Unitary Operators are Connected in a $C^*$-algebra Let $A$ be a unital $C^*$- algebra. Let $U = \{ u \in A : u^*u=uu^*=1\}$ be the unitary group of $A$.
Let $U'= \{ e^{ia_1}e^{ia_2} \cdots e^{ia_n} : a_k = a^*_k \in A, \text{for } 1\leq k \leq n \}$.
Show that $U'$ is the connected component of the identity in $U$.
If $A$ is commutative, show that $U' = \{ e^{ia} : a = a ^* \in A\}$.
I thought of using the theorems below from Banach Algebra Techniques in Operator Theory by Douglas.
Following the proof of 2.14, if $f = e^{ia} \in A$. Then $f\in U$. Consider $\phi: [0,1 ] \rightarrow e^(A)$ defined by $\phi(\lambda) = e^{i \lambda a}$, I don't see why $f \in U'$, and $e^A$ is contained in $U'$...
I also showed that if $\| u-1\| <2 $ , then we have $-1 \ne spec(u) $ and we can write $u = e^{ia}$ for some self-adjoint $a \in A$
Any help or suggestions will be appreciated.
Thank you!

 A: Any unitary sufficiently close to $1$ is the exponential of a self adjoint element. More precisely, if the distance from a unitary $u$ to $1$ is smaller than $2$, then $-1 \notin \mathrm{spec}(u)$, which allows you to define a logarithm for $u$ using continuous functional calculus.
Now if $u$ is connected to $1$ by a path $u_t$, $t \in [0,1]$, you can easily find
$$0 = t_0 < t_1 < \ldots < t_n = 1$$
such that the distance between $u_{t_i}$ and $u_{t_{i+1}}$ is always smaller than $2$.
Can you take it from there?
A: First of all, in the case that $A$ is commutative the result easily follows: Note that, when $ab=ba$, then $e^{b}\cdot e^{a}=e^{a}\cdot e^{b}=e^{a+b}$. Using this elementary result (which follows easily from the identity $e^x=\sum_{k=0}^\infty\frac{x^k}{k!}$), we have that
$$U'=\{e^{ia_1}\cdots e^{ia_n}: n\geq1,a_k\in A_{sa}\}=\{e^{i(a_1+\dots+a_n)}: n\geq1, a_k\in A_{sa}\}=\{e^{ia}:a\in A_{sa}\}$$
where $A_{sa}$ denotes the self-adjoint part of $A$.
Now forget about the commutative case and let $A$ be any unital $C^*$-algebra. Let's denote by $C$ the path connected component of $1_A$ in $U(A)$, i.e.
$$C=\{u\in U(A): \text{ there exists a continuous path from }u\text{ to }1_A\} $$
First of all, if $a\in A$ is a self-adjoint element, then the unitary $e^{ia}$ is path-homotopic to $1_A$: simply define $\alpha:[0,1]\to U(A)$ by setting $\alpha(t)=e^{ita}$. Then $\alpha(0)=1_A$, $\alpha(1)=e^{ia}$ and it is easy to verify that $\alpha$ is continuous. Therefore $e^{ia}\in C$ for all $a\in A_{sa}$. Moreover, if $u,v\in C$, then $uv\in C$. Indeed, if $\alpha,\beta:[0,1]\to U(A)$ are continuous paths from $u$ to $1_A$ and from $v$ to $1_A$ respectively, then $t\mapsto\alpha(t)\cdot\beta(t)$ is a continuous path from $uv$ to $1_A$, so $uv\in C$. These two results combined allow us to infer that, finite products of elemenets of the form $e^{ia}$ where $a\in A_{sa}$ belong to $C$. In other words, $U'\subset C$. Also, observe that $U'$ is closed under multiplication (obviously).
Now for the other inclusion, which is the hard one. Since $C$ is a path-connected component, it is connected, so if we show that $U'$ is a clopen subset of $C$, we are done.

*

*$U'$ is open in $C$: We show that $D(u,2)\cap C\subset U'$ for any $u\in U'$. Indeed, if $u\in U'$ and $v\in C$ satisfies $\|u-v\|<2$, then $\|u^*v-1_A\|<2$, so $-2\not\in\sigma(u^*v-1_A)$, so $-1\not\in\sigma(u^*v)$, therefore, as you explain in your post, we can define a logarithm which will give us $u^*v=e^{ih}$ for some self-adjoint element $h$, thus $v=ue^{ih}$, so $v\in U'$, (because $U'$ is closed under multiplication).


*$U'$ is closed in $C$: It suffices to show that $U'$ is closed in $U(A)$ (why?). So we will show that $U(A)\setminus U'$ is open. Note that, $(U(A), \cdot)$ is a group and $U'$ is a subgroup of $U(A)$. Let's pause for a moment and have a group-theory throwback:
Assume that $G$ is a group and $H\leq G$ is a subgroup. We have the relation $\sim$ on $G$ given by $x\sim y$ iff $y^{-1}x\in H$. This is an equivalence relation and $[x]=xH=\{xh: h\in H\}$. We take the quotient set $\{[x]: x\in G\}$ and we make a choice from each class (using of course the axiom of choice) and in this way we obtain a set $P\subset G$ so that, if $x,y\in P$ are distinct then $x\not\sim y$ and if $z\in G$ there exists a unique $x\in P$ so that $x\sim z$. From this we can conclude that
$$G=\bigcup_{x\in P}xH $$
Now if $x_0\in P$ is the element that satisfies $[x_0]=H$, we have that
$$G\setminus H=\bigcup_{x\in P-\{x_0\}}xH$$
So we have written the set $G\setminus H$ as a disjoint union of "copies" of $H$. But this will solve the problem, as we shall see:
Okay, back to our situation. We want to show that $U(A)\setminus U'$ is open; by the preceding remark, we have a set $P\subset U(A)$ so that
$$U(A)\setminus U'=\bigcup_{v\in P}vU'.$$
It is elementary to verify that the map $U'\mapsto vU'$, $u\mapsto vu$ is a topological homeomorphism for all $v\in P$. But, as we did in bullet 1, the set $U'$ is open in $U(A)$ (the exact same proof will work). Therefore, since every set $vU'$ is topologically homeomorphic to $U'$, every set $vU'$ is open in $U(A)$. Unions of open sets are open, so we have shown that $U(A)\setminus U'$ is open in $U(A)$, as we wanted.
I hope this is detailed enough. The idea of the proof is demonstrated in M. Rordam's book on K-theory of $C^*$-algebras, you might find Chapter 2 of this book very helpful.
