Let $D$ be an infinite UHF algebra, e.g. the infinite tensor product of the matrix algebra $M\_k(\mathbb{C})$. The permutation group $\Sigma\_n$ acts on the $n$-fold tensor product $D^{\otimes n}$ in a canonical way. 



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> What is known about the fixpoint algebra of this action? Is it simple? Is it again a UHF algebra? 
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fixpoint algebras of a permutation action
The answer is more interesting than I thought it would be. First, a clarification: you mean, that $k$ is fixed; the question still makes sense if $k$ is allowed to vary, and the analysis becomes more complicated (but can be done).


**Edit**: I made some serious (computational) errors in the original version, resulting in qualitative changes.


Just to give an idea of the method, let's do the case $n=2$ (outcome: the fp algebra is never UHF, but it is simple with unique trace).
We can rewrite $D \otimes D = \otimes (M\_k \otimes M\_k)$ (the first $\otimes$ on the right means the infinite tensor product), and the action is just $\otimes \pi$ where $\pi$ is the flip. The flip on $M\_k \otimes M\_k$ of course is implemented by an order two element of $M\_{k^2}$, that is, $\pi = \text{Ad } u$, where $u^2 = I$. It is not only product type, it is xerox (stationary). 


Thus the automorphism is of product type, namely $\otimes \text{Ad }\phi$, where $\phi$ is the representation ${\bf Z\_2 }\to M\_k \otimes M\_k$ given by $g \mapsto u$. Thus it comes within the purview of two papers of Handelman and Rossmann ([HR], Illinois J. Math. Volume 29, Issue 1 (1985), 51-95 *Actions of compact groups on AF C$^∗$-algebras* <http://projecteuclid.org/euclid.ijm/1256045841>, and the earlier one in the IUMJ). 


What is important is the character of $\phi$. This isn't entirely trivial to calculate (as I discovered after the first iteration). The trace of $u$ is $k$. Let $\chi$ denote the character of $\phi$. Then $\chi = {k \choose 2}\chi\_r + k \chi\_0$ where $\chi\_r$, $\chi\_0$ are respectively the regular rep char, and the trivial char.


Then from [HR], K$\_0$ of both the fixed point and crossed products is given as the direct limit of repeated multiplications by the character $\chi$, $\times\, \chi: R({\bf Z}\_2) \to R({\bf Z}\_2)$, where the $R(\cdot)$ is the representation ring of the enclosed group, the ring viewed as an ordered ring. Translating to a map
$\bf Z^2 \to \bf Z^2$, the matrix is (multn by $\chi$ on $R({\bf Z}\_2)$)
$$
\begin{pmatrix} {k \choose 2} + k & {k \choose 2} \\ {k \choose 2} & {k \choose 2} + k \\
\end{pmatrix}.
$$
[I'm not familiar with Latex, so used ams-tex. The matrix is circulant with top row ${k \choose 2} + k , {k \choose 2}$.]


Its determinant is $k^3\neq 0$, so the rank of K$\_0$ is two. Moreover, if $k' = 2k$, then we see that the matrix for $k'$ is the square of that for $k$. (Uniqueness of the trace follows from the matrix being primitive and the system being stationary.)


Now we can formalize this for any $n$. We write the action of $S\_n$ as a product type action on $D^{\otimes n} = \otimes (M\_k)^{\otimes n}$. Again the action on the little $n$-fold tensor product comes from a representation of $S\_n$, call the character of the representation $\chi$. We obtain the ordered K$\_0$ groups (actually ordered modules over $R(S\_n)$) as the direct limit $\times\, \chi: R(S\_n) \to R(S\_n)$--this formally inverts $\chi$, so we obtain the ring $R(S\_n) [1/\chi]$ (in case $\chi$ is a zero divisor—as occurs if it is the regular representation—this has to be interpreted as first factoring out the annihilator, then inverting). 


This will be rank 1 iff $\chi$ is a multiple of the regular representation (which forces $n!$ to divide $k^n$), and otherwise its rank is difficult to determine (at least for me) in general. So we obtain, the fixed point algebra is UHF iff $\chi$ is an integer multiple of the regular representation. Doing the computation even in the case that $n=3$ exhausted my patience, but I'm willing to bet that the thing is never a multiple of the reg rep.


At least, if $n!$ doesn't divide $k^n$, then $\chi$ is not a multiple of the reg rep char, so the fixed point algebra is a simple AF algebra with unique trace, and its K$\_0$-group has nonzero infinitesimals (and has rank at most the number of partitions of $n$). 


