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Let $A$ be a UHF-algebra of type $n^{\infty}$ and denote its unique and faithful trace by $\tau$. Let $L^2(A)$ be the Hilbert space of the GNS-representation associated to $\tau$. We have two commuting representations $L \colon A \to B(L^2(A))$ and $R \colon A^{\rm op} \to B(L^2(A))$ and by the universal property of the maximal tensor product and the nuclearity of $A$, we obtain a $*$-homomorphism $A \otimes A^{\rm op} \to B(L^2(A))$. I think the image of this is weakly dense, i.e. the von Neumann algebra completion of $A \otimes A^{\rm op}$ in this representation is type I and agrees with $B(L^2(A))$. Now consider the $*$-subalgebra $$ B = \left\{ \sum_{i} L_{a_i}R_{b_i} \ |\ \sum_{i} a_i b_i = 0\right\} $$ spanned by those operators corresponding to elements of $A \otimes A^{\rm op}$ that lie in the kernel of the multiplication map. Let $M$ be the weak closure of $B$ in $B(L^2(A))$.

What "is" this algebra $M$, more precisely: What is the type of $M$? Is it a factor?

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I see that $B$ is an $A-A$ bimodule, but why is $B$ a subalgebra? –  Dave Penneys May 19 at 15:22
    
@DavePenneys: The two actions commute and $R_aR_b = R_{ba}$, since it is a representation of the opposite algebra. If I now take two elements, say $\sum_i L_{a_i}R_{b_i}$ and $\sum_j L_{c_j}R_{d_j}$ of $B$ and multiply them, I end up with $\sum_{i,j} L_{a_ic_j}R_{d_jb_i}$, but $\sum_{i,j} a_ic_jd_jb_i$ should vanish since $\sum_j c_jd_j$ vanishes by assumption. –  Ulrich Pennig May 19 at 15:28
    
Whoops, to get a *-subalgebra, I probably have to demand that the sum $\sum_i b_ia_i$ vanishes as well in the definition of $B$. –  Ulrich Pennig May 19 at 17:22

1 Answer 1

up vote 5 down vote accepted

Let $\xi_0 \in L^2(A)$ denote the cyclic vector corresponding to the identity in $A$, and let $P_0$ denote the rank-one projection corresponding to $\xi_0$. Then we clearly have $xP_0 = P_0 x = 0$ for all $x \in M$, and hence $M \subset P_0^\perp B(L^2(A)) P_0^\perp$. I claim that we actually have equality. This should be a well known fact to experts in II$_1$ factors (one just needs that $A$ is a unital $*$-algebra which generates a II$_1$ factor), however I don't know a reference off hand so I'll give a proof instead.

To see that $P_0^\perp \in M$ note that if $u \in A$ is a unitary then the spectral projection of $1 - L_{u}R_{u^*}$ corresponding to $\mathbb C \setminus \{ 0 \}$ is contained in $M$. The supremum of these projections over all $u$ is equal to $P_0^\perp$ since $\tilde A := L(A)''$ is a factor.

Note that the representations $L$ and $R$ extend to normal commuting representations of $\tilde A$ (for which I will use the same notation), and it is then easy to show that in the definition of $B$ we may allow $a_i$ and $b_i$ to be in $\tilde A$.

Note also that if $A_0 \subset \tilde A$ is a von Neumann subalgebra, and if $Q$ denotes the projection onto the closure of $(A_0' \cap \tilde A) \xi_0$, then $Q \in \mathbb CP_0 \oplus M$. This follows from the observation that for $\eta \in L^2(A)$ we have that $Q\eta$ is the unique element of minimal norm in the convex closure of $\{ L_uR_{u^*} \eta \mid u \in \mathcal U(A_0) \}$. Hence, $Q$ is in the weak operator topology convex closure of the set $\{ L_uR_{u^*} \mid u \in \mathcal U(A_0) \} \subset \mathbb CP_0 \oplus M$.

In particular, if $p \in \mathcal P(\tilde A)$ is a projection and we set $A_0 = ( \mathbb Cp \oplus \mathbb Cp^\perp )' \cap \tilde A$, then since $\tilde A$ is a factor it follows that the rank one projection onto $(p - \tau(p)) \xi$ is contained in $M$.

Suppose now that we have a self-adjoint operator $T \in M' \cap P_0^\perp B(L^2(A)) P_0^\perp$. Then for $p \in \mathcal P(\tilde A)$ a non-zero projection, we have shown that there exists $\lambda_p \in \mathbb R$ so that $T( p - \tau(p) )\xi_0 = \lambda_p (p - \tau(p))\xi_0$. If $q \in \mathcal P(\tilde A)$ then we have \begin{align} \lambda_p \langle (p - \tau(p))\xi_0, (q - \tau(q))\xi_0 \rangle &= \langle T(p - \tau(p))\xi_0, (q - \tau(q))\xi_0 \rangle \\ &= \langle (p - \tau(p))\xi_0, T(q - \tau(q))\xi_0 \rangle \\ &= \lambda_q \langle (p - \tau(p))\xi_0, (q - \tau(q))\xi_0 \rangle. \end{align} Since $\tilde A$ is a factor, any pair of non-zero projections has a third projection so that these inner-products are non-zero. Hence, $\lambda := \lambda_p = \lambda_q$ for all non-zero projections $p, q \in \mathcal P(\tilde A)$. The span of projections is norm dense in $\tilde A$ by the spectral theorem, hence $T(x - \tau(x) ) \xi_0 = \lambda (x - \tau(x))\xi_0$ for all $x \in \tilde A$, and so $T = \lambda P_0^\perp$. Since $T$ was arbitrary, the double commutant theorem then gives the result.

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Nice proof. Thanks! –  Ulrich Pennig May 19 at 20:05

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