Although this is not research, I think the question is a little bit too specific for math.stackexchange

Let $G = \mathbb{R}$. By Stone's theorem, $U(t)\in\mathcal{B}(\mathcal{H})$ is generated by a self-adjoint operator $H$ (for which there is a resolution of the identity P(p), by the spectral theorem) $$ U(t) = e^{i t H} = \int_{\mathbb{R}} e^{i pt} d P(p)$$ One can thus define a spectral subspace, e.g. of eigenvectors of $H$ with eigenvalues in $[p_1,p_2]$, given by $(P(p_2)-P(p_1))\mathcal{H}$. Let's denote it $\mathcal{H}^U([p_1,p_2])$.

More generally, for a uniformly bounded action $\alpha$ of a locally compact group $G$ on a Banach space $X$, one can define the $\alpha$-spectrum $\mathrm{Sp}_{\alpha}(a)$ of $ x\in X$ as the set of maximal ideals of $L^1(G)$ containing $\mathcal{I}(a):=\{f\in L^1(G),\ \alpha_f(a) := \int_{G} f(x)\alpha_x(b) \operatorname{d}\! x \stackrel{!}{=}0\}$

By [8.2 p.218][2], maximal ideals are the kernels of the "characters" of $L^1(G)$ which are themselves in bijection with characters of the group via Fourier transform: $$ \forall\ \Phi: L^1(G) \rightarrow \mathbb{C}\ \text{ algebra morphism},\ \exists !\ \gamma\in\hat{G} \ \text{ s.t. }\ \Phi(f)= \int_G \overline{\gamma(x)} f(x)\operatorname{d}\!\mu (x)=: \hat{f}(\gamma) $$ The $\alpha$-spectrum can thus also be defined as a subset of the dual group $\hat{G}$ $$ \mathrm{Sp}_{\alpha}(a) := \{\gamma\in \hat{G},\ \hat{f}(\gamma) = 0,\ \forall\ f\in \mathcal{I}(a)\} $$

Now for a given subset $S\subset \hat{G}$ one can associate the following spectral subspaces $$ X^{\alpha}(S) := \overline{\{a\in X,\ \mathrm{Sp}_{\alpha}(a)\subseteq S\}} $$

$$ X^{\alpha}_0(S) := \overline{\mathrm{Span}\{\alpha_f(a),\ \forall\ a\in X,\ f\in L^1(G),\ \mathrm{supp}(\hat{f})\subseteq S\}} $$ (Relation between them: [Lemma 3.2.39 p.253][3])

[Bratteli, Robinson][3] say p.251 that it is "easy to derive"/ p.258 that it is "easily verified" that $$ \mathcal{H}([p_1,p_2]) = \mathcal{H}^U ([p_1,p_2])$$ where l.h.s. is given by the projections of the spectral theorem for the generator $H$ and r.h.s. is the abstract definition with $X:=\mathcal{H}$ and $\alpha:= U$

Question: how do we see this?

I have also seen, for $G= \mathbb{R}$ the following definition $$ \mathrm{Sp}_{\alpha}(a):=\{p\in\mathbb{R},\ \forall\ U\ni p\ \text{neighborhood},\ \exists\ f\in L^1(\mathbb{R}),\ \mathrm{supp}(\hat{f})\subset U \ , \enspace \alpha_f(a)\neq 0 \} $$ Which I find more accessible, and I guess it does agree with the one above by [3.2.40 (2) p.254][3]

[2]: "A course in functional analysis", Conway, (actually works even if $L^1(G)$ is not unital) ; or Thm. 34B p.136-137 for some algebra of measure containing $L^1(G)$, "Introduction to abstract harmonic analysis", Loomis

[3]: "Operator algebras and Quantum statistical mechanics vol. 1", Bratelli, Robinson

  • $\begingroup$ I see the following link: Let $f\in L^1(\mathbb R)$ and $x\in\mathcal H$ $$U_f(x)=\int_{\mathbb R}f(t)\ e^{it H}\cdot x\ dt=\int_{\mathbb R}\left(\int_{\mathbb{R}}f(t)\ e^{it p} dP(p)\right)\cdot x\ dt=\left(\int_{\mathbb R}\hat{f}(-p)\ dP(p)\right)\cdot x$$ If $x$ is an eigenvector of $H$ with eigenvalue $p$, then $U_f(x)=\hat{f}(-p)\ x.$ It is stunning that the relation group and generator $U= e^{itH}$ is the same as the character of the group $\mathbb R$. Maybe related to mathoverflow.net/questions/202653/…. $\endgroup$
    – Noix07
    Apr 12, 2015 at 11:35
  • $\begingroup$ I checked that strange minus sign: in Bratteli, Robinson (same pages as indicated previously), they actually write $U=e^{-i t H}$, as in physics. So there is no minus sign in the Fourier transform. $\endgroup$
    – Noix07
    Apr 12, 2015 at 12:59


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.