Spectral theorem for representations proof.

Question

Let $H$ be a separable Hilbert space, and $U$ a unitary representation of $\mathbb{Z}^d$ on $H$. Let $\chi_m$ be the characters of the Torus $T^d$, and $m$ the Haar measure on $T^d$. I would like to Show that if $U$ is cyclic with cyclic vector $v$, then there is a finite measure $\mu_v$ on $T^d$ such that $\langle U(m)v,v\rangle=\int_{T^d}\chi_m d\mu_v$. I would like to do this by considering the sequence of measure on $T^d$ defined by $d\mu_{v,N}=\frac{1}{(N+1)^d}||\sum_{m\in\mathbb{Z}^d, m_i\leqslant N}\chi_m^{-1} U(m)v||^2 dm$.

I guess we should pass to the Limit $T\rightarrow\infty$, but I am not too sure how to do it in Detail. Thanks for your help.

Answer 1

(note that you are using $m$ for two different things; I will just ignore the Haar measure as I don't think it is needed)

Since the characters $\chi_m$ evaluate at single points $m\in\mathbb Z^d\cap\mathbb T^d$ (which are the $d$-tuples consisting of $1$ and $-1$) , you have $$ \int_{\mathbb T^d}\chi_m\,d\mu_v=\mu_v(\{m\}). $$ And you want this to equal $\langle U(m)v,v\rangle$. So it is clear what the measure $\mu_v$ should be: $$ \mu_v(X)=\sum_{m\in X\cap\mathbb Z^d}\,\langle U(m)v,v\rangle\,\delta_m. $$ Then, for a fixed $m\in\mathbb Z^d\cap\mathbb T^d$, $$ \int_{\mathbb T^d}\chi_m\,d\mu_v=\mu_v(\{m\})=\langle U(m)v,v\rangle. $$

Answer 2

Let's fix some notations.

Denote the Haar measure on $\Bbb{T}^d$ by $\lambda$ instead of $m$.

The measures suggested should probably be normalized by $$ d\mu_{v,N}(y)=\frac{1}{(2N+1)^d}\left\|\sum_{m\in\mathbb{Z}^d, \lvert m_i \rvert \leq N}\chi_{-m}(y) U(m)v\right\|^2 d\lambda(y).$$

The characters of $\Bbb{T}^d$ are the functions $$\chi_m: \Bbb{T}^d \to C^*\\ y \mapsto e^{2\pi i m \cdot y}$$

Finally, introduce the shorthand notation for the truncated sums $$\sum_{m \le N} f(m) := \sum_{m\in\mathbb{Z}^d, \lvert m_i\rvert \leq N} f(m).$$

With this, we have \begin{align*} \left\| \sum_{m\le N} \chi_{-m}(y) U(m)v\right\|^2 & = \bigg\langle \sum_{m \le N} \chi_{-m}(y) U(m)v, \sum_{k\le N} \chi_{-k}(y) U(k)v\bigg\rangle \\ & = \sum_{m\le N} \sum_{k \le N} \big\langle\chi_{-m}(y) U(m)v,\chi_{-k}(y) U(k)v\big\rangle \\ & = \sum_{m\le N} \sum_{k \le N} \chi_{k-m}(y)\big\langle U(-k)U(m)v, v\big\rangle \\ & = \sum_{m\le N} \sum_{k \le N}e^{2\pi i (k-m) \cdot y}\big\langle U(m-k)v, v\big\rangle \\ \end{align*}

Now look at the terms $T_l = e^{2\pi i l \cdot y}\big\langle U(l)v, v\big\rangle$ that appear from the change of variables $l = m - k$. The peak contribution to the sum comes from $l = 0$ (with $(2N + 1)^d$ terms), and they decay as $\lvert l\rvert$ increases, until when some coordinate becomes greater than $2N$, when it vanishes.

Therefore, the mean defining $d\mu_{v,N}(y)$ corresponds to a Cèsaro sum of the Fourier series $$\sum_{l\le N} e^{2\pi i l \cdot y}\big\langle U(l)v, v\big\rangle.$$ Thus $d\mu_{v,N}(y)$ converges to a (tempered) distribution $T$.

A calculation shows that $\langle T, \chi_j\rangle = \langle U(j) v, v\rangle$, first integrating on the torus $\Bbb{T}^d$ to get only terms where $m-k = j$, and then taking the limit as $N\to\infty$. Let's see why: the definition of $\langle T, \chi_j\rangle$ is the limit as $N\to\infty$ of $$ \frac{1}{(2N+1)^d} \sum_{m\le N} \sum_{k \le N} \int_{\Bbb{T}^d} \chi_j (y) \chi_{k-m}(y) \big\langle U(m-k)v, v\big\rangle \, d\lambda(y). $$ The orthogonality of the characters $\chi_j$ and $\chi_{m-k}$ over $\Bbb{T}^d$ reduce the sum to only $$ \frac{1}{(2N+1)^d} \sum_{m,k\le N \, m-k = j} \big\langle U(m-k)v, v\big\rangle. $$ Now, a volume argument shows that for each fixed $j$, as $N\to\infty$, the number of solutions to $m-k = j$ with all coordinates of $m$ and $k$ smaller than $N$ is equal to $(2N+1)^d$ up to error terms of degree $d-1$ in $N$.

This establishes the functional calculus to the trigonometric polynomials, with the uniform bound $\lvert\langle T, P\rangle\rvert \leq \sup_{y\in \Bbb{T}^d} P(y)$. With this bound, we can both extend the functional calculus to all continuous funtions, and by the Riesz-Markov theorem this distribution is in fact a measure $\mu$ on $\Bbb{T}^d$.

To conclude, we note that the isometry between the spaces $L^2(\mu)$ and $H$ is given by the linear map $$ \mathcal{U}: L^2(\mu) \to H\\ P \mapsto P(U)v $$ constructed by the functional calculus above. This is an isometry since $$\langle P, Q\rangle_{L^2(\mu)} = \int_{\Bbb{T}^d} P(y) \bar{Q}(y) \, d\mu(y) = \langle P(U) Q(U)^* v, v\rangle_H = \langle Q(U)^* P(U) v, v\rangle_H = \langle P(U)v, Q(U)v\rangle_H. $$ (We used the commutativity of all $U(m)$ and $U^*(m)$, since they are unitary, to commute also $P(U)$ and $Q^*(U)$)

Since $v$ is cyclic, the function $\mathcal{U}$ is onto $H$, so indeed we have an isomorphism of Hilbert spaces.