normalizer of algebras and groups - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T13:17:43Z http://mathoverflow.net/feeds/question/22748 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/22748/normalizer-of-algebras-and-groups normalizer of algebras and groups Arnaud Brot 2010-04-27T17:30:00Z 2010-06-01T07:34:29Z <p>Hi, I am looking at inclusion of discrete groups $H\subset G$ such that $H$ is abelian and $(hgh^{-1},h\in H)$ is infinite if $g\in G-H$. If you have this, $LH\subset LG$ is a maximal abelian subalgebra of a finite von Neumann algebra. Suppose that $LH\subset LG$ is a Cartan subalgebra, i.e. the group of unitary of $LG$ that normalize the algebra $LH$ generates $LG$. Do we have necessarily that $H$ is a normal subgroup of $G$? Thanks for your help.</p> http://mathoverflow.net/questions/22748/normalizer-of-algebras-and-groups/22755#22755 Answer by HW for normalizer of algebras and groups HW 2010-04-27T17:51:27Z 2010-04-27T18:18:06Z <p>EDIT: On re-reading the question, I see that I misread 'at' as 'for' in the first line. This led me to read the first line as a question. Apologies! My answer is retracted.</p> http://mathoverflow.net/questions/22748/normalizer-of-algebras-and-groups/23050#23050 Answer by Owen Sizemore for normalizer of algebras and groups Owen Sizemore 2010-04-29T22:55:19Z 2010-04-29T22:55:19Z <p>So I don't think it is known in full generallity but there are some partial results. For example if additionally we assume that for any $c,d\in G\setminus H$ the stabilizer subgroups are either equal of noncommensurable then it is true. Much of the results rely on the Pukanzki invarient for $L(H)$ (and if $L(H)$ is cartan then the invarient is {1}), which in some cases you can calculate by the number of left-right cosets. </p> <p>This is mostly from memory, but Sinclair and Smith have a book "Finite von Neumann algebras and MASAS", and there is a chapter about the pukanzki invarient and masas coming from groups. So check that as well as references therein </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>