Can $xy$ and $yx$ lie in different connected components of the group of invertible elements of an algebra? What is  an example of  a  Banach or $C^{*}$ algebra $A$ which has two invertible elements $x, y$ such that $xy$ can not be connected to $yx$ in $G(A)$, the space of invertible elements of $A$.
A possible (weak) motivation for this question is that $K_{1}(A)$  is  an Abelian group.
Another motivation: Put  $F_{2}=$Free group on 2 generators $x,y$. then in the reduced $C^{*}-$algebra $C_{r}^{*} F_{2}$, $xyx^{-1}y^{-1}$ lies in the same connected component as of the identity.
 A: Yes, they can lie in different components. (Answer rewritten for clarity; conclusion the same.)
I'll give one example with a real Banach algebra and one with a complex $C^{\ast}$ algebra. Both rely on a result of [Samelson,H.,"Groups and spaces of loops, Comm.Math.Helv.,28,278-87 (1954)]
Theorem (Samelson) Define two maps $\lambda$ and $\rho: SU(2) \times SU(2) \to SU(2)$ by $\lambda(g,h)=gh$ and $\rho(g,h)=hg$. Then $\lambda$ and $\rho$ are not homotopic.
We will need to know that $\lambda$ and $\rho$ remain nonhomotopic if we embed $SU(2)$ into larger groups.
Lemma $\lambda$ and $\rho$ remain nonhomotopic if we view their target as the nonzero quaternions (embedding $SU(2)$ as the norm one quaternions.
Proof The map $q \mapsto q/|q|$ is a retraction from the nonzero quaternions onto $SU(2)$; any homotopy could be composed with this retraction to violate Samelson's result. $\square$
Lemma $\lambda$ and $\rho$ remain nonhomotopic if we view their target as $GL_2(\mathbb{C})$. 
Proof For a $2 \times 2$ matrix $X$, write $X^{\ast}$ for the conjugate transpose. Then $X \mapsto \sqrt{X X^{\ast}}^{-1} X$ is a retraction from $GL_2(\mathbb{C})$ onto $U(2)$, and $Y \mapsto Y \left( \begin{smallmatrix} \det(Y)^{-1} & 0 \\ 0 & 1 \end{smallmatrix} \right)$ is a retraction from $U(2)$ onto $SU(2)$. $\square$.

Now let $X = SU(2) \times SU(2)$ and let $H$ be either the quaternions or $\mathrm{Mat}_{2 \times 2}(\mathbb{C})$. Consider the algebra $A$ of continuous functions $X \to H$, equipped with the sup norm. If $H=\mathrm{Mat}_{2 \times 2}(\mathbb{C})$, this is a $C^{\ast}$ algebra, using the standard $C^{\ast}$ structure on $\mathrm{Mat}_{2 \times 2}(\mathbb{C})$. Let $x$ and $y \in A$ be the first and second projections from $X$ to $SU(2)$, followed by the obvious embeddings $SU(2) \to H$.
An element of $A$ is a unit if and only if it is valued in nonzero quaternions (if $H$ is the quaternions) or valued in $GL_2(\mathbb{C})$ (if $H$ is $\mathrm{Mat}_{2 \times 2}(\mathbb{C})$). So $xy$ and $yx$ han be joined by a path through units if and only if they give homotopic maps $X \to (\mathrm{nonzero quaternions})$ or $X \to GL_2(\mathbb{C})$ respectively -- which we showed they don't. $\square$
See these answers by Eric Wofsey and Achim Krause for more on the homotopy structure of maps $SU(2) \times SU(2) \to SU(2)$.
UPDATE 2/29/16 A very similar example appears in a 1973 paper of Yuen (Groups of Invertible Elements of Banach algebras, Yuen, Bull of the AMS Volume 79, Number 1 (1973), 82-84 Example 2); she credits it to E. Fadell. Also, Klaja and Ransford give an intriguing example of a different kind of noncommutativity -- a Banach algebra with elements $a$ and $b$ such that $1-ab$ is in the connected component of the identity but $1-ba$ is not!
