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The Gelfand Naimark Segal theorem says that any complex C* algebra $A$ is isometrically isomorphic to a C* sub-algebra of bounded operators on a Hilbert space.

Here we see that the set of all Invertible elements are connected in $B(H)$ for a Hilbert space $H$.

I wanted to ask if there is a category of non-commutative algebras for which the set of invertible elements is connected. Maybe a class of C*- algebras for which the Map in the Gelfand–Naimark-Segal construction is Surjective

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The group $K_1(A)$ is the invertibles in $\lim M_n(A)$ modulo the path component of the identity. (Here the embedding of $M_n(A)$ into $M_{n+1}(A)$ goes by adding an extra column and row of zeros.) So what you're asking for is something like triviality of $K_1(A)$, though not exactly the same. Anyway there are plenty of C*-algebras whose invertibles aren't connected, the simplest being $C(\mathbb{T})$ where the identity function $e^{it} \mapsto e^{it}$ is invertible but isn't connected by a path of invertibles to $e^{it} \mapsto 1$.

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  • $\begingroup$ Hi, thank you for the answer, so yea, what I am actually asking is, is there a characterization of non- commutative C*- algerbas $A$ for which $K_1(A)$ is trivial.? $\endgroup$
    – User93709
    Aug 4 at 17:12
  • $\begingroup$ I can't imagine any simpler condition than "$K_1(A)$ is trivial". $\endgroup$
    – Nik Weaver
    Aug 4 at 17:58
  • $\begingroup$ (Embedding $M_n(A)$ into $M_{n+1}(A)$ by adding an extra column and row of zeros and a $1$ in the $(n+1, n+1)$ entry.) $\endgroup$
    – Nik Weaver
    Aug 5 at 11:07
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Let me thoroughly answer this question in the commutative case. Hopefully this is helpful for the case when the $C^{*}$-algebra is not necessarily commutative.

If $B$ is a commutative $C^{*}$-algebra, then $B$ is isomorphic to $C(X)$ for some compact Hausdorff space $X$. Here, $X$ is simply the collection of all maximal $^{*}$-ideals of the $C^{*}$-algebra $B$.

Furthermore, if $B$ is generated by an element $a$, then $B$ is isomorphic to $C(\sigma(a))$ by an isomorphism that maps $a$ to the inclusion mapping $\iota:\sigma(a)\rightarrow\mathbb{C}$.

Let $X$ be a compact Hausdorff space. Suppose therefore that $g,h\in C(X)$ are invertible elements. Then there is a path from $g$ to $h$ in $C(X)^{\times}$ if and only if $g,h:X\rightarrow\mathbb{C}\setminus\{0\}$ are homotopic. Let $\pi:\mathbb{C}\setminus\{0\}\rightarrow S^{1}$ be the function defined by $\pi(z)=\frac{z}{|z|}$. Then $\pi$ is a homotopy equivalence, so there is a path from $g$ to $h$ in $C(X)^{\times}$ if and only if $\pi\circ g$ and $\pi\circ h$ are homotopic to each other.

If $X$ is a topological spaces, then observe that the collection of mappings $f:X\rightarrow S^{1}$ is an abelian group where $(f\cdot g)(x)=f(x)\cdot g(x)$. Homotopy equivalence is a congruence on this abelian group, so the collection of homotopy classes $[X,S^{1}]$ of mappings from $X$ to $S^{1}$ is also an abelian group. One might suspect that there is a relationship between $[X,S^{1}]$ and some sort of homology, cohomology, or homotopy groups, and there is.

It turns out that if $X$ is homotopy equivalent to a CW complex, then the group $[X,S^{1}]$ is isomorphic to the singular cohomology group $H^{1}(X,\mathbb{Z})$. If $X$ is not homotopy equivalent to a CW complex, then we will have to use a different notion of cohomology. Fortunately, it has been shown by Peter Huber [1] that for paracompact Hausdorff spaces $X$, the group $[X,S^{1}]$ is isomorphic to the Cech-cohomology group $\check{H}^{1}(X,\mathbb{Z})$. These are actually special cases of more general isomorphisms $H^{n}(X,G)\simeq[X,K(G,n)]$ where $K(G,n)$ is the Eilenberg-MacLane spaces where $X$ is homotopy equivalent to a CW-complex or the isomorphism $\check{H}^{n}(X,G)\simeq[X,K(G,n)]$ where $X$ is paracompact and Hausdorff and $G$ is countable. The special case follows since $S^{1}=K(\mathbb{Z},1)$.

  1. Huber, P.J. Homotopical Cohomology and čech Cohomology. Math. Ann. 144, 73–76 (1961). https://doi.org/10.1007/BF01396544
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  • $\begingroup$ I am just learning about operator K-theory. $\endgroup$ Aug 4 at 16:47
  • $\begingroup$ Can you tell what $C(X)^{\times}$ stands for? Can you give a reference for the result: There is a path from $g$ to $h$ in $C(X)^{\times}$ if and only if $g,h:X\rightarrow\mathbb{C}\setminus\{0\}$ are homotopic. $\endgroup$
    – User93709
    Aug 4 at 17:19
  • $\begingroup$ $C(X)^{\times}$ is the set of all invertible elements in $C(X)$. Here $C(X)$ is the space of all bounded continuous function $f:X\rightarrow\mathbb{C}$ (but since $X$ is compact, all such functions are bounded) with the compact open topology or equivalently, the topology given by the $\|\cdot\|_{\infty}$ norm. $\endgroup$ Aug 4 at 17:36
  • $\begingroup$ The correspondence between paths and homotopies is a special case of the following well-known fact that is mentioned in many general topology and algebraic topology textbooks: Suppose that $f:X\times Q\rightarrow Y,F:Q\rightarrow C(X,Y)$ are functions defined by $F(t)(x)=f(x,t)$ where $X,Q$ are compact and $C(X,Y)$ is the space of all continuous functions $f:X\rightarrow Y$ with the compact open topology. Then $f$ is continuous iff $F$ is continuous. $\endgroup$ Aug 4 at 17:36

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