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For what type of $C^{*}$ algebras $A$, the group of invertible elements of $C(S^{2}) \otimes A$ is a connected group?

All finite dimensional $A$ satisfy this property.

Is it true to say that $A$ satisfy the above property if and only if the group of invertible elements of $A$ is connected?

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  • $\begingroup$ You can generally replace "invertible" with "unitary" up to homotopy when working with C* algebras, so by Bott periodicity I think you just want to look at $K_1(A)$. $\endgroup$
    – Paul Siegel
    Commented Nov 2, 2015 at 16:15
  • $\begingroup$ @PaulSiegel I think that the following two, are not equivalent: 1.unitary group of A is connected. 2. $K_{1}(A)$ is trivial, right? $\endgroup$
    – Ali Taghavi
    Commented Nov 2, 2015 at 16:26
  • $\begingroup$ Perhaps I am missing something, but isn't it the case that for a Banach algebra $B$, $GL(B)$ is connected iff it is path-connected? I think it's then clear that $GL(C(S^2)]\otimes A)) = C(S^2, GL(A))$ is path-connected iff $GL(A)$ is? $\endgroup$
    – user75274
    Commented Nov 8, 2015 at 19:05
  • $\begingroup$ @user75274 Is it obvious that the connected component of GL(A) has laways trivial second homotopy? $\endgroup$
    – Ali Taghavi
    Commented Nov 9, 2015 at 15:17
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    $\begingroup$ @Ali: If A is stabilized by the compact Operators, i.e. $A=(B \otimes K)^+$, then the answer is true if and only if $K_1(C(S^2) \otimes A)=0$. Bott periodicity Needs $C(R^2)$ instead of $C(S^2)$. $\endgroup$
    – hänsel
    Commented Mar 30, 2016 at 17:18

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