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 $B(L^2(A))$.



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> What "is" this algebra $M$, more precisely: What is the type of $M$? Is it a factor?
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von Neumann algebras generated by commutators
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.


