User jesse peterson - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T03:30:24Z http://mathoverflow.net/feeds/user/6460 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/115875/which-von-neumann-algebras-have-inner-permutation-of-tensor-factors/115936#115936 Answer by Jesse Peterson for Which von Neumann algebras have inner permutation of tensor factors? Jesse Peterson 2012-12-10T00:46:20Z 2013-03-21T04:27:16Z <p>Here's an argument showing that in the ${\rm II}_1$ case the flip automorphism is never inner. </p> <p>Let $M$ be a type ${\rm II}_1$ factor and $\tau$ its trace, so that $M \subset L^2(M, \tau)$, and $M$ acts standardly on $L^2(M, \tau)$. Suppose that the flip automorphism is implemented by a unitary $U \in \mathcal U(M \overline \otimes M)$. I'll reach a contradiction by showing that $U$ is orthogonal to every vector in the dense subspace of $L^2(M \overline \otimes M, \tau \otimes \tau)$ spanned by vectors of the form $v \otimes w$ where $v, w \in \mathcal U(M)$.</p> <p>Fix $v, w \in \mathcal U(M)$, and $\varepsilon > 0$. Take $n \in \mathbb N$ such that $2^{-n} &lt; \varepsilon$, and take a partition of unity <code>$\{ p_k \}_{k = 1}^{2^n} \subset M$</code> such that each $p_k$ is a projection of trace $2^{-n}$. Then $$| \langle U, v \otimes w \rangle | = \left| \sum_{k = 1}^{2^n} (\tau \otimes \tau) ((p_k \otimes 1)(v^* \otimes w^*) U (p_k \otimes 1) ) \right|$$ $$\leq \sum_{k = 1}^{2^n} | (\tau \otimes \tau) ((v^* \otimes w^*) U (p_k \otimes vp_kv^*) ) |$$ $$\leq \sum_{k = 1}^{2^n} (\tau \otimes \tau)(p_k \otimes vp_kv^*)$$ $$= \sum_{k = 1}^{2^n} \tau(p_k)^2 = 2^{-n} &lt; \varepsilon.$$</p> <p><b>Update</b>: I've recently come across the 1975 paper of Sakai "Automorphisms and tensor products of operator algebras" where he proves that the flip automorphism for a von Neumann algebra $M$ is inner if and only if $M$ is a type ${\rm I}$ factor. His proof is roughly as follows:</p> <p>For the type ${\rm II}_1$ case he proceeds as I did above by showing that the unitary $U$ would have to be orthogonal to every vector of the form $v \otimes w$. His argument for this is not as direct as the one above, but the argument I gave above is based on techniques of Popa which came later.</p> <p>For the type ${\rm II}_\infty$ case he writes $M$ as $N \overline \otimes \mathcal B(\mathcal H)$ where $N$ is a type ${\rm II}_1$ factor and then shows with a simple argument that if the flip automorphism is inner on $M$ then it must also be inner on $N$ which it cannot be by the arguments above.</p> <p>For the type ${\rm III}$ case he first writes $M$ as $N \overline \otimes \mathcal B(\mathcal H)$ where $N$ is type ${\rm III}$ and countably decomposable. Next he shows that if the flip is inner on $N$ then $N$ has trivial outer automorphism group. Indeed, if $\sigma$ denotes the flip and $\rho \in {\rm Aut}(N)$ then since $\sigma$ is inner, and ${\rm Inn}(M)$ is a normal subgroup, we must have $\tilde \rho = \sigma (\rho^{-1} \otimes {\rm id}) \sigma (\rho \otimes {\rm id})$ is also inner. Restricting $\tilde \rho$ to $N \otimes \mathbb C$ we then see then that there is a unitary $V \in N \overline \otimes N$ such that $(a \otimes 1)V = V(\rho(a) \otimes 1)$ for all $a \in N$. If we then consider the normal conditional expectation $E$ from $N \overline \otimes N$ to $N \otimes \mathbb C$, then there exists some operator $x \in \mathbb C \otimes N$ such that $E(Vx) \not= 0$, and we then have $a E(Vx) = E(Vx) \rho(a)$ for all $a \in N$. By conjugating this formula it then follows easily that $E(Vx)E(Vx)^* \in \mathcal Z(N) = \mathbb C$ and also $E(Vx)^*E(Vx) \in \mathbb C$, hence $E(Vx)$ is a non-zero scalar multiple of a unitary showing that $\rho$ is inner. Tomita-Takesaki theory though gives continuum many outer automorphism of $N$, a contradiction.</p> http://mathoverflow.net/questions/103971/how-well-do-we-know-relative-commutants-in-l-mathbbf-infty/103980#103980 Answer by Jesse Peterson for How well do we know relative commutants in $L(\mathbb{F}_\infty)$? Jesse Peterson 2012-08-04T22:08:48Z 2012-08-04T22:08:48Z <p>Given any two tracial von Neumann algebras $(N_1, \tau_1)$ and $(N_2, \tau_2)$ the $L^2$ space of the free product $(N_1 * N_2, \tau)$ canonically decomposes as $$L^2(N_1 * N_2, \tau) = \mathbb C \oplus_{n \in \mathbb N} \bigoplus_{i_1 \not= i_2, i_2 \not= i_3, \ldots, i_{n - 1} \not= i_n } \overline {\otimes}_{k = 1}^n L^2_0(N_{i_k}, \tau_{i_k}),$$ where $L^2_0(N_i, \tau_i)$ is the orthogonal complement of the scalars.</p> <p>From this decomposition it's not hard to see that as an $N_1$ bimodule, $L^2(N_1 * N_2, \tau)$ decomposes as a direct sum of one copy of the trivial bimodule $L^2(N_1, \tau_1)$ and copies of the coarse bimodule $L^2(N_1 \overline \otimes N_1, \tau_1 \otimes \tau_1)$. Specifically, if $\xi \in L^2(N_1 * N_2, \tau)$ is a unit vector which is an elementary word'' starting and ending with vectors in $L^2_0(N_2, \tau_2)$ then $x \xi y \mapsto x \otimes y$ extends to an isomorphism from the $N_1$ bimodule generated by $\xi$ and $L^2(N_1 \overline \otimes N_1, \tau_1 \otimes \tau_1)$.</p> <p>Thus $N_1' \cap (N_1 * N_2) \not= \mathcal Z(N_1)$ if and only if the coarse $N_1$ bimodule has non-zero central vectors, which, by identifying $L^2(N_1 \overline \otimes N_1, \tau_1 \otimes \tau_1)$ with Hilbert-Schmidt operators and then taking a spectral projection, is if and only if $L^2(N_1, \tau_1)$ has a finite dimensional $N_1$-sub-bimodule, which is if and only if $N_1$ has a non-trivial finite dimensional direct summand.</p> <p>In the case you're looking at $N_1$ is a ${\rm II}_1$ factor and so $N_1' \cap (N_1 * N_2) = \mathcal Z(N_1) = \mathbb C$.</p> http://mathoverflow.net/questions/101321/injective-von-neumann-algebra/101340#101340 Answer by Jesse Peterson for Injective von Neumann algebra Jesse Peterson 2012-07-04T20:33:27Z 2012-07-04T20:33:27Z <p>Here is an adaptation of the standard proof that $G$ is amenable if $LG$ is injective. (I believe for instance that it is contained in the book of Brown and Ozawa).</p> <p>Suppose $p \in LG$ is a non-zero central projection such that $p LG$ is injective. Thus, there exists a conditional expectation $E: \mathcal B(p \ell^2 G) \to p LG$. If we view $\ell^\infty G \subset \mathcal B(\ell^2 G)$ as diagonal multiplication operators (for $f \in \ell^\infty G$ and $\xi \in \ell^2 G$ we set $(M_f \xi)(\gamma) = f(\gamma) \xi(\gamma)$), and if we denote by $\tau$ a tracial state on $pLG$ then we can construct a state $\varphi$ on $\ell^\infty G$ by the formula $\varphi(f) = \tau \circ E(p M_f p)$. If $\gamma \in G$ then we have $$\varphi( f \circ \gamma) = \tau \circ E(p M_{f \circ \gamma} p)$$ $$= \tau \circ E(p \lambda_{\gamma^{-1}} M_f \lambda_{\gamma} p) = \tau( (p\lambda_{\gamma^{-1}}p) E(p M_f p) (p \lambda_{\gamma}p) ) = \varphi(f).$$ Thus $\varphi$ is an invariant mean for $G$ and so $G$ is amenable.</p> http://mathoverflow.net/questions/96068/ergodicity-of-the-group-of-transformations-preserving-a-partition/96075#96075 Answer by Jesse Peterson for ergodicity of the group of transformations preserving a partition Jesse Peterson 2012-05-05T16:21:51Z 2012-05-05T18:19:54Z <p>It's easy to see that the action is always ergodic since $\mathcal G$ contains the group of finite permutations on the indices, which acts ergodically. In fact, the group $\mathcal G$ (which equals $\mathcal H$ in the case $m = \mu^{\mathbb N}$ with $\mu(0) = 1/2$.) that you are describing is the full group of the ergodic hyperfinite measurable equivalence relation. It, and other full groups are discussed in Sections I.3 and I.4 in the book by Alexander Kechris: Global aspects of ergodic group actions, Mathematical Surveys and Monographs, 160, American Mathematical Society, 2010. </p> http://mathoverflow.net/questions/92922/abelian-sub-w-algebras/92947#92947 Answer by Jesse Peterson for Abelian sub-W*-algebras Jesse Peterson 2012-04-03T00:38:01Z 2012-04-03T05:55:28Z <p>This is not true. Popa showed in "<a href="http://www.ams.org/mathscinet-getitem?mr=703810" rel="nofollow">Orthogonal pairs of ∗-subalgebras in finite von Neumann algebras</a>" (1983), that if $F$ is a free group with arbitrary cardinality than any abelian von Neumann subalgebra of the group von Neumann algebra $LF$ must have separable predual.</p> <p>Edit: This doesn't even hold when $M$ is abelian since $\ell^\infty(\mathbb R)$ has no faithful state and hence does not embed into any $\sigma$-finite von Neumann algebra.</p> http://mathoverflow.net/questions/89322/non-isomorphic-groups-with-the-same-oriented-cayley-graph/90028#90028 Answer by Jesse Peterson for Non-isomorphic groups with the same oriented Cayley graph Jesse Peterson 2012-03-02T11:42:08Z 2012-03-02T11:42:08Z <p>If we consider the dihedral group of order 12, $G = \langle a, b \mid a^2 = b^2 = (ab)^6 = e \rangle$, then the Cayley graph corresponding to <code>$\{ a, b \}$</code> is the cyclic graph on 12 vertices with edges labeled alternately by $a$ and $b$. We may then consider $H_1 = G \times \mathbb Z/2 \mathbb Z$ and $H_2 = G \rtimes \mathbb Z/2 \mathbb Z$ where the copies of $\mathbb Z/ 2 \mathbb Z$ are generated by $c$ and $d$ respectively, and where $d$ acts by $d a d = b$, $d b d = a$. Then the Cayley graph of $H_1$ with respect to <code>$\{ a, b, c \}$</code> consists of two copies of cyclic graphs of order 12 with edges labeled by $c$ connecting the two graphs. Since $d$ just interchanges $a$ and $b$ we have that the Cayley graph of $H_2$ with respect to <code>$\{ a, b, d \}$</code> can be obtained from the Cayley graph of $H_1$ by just relabeling the second copy of the cyclic graph, so that the two Cayley graphs will be isomorphic.</p> <p>Unfortunately, the generating set for $H_2$ is not minimal since by the definition of $d$ we have $b \in \langle a, d \rangle$. This can be fixed however by taking two automorphisms $c$ and $d$ which are complicated enough so that this doesn't occur.</p> <p>Specifically, we can let $c$ act on $G$ by $cac = ababa$, and $cbc = babab$, and we can let $d$ act on $G$ by applying $c$ and then interchanging $a$ and $b$, i.e., $dad = babab$, and $dbd = ababa$. It's not hard to see that these indeed define order two automorphisms of $G$, and for the same reason as above we have that the Cayley graphs of $H_1$ and $H_2$ will be isomorphic.</p> <p>It is also not hard to check that we now have $| \langle a, c \rangle | = | \langle b, c \rangle | = 12 &lt; 24$ and $| \langle a, d \rangle | = | \langle a, d \rangle | = 8 &lt; 24$ so that the generating sets are now minimal. Moreover, the groups $H_1$ and $H_2$ will not be isomorphic, this can be seen for instance by counting the number of elements of order 2, (I counted 15 for $H_1$ and 9 for $H_2$, but I've omitted the tedious details).</p> http://mathoverflow.net/questions/86472/left-mean-values-vs-right-mean-values/86479#86479 Answer by Jesse Peterson for Left mean values vs right mean values Jesse Peterson 2012-01-23T20:14:59Z 2012-01-23T20:14:59Z <p>For a given amenable group $G$, these sets will coincide for all $f \in \ell^\infty(G)$ if and only if the sets $L$, $R$, and $I$ coincide. Just notice that if $f \in \ell^\infty(G)$, and $g \in G$ then we have <code>$M_L( f - \lambda_g(f) ) = \{ 0 \}$</code>.</p> http://mathoverflow.net/questions/85873/reference-for-embedding-an-infinite-direct-product-of-matrix-algebras-into-the-hy/85878#85878 Answer by Jesse Peterson for Reference for embedding an infinite direct product of matrix algebras into the hyperfinite $II_1$ factor Jesse Peterson 2012-01-17T03:00:21Z 2012-01-17T04:49:32Z <p>Murray and von Neumann showed that if $p \in \mathcal R$ is a non-zero projection then $p\mathcal R p \cong \mathcal R$, i.e., the fundamental group of $\mathcal R$ is all positive reals (You should be able to find this in most books that discuss the hyperfinite II$_1$ factor. Also, note that this is easy to see if $p$ has rational trace by viewing $\mathcal R$ as an infinite tensor product of matrices). Thus, if <code>$\{ p_n \}_{n \in \mathbb N}$</code> is a partition of $1$ by non-zero projections then we obtain an embedding <code>$$\oplus_{n \in \mathbb N} \mathbb M_n(\mathbb C) \subset \oplus_{n \in \mathbb N} \mathcal R \cong \oplus_{n \in \mathbb N} p_n \mathcal R p_n \subset \mathcal R.$$</code></p> <p>Note also that since <code>$L^\infty([0, 1], \lambda)$</code> embeds into $\mathcal R$ (e.g., as tensor products of diagonal matrices), it follows that <code>$\mathbb M_n(\mathbb C) \overline \otimes L^\infty([0, 1], \lambda) \subset \mathbb M_n(\mathbb C) \overline \otimes \mathcal R \cong \mathcal R$</code>. Thus, since any separable abelian von Neumann algebra <code>$A_n$</code> is a von Neumann subalgebra of <code>$L^\infty([0, 1], \lambda)$</code> it follows from above that <code>$\oplus_{n \in \mathbb N}( A_n \overline \otimes \mathbb M_n(\mathbb C) )$</code> embeds into <code>$\mathcal R$</code>. All separable finite type I von Neumann algebras are of this form where some of the $A_n \overline \otimes \mathbb M_n(\mathbb C)$ terms may be omitted (this can be found for instance in Takesaki, vol. 1, chapter 1). Thus all separable finite type I von Neumann algebras embed into <code>$\mathcal R$</code>.</p> http://mathoverflow.net/questions/85092/relative-commutants-of-abelian-von-neumann-algebras Relative commutants of abelian von Neumann algebras Jesse Peterson 2012-01-06T22:13:41Z 2012-01-06T22:13:41Z <p>This question arose from the discussion over at the question <a href="http://mathoverflow.net/questions/84863/centralizers-in-c-algebra" rel="nofollow">Centralizers in $C^*$-algebras</a>.</p> <p>Which von Neumann algebras $N$ satisfy the property that $A' \cap N = B' \cap N \implies A = B$, for all commutative von Neumann subalgebras $A, B \subset N$?</p> <p>Note that $N = \mathcal B(\mathcal H)$ has this property by von Neumann's double commutant theorem, and perhaps this property characterizes $\mathcal B(\mathcal H)$. It is clear that $N$ must be a factor by considering $A = \mathbb C$ and $B = \mathcal Z(N)$. If $\mathbb F_2 = \langle a, b \rangle$ is the free group on two generators, then by considering the Fourier expansion of elements in $L\mathbb F_2$ it is not hard to see that for $A = L\langle a \rangle$ and $B = L\langle a^2 \rangle$ we have $A' \cap L\mathbb F_2 = B' \cap L\mathbb F_2$ thus $L\mathbb F_2$ does not have this property. </p> <p>Also note that if we were to consider the case when $A$ and $B$ are allowed to be non-commutative then relevant is Corollary 4.1 in Popa's paper <a href="http://www.ams.org/mathscinet-getitem?mr=641131" rel="nofollow">On a Problem of R.V. Kadison on Maximal Abelian $*$-Subalgebras in Factors</a> which shows that every type $II$ factor $N$ contains a hyperfinite subfactor $R$ such that $R' \cap N = \mathbb C$.</p> http://mathoverflow.net/questions/84120/fubinis-theorem-and-unique-mean-value/84126#84126 Answer by Jesse Peterson for Fubini's theorem and unique mean value Jesse Peterson 2011-12-22T23:09:09Z 2011-12-22T23:09:09Z <p>Let $g$ be the characteristic function on the positive odd numbers and let $f(x) = g(x) - g(x + 1)$, so that $f$ has a unique mean value. If $\mu$ is non-principle and supported on the positive odd numbers and $\nu$ is non-principle and supported on the negative even numbers then we have $\iint f(x + y) \; d\mu d\nu = 1$, while $\iint f(x + y) \; d\nu d\mu = 0$.</p> http://mathoverflow.net/questions/83336/induced-representations-of-topological-groups/83390#83390 Answer by Jesse Peterson for Induced representations of topological groups Jesse Peterson 2011-12-14T04:06:58Z 2011-12-14T07:34:40Z <p>If $G$ is compact this is the Frobenius Reciprocity Theorem, see e.g., Section 6.2 in Folland's A Course in Abstract Harmonic Analysis for a proof. When $G$ is not compact then this fails already for $H$ the trivial group and $U$, and $V$ trivial representations. Indeed, in this case Ind$_H^G(1_H) = L^2G$ the left regular representation, and since constant functions are not square integrable we have Hom<code>$_G( L^2G, 1_G ) = \{ 0 \}$</code>, while Hom<code>$_H(1_H, 1_H) \not= \{ 0 \}$</code>.</p> http://mathoverflow.net/questions/80391/topological-weak-mixing-and-omega-linearly-independent-sequences-generated-by/80457#80457 Answer by Jesse Peterson for Topological weak mixing and $\omega$-linearly-independent sequences generated by composition operators Jesse Peterson 2011-11-09T07:11:30Z 2011-11-09T07:11:30Z <p>For each $k > 0$, <code>$F_k = \{ x \in X \mid T^k(x) = x \}$</code> is closed, hence if $T$ is not periodic then for each $N > 0$, $G_N = X \setminus (\cup_{k \leq N} F_k)$ is open, and non-empty.</p> <p>If we are given $(a_n)$ absolutely summable, (for which we will assume $a_0 = 1$ by scaling and shifting, which is fine since $T$ is not periodic), take $N > 0$ such that $\sum_{n > N} |a_n| &lt; 1$. Since $G_N$ is open and non-empty there exists a non-empty, open subset $O \subset G_N$ such that $T^n(O) \cap O = \emptyset$ for all $1 \leq n \leq N$.</p> <p>If we consider a continuous function $f$ such that $\| f \|_\infty \leq 1$, $f$ has support contained in $O$, and $f$ obtains the values $\pm 1$ within $O$, then <code>$\sum_{n = 0}^N a_nf \circ T^n$</code> agrees with $f$ on $O$. Hence <code>$\sum_{n = 0}^\infty a_n f \circ T^n$</code> has both positive and negative values and is therefore not constant.</p> http://mathoverflow.net/questions/77699/uniqueness-of-free-complements/77706#77706 Answer by Jesse Peterson for Uniqueness of free complements Jesse Peterson 2011-10-10T16:43:22Z 2011-10-10T16:43:22Z <p>If you want uniqueness up to equality, then the answer is no since if $B$ is a free complement for $A$ in $M$ then so is $u B u^*$ for any unitary $u \in A$. This is the case also in the specific example you give above.</p> <p>If you want uniqueness up to isomorphism then this is more subtle since showing that II$_1$ factors are (non)isomorphic is usually quite difficult. I can't think of an example for this situation in general, but I do have an example if you assume that the free group factors $L\mathbb F_2$ and $L\mathbb F_3$ are isomorphic.</p> <p>Indeed, Dykema and Radulescu showed in (http://www.ams.org/mathscinet-getitem?mr=1955269) that if the free group factors are isomorphic then for any II$_1$ factors $M$ and $N$ we have that $M * N \cong M * N * L\mathbb F_\infty$. In particular, this would hold if $M$ were a II$_1$ factor with property (T), and $N = R$ were the hyperfinite II$_1$ factor. But in this case by a result of Ioana, Popa, and myself (http://www.ams.org/mathscinet-getitem?mr=2386109) there would then also exist an isomorphism $\theta: M * R \to M * R * L\mathbb F_\infty$ such that $\theta(M) = M$. This would give a situation where $R$ and $R * L\mathbb F_\infty$ are both free complements of $M$, but $R * L\mathbb F_\infty$ is not hyperfinite and hence not isomorphic to $R$.</p> http://mathoverflow.net/questions/76515/how-big-are-the-ultrapowers-of-the-hyperfinite-ii-1-factor/76524#76524 Answer by Jesse Peterson for How big are the ultrapowers of the hyperfinite $II_1$-factor? Jesse Peterson 2011-09-27T15:25:21Z 2011-09-27T15:25:21Z <p>This is not true. If we denote by $\mathbb F_{\mathbb R}$ the free group with generators indexed by $\mathbb R$ then any separable von Neumann subalgebra $N \subset L\mathbb F_{\mathbb R}$ must necessarily be contained in the von Neumann subalgebra $L\mathbb F_I$ for some countably infinite subset $I \subset \mathbb R$. (This is a direct consequence of applying Parseval's identity to vectors in $L^2N \subset \ell^2 \mathbb F_{\mathbb R}$.) </p> <p>The separable $II_1$ factor $L ({\rm PSL}(3, \mathbb Z))$ embeds into $R^\omega$ since the group is residually finite, but does not embed into $L\mathbb F_I$ by a result of Connes and Jones since it has property (T). Hence, $L\mathbb F_{\mathbb R}$ cannot contain $R^\omega$.</p> http://mathoverflow.net/questions/71362/trace-norm-inequality-for-positive-matrices/71370#71370 Answer by Jesse Peterson for trace norm inequality for positive matrices Jesse Peterson 2011-07-27T00:21:52Z 2011-07-27T01:02:56Z <p>If $A$ are $B$ are projections which are not orthogonal, but are close to being orthogonal so that $ABA \not= 0$ but has only small eigenvalues then we have <code>$\| A B A \|_\text{tr} \ll \| (A B A)^{1/2} \|_\text{tr} = \| A^2 B \|_\text{tr}$</code>. Hence, no such constant $C_n$ exists.</p> <p>For a specific example, if <code>$A = \left( \begin{array}{cc} 1 &amp; 0 \\ 0 &amp; 0 \end{array} \right)$</code> and <code>$B = \left( \begin{array}{cc} \sin^2 t &amp; - \sin t \cos t \\ - \sin t\cos t &amp; \cos^2 t \end{array} \right)$</code>, then as $t \to 0$ we have <code>$\| ABA \|_\text{tr} / \| A^2 B \|_\text{tr} = | \sin t | \to 0$</code>.</p> http://mathoverflow.net/questions/42685/pro-discrete-locally-compact-and-open-normal-subgroups-have-trivial-intersectio/70798#70798 Answer by Jesse Peterson for pro-discrete = locally compact and open normal subgroups have trivial intersection? Jesse Peterson 2011-07-20T07:12:32Z 2011-07-20T07:12:32Z <p>Let $K$ be an infinite profinite group and let $K_n &lt; K$ be a decreasing family of open subgroups with trivial intersection. Thus, $K$ acts continuously on the discrete space $X = \sqcup_n K/K_n$ and this in turn gives rise to a continuous action of $K$ on the free abelian group $\mathbb Z[X]$. Define $G$ to be the semidirect product $G = K \ltimes \mathbb Z[X]$.</p> <p>$G$ is a locally compact group and if we denote $X_n = \sqcup_{m \geq n} K/K_m \subset X$ then since $K_n$ acts trivially on $X \setminus X_n$ we have that $K_n \ltimes \mathbb Z[X_n]$ is a family of open normal subgroups with trivial intersection in $G$. Also, $K &lt; G$ is an open subgroup but contains no nontrivial normal subgroups since the action of $K$ on $X$ is faithful.</p> http://mathoverflow.net/questions/65477/actions-orbit-equivalent-to-profinite-ones/65538#65538 Answer by Jesse Peterson for Actions orbit equivalent to profinite ones Jesse Peterson 2011-05-20T13:02:26Z 2011-05-29T18:11:35Z <p>If a group has a free profinite action then in particular it must be residually finite. On the other hand, any measure preserving action of an amenable group is weakly compact. Thus, any free measure preserving action of infinite amenable group $G$ which is not residually finite cannot be orbit equivalent to a profinite action of $G$.</p> <p>Of course by Ornstein and Weiss' theorem all such orbit equivalence relations are hyperfinte and so this doesn't work if you allow yourself to change the group $G$. </p> <p><b>Update:</b></p> <p>For residually finite groups an example will have to be more sophisticated since this will not work with amenable groups. However, there are enough orbit equivalence rigidity theorems around that this is still possible. The relevant theorem here is Ioana's rigidity theorem for profinite actions of property (T) groups (http://arxiv.org/abs/0805.2998):</p> <p><b>Theorem</b>[Adrian Ioana]: Let $\Gamma$, and $\Lambda$ be countable discrete groups and suppose that $\Gamma$ has property (T). Let $\Gamma \curvearrowright X$ be a free, ergodic, profinite, probability measure preserving action and suppose that $\Lambda \curvearrowright Y$ is a free probability measure preserving action which is orbit equivalent to the action $\Gamma \curvearrowright X$. Then there exists finite index subgroups <code>$\Gamma_1 &lt; \Gamma$</code> and <code>$\Lambda_1 &lt; \Lambda$</code> such that restricting to an <code>$\Gamma_1$</code> (resp. <code>$\Lambda_1$</code>) ergodic component <code>$X_1$</code> (resp. <code>$Y_1$</code>) we have that the actions <code>$\Gamma_1 \curvearrowright X_1$</code> and <code>$\Lambda_1 \curvearrowright Y_1$</code> are conjugate.</p> <p>In particular, this shows that for orbit equivalent actions of property (T) groups if one action is profinite then so is the other. This is because a measure preserving action $\Gamma \curvearrowright X$ is profinite if and only if the weak closure $\overline \Gamma$ of $\Gamma$ in Aut$(X)$ is a profinite group, and $\overline \Gamma$ is invariant under conjugation. Thus one just needs to construct a weakly compact action of a residually finite property (T) group which is not profinite. </p> <p>One such example is given by Dennis Sullivan (http://www.ams.org/mathscinet-getitem?mr=590825) who showed that for $n > 3$ there is a countable dense subgroup <code>$\Gamma_n$</code> of $O(n + 1)$ which has property (T). <code>$\Gamma_n$</code> will be residually finite by a classical result of Malcev since it is finitely generated and linear. Also, the Haar measure preserving action <code>$\Gamma_n \curvearrowright O(n + 1)$</code> given by left multiplication will be free (since <code>$\Gamma_n$</code> is a subgroup), ergodic (since <code>$\Gamma_n$</code> is dense), and compact (since $O(n + 1)$ is compact), and hence also weakly compact. However, <code>$\Gamma_n \curvearrowright O(n + 1)$</code> is not profinite since <code>$O(n + 1) = \overline{\Gamma_n}$</code> is not profinite.</p> <p>Note that this gives an example if you consider orbit equivalence with other groups $\Lambda$ as well.</p> http://mathoverflow.net/questions/66343/commutators-in-the-reduced-c-algebra-of-the-free-group/66348#66348 Answer by Jesse Peterson for Commutators in the reduced C*-algebra of the free group Jesse Peterson 2011-05-29T09:37:47Z 2011-05-29T09:37:47Z <p>This is true. Perhaps it is known whether this property for a group $G$ is equivalent to <code>$C^*_\lambda G$</code> having a unique trace? In any case, the same proof of Powers which shows unique trace for free groups, can be adapted to this question.</p> <p>First note that if <code>$F = F(x_1, x_2, \ldots, x_n)$</code> is the free group on $n$ generators, then by a trick of Bob Powers, in <code>$\mathcal B(\ell^2 F)$</code> we have <code>$\| \Sigma_{i = 1}^n \lambda(x_i) \| \leq 2 \sqrt{n}$</code>. Indeed, if we denote by <code>$P_i$</code> the projection in <code>$\mathcal B(\ell^2 F)$</code> onto the subspace generated by Dirac functions on all words which in reduced form begin with the letter <code>$x_i$</code>, then we have <code>$(1 - P_i) \lambda(x_i) (1 - P_i) = 0$</code> and hence <code>$$\Sigma_{i = 1}^n \lambda(x_i) = \Sigma_{i = 1}^n P_i \lambda(x_i) + ( \Sigma_{i = 1}^n P_i \lambda(x_i) (1 - P_i))^*.$$</code> Since <code>$P_i$</code> have orthogonal ranges we have that <code>$\| \Sigma_{i = 1}^n P_i \lambda(x_i) \| \leq \sqrt{n}$</code> and <code>$\| \Sigma_{i = 1}^n P_i \lambda(x_i) (1 - P_i) \| \leq \sqrt{n}$</code>.</p> <p>Next note that the limits of sums of commutators forms a subspace and hence it is enough in the reduced group <code>$C^*$</code>-algebra to show that the non-trivial group elements can be written as a limit of a sums of commutators. Moreover, in a free non-abelian group, every element has an element which is free from it, and hence by restricting to a subgroup it is enough to show that in the free group on 2 generators $F(a, b)$ we can write $\lambda(a)$ as a limit of sums of commutators.</p> <p>For this we consider an arbitrary $n \in \mathbb N$ and easily verify the formula <code>$$\frac{1}{n} \Sigma_{i = 0}^{n -1} \lambda( b^{-i} a b^i ) = \lambda(a) - \frac{1}{n}\Sigma_{i = 1}^{n - 1} [ \lambda(b^i), \lambda(b^{-i} a) ].$$</code></p> <p>Since <code>$x_i \mapsto b^{-(i - 1)} a b^{(i - 1)}$</code> extends to an isomorphism between <code>$F(x_1, \ldots, x_n)$</code> to the subgroup $\langle a, b^{-1} a b, \ldots, b^{-(n - 1)}a b^{(n - 1)} \rangle$ it follows from Powers trick that <code>$$\| \lambda(a) - \frac{1}{n}\Sigma_{i = 1}^{n - 1} [ \lambda(b^i), \lambda(b^{-i} a) ] \| \leq 2/ \sqrt{n}.$$</code> Since $n$ was arbitrary, this finishes the proof. </p> http://mathoverflow.net/questions/62360/is-independence-meaningful-for-commutative-c-algebras/62369#62369 Answer by Jesse Peterson for Is independence meaningful for commutative $C^*$-algebras? Jesse Peterson 2011-04-20T04:33:01Z 2011-04-20T04:33:01Z <p>If $X$ is a separable compact Hausdorff space then for any Radon measure $\mu$ with dense support we have $C(X) \subset L^\infty(X, \mu)$, so the answer to your first question is yes.</p> <p>Independence will depend on the measure you take, consider for instance the difference between taking Lebesgue measure on $[0, 1] \times [0,1]$ versus taking a measure which gives most of its weight to the diagonal. In the former case functions of $x$ will be independent from functions of $y$, while this is not the case in the latter.</p> <p>These types of results would be covered in any introductory book on operator algebras. For instance R. Douglas "Banach algebra techniques in operator theory" (http://www.ams.org/mathscinet-getitem?mr=1634900).</p> http://mathoverflow.net/questions/61151/property-t-for-pseudogroups/61352#61352 Answer by Jesse Peterson for Property (T) for pseudogroups Jesse Peterson 2011-04-12T00:48:36Z 2011-04-14T16:43:44Z <p>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.</p> <p>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:</p> <p>(i) ${\rm id}_{|S} \in \mathcal G$, for any subset $S \subset X$.</p> <p>(ii) If $\gamma \in \mathcal G$ then $\gamma^{-1} \in \mathcal G$.</p> <p>(iii) If $\gamma, \delta \in \mathcal G$ then $\delta \circ \gamma \in \mathcal G$, where $\delta \circ \gamma$ is understood to have domain $\gamma^{-1}(r(\gamma) \cap s(\delta))$.</p> <p>(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 $\gamma \in \mathcal G$.</p> <p>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 $\gamma \in \mathcal G$. </p> <p>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:</p> <p>(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 $\mathcal B(K)$.</p> <p>(ii) If $v \in \mathcal H$ then $v^* \in \mathcal H$.</p> <p>(iii) If $v, w \in \mathcal H$ then $w \cdot v \in \mathcal H$ where $w \cdot v$ is the partial isometry <code>$w( w^*w \wedge vv^* ) v$</code>, 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 $w( K_2 \cap K_3)$. </p> <p>(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 $v \in \mathcal H$.</p> <p>A nice example is the set of all partial isometries $S(K)$ described by Mark above.</p> <p>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 $v_\gamma(\xi) = \xi \circ \gamma^{-1}$.</p> <p>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 $v = v_\gamma$. </p> <p>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 $\gamma \in \mathcal G$. </p> <p>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.</p> <p><b>Proposition.</b> 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.</p> <p><b>Proof.</b> 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 $$\pi(1_S) \wedge (1 - \pi(1_T)) \leq \pi(1_S) \vee \pi(1_T) - \pi(1_T) \leq 1 - \pi(1_T).$$<br> 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. $\square$</p> <p>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 $\gamma \in \mathcal H$).</p> <p>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.</p> <p>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)?).</p> <p>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. </p> <p>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 <code>$\phi(v^*v) = \phi(vv^*)$</code> for all $v \in \mathcal H$. This definition is consistent with the definition in the standard setting.</p> <p>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.</p> <p>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?</p> http://mathoverflow.net/questions/61151/property-t-for-pseudogroups/61156#61156 Answer by Jesse Peterson for Property (T) for pseudogroups Jesse Peterson 2011-04-09T18:07:56Z 2011-04-09T18:37:01Z <p>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. </p> <p>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 (<a href="http://www.ams.org/mathscinet-getitem?mr=630773" rel="nofollow">Amer. J. Math. 103 (1981), no. 5, 937–951</a>). </p> <p>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 (<a href="http://www.ams.org/mathscinet-getitem?mr=2158394" rel="nofollow">Ergodic Theory Dynam. Systems 25 (2005), no. 4, 977–1013</a>). </p> <p><b>Edit</b>: 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 <a href="http://en.wikipedia.org/wiki/Pseudogroup" rel="nofollow">pseudogroup</a>, and (at least in the measurable setting) this is the natural thing to use to define representations of groupoids. </p> http://mathoverflow.net/questions/58186/positive-element-in-c-tensor-product/58200#58200 Answer by Jesse Peterson for positive element in C* tensor product Jesse Peterson 2011-03-11T21:43:00Z 2011-03-11T21:43:00Z <p>The same answer as before, the matrix <code>$$a=\begin{bmatrix} 1 &amp; 0 &amp; 0 &amp; 1 \\ 0 &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; 0 \\ 1 &amp; 0 &amp; 0 &amp; 1 \end{bmatrix}$$</code> in $M_2(\mathbb{C})\otimes M_2(\mathbb{C})$, also works here since it is twice a rank one projection and so any smaller positive matrix must be a scalar multiple of $a$.</p> http://mathoverflow.net/questions/45102/endomorphism-of-factor-can-it-be-idempotent-up-to-congugacy/45132#45132 Answer by Jesse Peterson for endomorphism of factor: can it be idempotent up to congugacy? Jesse Peterson 2010-11-07T02:41:05Z 2010-11-07T02:41:05Z <p>This is not possible. If it were, then using the notation above, given any $x \in \phi(M)$, we would have <code>$x u^* = u^* \phi(x)$</code>, and $\phi(x) u = u x$. Hence, for any $x \in \phi(M)$ we have <code>$$x u^* \phi(u^*) u^2 = u^* \phi(x u^*) u^2$$</code> <code>$$= u^* \phi(u^*) \phi \circ \phi (x) u^2 = u^* \phi(u^*) u^2 x.$$</code></p> <p>Hence <code>$u^* \phi(u^*)u^2 \in \phi(M)' \cap M = \mathbb C$</code> and so <code>$\phi(u^*) \in \mathbb C \cdot u^*$</code>. Then, for any $y \in M$ we would have that <code>$$\phi \circ \phi (y) = u \phi(y) u^*$$</code> <code>$$= \phi(u y u^*).$$</code> Since $\phi$ is injective we then have <code>$\phi(y) = u y u^*$</code>, and hence $\phi$ is invertible.</p> <p>If you don't require that $\phi(M)$ be irreducible then this is possible.</p> http://mathoverflow.net/questions/42928/orders-and-length-functions-on-finitely-generated-groups/42953#42953 Answer by Jesse Peterson for orders and length functions on finitely generated groups Jesse Peterson 2010-10-20T21:39:13Z 2010-10-20T21:39:13Z <p>If $G$ is a finitely generated infinite group and $\leq$ is a linear word order, then for each $a, c \in G$ there are only finitely many elements $b \in G$ such that $a \leq b \leq c$. From this it follows that $(G, \leq)$ is order isomorphic to $\mathbb Z$. If $\leq$ is also left invariant, then this isomorphism must be a group isomorphism as well.</p> http://mathoverflow.net/questions/32147/examples-of-groups-without-the-n-positive-approximation-property Examples of groups without the n-positive approximation property Jesse Peterson 2010-07-16T10:46:59Z 2010-07-17T00:24:13Z <p>Let $G$ be a locally compact group and let $A(G)$ be the Fourier algebra of $G$, which we view as the predual of the group von Neumann algebra $\mathcal M(G)$. Let $MA(G)$ be the space of multipliers of $A(G)$, i.e., $\varphi \in MA(G)$ if and only if $\varphi \psi \in A(G)$ for all $\psi \in A(G)$. Then $\varphi \in MA(G)$ induces a bounded operator $m_\varphi: A(G) \rightarrow A(G)$, and hence also a bounded operator <code>$M_\varphi = m_\varphi^*$</code> on $\mathcal M(G)$. </p> <p><code>$M_\varphi$</code> is completely bounded if <code>$\| M_\varphi \|_{CB} = \sup_n \| M_\varphi \otimes {\rm id}_n \| &lt; \infty$</code>, where <code>${\rm id}_n$</code> is the identity operator on the $n \times n$ matrices <code>$\mathbb M_n(\mathbb C)$</code>. <code>$M_\varphi$</code> is $n$-positive if <code>$M_\varphi \otimes {\rm id}_n$</code> takes the positive cone <code>$\mathcal M(G)_+$</code> into itself, or equivalently <code>$\| M_\varphi \otimes {\rm id}_n \| = \varphi(e)$</code>. <code>$M_\varphi$</code> is completely positive if it is $n$-positive for every $n \in \mathcal N$.</p> <p>A well known result is that $G$ is amenable if and only if $A(G)$ has an approximate unit <code>$\{ \varphi_k \}_k$</code> such that <code>$M_{\varphi_k}$</code> is completely positive for all $k$. Haagerup showed that <code>$SL_2(\mathbb R)$</code>, and all of its lattices have the completely bounded approximation property: For these groups, $A(G)$ has an approximate unit <code>$\{ \varphi_k \}_k$</code> such that <code>$\sup_k \| M_{\varphi_k} \|_{CB} &lt; \infty$</code>, (in fact he showed that this supremum can be 1 for <code>$SL_2(\mathbb R)$</code>, and all of its lattices). To contrast, he also showed that <code>$SL_m(\mathbb R)$</code>, and all of its lattices do not have the completely bounded approximation property whenever $m \geq 3$. De Canniere and Haagerup have also shown that free groups have the $n$-positive approximation property for every $n \in \mathbb N$: For these groups, given any $n \in \mathbb N$, $A(G)$ has an approximate unit <code>$\{ \varphi_k \}$</code> of compactly supported functions such that <code>$\varphi_k$</code> is $n$-positive.</p> <p>Recently, I was at a conference and Mikael de la Salle asked me if I knew of any examples of groups which do not have the $n$-positive approximation property. I do not and so I thought I would ask here.</p> <p>1) What is an example of a group which for some $n$ does not have the $n$-positive approximation property?</p> <p>2) What is an example of a group which for any $n$ does not have the $n$-positive approximation property?</p> <p>3) Are there groups which have the $n$-positive approximation property for some $n$, but not for $n + 1$?</p> http://mathoverflow.net/questions/31278/does-a-crossed-product-r-f-n-of-the-hyperfinite-factor-of-type-ii-1-and-a-fre/31282#31282 Answer by Jesse Peterson for Does a crossed product R⋊_α F_n of the hyperfinite factor of type II_1 and a free group have the QWEP? Jesse Peterson 2010-07-10T09:14:43Z 2010-07-10T09:14:43Z <p>Yes. If $a$ and $b$ are generators of $\mathbb F_2$ then $\mathcal R \rtimes_\alpha \mathbb F_2$ decomposes as an amalgamated free product of $(\mathcal R \rtimes_\alpha \langle a \rangle)$ and $(\mathcal R \rtimes_\alpha \langle b \rangle)$ over $\mathcal R$, where each of these are hyperfinite. Brown, Dykema, and Jung showed in <a href="http://arxiv.org/abs/math/0609080" rel="nofollow">http://arxiv.org/abs/math/0609080</a> that for separable finite von Neumann algebras being embeddable into $\mathcal R^\omega$ is stable under amalgamated free products over a hyperfinite von Neumann algebra. Thus $\mathcal R \rtimes_\alpha \mathbb F_2$ is embeddable into $\mathcal R^\omega$, which is equivalent to QWEP. Induction then gives the case when $2 \leq n &lt; \infty$, and the case $n = \infty$ then follows since QWEP is preserved under (the weak-closure of) increasing unions.</p> <p>Related to this, Collins and Dykema in <a href="http://arxiv.org/abs/1003.1675" rel="nofollow">http://arxiv.org/abs/1003.1675</a> have recently shown that the class of Sophic groups is stable under taking amalgamated free products over amenable groups.</p> <p>I believe this is an open problem however if we consider arbitrary residually finite groups instead of only $\mathbb F_n$.</p> http://mathoverflow.net/questions/29605/group-action-with-a-fixed-point-property/29725#29725 Answer by Jesse Peterson for Group Action with a Fixed-Point Property Jesse Peterson 2010-06-27T18:34:51Z 2010-06-27T18:57:20Z <p>One way to get an example is to let $F_2$ be the free group on two generators, then consider the collection of all maximal cyclic subgroups and choose one representative $H_n$ for each conjugacy class. $F_2$ then acts by left multiplication on the disjoint union of the left coset spaces $F_2/H_n$. Since maximal cyclic subgroups in $F_2$ are malnormal, each non-trivial element will have exactly one fixed point. Hence we can do as Tom Goodwillie suggests and add a global fixed point.</p> <p>I'm not sure how one would get a transitive action, but one thing which might help in this direction is that Ashot Minasyan, building on the work of Denis Osin, shows in Lemma 3.4 of <a href="http://eprints.soton.ac.uk/54841/" rel="nofollow">http://eprints.soton.ac.uk/54841/</a> that there is a countable torsion free group containing a malnormal copy of $\mathbb Z$ and having every element being conjugate to an element in $\mathbb Z$. It follows that this group acts transitively on a set such that each non-trivial element has exactly one fixed point.</p> http://mathoverflow.net/questions/27521/injection-between-non-isomorphic-irreducible-hilbert-space-reps/27528#27528 Answer by Jesse Peterson for Injection between non-isomorphic irreducible Hilbert space reps? Jesse Peterson 2010-06-08T21:47:04Z 2010-06-08T21:47:04Z <p>No conditions are needed on the group $G$, or on the continuity of the representation, you do need the assumption that $i$ is continuous however. Since $i:V \to W$ is continuous (equivalent to being bounded) it has a continuous adjoint $i^* : W \to V$ which is also $G$ equivariant, hence $i^* i:V \to V$ is $G$ equivariant and since $G$ acts on $V$ irreducibly must be a scalar multiple of the identity (if not the spectrum would contain more than one point and you could take a spectral projection of $i^* i$ and obtain a closed non-trivial $G$-invariant subspace). This shows that $i$ is a scalar multiple of an isometry and hence the image must be closed. If $G$ also acts irreducible on $W$ then $ii^*$ is also a scalar multiple of the identity and hence if $i \not= 0$ it must be a non-zero scalar multiple of a unitary between $V$ and $W$.</p> http://mathoverflow.net/questions/22748/normalizer-of-algebras-and-groups/26638#26638 Answer by Jesse Peterson for normalizer of algebras and groups Jesse Peterson 2010-06-01T00:10:16Z 2010-06-01T07:34:29Z <p>This is true, and in fact more has been shown in the recent preprint <a href="http://arxiv.org/abs/1005.3049" rel="nofollow">http://arxiv.org/abs/1005.3049</a> of Fang, Gao, and Smith. One can also give the following alternative argument based on ideas of Popa:</p> <p>If $LH \subset LG$ is a MASA then it follows from the condition $( hgh^{-1} \ | \ h \in H ) = \infty$ for all $g \in G \setminus H$, that the normalizer of $H$ in $G$ is the same as the set of elements $g \in G$ such that $[H: H \cap gHg^{-1}] &lt; \infty$. (This set is not in general closed under inversion but in this case it is since it coincides with the normalizer.)</p> <p>Suppose we fix $g \in G$ such that $[H: H \cap gHg^{-1}] = \infty$ and let's show that $u_g$ is orthogonal to $\mathcal N_{LG}(LH)''$. Since $\mathcal N_{LG}(LH)''$ is spanned by $\mathcal N_{LG}(LH)$ it is enought to show that $u_g$ is orthogonal to this set and so let's fix $v \in \mathcal N_{LG}(LH)$.</p> <p>Before we show that $u_g$ and $v$ are orthogonal let's rewrite the condition $[H: H \cap gHg^{-1}] = \infty$ in a more von Neumann algebraic friendly context which states that there are always "large" subalgebras of $LH$ which are almost moved orthogonal to $LH$.</p> <p><strong>Lemma:</strong> For all $n \in \mathbb N, \delta > 0$ there exists a finite dimensional subalgebra $A_0 \subset LH$ such that if $p$ is any minimal projection in $A_0$ then $\tau(p) = 1/2^n$ and $| \langle x, u_g^* p u_g - \tau(p) \rangle | &lt; \delta \|x \|_2$ for all $x \in LH$.</p> <p><strong>Proof.</strong> This essentially follows from Popa's intertwining techniques since the condition $[H: H \cap gHg^{-1}] = \infty$ translates in this context to $LH \not\prec_{LH} L(H \cap gHg^{-1})$ (See Popa's paper <a href="http://www.ams.org/mathscinet-getitem?mr=2231961" rel="nofollow">http://www.ams.org/mathscinet-getitem?mr=2231961</a>). </p> <p>Let's show this by induction on $n$. For the case when $n = 1$ consider the group $\mathcal G = ( u \in \mathcal U(LH) \ | \ u = 1 - 2p, p \in \mathcal P(LH), \tau(p) = 1/2 ) \cup (1)$. Since $\mathcal G$ generates $LH$ as a von Neumann algebra and since $LH \not\prec_{LH} L(H \cap gHg^{-1})$ it follows from Popa's intertwining Theorem that there exists a sequence $p_k \in \mathcal P(LH)$ with $\tau(p_k) = 1/2$ such that $\lim_{k \to \infty} \| E_{L(H \cap gHg^{-1})}(1 - 2p_k ) \|_2 = 0$ (see Popa, op. cit.). In particular, for some $k$ this is less than $2\delta$ and so if $x \in LH$, $\| x \|_2 &lt; 1$ we have $| \langle x, u_g^*p_ku_g - \tau(p) \rangle | \leq \| E_{LH}(u_g^* p u_g - \tau(p) ) \|$ $<em>2 = \| E</em>{L(H \cap gHg^{-1})} (p_k - 1/2) \|_2 &lt; \delta$. The same inequality holds for the other minimal projection $1 - p_k$.</p> <p>Once we have produced such an $A_0$ for $1/2^n$ then given any minimal projection $p \in A_0$ we again have that $pLH \not\prec_{pLH} pL(H \cap gHg^*)$ and so the argument above shows that there exists $p_1$ and $p_2$ in $\mathcal P(LH)$ such that $p_1 + p_2 = p$, each has half the trace and $| \langle x, u_g^* p_j u_g - \tau(p_j) \rangle | &lt; \delta$. This proves the induction step. QED</p> <p>Now that we have established the above lemma, the fact that $u_g$ and $v$ are orthogonal follows from a lemma of Popa's in <a href="http://www.ams.org/mathscinet-getitem?mr=703810" rel="nofollow">http://www.ams.org/mathscinet-getitem?mr=703810</a>. Let's give the proof here.</p> <p>Let $\varepsilon > 0$ be given and take $n \in \mathbb N$ such that $1/2^n &lt; \varepsilon/2$. From the above lemma let's consider a finite dimensional subalgebra $A_0 \subset LH$ such that if $p$ is any minimal projection in $A_0$ then $\tau(p) = 1/2^n$ and $| \langle x, u_g^*pu_g - \tau(p) \rangle | &lt; \| x \|_2 \varepsilon/2^{n + 1}$. Let's denote the minimal projections in $A_0$ by $p_k$ where $1 \leq k \leq 2^n$. Denote by $B_0$ the commutant of $A_0$ in $LG$.</p> <p>Since $v \in \mathcal N_{LG}(LH)$ we have that $vLHv^* = LH$, hence $v^* p_k v \in LH$ for each $k$. Therefore $| \langle v, u_g \rangle |^2 \leq \| E_{B_0} ( vu_g^*) \|_2^2$ $= \|$ $\Sigma_k$ $p_k v u_g^* p_k \|_2^2 = \Sigma_k \langle v^* p_k v, u_g^* p_k u_g \rangle &lt; (\Sigma_k \tau(p_k)^2 ) + \Sigma_k \varepsilon/2^{n + 1} &lt; \varepsilon$.</p> <p>Since $\varepsilon$ was arbitrary we conclude that $u_g$ and $v$ are orthogonal. Hence since $v$ was arbitrary we conclude that $\mathcal N_{LG}(LH)'' = L(\mathcal N_G(H))$.</p> http://mathoverflow.net/questions/946/operator-valued-weights/26632#26632 Answer by Jesse Peterson for Operator Valued Weights Jesse Peterson 2010-05-31T22:23:13Z 2010-05-31T22:23:13Z <p>Since $T(m_T)$ is a non-trivial two sided ideal in $N$, it follows from spectral calculus that $T(m_T)$ contains a non-zero projection. $T(m_T)$ then contains all subprojections, hence there exists $x \in m_T$ such that $p = T(x)$ is a projection with trace Tr$(p) = 1/n$ for some natural number $n$. </p> <p>We may assume that $pxp = x$, and by considering $(x + x^*)/2$ we may assume that $x$ is self-adjoint. Since $m_T$ is spanned by its positive part it follows that we can write $x$ as $x_1 - x_2$ where $x_j \in m_T$ are positive. Hence $T(x_1) \geq p$ and so by considering the element $(pT(x_1)^{-1/2}p)x_1(pT(x_1)^{-1/2}p) \in m_T$ we may assume that $x \geq 0$.</p> <p>If $N$ is a factor, then since Tr$(p) = 1/n$ we may find a (finite, if $N$ is finite) sequence of partial isometries $v_k$ such that $v_k^*v_k = p$ for each $k$ and $\Sigma_k v_kv_k^* = 1$. Then $x_m = \Sigma_{k = 1}^m v_k x v_k^*$ is a bounded increasing sequence in $m_T$ and we have that $T(x_m)$ increases to $1$. Since the weight is normal we have that $x_\infty = \Sigma_k v_k x v_k^* \in m_T \cap M_+$ and $T(x_\infty) = 1$. In particular, since $T(m_T)$ is an ideal we have $N \subset T(m_T)$.</p> <p>If $N$ has infinite dimensional center then as Dmitri explained there are many examples (even bounded ones) for which $1 \not\in T(M_+)$. I imagine that for finite dimensional center the argument above should still work.</p> http://mathoverflow.net/questions/129067/metrics-and-completions-on-the-direct-limit-of-matrices-of-all-sizes-over-arbitra Comment by Jesse Peterson Jesse Peterson 2013-05-01T16:59:02Z 2013-05-01T16:59:02Z For $A \in M_n$, the two norm is $\| A \|_2 = (\frac{1}{n} {\rm Tr}(A^*A) )^{1/2}$. This is different than the uniform norm/maximum singular value $\| A \|$. Although, one has the inequality $\| A \|_2 \leq \| A \|$. One way to construct the hyperfinite II$_1$ factor is to complete each uniform ball in the two norm and then take the union. But if you don't restrict to the uniform balls then you will not get an algebra since multiplication is not jointly continuous in the two norm http://mathoverflow.net/questions/129067/metrics-and-completions-on-the-direct-limit-of-matrices-of-all-sizes-over-arbitra Comment by Jesse Peterson Jesse Peterson 2013-04-30T16:08:49Z 2013-04-30T16:08:49Z To construct the hyperfinite II$_1$ factor in this way you also need to use the uniform norm. You need to restrict to uniformly bounded subsets when you take the completion so that multiplication is continuous in the two norm. http://mathoverflow.net/questions/128667/weights-on-von-neuman-factors Comment by Jesse Peterson Jesse Peterson 2013-04-28T04:00:03Z 2013-04-28T04:00:03Z In either case, if $H = \mathbb C^2$, and $A = \mathbb M_2(\mathbb C)$. Then requiring $A$ to be isomorphic to $\mathbb C_{H_2} \otimes B(H_1)$ means that $H_1$ is two dimensional. Since $H$ is isomorphic to $H_2 \otimes H_1$ you then know that $H_2$ is one dimensional. Thus $P_{H_2} \not= 1$. http://mathoverflow.net/questions/128667/weights-on-von-neuman-factors Comment by Jesse Peterson Jesse Peterson 2013-04-28T03:52:44Z 2013-04-28T03:52:44Z Perhaps I am still confused about your question. When you ask for $H$ to be isomorphic to $H_2 \otimes H_2$ such that $A$ is isomorphic to $\mathbb C_{H_2} \otimes B(H_1)$, am I correct that you mean for the second isomorphism to be implemented by the first? http://mathoverflow.net/questions/128667/weights-on-von-neuman-factors Comment by Jesse Peterson Jesse Peterson 2013-04-25T03:35:07Z 2013-04-25T03:35:07Z In that case a simple counter example is to take $A = B(H)$. If the dimension of $H$ is at least 2 then the right hand side of the formula cannot be faithful. Just consider $T = 1 - P_{H_2}$. http://mathoverflow.net/questions/128667/weights-on-von-neuman-factors Comment by Jesse Peterson Jesse Peterson 2013-04-25T00:25:10Z 2013-04-25T00:25:10Z It's a bit unclear to me how you are making sense of the formula. Does $P_{H_2}$ denote the orthogonal projection onto $H_2$? If so, what is the domain, and where does $T$ live? http://mathoverflow.net/questions/125693/stoneech-compactification-of-mathbbz-with-furstenberg-topology Comment by Jesse Peterson Jesse Peterson 2013-03-27T18:33:48Z 2013-03-27T18:33:48Z I'm just joining the conversation here and I must say that I am very confused since it appears that the notation $\beta \mathbb Z$ is being used by some to denote the Stone-Cech compactification of $\mathbb Z$ with the discrete topology, and by others to denote the Stone-Cech compactification with respect to the topology $\mathcal T$. Perhaps writing $\beta (\mathbb Z, \mathcal T)$ to denote the latter would help? http://mathoverflow.net/questions/122353/not-measure-equivalent-icc-groups-g-and-h-but-lg-cong-lh Comment by Jesse Peterson Jesse Peterson 2013-02-21T01:08:37Z 2013-02-21T01:08:37Z This problem is still open as far as I know. http://mathoverflow.net/questions/121741/compact-subgroups-of-the-unitary-group-of-operators-in-a-hilbert-space Comment by Jesse Peterson Jesse Peterson 2013-02-13T23:42:44Z 2013-02-13T23:42:44Z @Andras Batkai: Actually, the unitaries with the weak topology do not form a compact group. In fact, the weak and strong topologies give the same relative topology on the space of unitaries. http://mathoverflow.net/questions/115875/which-von-neumann-algebras-have-inner-permutation-of-tensor-factors/115936#115936 Comment by Jesse Peterson Jesse Peterson 2012-12-10T01:17:23Z 2012-12-10T01:17:23Z In fact, many of the examples of ${\rm II}_1$ factors without ourter automorphisms can also be shown to be prime, i.e., they are not isomorphic to any tensor product of other ${\rm II}_1$ factors. http://mathoverflow.net/questions/115875/which-von-neumann-algebras-have-inner-permutation-of-tensor-factors/115901#115901 Comment by Jesse Peterson Jesse Peterson 2012-12-10T00:51:03Z 2012-12-10T00:51:03Z For the hyperfinte ${\rm II}_1$ factor the flip is approximately inner which can be seen by restricting to finite dimensional subalgebras, but the unitaries you get won't converge. http://mathoverflow.net/questions/113984/tensor-product-of-von-neumann-algebras Comment by Jesse Peterson Jesse Peterson 2012-11-21T08:38:28Z 2012-11-21T08:38:28Z @Yemon: Yes, my comment was about normality. If $A = L^\infty[0, 1]$ then the map will not extend to a normal map on $A \overline \otimes A \cong L^\infty([0, 1] \times [0, 1])$. http://mathoverflow.net/questions/113984/tensor-product-of-von-neumann-algebras Comment by Jesse Peterson Jesse Peterson 2012-11-21T06:58:30Z 2012-11-21T06:58:30Z Which topology on $A \otimes A$ are you considering? In any case, try $A = L^\infty[0, 1]$, if continuity fails here then it also fails for any von Neumann algebra which contains a copy of $L^\infty[0, 1]$. http://mathoverflow.net/questions/108987/non-invariant-subspaces-for-subfactors Comment by Jesse Peterson Jesse Peterson 2012-10-07T12:02:44Z 2012-10-07T12:02:44Z You probably want to exclude the case $\mathcal N = \mathcal M$ as well. http://mathoverflow.net/questions/106190/the-set-of-upper-semi-continuous-functions-as-a-ring Comment by Jesse Peterson Jesse Peterson 2012-09-02T21:55:43Z 2012-09-02T21:55:43Z Are you sure that is the correct topology to use on $\mathbb R$? This seems to be stronger than continuity, rather than weaker. Also, Wikipedia at least suggests the name Sorgenfrey rather than surgenfery.