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I'm trying to make an example for spectral graph theory over infinite graphs. Thus, I took the example of the graph $G = (\mathbb{Z},E)$ with $E = \{ (n,n+1)\;| \; n\in \mathbb{Z}\}$ with the adjacency operator $A: \ell^2(\mathbb{Z})\rightarrow \ell^2(\mathbb{Z})$ such that $Af(n) = f(n+1)$; the translation operator. How I prove that the spectrum of this operator is $S^1$?

温泽海
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user_sion
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  • A lot of information here https://math.stackexchange.com/questions/617601/how-to-compute-the-spectrum-of-shift-operators-in-ell2-mathbb-n. Does this help? – HackR May 25 '25 at 18:34
  • @HackR that is useful but not exactly the same. The left shift operator on $\ell^2(\mathbb N)$ has nontrivial kernel while the on on $\ell^2(\mathbb Z)$ is unitary. – Just a user May 26 '25 at 02:01
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    I'm voting this question up to defend against those who again try to vote down and close a question perfectly suitable for this site with no good reason. Not only this question hasn't shown up on the site yet (as far as I can see), both answers below are very good for the technique and insight. Plus, the study of Cayley graphs of groups (here $G$ is just the Cayley graph of $\mathbb Z$) is an important subject in modern group theory and operator algebra. – Just a user May 26 '25 at 05:10

2 Answers2

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The operator $A:\ell^p(\mathbb{Z})\to \ell^p(\mathbb{Z})$ satisfies $\|Af\|_p=\|f\|_p$ and $\|A^{-1}f\|_p$ for any $1\le p\le\infty.$ Hence its spectrum $\sigma_p(A)$ is contained in $ S^1.$

Solution 1 For any $z\in S^1$ and the isometry operator $(U_zf)(n)=z^nf(n)$ we have $U_z^{-1}AU_z= zA.$ Therefore $$\sigma_p(A)=\sigma_p( U_z^{-1}AU_z)=\sigma_p(zA)=z\sigma_p(A)$$ Thus $\sigma_p(A)$ is invariant under all rotations. Hence $\sigma_p(A)=S^1.$

Solution 2 Let $\{\delta_n\}_{n=-\infty}^\infty $ denote the standard basis in $\ell^p(\mathbb{Z}).$ Then $A\delta_n=\delta_{n-1}.$ For fixed $N$ and $|z|=1$ let $f_N=\sum_{n=-N}^Nz^n\delta_n.$ Then $$ Af_N-zf_N=z^N\delta_{N-1}-z^{N+1}\delta_N$$ Thus for $1\le p<\infty$ we get $$ {\|Af_N-zf_N\|_p\over \|f_N\|_p}={2^{1/p}\over (2N+1)^{1/p}}\underset{N}{\longrightarrow} 0$$ Therefore the number $z$ is an approximate eigenvalue of $A.$ In particular $z\in \sigma_p(A).$ For $p=\infty $ we have $$A\left (\sum_{n=-\infty}^\infty z^n\delta_n\right )=z\sum_{n=-\infty}^\infty z^n\delta_n$$ Thus $z$ is an eigenvalue of $A,$ i.e. $z\in \sigma_\infty(A).$

Ryszard Szwarc
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A hint:

While the group $\mathbb{Z}$ is certainly not isomorphic to $S^1$ ( its group of characters), $\ell^2(\mathbb{Z})$ is isometric to $L^2(S^1)$ under the Fourier transform. The translation operator "becomes" the following operator on $L^2(S^1)$: $\phi(z) \mapsto z\cdot \phi(z)$. Not that it simplifies the problem too much, but it's an alternative way to see it.

orangeskid
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