Let $H$ be a Hilbert space, $S(H)$ be the inverse semigroup (pseudogroup) of linear maps between (closed) subspaces of $H$ preserving the dot product (the operation is composition of partial maps). Now consider an arbitrary inverse semigroup $A$ and all possible homomorphisms $A\to S(H)$. An inverse semigroup, by definition, is a semigroup with unary operation $^{-1}$ such that $(a^{-1})^{-1}=a, aa^{-1}a=a, aa^{-1} bb^{-1}=bb^{-1} aa^{-1}$ (by Wagner's theorem, similar to the Cayley theorem for groups) these are precisely the semigroups of partial bijections between subsets of a set under composition).

  1. Is it possible to define property (T) for inverse semigroups using these representations?

  2. Let $I_n(\mathbb Z)$ be the set of restrictions of all operators from $SL_n(\mathbb Z)$ onto all subspaces of $\mathbb{R}^n$, $n\ge 3$. It is an inverse semigroup. Does it have property (T)?

Note that amenability of inverse semigroups (pseudogroups) has been considered, for example, here.

Update Since there is a confusion about what partial maps are, here is a formal definition. A partial map $X\to X$ is a map from a subset $Y$ of $X$, called domain, to another subset $Z\subseteq X$, called range. If $X$ is a metric space (say, a Hilbert space), a partial isometry is an isometry between subspaces $Y$ and $Z$. The composition of a partial map $f: X\to X$ and another partial map $g: X\to X$ is a partial map $fg$ defined of all $x$ such that $x$ is in the domain of $g$ and $g(x)$ is in the domain of $f$, $fg(x)=f(g(x))$. If $f:Y\to Z$ is a (partial) bijection $X\to X$, then the map $f^{-1}f$ is the identity map on the domain of $f$, and it is an idempotent (obviously). If $f$ is an idempotent partial bijection with damain (=range) $Y$, and $g$ is an idempotent with domain (=range) $Z$, then $fg=gf$ is the idempotent with domain (=range) $Y\cap Z$, so idempotents commute. It is obvious that any set of partial bijections closed under products and taking $^{-1}$ is an inverse semigroup. The question is about representations of inverse semigroups into the inverse semigroup of partial (!) bijective unitary (=preserving the dot product) operators of a Hilbert space.

  • $\begingroup$ Your second example does not satisfy the axiom $aa^{-1}bb^{-1} = bb^{-1}aa^{-1}$. Indeed $aa^{-1}$ is just the projection onto the image of $a$, which does not necessarily commute with the projection onto the image of $b$. $\endgroup$ – Andreas Thom May 5 '11 at 7:40
  • $\begingroup$ @Andreas: No, it does. $aa^{-1}$ is the partial trivial isometry of a subspace onto itself. That is domain$(aa^{-1})=H=$range$(aa^{-1})$, where $H$ is a subspace, and the action is identity. $\endgroup$ – user6976 May 5 '11 at 11:14
  • $\begingroup$ Therefore if $aa^{-1}$ corresponds to suspace $H$, $bb^{-1}$ corresponds to $H'$, then their product (in any order) corresponds to the intersection $H\cap H'$. $\endgroup$ – user6976 May 5 '11 at 11:16

I can perhaps say something about the first question. $S(H)$ will not be a groupoid under composition since the composition of two projections need not preserve the dot product. However if one first fixes a standard probability space $(X, \mu)$ then one can consider the set of partial isometries $V$ on $L^2(X, \mu)$ such that the range and support are characteristic functions on $X$, i.e., there exists measurable subsets $A, B \subset X$ such that $V^*V(f) = f_{|A}$ and $VV^*(f) = f_{|B}$ for all $f \in L^2(X, \mu)$. This is a groupoid since the range and source projections will always commute.

In this setting it is then natural to define property (T) for measured groupoids. This was first done for probability measure preserving group actions and measured equivalence relations by Zimmer (Amer. J. Math. 103 (1981), no. 5, 937–951).

This definition, as well as analogues of several of the well known characterizations of property (T) for groups (for example the cohomological characterization by Delorme and Guichardet) can be found in the more recent paper of Anantharaman-Delaroche (Ergodic Theory Dynam. Systems 25 (2005), no. 4, 977–1013).

Edit: Actually, I suppose $S(H)$ is a groupoid if you only allow composition when the corresponding subspaces are equal. Rather what I mean is that it is not a pseudogroup, and (at least in the measurable setting) this is the natural thing to use to define representations of groupoids.

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  • $\begingroup$ @Jesse: Why is the composition not preserve the dot product? If $\phi$ is a map $A\to B$ and $\psi$ is a map $C\to D$, both preserving the dot product, then $\phi\psi$ ($\phi$ acts first) is a map from $\phi^{-1}(B\cap C)$ to $\psi(B\cap C)$ and for every $a,b$ in the domain $< a,b >=< \psi\phi(a), \psi\phi(b) >$ . Right? $\endgroup$ – user6976 Apr 9 '11 at 18:42
  • $\begingroup$ What I mean is that composition of a partial isometry from $A$ to $B$ with a partial isometry from $C$ to $D$ will not be a partial isometry from $\phi^{-1}(B \cap C)$ to $\psi(B \cap C)$ in general. Consider for example $\phi$ the projection in $\mathbb C^2$ onto the subspace spanned by $e_1 + e_2$ and $\psi$ the projection onto the subspace spanned by $e_1$, where $e_j$, $j = 1,2$ form the standard basis. $\endgroup$ – Jesse Peterson Apr 9 '11 at 18:53
  • $\begingroup$ You can of course restrict the composition to the subspace $\phi^{-1}(B \cap C)$ but I feel like this would not be desirable for looking at representations since this is giving up the best feature of Hilbert space, that you have angles between vectors. $\endgroup$ – Jesse Peterson Apr 9 '11 at 18:58
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    $\begingroup$ Another possibly relevant reference (it is also in the measurable setting) is Popa, Vaes: "Cocycle and orbit superrigidity for lattices in $SL(n, \mathbb R)$ acting on homogeneous spaces" (arxiv.org/abs/0810.3630). $\endgroup$ – Jesse Peterson Apr 9 '11 at 20:01
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    $\begingroup$ I guess, there is a problem with terminology. The word "groupoid" has different meaning in different subject. The object I am talking about it defined in the question. It consists of all partial isometries between subspaces of $H$ with operation: the natural composition of partial isometries. It is an inverse semigroup or pseudo-group in the sense of Cartan: en.wikipedia.org/wiki/Pseudogroup . The question is whether one can define (T) of abstract inverse semigroups using representations in $S(H)$. $\endgroup$ – user6976 Apr 9 '11 at 21:21

After looking at the paper of Ceccherini-Silberstein, Grigorchuk, and de la Harpe linked to in the question I think I have a better idea of what is being asked, so let me record some further thoughts here in a separate answer.

Given a set $X$ and a bijection $\gamma : S \to T$ between subsets of $X$ let me denote by $s(\gamma) = S$ (the support) and $r(\gamma) = T$ (the range). As in the above paper let me define a (standard) pseudogroup $\mathcal G$ of transformations on $X$ to be a set of bijections $\gamma : S \to T$ between subsets $S, T \subset X$ such that:

(i) ${\rm id}_{|S} \in \mathcal G$, for any subset $S \subset X$.

(ii) If $\gamma \in \mathcal G$ then $\gamma^{-1} \in \mathcal G$.

(iii) If $\gamma, \delta \in \mathcal G$ then $\delta \circ \gamma \in \mathcal G$, where $\delta \circ \gamma$ is understood to have domain $\gamma^{-1}(r(\gamma) \cap s(\delta))$.

(iv) If $\gamma: S \to T$ is a bijection of subsets $S, T \subset X$ and $S = \sqcup_{1 \leq j \leq n} S_j$ is a finite partition with $\gamma_{|S_j} \in \mathcal G$ for all $1 \leq j \leq n$, then $\gamma \in \mathcal G$.

I will only consider the case when $X$ a discrete space with counting measure, although much of this should work in the setting where $X$ is is a standard Borel space with a (possibly infinite) measure $\lambda$ on $X$ such that $\lambda(s(\gamma)) = \lambda(r(\gamma))$ for all $\gamma \in \mathcal G$.

Let me make now a "quantum" definition. A pseudogroup $\mathcal H$ of quantum transformations on a Hilbert space $K$ is a set of partial isometries on $K$ such that:

(i) $1 = {\rm id}_{K} \in \mathcal H$, and the von Neumann algebra generated by the set of projections in $\mathcal H$ contains a maximal abelian self-adjoint subalgebra (MASA) of $\mathcal B(K)$.

(ii) If $v \in \mathcal H$ then $v^* \in \mathcal H$.

(iii) If $v, w \in \mathcal H$ then $w \cdot v \in \mathcal H$ where $w \cdot v$ is the partial isometry $w( w^*w \wedge vv^* ) v$, e.g., if $v$ is a partial isometry from $K_1$ to $K_2$ and $w$ is a partial isometry from $K_3$ to $K_4$ then $w \cdot v$ is a partial isometry from $v^*( K_2 \cap K_3 )$ to $w( K_2 \cap K_3)$.

(iv) If $v \in \mathcal B(K)$ is a partial isometry and $v^*v = \Sigma_{1 \leq j \leq n} p_j$ is a finite partition of projections with $v p_j \in \mathcal H$ for all $1 \leq j \leq n$, then $v \in \mathcal H$.

A nice example is the set of all partial isometries $S(K)$ described by Mark above.

A pseudogroup of transformations on a set $X$ has a natural representation (which I will call the regular representation) as a pseudogroup of quantum transformations on $\ell^2X$, by viewing a bijection of subsets $\gamma: S \to T$ as the partial isometry $v_\gamma : \ell^2S \to \ell^2T$ given by $v_\gamma(\xi) = \xi \circ \gamma^{-1}$.

Conversely, if $\mathcal H$ is a pseudogroup of quantum transformations on $K$ such that the set of projections in $\mathcal H$ generates a purely atomic MASA then by letting $X$ be the set of rank one projections in $\mathcal H$, we have a natural identification $K = \ell^2X$ and we have that each $v \in \mathcal H$ induces a bijection $\gamma : S \to T$ of subsets of $X$ such that $v = v_\gamma$.

A homomorphism between (quantum) pseudogroups $\mathcal G$ and $\mathcal H$ is a map $\pi: \mathcal G \to \mathcal H$ which preserves the structures $(i)-(iv)$. If both $\mathcal G$ and $\mathcal H$ are pseudogroups of transformations on $X$ and $Y$ then homomorphisms are somewhat rigid, if we restrict to characteristic functions on points we obtain a map $f: X \to Y$ and we then can check that $\pi(\gamma) \circ f = f \circ \gamma$, for all $\gamma \in \mathcal G$.

Given a homomorphism between standard pseudogroups $\mathcal G$ and $\mathcal H$ we obtain a representation of $\mathcal G$ by composition with the regular representation for $\mathcal H$. One could hope that by viewing a pseudogroup of transformatins $\mathcal G$ as a pseudogroup of quantum transformations then maybe the homomorphism (and hence also the representation) structure would be richer. This however does not give a larger class.

Proposition. Let $\mathcal G$ be a pseudogroup of transformations on a set $X$. If $\pi: \mathcal G \to S(K)$ is a representation then $\pi(2^X)$ generates an abelian von neumann subalgebra of $\mathcal B(K)$, hence every representation of $\mathcal G$ is given by composing a homomorphism into a standard pseudogroup with the regular representation.

Proof. Since $\pi$ preserves (ii) we have that $\pi(1_S 1_T) = \pi(1_S) \wedge \pi(1_T)$, for all $S, T \subset X$. Since $\pi(1_X) = 1$ and since $\pi$ preserves (iv) we have that $\pi( 1_X - 1_S ) = 1 - \pi(1_S)$ for all $S \subset X$. Hence by again using the fact that $\pi$ preserves (iv) we have that $\pi( 1_{S \cup T}) = \pi( 1_S ) \vee \pi( 1_T)$ and $$ \pi(1_S) - \pi(1_S) \wedge \pi(1_T) = \pi(1_S - 1_{S \cap T}) $$ $$ = \pi(1_{S \cup T} - 1_T) = \pi(1_S) \vee \pi(1_T) - \pi(1_T). $$ A standard fact from operator algebras then shows that $\pi(1_T)$ and $\pi(1_S)$ commute. Indeed, the left hand side $\pi(1_S) - \pi(1_S) \wedge \pi(1_T)$ is the projection onto the closure of the range of $\pi(1_S)(1 - \pi(1_T))$ and satisfies $$ \pi(1_S) \wedge (1 - \pi(1_T)) \leq \pi(1_S) - \pi(1_S) \wedge \pi(1_T) \leq \pi(1_S), $$ while the right hand side $\pi(1_S) \vee \pi(1_T) - \pi(1_T)$ is the projection onto the closure of the range of $(1 - \pi(1_T))\pi(1_S)$ and satisfies $$ \pi(1_S) \wedge (1 - \pi(1_T)) \leq \pi(1_S) \vee \pi(1_T) - \pi(1_T) \leq 1 - \pi(1_T). $$
Hence both sides equal $P = \pi(1_S) \wedge (1 - \pi(1_T))$ and we have $$ \pi(1_S)(1 - \pi(1_T)) = P \pi(1_S)(1 - \pi(1_T)) $$ $$ = P = P (1 - \pi(1_T))\pi(1_S) = (1 - \pi(1_T))\pi(1_S), $$ which shows that $\pi(1_T)$ and $\pi(1_S)$ commute. Since $S$ and $T$ were arbitrary this gives the result. $\square$

One could certainly define property (T) for pseudogroups of transformations in this setting by requiring that any representation which almost contains invariant vectors must contain a non-trivial invariant vectors. But given the above proposition it is equivalent to say that $\mathcal G$ has property (T) if and only if any quotient of $\mathcal G$ which is amenable must be finite. (Recall that a pseudogroup $\mathcal H$ of transformations on $X$ is amenable if there is a state $\varphi$ on $\ell^\infty X$, such that $\varphi(\gamma \gamma^{-1}) = \varphi(\gamma^{-1} \gamma)$, for all $\gamma \in \mathcal H$).

This definition seems like an interesting property, but it lacks a bit in functoriality. For instance, if $\Gamma$ is a countable group and we consider the pseudogroup $\mathcal G(\Gamma)$ of all maps on subsets of $\Gamma$ which arise from the action of $\Gamma$ on itself by left multiplication. Then I believe that the property that $\mathcal G(\Gamma)$ has no quotient onto an infinite amenable pseudogroup, can be restated in terms of group actions as: Every transitive action of $\Gamma$ on an infinite set $Y$ has no invariant mean.

If $\Gamma$ has property (T) then this condition is satisfied, but this condition will also be satisfied for groups which do not have property (T), although it seems to be non-trivial to show this. If $\Gamma$ is an infinite product of infinite simple property (T) groups then this should be an example. This condition is also satisfied for Tarski monsters (are there Tarski monsters which do not have property (T)?).

Coming to question 2, $I_n(\mathbb Z)$ in this setting is not a pseudogroup of transformations on a set, but rather a pseudogroup of quantum transformations in $M_n(\mathbb R) \subset M_n(\mathbb C)$. Finite dimensional algebras correspond to finite sets in the standard setting and so I think that $I_n(\mathbb Z)$ in this setting should be considered as "finite", and hence will have property (T) but for trivial reasons.

For instance, the definition of amenability for a pseudogroup $\mathcal H$ of quantum transformations on $H$ should be that there exists a state $\phi$ on $\mathcal B(K)$ such that $\phi(v^*v) = \phi(vv^*)$ for all $v \in \mathcal H$. This definition is consistent with the definition in the standard setting.

There may be another perspective in which $I_n(\mathbb Z)$ can be seen to be "rigid", but I'm not sure what it would be.

One could also fix a dimension $d$ and consider the action of $SL_n(\mathbb Z)$ on the Grassmannian $Gr(d, \mathbb R^n)$ and ask if the generated pseudogroup of transformations on $Gr(d, \mathbb R^n)$ does not have any non-trivial amenable quotients ($d \not= 0, n$). (This is no longer the discrete setting.) Using the property (T) of $SL_n(\mathbb Z)$ it should be enough to check the following property (perhaps this is well known or is discussed in the paper of Popa and Vaes that I referred to before, I am not sure): If $Y$ is a non-finite compact Hausdorff space which has a continous action of $SL_n(\mathbb Z)$ and such that there is a continuous $SL_n(\mathbb Z)$-invariant surjection $\pi: Gr(d, \mathbb R^n) \to Y$, can $Y$ have a $SL_n(\mathbb Z)$-invariant probability measure?

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  • $\begingroup$ @Jesse: I certainly do not understand some parts of your answer. But $I_n(\mathbb{Z})$ should not be considered finite, it is a "lattice" in $I_n(\mathbb{R})$, and the proof of (T) for it cannot be muxch simpler than for $SL_n(\mathbb{Z})$. By definition a pseudogroup (inverse semigroup) of transformations of any set $X$ is any set of partial injective maps $X\to X$ closed under composition and taking inverses. We do not necessarily require that it is closed under restrictions to subsets. $\endgroup$ – user6976 Apr 15 '11 at 16:50
  • $\begingroup$ @Mark: My answer is only relevant to pseudogroups of transformations on a set which are closed under restricting to subsets, this is the definition which is taken in the paper of Ceccherini-Silberstein, Grigorchuk, and de la Harpe. For the case when it is not closed under restriction one can still define amenability and property (T), but the proposition I wrote above will not be true in that case. $\endgroup$ – Jesse Peterson Apr 15 '11 at 21:08
  • $\begingroup$ @Mark cont.: $I_n(\mathbb Z)$ is not defined as partial transformations on a set but rather partial transformations on a quantum set (= Hilbert space), my answer is a possible way to extend the framework of Ceccherini-Silberstein, Grigorchuk, and de la Harpe to this setting. There very well may be another way. Do you know if amenability has been defined for inverse semigroups which are not pseudogroups of transformations on a set? $\endgroup$ – Jesse Peterson Apr 15 '11 at 21:08

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