# Connectedness of Invertible elements in a non- commutative C*- algebra

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

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 $$e^{it} \mapsto 1$$.

• 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.? Aug 4 at 17:12
• I can't imagine any simpler condition than "$K_1(A)$ is trivial". Aug 4 at 17:58
• (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.) Aug 5 at 11:07

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
• I am just learning about operator K-theory. Aug 4 at 16:47
• 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. Aug 4 at 17:19
• $C(X)^{\times}$ 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. Aug 4 at 17:36
• 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. Aug 4 at 17:36