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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.

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Arnaud, I think you had H and G backwards in the first inclusion, so I edited it. –  Noah Snyder Apr 27 '10 at 17:34
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I also think the title should perhaps be edited to make it clear that the question is (probably) specifically focused on von Neumann algebras -- the fact that they have big unitary groups seems like it would be relevant to the difficulty of the question –  Yemon Choi Apr 27 '10 at 17:41
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3 Answers

up vote 6 down vote accepted

This is true, and in fact more has been shown in the recent preprint http://arxiv.org/abs/1005.3049 of Fang, Gao, and Smith. One can also give the following alternative argument based on ideas of Popa:

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 $[H: H \cap gHg^{-1}] < \infty$. (This set is not in general closed under inversion but in this case it is since it coincides with the normalizer.)

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 $v \in \mathcal N_{LG}(LH)$.

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 $LH$.

Lemma: 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 $| \langle x, u_g^* p u_g - \tau(p) \rangle | < \delta \|x \|_2$ for all $x \in LH$.

Proof. 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 http://www.ams.org/mathscinet-getitem?mr=2231961).

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 $x \in LH$, $\| x \|_2 < 1$ we have $| \langle x, u_g^*p_ku_g - \tau(p) \rangle | \leq \| E_{LH}(u_g^* p u_g - \tau(p) ) \|$ $_2 = \| E_{L(H \cap gHg^{-1})} (p_k - 1/2) \|_2 < \delta$. The same inequality holds for the other minimal projection $1 - p_k$.

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 $| \langle x, u_g^* p_j u_g - \tau(p_j) \rangle | < \delta$. This proves the induction step. QED

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 http://www.ams.org/mathscinet-getitem?mr=703810. Let's give the proof here.

Let $\varepsilon > 0$ be given and take $n \in \mathbb N$ such that $1/2^n < \varepsilon/2$. 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 $| \langle x, u_g^*pu_g - \tau(p) \rangle | < \| x \|_2 \varepsilon/2^{n + 1}$. 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 $LG$.

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 < (\Sigma_k \tau(p_k)^2 ) + \Sigma_k \varepsilon/2^{n + 1} < \varepsilon$.

Since $\varepsilon$ was arbitrary we conclude that $u_g$ and $v$ are orthogonal. Hence since $v$ was arbitrary we conclude that $\mathcal N_{LG}(LH)'' = L(\mathcal N_G(H))$.

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Apparently I can not get some to the tex to compile correctly. I'm not sure how to fix this though. –  Jesse Peterson Jun 1 '10 at 0:16
    
Thanks Jesse. It is nice to have this direct proof in a Popa's way. –  Arnaud Brot Jun 2 '10 at 1:00
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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.

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

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"Pukanzki" ---> "Pukanszky" in most sources I know of, just for reference. –  Yemon Choi Apr 30 '10 at 3:57
    
Thanks Owen for your answer. Unfortunately, in the book of Sinclair and Smith they are proving what you are talking about but don't discuss more about. Do you have an idea of an other reference for this topic? –  Arnaud Brot Apr 30 '10 at 20:32
    
Hey Arnaud, unfortunately I don't know any other good references, and in fact I don't think much else is known. –  Owen Sizemore May 6 '10 at 3:25
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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.

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At a quick glance, the subtlety/difficulty is that a priori the Cartan subalgebra condition doesn't tell us that individual elements of $G$ normalize $LH$ (but I may have misunderstood) –  Yemon Choi Apr 27 '10 at 18:06
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