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Let $F_{2}$ be the free group with two generators.

Then $F_{2}=\{\text{odd words}\}\sqcup\{\text{even words}\}$. This gives us a $Z_{2}$ graded structure for $C^{*}_{red} (F_{2})$, in a natural way.

My question:

Is there another $Z_{2}$ graded structure for $C^{*}_{red} (F_{2})$ which is not graded isomorphic to the above $Z_{2}$- graded structure?

Please consider the same question for $C^{*}_{red} (F_{1}) \sim C(\mathbb{S}^{1})$.

Namely: is there a $Z_{2}$ graded structure for $C(\mathbb{S}^{1})$ which is not graded isomorphism to the standard grading structure(decomposition to even and odd continuous functions)?

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    $\begingroup$ Have you considered the $\mathbb{Z}_2$-grading associated with the involutive automorphism of $F_2$ obtained by flipping the generators? $\endgroup$
    – Alain Valette
    Commented Dec 23, 2013 at 22:34
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    $\begingroup$ For $C(S^1)$, you also have the grading associated to the automorphism $z\mapsto z^{-1}$ of $S^1$ (viewed as the group of complex numbers of modulus one). The sub-algebra of elements of degree 0, is isomorphic to $C[-1,1]$, proving that this grading is not isomorphic to the one by even/odd functions (which is associated to $z\mapsto -z$). $\endgroup$
    – Alain Valette
    Commented Dec 23, 2013 at 22:47
  • $\begingroup$ @AlainValette Thank you very much for your comments. I consider the grading structure for $C^{*}_{red}( F_{2})= A_{0} \oplus A_{1}$ where $A_{0}$ is the banach space generated by even words and $A_{1}, with odd words. $\endgroup$
    – Ali Taghavi
    Commented Dec 24, 2013 at 5:50
  • $\begingroup$ In particular, does the graded structure which you mentioned is graded isomorphic to this structure which I mentioned? $\endgroup$
    – Ali Taghavi
    Commented Dec 24, 2013 at 6:01
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    $\begingroup$ Is the tag 'graded-banach-algebras' not overly specific? $\endgroup$
    – Ricardo Andrade
    Commented Apr 8, 2014 at 22:03

1 Answer 1

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It is well-known that $K_1(C^*_{\rm red}(F_2))={\mathbb Z^2}$ with generators given by $[u]$ and $[v]$, where $F_2=\langle u,v\rangle$.

Now, the automorphism of order two associated with the even-odd grading is multiplying each generator by $-1$, which is homotopic to the multiplication by $1$. Hence it is trivial on $K_1$.

On the other side, the automorphism of order two that comes mapping $u$ to $v$ and $v$ to $u$ gives rise to a non-trivial action on $K_1$. Thus, the two automorphisms of order two cannot be conjugate - and hence the associated graded algebras cannot be isomorphic.

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  • $\begingroup$ thank you very much for the answer. I think that I can get the answer to my question from first version of your answer: two "Flipping" and "antipodal" automorphism of $\mathbb{Z}^{2}$ are not conjugate. I think in the revised version, some thing is missing:the two C* automorphisms are not homotopic because their induced K group automorphism are not equal. Thanks again for your help on my question. $\endgroup$
    – Ali Taghavi
    Commented Apr 9, 2014 at 8:24
  • $\begingroup$ Just to clarify, the first version contained a mistake and the current version is correct. $\endgroup$
    – Andreas Thom
    Commented Apr 9, 2014 at 10:03
  • $\begingroup$ what is the homotopy which you mentioned in the current version? $\endgroup$
    – Ali Taghavi
    Commented Apr 9, 2014 at 11:12
  • $\begingroup$ for the homotopy, are you considering a curve in $\mathbb{C}-\{0\})$ which connect 1 to -1? $\endgroup$
    – Ali Taghavi
    Commented Apr 9, 2014 at 13:08
  • $\begingroup$ Yes, exactly, along the unit circle. $\endgroup$
    – Andreas Thom
    Commented Apr 9, 2014 at 21:10

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