1
$\begingroup$

Consider the space $l^2(\mathbb{Z}) \otimes l^2(\mathbb{Z})$ with distinguished computational basis $e_i \otimes e_j $ and a group of translations $T_a$ defined by $T_a e_i \otimes e_j = e_{i+a} \otimes e_{j+a}$.

Suppose that I have a bounded operator $A: l^2(\mathbb{Z}) \otimes l^2(\mathbb{Z}) \to l^2(\mathbb{Z}) \otimes l^2(\mathbb{Z}) $ which is invariant under these translations, i.e. $T_a A T_{-a} = A$. Further, $A$ has the property that $A \left( e_j \otimes e_k \right) = 0 $ for $\vert j-k \vert > R$ for some $R$. Now, analogously with translation invariance in $ l^2(\mathbb{Z})$, I'd like to do some kind of Fourier-transformation to find a basis where the operator is nicer.

I suggest $\mathcal{F}: l^2(\mathbb{Z}) \otimes l^2(\mathbb{Z}) \to \oplus_{r \in \mathbb{Z}} L^2 \left( \lbrack 0,2\pi \rbrack \right)_r$ defined by $(\mathcal{F}\psi)(k,r) = \sum_{x \in \mathbb{Z}} e^{-ikx} \psi(x,x-r)$ and inverse $(\mathcal{F}^{-1}\phi)(x,y) = \int_{0}^{2\pi} e^{ikx} \phi(k,x-y) dk $ with some normalization constants.

Does this make sense, and if yes could you point me to some literature about it? Further, if this does make sense how would one go about showing that $\mathcal{F}$ is unitary? I am ultimately interested in the spectrum of $A$.

Improve this question
$\endgroup$
5
  • $\begingroup$ Why does $\mathcal F^{-1}\phi$ live in $\ell^2(\mathbb Z) \otimes \ell^2(\mathbb Z)$? Is that meant to be some sort of completed tensor product? (This probably has to do with what you mean by 'basis', too.) $\endgroup$
    – LSpice
    Commented Mar 18, 2020 at 19:52
  • $\begingroup$ Well, I might be wrong but here is my take. The norm of a vector in $l^2(\mathbb{Z}) \otimes l^2(\mathbb{Z}) $ I calculate as $\vert\vert a \vert \vert = \sum_{k,j} \vert a_{k,j} \vert^2$ and hence $ \vert \vert \mathcal{F}^{-1} \phi \vert \vert = \sum_{x,y} \vert \int_0^{2 \pi} e^{ikx} \phi(k, x-y) dk \vert^2 = \sum_{x,y} \int_0^{2\pi} \int_0^{2\pi} dk dk' e^{i(k-k')x} \vert \phi(k,x-y) \vert^2 $. Now $\sum_{y} \vert \phi(k,x-y) \vert^2 $ is independent of $x$ and then the sum over $x$ of $e^{i(k-k')x}$ collapses one into to yield $\int_0^{2\pi} dk \sum_{n} \vert \phi(k,n) \vert^2 $. $\endgroup$
    – Frederik Ravn Klausen
    Commented Mar 18, 2020 at 20:25
  • $\begingroup$ And this is supposed to be the norm of $\phi$ in the space $\otimes_{r \in \mathbb{Z}} L^2( \lbrack 0, 2 \pi \rbrack) $. $\endgroup$
    – Frederik Ravn Klausen
    Commented Mar 18, 2020 at 20:27
  • $\begingroup$ I believe that $\mathcal F^{-1}\phi$ lies in $\ell^2(\mathbb Z \times \mathbb Z)$, but that properly contains the ordinary vector-space product of $\ell^2(\mathbb Z) \otimes \ell^2(\mathbb Z)$. Did you mean some kind of completed vector product? $\endgroup$
    – LSpice
    Commented Mar 18, 2020 at 20:28
  • $\begingroup$ Yes, we complete with respect to the norm coming from the inner product in the tensor product. $\endgroup$
    – Frederik Ravn Klausen
    Commented Mar 19, 2020 at 10:45

1 Answer 1

Reset to default
1
$\begingroup$

Fine question. The first thing to notice is that your problem becomes simpler if you work in the $(1,1)$, $(0,1)$ basis of $\mathbb{Z}\times\mathbb{Z}$. I.e., we have an isomorphism from $\mathbb{Z}\times \mathbb{Z}$ to itself given by the matrix $ \left[\begin{matrix}1&0\cr 1&1 \end{matrix}\right]$, and conjugating by this turns your operator into one which commutes with horizontal translations.

Then we can take the Fourier transform in the first variable to go to $L^2(\mathbb{T})\otimes l^2(\mathbb{Z})$ --- these two steps, change basis and Fourier transform are effectively what you have done. (BTW you can see directly that it's unitary by noticing that it takes the orthonormal basis $(e_i\otimes e_j)$ to an orthonormal basis of the range.) You now have an operator on this Hilbert space which commutes with multiplication by the first variable, which means it commutes with all of the spectral projections of that multiplication operator, which means the subspace $L^2(S)\otimes l^2(\mathbb{Z})$ is invariant, for any measurable $S \subseteq \mathbb{T}$.

Actually, the band-limiting condition means it's effectively an operator from $L^2(\mathbb{T})\otimes l^2[-R,R]$ to $L^2(\mathbb{T})\otimes l^2(\mathbb{Z})$.

I can't say how that will help with your problem without knowing more about $A$, but I think it cleanly packages the information you've given us. I guess you can write $A$ as a direct integral over $\mathbb{T}$ of a family of operators on $l^2(\mathbb{Z})$, and you can relate the spectrum of $A$ to the spectra of these operators, but I somehow doubt that would really help. I can give references for that last comment if you like, the rest is just standard functional analysis.

(Actually, a better way to say this is that after the transformation your operator belongs to $B(l^2(\mathbb{Z}))\otimes L^\infty(\mathbb{T})$.)

Improve this answer
$\endgroup$
3
  • $\begingroup$ I haven't worked through the calculation properly but does the band-limiting condition mean you land in the uniform Roe algebra of ${\bf Z}$ with coefficients in $L^\infty({\bf T})$? $\endgroup$
    – Yemon Choi
    Commented Mar 19, 2020 at 19:19
  • $\begingroup$ @YemonChoi Not exactly, I don't think. The condition isn't that we can't move $e_{ij}$ more than $R$ units away from where it starts, it's that the operator kills $e_{ij}$ if $|i-j| >R$. $\endgroup$
    – Nik Weaver
    Commented Mar 19, 2020 at 20:42
  • $\begingroup$ Thanks, I used the trick here: arxiv.org/abs/2206.09879 $\endgroup$
    – Frederik Ravn Klausen
    Commented Jun 22, 2022 at 18:30

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .