gr.group theory - 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 $LH notprec_{LH} L(H cap gHg^{-1})$ (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 $LH notprec_{LH} L(H cap gHg^{-1})$ 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 $2delta$ 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 $pLH notprec_{pLH} pL(H cap gHg^*)$ 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 $mathcal N_{LG}(LH)'' = L(mathcal N_G(H))$.