I'm working through Li's and Barlak's Cartan Subalgebras and the UCT Problem but I'm stuck at one of the simpler proofs of the paper. On page 9 they deal with masas (maximal abelian subalgebras) of a C*-algebra $A$ and look at the crossed products $A \rtimes_\alpha \mathbb{Z}_n$ where $\alpha\in Aut(A)$ with $\alpha^n=id$. I now have two questions:

1.) Before prop. 4.1 the paper says "let $t$ be the unitary in (the multiplier algebra of) $A\rtimes_\alpha \mathbb{Z}_n$ implementing the $\mathbb{Z}_n$-action." I'm not quite familiar with theory of crossed products. What does this mean and why does such a unitary exist? (Do you have literature recommendations?)

2.) In the proof they compute $[t^k x_k, b]$ and the result seems to be $t^k(x_kb-\alpha^{-k}(b)x_k)$. By construction of crossed products I would've guessed $[t^k x_k, b]=t^k\left(x_{k}\alpha^{k}\left(b\right)-bx_{k}\right)$. Where's my mistake?


The unitary exists by the definition of the crossed product. In the case of a $\mathbb{Z}/n\mathbb{Z}$-action you can think of an element of the crossed product as a linear combination $$ \sum_{n = 0}^{n-1} a_n t^n $$ with $a_n \in A$ and where $t$ is a unitary (i.e. $t^* = t^{-1}$), such that the following relations hold $$ tat^* = tat^{-1} = \alpha(a) $$ In particular $t^kat^{-k} = \alpha^k(a)$ and $t^{-k}at^{k} = \alpha^{-k}(a)$. In the case $A = \mathbb{C}$ and $\alpha$ trivial, you obtain the group $C^*$-algebra of $\mathbb{Z}/n\mathbb{Z}$ this way, which you might be more familiar with. Concerning the second question we have $$ [t^kx_k, b] = t^kx_kb - bt^kx_k = t^kx_kb - t^k(t^{-k}bt^k)x_k = t^k(x_kb - \alpha^{-k}(b)x_k)\ . $$ Therefore their calculation seems to be correct. A good source of information about crossed products is this book by Dana P. Williams.

Here are some more details about crossed products. Formally the crossed product is defined as follows: Let $G = \mathbb{Z}/n\mathbb{Z}$. Consider the algebra $C(G,A)$ of functions on $G$ with values in $A$ equipped with the following twisted multiplication $$ (f \cdot h)(x) = \sum_{g \in G} f(g)\alpha_g(h(g^{-1}x)) $$ and the $*$-operation $$ f^*(x) = \alpha_x(f(x^{-1})^*). $$ This $*$-algebra has several interesting faithful representations, for example a regular and a maximal one. These correspond to (possibly different) completions to $C^*$-algebras, e.g. the reduced or maximal crossed product.

To obtain the picture described above, we can then identify the unitary $t$ with the function $$ t(x) = \begin{cases} 1_A & if x = 1 \\ 0 & else \end{cases} $$ where $1 \in \mathbb{Z}/n\mathbb{Z}$ refers to the generator on the right hand side and $1_A$ is the unit of $A$. Strictly speaking this only works if $A$ is unital. Otherwise $t$ will not be an element in the crossed product, but in its multiplier algebra. Then we have $$ t\cdot a \cdot t^*(x) = \sum_{g,h} t(g)\alpha_g(a(h)\alpha_h(t^{-1}((gh)^{-1}x))) = \alpha(a(x)) $$ which corresponds to the relation mentioned above.

The crossed product therefore is a $C^*$-algebra that contains the original algebra $A$ (in the unital case) and in which the action becomes inner.

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  • $\begingroup$ Thanks for your answer! I see that if we have a $t$ such that $tat^*=\alpha(a)$ elements of the above polynomial form are exactly the elements of the crossed product. But why is every automorphism of this form? Is this a general property of the multiplier algebra? $\endgroup$ – worldreporter14 Aug 20 '17 at 15:54
  • $\begingroup$ Note that $t$ is not an element of $A$, i.e. the automorphism does not have to be inner from the start. However, the crossed product is defined in such a way that the automorphism becomes inner by adding another element $t$ that satisfies $tat^* = \alpha(a)$. $\endgroup$ – Ulrich Pennig Aug 20 '17 at 15:58
  • $\begingroup$ All right! I think I get it now. Tanks a lot! $\endgroup$ – worldreporter14 Aug 20 '17 at 16:19

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