\end{array}\right).$$So it would be interesting to know whether the theorem extends to such a case. Or perhaps there are no interesting non-Lie groups for this theorem, which would be a bit of a let down.

A word about the relation to ergodicity: By the ergodic theorem the left shift $\tau_g:x\mapsto gx$ is ergodic if and only if for almost every $x$ the sequence $g^n x$ equidistributes; on the other hand $\tau_g$ is

*uniquely ergodic,*i.e., the only $\tau_g$-invariant measure is the given one, if and only if for*every*$x$ the sequence $g^n x$ equidistributes. Thus to prove the theorem above we must not only prove that $\tau_g$ is ergodic but that it is uniquely ergodic. Fortunately one can prove these two properties are equivalent in this case.Lemma: If $\tau_g:G/\Gamma\to G/\Gamma$ is ergodic then it's uniquely ergodic.Proof: (Furstenberg) By the ergodic theorem the set $A$ of $\mu_{G/\Gamma}$-generic points, in other words points $x$ for which$$\frac{1}{N}\sum_{n=0}^{N-1} f(g^n x) \to \int f \,d\mu_{G/\Gamma}$$for every $f\in C(G/\Gamma)$, has $\mu_{G/\Gamma}$-measure $1$, and clearly if $x\in A$ and $c\in G_s$ then $xc\in A$, so $A = p^{-1}(p(A))$, where $p$ is the projection of $G/\Gamma$ onto $G/G_s\Gamma$.

Now let $\mu'$ be any $\tau_g$-invariant ergodic measure. By induction we may assume that $\tau_g:G/G_s\Gamma\to G/G_s\Gamma$ is uniquely ergodic, so we must have $p_*\mu' = p_*\mu_{G/\Gamma}$, so$$\mu'(A) = p_*\mu'(p(A)) = p_*\mu_{G/\Gamma}(p(A)) = \mu_{G/\Gamma}(A) = 1.$$But by the ergodic theorem the set of $\mu'$-generic points must also have $\mu'$-measure $1$, so there must be some point which is both $\mu_{G/\Gamma}$- and $\mu'$-generic, and this implies that $\mu'=\mu_{G/\Gamma}$.$\square$

We need one more preliminary lemma about topological groups before we really get started on the proof.

Lemma: If $H$ and $K$ are connected subgroups of some ambient topological group then $[H,K]$ is also connected.Proof: Since $(h,k)\mapsto [h,k]=h^{-1}k^{-1}hk$ is continuous certainly $C = \{[h,k]:h\in H,k\in K\}$ is connected, so $C^n = CC\cdots C$ is also connected, so because $1\in C^n$ for all $n$ we see that $[H,K]=\bigcup_{n=1}^\infty C^n$ is connected.$\square$

Thus if $G$ is connected then every term $G_1,G_2,G_3,\dots$ in the lower central series of $G$ is connected.

Proof of Leon Green's theorem: As noted it suffices to prove that $\tau_g$ acts ergodically on $G/\Gamma$ whenever it acts ergodically on $G/G_2\Gamma$. By induction we may assume that $\tau_g$ acts ergodically on $G/G_s\Gamma$. So suppose that $f\in L^2(G/\Gamma)$ is $\tau_g$-invariant. By decomposing $L^2(G/\Gamma)$ as a $\overline{G_s\Gamma}/\Gamma$-space we may assume that $f$ obeys$$f(cx)=\gamma(c)f(x)\quad(c\in G_s, x\in G/\Gamma)$$for some character $\gamma:G_s\Gamma/\Gamma\to S^1$. In particular $|f|$ is both $G_s$-invariant and $\tau_g$-invariant, so it factors through a $\tau_g$-invariant function $G/G_s\Gamma\to\mathbf{R}$, so it must be constant, say $1$. Moreover for every $b\in G_{s-1}$ the function$$\Delta_bf(x) = f(bx)\overline{f(x)}$$is $G_s$-invariant, and also a $\tau_g$ eigenvector:$$\Delta_bf(gx) = \gamma([b,g])\Delta_bf(x).$$By integrating this equation we find that either $\gamma([b,g])=1$, so $\Delta_bf$ is constant, or $\int\Delta_bf \,d\mu_{G/\Gamma}= 0$, so either way we have$$\int \Delta_bf\,d\mu_{G/\Gamma}\in \{0\}\cup S^1.$$But since $\int\Delta_bf\,d\mu_{G/\Gamma}$ is a continuous function of $b$ and equal to $1$ when $b=1$ we must have $\gamma([b,g])=1$ for all sufficiently small $b$, and thus for all $b$ by connectedness of $G_{s-1}$ and the identity$$[b_1b_2,g]=[b_1,g][b_2,g].$$Thus setting $\gamma(b)=\Delta_bf$ extends $\gamma$ to a homomorphism $G_{s-1}\to S^1$. In fact we can extend $\gamma$ still further to a function $G\to D_1$, where $D_1$ is the unit disc in $\mathbf{C}$, by setting$$\gamma(a) = \int \Delta_af\,d\mu_{G/\Gamma}.$$Now if $a\in G$ and $b\in G_{s-1}$ then$$\gamma(ba) = \int f(bax) \overline{f(x)}\,d\mu_{G/\Gamma} = \int \gamma(b)f(ax)\overline{f(x)}\,d\mu_{G/\Gamma}=\gamma(b)\gamma(a),$$and$$\gamma(ab) = \int f(abx)\overline{f(x)}\,d\mu_{G/\Gamma} = \int f(ax) \overline{f(b^{-1}x)}\,d\mu_{G/\Gamma} = \int f(ax) \overline{\gamma(b^{-1})}\overline{f(x)}\,d\mu_{G/\Gamma} = \gamma(b)\gamma(a),$$so$$\gamma(b)\gamma(a)=\gamma(ab)=\gamma(ba[a,b]) = \gamma(ba)\gamma([a,b])=\gamma(b)\gamma(a)\gamma([a,b]).$$Since $|\gamma(b)|=1$ we can cancel $\gamma(b)$, so$$\gamma(a)(\gamma([a,b])-1) = 0.$$Finally observe that $\gamma(a)$ is a continuous function of $a$, and $\gamma(1)=1$, so we must have $\gamma([a,b])=1$ for all sufficiently small $a$, and thus by connectedness of $G$ and the identity$$[a_1a_2,b]=[a_1,b][a_2,b]$$we must have $\gamma([a,b])=1$ identically. But this implies that $\gamma$ vanishes on all $s$-term commutators and thus on all of $G_s$, so in fact $f$ factors through $G/G_s\Gamma$, so it must be constant.$\square$

A remark is in order about the possibility that some of the groups $G_i$ and $G_i\Gamma$ are not closed. This should not matter. One could either read the above proof as it is written, noting carefully that I never said groups should be Hausdorff, or, what's similar, instead modify it so that whenever you quotient by a group $H$ you instead quotient by the group $\overline{H}$.

Embarrassingly, it's difficult to come up with a non-Lie group to which this generalised Leon Green's theorem applies. It seems that many natural candidates have the property that $G$ is not connected but $G/\Gamma$ is: for example consider$$\left(\begin{array}{ccc}1&\mathbf{R}\times\mathbf{Q}_2&\mathbf{R}\times\mathbf{Q}_2\\0&1&\mathbf{R}\times\mathbf{Q}_2\\0&0&1\end{array}\right)/\left(\begin{array}{ccc}1&\mathbf{Z}[1/2]&\mathbf{Z}[1/2]\\0&1&\mathbf{Z}[1/2]\\0&0&1\end{array}\right).$$So it would be interesting to know whether the theorem extends to such a case. Or perhaps there are no interesting non-Lie groups for this theorem, which would be a bit of a let down.

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