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Let $A$ be a $C^*$-algebra, let $p$ and $q$ be Murray-von Neumann equivalent projections in $A$, i.e. there is a partial isometry $v$ such that $v^*v = p$ and $vv^* = q$. Suppose $\alpha \in Aut(A)$ is an automorphism of $A$, such that $\alpha(p) = p$ and $\alpha(q) = q$.

Under what condition is it true that $\alpha$ fixes $v$ as well?

For example, if we assume that $\lVert p - q \rVert < 1$, then $p$ and $q$ are Murray-von Neumann equivalent via a partial isometry explicitly constructed from $p$ and $q$ only - using continuous functional calculus. Thus, $\alpha$ fixes $v$ as well in this case.

If it helps, you may assume $A$ to be a unital Kirchberg algebra.

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    $\begingroup$ Well, in the very special case $p=q=1$, this is just the question "which unitaries are fixed by $\alpha$?" What sort of condition do you have in mind? Or should this be read as "when does there exist a $v$ fixed by $\alpha$"? $\endgroup$ – Michael Jul 29 '13 at 19:33
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    $\begingroup$ I am wondering more about the case, where $p$ and $q$ are arbitrary. The question is: Is there a partial isometry $v$ that mediates the equivalence and is fixed by the automorphism. In the example I gave, there is such a $v$. $\endgroup$ – Ulrich Pennig Jul 29 '13 at 19:38
  • $\begingroup$ I think, the argument I sketched also works, if I can find a path from $p$ to $q$, which is fixed under pointwise application of $\alpha$. $\endgroup$ – Ulrich Pennig Jul 29 '13 at 19:40
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    $\begingroup$ Yes, if you can find a fixed path from $p$ to $q$, you just have to exhibit $p=p_0,p_1,\ldots,p_n=q$ all fixed with $\|p_i-p_{i+1}\|<1$. Then at each step $p_i$ and $p_{i+1}$ are unitarily equivalent via $u_i:=(p_i+p_{i+1}-1)|p_i+p_{i+1}-1|^{-1}$ wich is fixed. So $u=u_1\cdots u_n$ is a fixed unitary such that $q=u^*pu$ and you just have to set $v:=pu$ to get your fixed partial isometry. But this strategy requires at least that $p$ and $q$ be homotopic, which is not guaranteed if they are just MvN equivalent in a purely infinite $C^*$ algebra. $\endgroup$ – Julien Jul 29 '13 at 21:17
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    $\begingroup$ Because of your comment, your question really is: Which conditions do you have to impose on $\alpha$ or $A$ such that MvN equivalence of any two projections $p,q\in A^\alpha$ inside $A$ is already a MvN equivalence inside $A^\alpha$? $\endgroup$ – Gabor Szabo Jul 29 '13 at 21:40
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This isn't really an answer to the question, but I thought I'd add a simple-minded example demonstrating how things can break down.

Let $A$ be the full algebra of bounded operators on the Hilbert space $L^2(\mathbb{T})$. Let $u \in A$ be multiplication by $z \mapsto z$. Let $\alpha \in \mathrm{Aut}(A)$ be conjugation by $u$. The fixed subalgebra is all elements commuting with $u$. But, $u$ generates all of $C(\mathbb{T})$ and the commutant of $C(\mathbb{T})$ in $A$ is $L^\infty(\mathbb{T})$. Certainly there are plenty of pairs of projections in $L^\infty(\mathbb{T})$ that are MvN equivalent in $A$. But, only trivial equivalences are possible the (commutative) fixed subalgebra.

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    $\begingroup$ Another simple-minded example in this vein would be almost anything involving $A=\mathcal{O}_2$. There are many automorphism whose fixed point algebra has non-trivial K-theory. While every pair of non-trivial projections in $\mathcal{O}_2$ is MvN equivalent, this is certainly not true for every pair of non-trivial projections in most fixed point algebras you encounter. $\endgroup$ – Gabor Szabo Jul 30 '13 at 15:21
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I don't think you'll be able to prove that $v$ itself is fixed in general, even if $p$ and $q$ are MvN equivalent in $A^\alpha$: you could perturb a fixed partial isometry to get another partial isometry from $p$ to $q$ that is no longer fixed.

It seems like you are asking for sufficient (and necessary?) conditions for the inclusion $A^\alpha\hookrightarrow A$ to induce an injective map $K_0(A^\alpha)\to K_0(A)$. The naive thing to do is to simply average the partial isometry $v$ to get something that is fixed and almost a partial isometry with the right source and range projections, and then use functional calculus in the fixed point algebra. However, this does not in general work since you don't in general get an approximate partial isometry (even if the group is compact, in which case you can take the conditional expectation).

In some cases you will get this conclusion for trivial reasons. For example, if $\alpha$ is a pointwise outer automorphism of $\mathcal{O}_2$, then $\mathcal{O}_2 \rtimes_\alpha \mathbb{Z}$ has trivial $K$-theory by the Pimsner-Voiculescu exact sequence, and this crossed product is moreover Morita equivalent to the fixed point algebra. In this case you get that $K_0(A^\alpha)\to K_0(A)$ is injective (and even an isomorphism!) because both $K$-groups are trivial. If you look at non-pointwise outer automorphisms of $\mathcal{O}_2$ (specifically, of finite order), than the examples that Gabor mentioned show that this can fail for non-pointwise outer automorphisms.

If you're working (or willing to work) with automorphisms on Kirchberg algebras of finite order (equivalently, actions of $\mathbb{Z}_n$), then pointwise outerness will of course not be enough, and the condition you need is probably the Rokhlin property (though this may be overkill). If $\alpha\colon \mathbb{Z}_n\to \mbox{Aut}(A)$ is an action with the Rokhlin property on any unital $C^*$-algebra (Kirchberg or not), then $A^\alpha\hookrightarrow A$ induces an injective map $K_\ast(A^\alpha)\to K_\ast(A)$ (you also get this for $K_1$).

There is also a definition of the Rokhlin property for automorphisms ($\mathbb{Z}$-actions), but I don't know if you get the same conclusion in this case. In the case of Kirchberg algebras, the Rokhlin property for an automorphism is equivalent to it being pointwise outer.

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