## Notes of Henrik Petersen’s lecture

Quasiisometries of nilpotent groups

Joint work (in progress) with David Kyed. The paper is not fully written, so be careful.

1. Nilpotent groups

Let ${\Gamma}$ be finitely generated torsion free nilpotent. Mal’cev’s theorem states that there exists a unique connected Lie group ${G=\Gamma\otimes{\mathbb R}}$ where ${\Gamma}$ sits as a discrete cocompact subgroup.

This is not hard for the free step ${d}$ nilpotent group, the general case follows. See Baumslag’s notes.

Fact. Any two cocompact lattices in a Lie group are quasiisometric. Therefore nilpotent group having the same Mal’cev completion are quasiisometric.

Question. Is the converse true ?

2. Earlier results

In 1989, Pansu showed that quasiisometric nilpotent groups have the same graded Lie algebras. Given a Lie algebra ${\mathfrak{g}}$, let ${\mathfrak{g}_{[i]}}$ denote the descending central series. Then

$\displaystyle \begin{array}{rcl} gr(\mathfrak{g})=\bigoplus_i \mathfrak{g}_{[i]}/\mathfrak{g}_{[i+1]}, \end{array}$

with the induced Lie bracket.

In 2002, Shalom showed that quasiisometric nilpotent groups have the same usual Betti numbers. Apart from degree one, this does not follow from the previous result.

2.1. Sketch of Shalom’s argument

Two groups ${\Gamma}$ and ${\Lambda}$ are quasiisometric iff they are uniformly measure equivalent.

Definition 1 ${\Gamma}$ and ${\Lambda}$ are uniformly measure equivalent if they have commuting measure preserving actions on some measure space ${\Omega}$, and both action are co-finite. Furthermore, pick a fundamental domain ${X}$ with resulting cocycle ${\omega:\Gamma\times X\rightarrow \Lambda}$; one requires that for fixed ${\gamma\in\Gamma}$, ${\omega(\gamma,\cdot):X\rightarrow \Lambda}$ is bounded.

In this case, cohomology of the module ${L^2(X)}$ can be transferred from ${\Gamma}$ to ${\Lambda}$.

Next, Shalom uses Property ${H_T}$ (see below). This implies that

$\displaystyle H^n(\Gamma,{\mathbb R})\simeq H^n(\Gamma,L^2(X))\simeq H^n(\Lambda,L^2(X))\simeq H^n(\Lambda,{\mathbb R}).$

3. Cohomology and higher order cohomology

3.1. Cohomology

Definition 2 A continuous ${G}$-module is relatively injective if for every exact sequence

$\displaystyle \begin{array}{rcl} 0\rightarrow A\rightarrow B \end{array}$

of continuous ${G}$-modules, which admits a continuous linear section ${\sigma}$ (when this is the case, we speak of a strengthened morphism), also admits a ${G}$-equivariant continuous linear section.

Proposition 3 For every ${G}$-module ${E}$, the module ${C(G,E)}$ of continuous equivariant maps ${G\rightarrow E}$ is relatively injective.

A strengthened resolution of ${E}$ follows:

$\displaystyle \begin{array}{rcl} 0\rightarrow E\rightarrow C(G,E)\rightarrow C(G\times G,E)\rightarrow\cdots. \end{array}$

Definition 4 Let ${E}$ be a ${G}$-module. Pick a relatively injective resolution. The continuous cohomology of ${E}$ is the cohomology of the subcomplex of ${G}$-invariant vectors of the resolution.

Note that we do not take closures.

3.2. Higher order invariants

Definition 5 For ${\xi\in E}$ and ${g\in G}$, denote by

$\displaystyle \begin{array}{rcl} \partial_g(\xi):=g\xi-\xi. \end{array}$

For instance, invariant vectors

$\displaystyle \begin{array}{rcl} E^G=\{\xi\in E\,;\,\forall g\in G,\,\partial_g \xi=0\}. \end{array}$

Definition 6 For ${\xi\in E}$ and ${g\in G}$, denote by

$\displaystyle \begin{array}{rcl} E^{G(d)}:=\{\xi\in E\,;\,\forall g_1,\ldots,g_d\in G,\,\partial_{g_1}\cdots\partial_{g_d}\xi=0\}. \end{array}$

The functor ${E\mapsto E^{G(d)}}$ is left-exact. As for all left-exact functors, one can define its polynomial cohomology.

Definition 7 Let ${E}$ be a ${G}$-module. Pick an arbitrary relative injective resolution

$\displaystyle \begin{array}{rcl} 0\rightarrow E\rightarrow E_0\rightarrow E_1\rightarrow\cdots. \end{array}$

Its polynomial cohomology in degree ${d}$, ${H^{n}_{(d)}(G,E)}$, is the cohomology of the resulting complex of order ${d}$ invariants ${\cdots\rightarrow E_i^{G(d)}\rightarrow\cdots}$.

3.3. Computation

As an exercise, let us compute ${C(G,{\mathbb R})^{G(2)}}$. ${f\in C(G,{\mathbb R})^{G(2)}}$ iff for all ${g_1}$ and ${g_2\in G}$,

$\displaystyle \begin{array}{rcl} f(g_1g_2)-f(g_1)-f(g_2)+f(e)=0. \end{array}$

I.e. ${f}$ is a homomorphism plus a constant.

Elements of ${C(G,{\mathbb R})^{G(3)}}$ are known as bi-characters

${C({\mathbb R}^n,{\mathbb R})^{{\mathbb R}^n(d)}=\{}$polynomials of degree ${\leq d-1\}}$. Whence the name polynomial cohomology.

3.4. Polynomial maps

More generally, Lazard (’50s), Leibman (2002) call elements of ${C(G,{\mathbb R})^{G(d)}}$ polynomial maps on ${G}$. Note that polynomials can be multiplied together.

Proposition 8

$\displaystyle \begin{array}{rcl} H^1_{(d)}(G,{\mathbb R})=Pol_d(G)/Pol_{d-1}(G). \end{array}$

Now I describe polynomials in the nilpotent Lie group case. Fix a Mal’cev basis, i.e. a linear basis ${\{X_{ij}\}}$ adapted to the descending central filtration. Denote ${g_{ij}:=\exp(X_{ij})}$. Then the map

$\displaystyle \begin{array}{rcl} (t_{ij})\mapsto g=\prod g_{ij}^{t_{ij}} \end{array}$

is a diffeomorphism. Each coordinate ${\xi_{ij}:g\mapsto t_{ij}}$ becomes a function on ${G}$, this is a polynomial map of degree ${i}$.

Theorem 9 ${G}$ nilpotent Lie group. Then ${Pol(G)}$ is generated as an algebra by functions ${\xi_{ij}}$.

Example. ${G=}$ Heisenberg group. Then

$\displaystyle \begin{array}{rcl} H^1_{(2)}(G,{\mathbb R})=span\langle \xi_x^2,\xi_y^2,\xi_x\xi_y,\xi_z\rangle. \end{array}$

4. Main result

Theorem 10 (Kyed-Petersen) Let ${G}$, ${H}$ be nilpotent Lie groups. Assume they are uniformly measure equivalent. Then for all ${n}$ and ${d}$,

$\displaystyle \begin{array}{rcl} H^n_{(d)}(G,{\mathbb R})\simeq H^n_{(d)}(H,{\mathbb R}). \end{array}$

Let ${m:G\times G\rightarrow G}$ be the group multiplication. One can show that it induces ${m^*:Pol(G)\rightarrow Pol(G\times G)}$ which is a complete invariant of ${G}$. Our strategy is to express ${m^*}$ in terms of polynomial cohomology. In this way, we hope to be able to prove that quasiisometric nilpotent Lie groups must be isomorphic.

5. Proof

Recall that uniform measure equivalence means there exists a measure space ${(\Omega,\mu)}$ with a ${G\times H}$ action, isomorphisms to ${G\times Y}$ and ${G\times X}$ respectively, with ${X}$ and ${Y}$ compact.

Theorem 11 (Reciprocity theorem, inspired by Monod-Shalom)

$\displaystyle H^n_{(d)}(G,L^2_{loc}(\Omega, E)^{H(d')})\simeq H^n_{(d')}(H,L^2_{loc}(\Omega, E)^{G(d)}).$

Then Property ${H_T}$ leads to

$\displaystyle \begin{array}{rcl} H^n_{(d)}(G,{\mathbb R})\simeq H^n_{(d)}(G,L^2(X))\simeq H^n(H,L^2(Y,Pol_{d-1}(G))). \end{array}$

which relates to the higher cohomology of ${Pol_{d-1}(H)}$. Then recursion.