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.



normalizer of algebras and groups
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 $\mathcal N\_{LG}(LH)'' = L(\mathcal N\_G(H))$.


