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 $\{e^{itA} : t\in\mathbb R\}$.