A unital $C^{*}$ algebra is called a "Path connected algebra" if the spectrum of all its elements is a path connected subset of $\mathbb{C}$.

What is an example of a non commutative path connected $C^*$ algebra? What is an example of a simple path connected $C^{*}$ algebra? In particular does $C^{*}_{red} F_{2}$ satisfiy this property?

As another question: Is the tensor product of two path connected algebra, a path connected algebra?(For spatial norm).

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    $\begingroup$ Oh you meant simple as in simple not as in easy... $\endgroup$ Feb 7, 2015 at 0:31
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    $\begingroup$ Well $\mathbb{C}$ is simple. But I assume you want something less trivial. A stab in the dark: One could imagine putting some group action on $C(X)$ such that none of its non-trivial ideals are $G$-invariant and hope that $C(X)\rtimes G$ is simple. $\endgroup$ Feb 7, 2015 at 0:37
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    $\begingroup$ The spectrum of a nontrivial projection (i.e., nonzero and not the unit) is disconnected. Thus, a unital C*-algebra with the property you consider cannot contain any nontrivial projections. Conversely, if a C*-algebra contains a normal element with disconnected spectrum, then it contains a nontrivial projection. Thus, in a unital, simple C*-algebras without nontrivial projections (such as the Jiang-Su algebra), at least the spectrum of every normal element is connected. $\endgroup$ Feb 25, 2015 at 18:46
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    $\begingroup$ Hannes' observation can be improved: all elements of a C*-algebra with no non-trivial projections have connected spectrum. Indeed given an element with disconnected spectrum the holomorphic functional calculus gives a non-trivial idempotent. This will be similar to some self-adjoint idempotent (see Blackadar's K-Theory book) which will be non-trivial. $\endgroup$ Mar 4, 2015 at 15:13
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    $\begingroup$ Example of a connected C∗ algebra : See B.E. Blackadar. A simple unital projectionless C∗-algebra. J. Operator Theory, 5:63–71, 1981. By the comments of Hannes Thiel and Sam Evington this algebra is an example of a connected C* algebra. But is it also path connected (remark by Ali Taghavi) ? $\endgroup$
    – jjcale
    Nov 7, 2015 at 16:29


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