# How to see that group of unitary operators with strong operator topology is connected?

This question stems from a remark on page 39 of Folland's book "Quantum Field Theory: A Tourist Guide for Mathematicians," in which Folland writes:

"The group of unitary operators [on a Hilbert space] is connected because, by spectral theory, every unitary operator belongs to a continuous one-parameter group of unitary operators and so is connected to the identity by a continuous path."

I know by spectral theory that, as a normal operator, a unitary operator $U$ has a unique resolution of the identity $E$ such that $U = \int_{\sigma(U)}\lambda dE(\lambda)$, where $\sigma(U)$ is the spectrum of $U$. I also know that $I = \int_{\sigma(U)}dE(\lambda)$, so perhaps $U_t = \int_{\sigma(U)}\lambda^{t}dE(\lambda)$, $t \in [-1,1]$ is the one-parameter group he's talking about? How would I show that each $U_t$ is unitary and that the path $\gamma(t) = U_t$ is continuous when the unitary group is given the topology inherited from the strong operator topology on all bounded operators on a Hilbert space?

Any help is appreciated.

You can see this as follows: $$\|U_tx\|^2 = \left\|\int\lambda^t\,dEx\right\|^2 = \int|\lambda|^{2t}d\|Ex\|^2 = \int\,d\|Ex\|^2 = \|E(\sigma(U))x\|^2 = \|x\|^2.$$ The continuity of $t\mapsto U_t$ follows from the uniform Lipschitz continuity of $t\mapsto\lambda^t$ for $\lambda$ on the unit circle.
But I prefer another way. One can show that $U = e^{iA}$ with a bounded selfadjoint operator $A$ and that for each bounded selfadjoint operator $B$ the operator $e^{iB}$ is unitary. Hence, $U$ belongs to the continuous one-parameter group $\{e^{itA} : t\in\mathbb R\}$.
• @amsmath , and how can show the group is connected, or the same the function $f(t)=e^{itA}$ is continuous – user89940 Nov 29 '18 at 1:07
• @user89940 You can use the properties of the continuous functional calculus. For $s,t\in\mathbb R$ define the function $f(x) = e^{itx} - e^{isx}$. Then $f$ is continuous and $\|f(A)\| = \sup\{|f(x)| : x\in\sigma(A)\}\le\sup\{|f(x)| : x\in[-\pi,\pi]\}$ because $\sigma(A)\subset [-\pi,\pi]$, – amsmath Jan 3 '19 at 21:58