Spatio-spectral limiting on discrete tori: adjacency invariant spaces

Abstract

Discrete tori are ${\mathbb {Z}}_m^N$ thought of as vertices of graphs ${\mathcal {C}}_m^N$ whose adjacencies encode the Cartesian product structure. Space-limiting refers to truncation to a symmetric path neighborhood of the zero element and spectrum-limiting in this case refers to corresponding truncation in the isomorphic Fourier domain. Composition spatio-spectral limiting (SSL) operators are analogues of classical time and band limiting operators. Certain adjacency-invariant spaces of vectors defined on ${\mathbb {Z}}_m^N$ are shown to have bases consisting of Fourier transforms of eigenvectors of SSL operators. We show that when $m=3$ or $m=4$, all eigenvectors of SSL arise in this way. We study the structure of corresponding invariant spaces when $m\ge 5$ and give an example to indicate that the relationship between eigenvectors of SSL and the corresponding adjacency-invariant spaces should extend to $m\ge 5$.

Appendices

Proofs of Auxiliary lemmas for Theorem 22

Proof of Lemma 18

The basic idea is similar to that of the proof of Lemma 3. We begin with a computation of $[A_{+,\ell },A_{-,\ell '}]$ for $1\le \ell ,\ell '\le M$.

$$\begin{aligned} (A_{+,\ell }A_{-,\ell '}f)(v)= & {} \sum _{d_\nu (v)=\ell } (A_{-,\ell '}f)(v_\nu ^-)=\sum _{d_\nu (v)=\ell } \sum _{d_\mu (v_\nu ^-)=\ell '-1} f((v_\nu ^-)_\mu ^+)\\= & {} \sum _{d_\nu (v)=\ell } \Bigl [ f((v_\nu ^-)_\nu ^+){\varvec{1}}_{\ell '=\ell }+\sum _{d_\mu (v)=\ell '-1} f((v_\nu ^-)_\mu ^+) {\varvec{1}}_{\mu \ne \nu }\Bigr ] \\= & {} \sum _{d_\nu (v)=\ell } f((v_\nu ^-)_\nu ^+){\varvec{1}}_{\ell '=\ell }+\sum _{d_\mu (v)=\ell '-1}\sum _{d_\nu (v)=\ell } f((v_\mu ^+)_\nu ^-) {\varvec{1}}_{\mu \ne \nu }\\= & {} \sum _{d_\nu (v)=\ell } f((v_\nu ^-)_\nu ^+){\varvec{1}}_{\ell '=\ell }+\sum _{d_\mu (v)=\ell '-1}\sum _{d_\nu (v_\mu ^+)=\ell } f((v_\mu ^+)_\nu ^-) {\varvec{1}}_{\mu \ne \nu }\\= & {} \sum _{d_\nu (v)=\ell } f((v_\nu ^-)_\nu ^+){\varvec{1}}_{\ell '=\ell }\\&+\sum _{d_\mu (v)=\ell '-1}\Bigl [ - f((v_\mu ^+)_\mu ^-){\varvec{1}}_{\ell '=\ell } +\sum _{d_\nu (v_\mu ^+)=\ell } f((v_\mu ^+)_\nu ^-)\Bigr ]\\= & {} \sum _{d_\nu (v)=\ell } f((v_\nu ^-)_\nu ^+){\varvec{1}}_{\ell '=\ell }\\&+\sum _{d_\mu (v)=\ell '-1} \Bigl [ - f((v_\mu ^+)_\mu ^-){\varvec{1}}_{\ell '=\ell } +(A_{+,\ell } f)(v_\mu ^+)\Bigr ]\\= & {} \left[ \sum _{d_\nu (v)=\ell } f((v_\nu ^-)_\nu ^+) - \sum _{d_\mu (v)=\ell '-1} f((v_\mu ^+)_\mu ^-) \right] {\varvec{1}}_{\ell '=\ell } +(A_{-,\ell '}A_{+,\ell } f)(v). \end{aligned}$$

To get from the first line to the second we used the condition that if $d_\nu (v)=\ell$ and $d_\mu (v_\nu ^-)=\ell '-1$ then either $\nu =\mu$ and $\ell '=\ell$ or else $\mu \ne \nu$ and $d_\mu (v)=\ell '-1$. To get from the third to fourth lines we used that if $\mu \ne \nu$ then $d_\nu (v)=d_\nu (v_\mu ^+)$ and to get from the fourth to fifth line we used that if $d_\mu (v)=\ell '-1$ and $d_\nu (v_\mu ^+)=\ell$ then either $\mu =\nu$ and $\ell '=\ell$ or $\mu \ne \nu$. We conclude that

$$\begin{aligned} ([ A_{-,\ell '}, \, A_{+,\ell }]f)(v)=\left[ \sum _{d_\mu (v)=\ell '-1} f((v_\mu ^+)_\mu ^-) -\sum _{d_\nu (v)=\ell } f((v_\nu ^-)_\nu ^+) \right] {\varvec{1}}_{\ell '=\ell }. \end{aligned}$$

(12)

In particular, if $\ell '\ne \ell$ then $[ A_{-,\ell '}, \, A_{+,\ell }]=0$, which is Lemma 18.i.

For the next set of identities in Lemma 18.ii, first consider the case $\ell =1$. In this case if $d_\mu (v)=\ell -1=0$ then using the convention above, $f((v_\mu ^+)_\mu ^-) =2f(v)$ while if $d_\nu (v)=\ell =1$ then $f((v_\nu ^-)_\nu ^+)=f(v)+f({\tilde{v}}_\nu )$ and

$$\begin{aligned}&\sum _{d_\mu (v)=0} f((v_\mu ^+)_\mu ^-) -\sum _{d_\nu (v)=1} f((v_\nu ^-)_\nu ^+) \\&\quad =f(v)\bigl (2\#\{\mu :d_\mu (v)=0\}- \#\{\nu :d_\nu (v)=1\}\bigr )\\&\qquad -\sum _{d_\nu (v)=1} f({\tilde{v}}_\nu ) =(2q_0-q_1) f(v) -(R_1f)(v)\, . \end{aligned}$$

This establishes the case $\ell =1$ of Lemma 18.ii.

In the case $\ell =M$, again using the conventions above, if $d_\nu (v)=M$ then $f((v_\nu ^-)_\nu ^+) =2f(v)$ while if $d_\mu (v)=M-1$ then $f((v_\mu ^+)_\mu ^-)=f(v)+f({\tilde{v}}_\mu )$ so

$$\begin{aligned}&\sum _{d_\mu (v)=M-1} f((v_\mu ^+)_\mu ^-) -\sum _{d_\nu (v)=M} f((v_\nu ^-)_\nu ^+) \\&\quad = f(v)\bigl (\#\{\mu :d_\mu (v)=M-1\} - 2 \#\{\nu :d_\nu (v)=M\}\bigr )+\sum _{d_\mu (v)=M-1} f({\tilde{v}}_\mu ) \\&\quad =(q_{M-1}-2q_M) f(v) +(R_{M-1}f)(v)\, . \end{aligned}$$

This establishes the case $\ell =M$ of Lemma 18.ii.

Finally, let $1<\ell <M$. In this case, $(v_\mu ^+)_\mu ^-=(v_\nu ^-)_\nu ^+=v$ for each $\mu ,\nu$ so

$$\begin{aligned}&\sum _{d_\mu (v)=\ell '-1} f((v_\mu ^+)_\mu ^-) -\sum _{d_\nu (v)=\ell } f((v_\nu ^-)_\nu ^+) \\&\quad =f(v)\bigl (\#\{\mu :d_\mu (v)=\ell -1\}-\#\{\nu :d_\nu (v)=\ell \}\bigr ) =(q_{\ell -1}-q_\ell ) f(v). \end{aligned}$$

This completes the proof of Lemma 18.ii.

For Lemma 18.iii, a simple calculation shows that $([R_\ell ,A_{+,\ell }]f)(v)=\sum _{d_\nu (v)=\ell } f((v_\nu ^-)_\nu ^\sim )$. We then compute $[A_{-,\ell }, [R_\ell ,A_{+,\ell }]]$ as follows.

$$\begin{aligned} (A_{-,\ell } [R_\ell ,A_{+,\ell }]f)(v)= & {} \sum _{d_\mu (v)=\ell -1} ( [R_\ell ,A_{+,\ell }]f)(v_\mu ^+)= \sum _{d_\mu (v)=\ell -1}\sum _{d_\nu (v_\mu ^+)=\ell } f(((v_\mu ^+)_\nu ^-)_\nu ^\sim )\\= & {} \sum _{d_\mu (v)=\ell -1}\left[ f(((v_\mu ^+)_\mu ^-)_\mu ^\sim )+\sum _{d_\nu (v)=\ell } f(((v_\mu ^+)_\nu ^-)_\nu ^\sim )\right] \\= & {} \sum _{d_\mu (v)=\ell -1} f({\tilde{v}}_\mu )+\sum _{d_\nu (v)=\ell }\sum _{d_\mu (v)=\ell -1} f(((v_\nu ^-)_\nu ^\sim )_\mu ^+){\varvec{1}}_{\mu \ne \nu }\\= & {} (R_{\ell -1} f)(v)-\sum _{d_\nu (v)=\ell }\left[ f(((v_\nu ^-)_\nu ^\sim )_\nu ^+)-\sum _{d_\mu (v_\nu ^-)=\ell -1} f(((v_\nu ^-)_\nu ^\sim )_\mu ^+)\right] \\= & {} (R_{\ell -1} f)(v)-(R_\ell f)(v) +\sum _{d_\nu (v)=\ell }(A_{-,\ell }f)((v_\nu ^-)_\nu ^\sim )\\= & {} (R_{\ell -1} f)(v) -(R_\ell f)(v) +[R_\ell ,A_{+,\ell }] A_{-,\ell }f)(v) \, . \end{aligned}$$

Subtracting $([R_\ell ,A_{+,\ell }] A_{-,\ell }f)(v)$ from both sides gives Lemma 18.iii.

To prove Lemma 18.iv, let $\ell \ne \ell '$. Recalling that $([R_\ell ,A_{+,\ell }]f)(v)=\sum _{d_\nu (v)=\ell } f((v_\nu ^-)_\nu ^\sim )$,

$$\begin{aligned} (A_{-,\ell '}[R_\ell ,A_{+,\ell }]f)(v)= & {} \sum _{d_\nu (v)=\ell '-1} ([R_\ell , A_{+,\ell }]f)(v_\nu ^+)\\= & {} \sum _{d_\nu (v)=\ell '-1} \sum _{d_\mu (v_\nu ^+)=\ell } f(((v_\nu ^+)_\mu ^-)_\mu ^\sim )\\= & {} \sum _{d_\nu (v)=\ell '-1} \sum _{d_\mu (v)=\ell } f(((v_\nu ^+)_\mu ^-)_\mu ^\sim )\\= & {} \sum _{d_\mu (v)=\ell } \sum _{d_\nu (v)=\ell '-1} f(((v_\mu ^-)_\mu ^\sim )_\nu ^+)\\= & {} \sum _{d_\mu (v)=\ell } \sum _{d_\nu (v_\mu ^-)=\ell '-1} f(((v_\mu ^-)_\mu ^\sim )_\nu ^+) =\sum _{d_\mu (v)=\ell } A_{-,\ell '}f((v_\mu ^-)_\mu ^\sim ) \\= & {} [R_\ell ,A_{+,\ell }] A_{-,\ell '}f(v)\, . \end{aligned}$$

Here we observed that if $d_\nu (v)=\ell '-1$ and $d_\mu (v_\nu ^+)=\ell$ then $\mu \ne \nu$ since $\ell \ne \ell '$. Similarly, if $d_\mu (v)=\ell$ and $d_\nu (v_\mu ^-)=\ell '-1$ then $\mu \ne \nu$. Subtracting $[R_\ell ,A_{+,\ell }] A_{-,\ell '}f$ from both sides gives Lemma 18.iv and completes the proof of the lemma. $\square$

Proof of Lemma 19

Since we finish with a linear term to the left of $A_{+,\ell }$ in each case, it suffices to assume that $p({\mathbf {R}})$ is a monomial of the form $p({\varvec{R}})=R_1^{\nu _1}\cdots R_{M-1}^{\nu _{M-1}}$. First consider the case $\ell =1$. We wish to show that $p({\varvec{R}})A_{+,1}=A_{+,1}p_{+,1}({\varvec{R}})$ for an appropriate $p_{+,1}$. Since $[R_j,A_{+,1}]=0$ for each $j>1$ we evidently have $p({\varvec{R}})A_{+,1}=R_1^{\nu _1}A_{+,1}R_2^{\nu _2}\cdots R_{M-1}^{\nu _{M-1}}$. By iterating Lemma 20.i we have $R_1^{\nu _1}A_{+,1}=A_{+,1}(I+R_1)^{\nu _1}$ so we may take $p_{+,1}=(I+R_1)^{\nu _1}R_2^{\nu _2}\cdots R_{M-1}^{\nu _{M-1}}$. The case $\ell =M-1$ is similar yielding $p({\varvec{R}})A_{+,M}= A_{+,1}R_1^{\nu _1}\cdots R_{M-2}^{\nu _{M-2}}(R_{M-1}-I)^{\nu _{M-1}}$. Now consider the case $1<\ell <M$. By Lemma 20.ii for the same $p({\varvec{R}})$ we can write

$$\begin{aligned} p({\varvec{R}})A_{+,\ell }= & {} R_{\ell -1}^{\nu _{\ell -1}}R_{\ell }^{\nu _{\ell }} A_{+,\ell }\prod _{j\ne \ell ,\ell -1}R_j^{\nu _j} =(R_{\ell -1}+R_\ell -R_\ell )^{\nu _{\ell -1}}R_{\ell }^{\nu _{\ell }} A_{+,\ell }\prod _{j\ne \ell ,\ell -1}R_j^{\nu _j}\\= & {} \sum _{j=0}^{\nu _{\ell -1}} (-1)^j\left( {\begin{array}{c}\nu _{\ell -1}\\ j\end{array}}\right) (R_\ell )^{\nu _\ell +\nu _{\ell -1}-j}A_{+,\ell }(R_{\ell -1}+R_\ell )^j \prod _{j\ne \ell ,\ell -1}R_j^{\nu _j}\, . \end{aligned}$$

Splitting this sum into its monomial pieces then reduces the problem to showing that, for each $\nu =1,2,\dots$ we can write

$$\begin{aligned} (R_\ell )^{\nu }A_{+,\ell }=aA_{+,\ell }p_{+,\ell }+bR_\ell A_{+,\ell }q_{+,\ell } \end{aligned}$$

for appropriate p and q. We can proceed by induction. Assuming such an expression is valid for $\nu$, we have

$$\begin{aligned} (R_\ell )^{\nu +1}A_{+,\ell }=aR_\ell A_{+,\ell }p_{+,\ell }+bR_\ell ^2A_{+,\ell } q_{+,\ell }\, . \end{aligned}$$

(13)

By Lemma 20.ii we can write

$$\begin{aligned} R_\ell ^2 A_{+,\ell }&= [R_\ell , [R_\ell ,A_{+,\ell }]]+2[R_\ell ,A_{+,\ell }]R_\ell +A_{+,\ell }R_\ell ^2\\&=A_{+,\ell }+2R_\ell A_{+,\ell }R_\ell -A_{+,\ell }R_\ell ^2 \, . \end{aligned}$$

Substituting this into (13) and gathering like terms then proves the induction step and completes the proof of Lemma 19$\square$

Proof of Lemma 20

To prove Lemma 20.i in the case $\ell =1$ we have

$$\begin{aligned} (R_1A_{+,1} f)(v)= & {} \sum _{d_\mu (v)=1 }(A_{+,1} f)({\tilde{v}}_\mu )=\sum _{d_\mu (v)=1 }\sum _{d_\nu (v)=1}f(({\tilde{v}}_\mu )_\nu ^-)\\= & {} \sum _{d_\nu (v)=1}\sum _{d_\mu (v)=1 }f((v_\nu ^-)_\mu ^\sim ) =\sum _{d_\nu (v)=1} \left[ f((v_\nu ^-)_\nu ^\sim ) +\sum _{d_\mu (v_\nu ^-)=1}f((v_\nu ^-)_\mu ^\sim )\right] \\= & {} \sum _{d_\nu (v)=1}f(v_\nu ^-) +(R_1f((v_\nu ^-)) =A_{+,1}f(v)+(A_{+,1}R_1f)(v)\, . \end{aligned}$$

Here we use the fact that if $d_\nu (v)=1$ then $(v_\nu ^-)_\nu ^\sim =v_\nu ^-$. The proof that $[R_{M-1},A_{+,M}]=-A_{+,M}$ is essentially the same. This proves Lemma 20.i. To prove Lemma 20.ii,

$$\begin{aligned} ((R_{\ell -1}+R_{\ell }) A_{+,\ell }f)(v)= & {} \left( \sum _{d_\nu (v)=\ell -1}+\sum _{d_\nu (v)=\ell }\right) (A_{+,\ell }f)({\tilde{v}}_\nu )\\= & {} \left( \sum _{d_\nu (v)=\ell -1}+\sum _{d_\nu (v)=\ell }\right) \sum _{d_\mu ({\tilde{v}}_\nu )=\ell } f(({\tilde{v}}_\nu )_\mu ^-)\\= & {} \sum _{d_\mu (v)=\ell } \left( \sum _{d_\nu (v)=\ell -1}+\sum _{d_\nu (v)=\ell }\right) f((v_\mu ^-)_\nu ^\sim ) \\= & {} \sum _{d_\mu (v)=\ell } \left( \sum _{d_\nu (v_\mu ^-)=\ell -1}+\sum _{d_\nu (v_\mu ^-)=\ell }\right) f((v_\mu ^-)_\nu ^\sim )\\= & {} \sum _{d_\mu (v)=\ell } (R_{\ell -1}f+R_{\ell }f)(v_\mu ^-) \\= & {} (A_{+,\ell } (R_{\ell -1}f+R_{\ell })f)(v)\, . \end{aligned}$$

Here we observed that if $d_\mu (v)=\ell$ then $\{\nu :d_\nu (v)=\ell \}\cup \{\nu :d_\nu (v)=\ell -1\} =\{\nu :d_\nu (v_\mu ^-)=\ell \}\cup \{\nu :d_\nu (v_\mu ^-)=\ell -1\}$ since $d_\mu (v_\mu ^-)=\ell -1$ and $d_\nu (v_\mu ^-)=d_\nu (v)$ if $\mu \ne \nu$.

To prove Lemma 20.iii, using again that $([R_\ell ,A_{+,\ell }]f)(v)=\sum _{d_\nu (v)=\ell } f((v)_\nu ^-)_\nu ^\sim )$,

$$\begin{aligned} (R_\ell , [R_\ell ,A_{+,\ell }]f)(v)= & {} \sum _{d_\mu (v)=\ell } ( [R_\ell ,A_{+,\ell }]f)({\tilde{v}}_\mu )= \sum _{d_\mu (v)=\ell } \sum _{d_\nu ({\tilde{v}}_\mu )=\ell }f((({\tilde{v}}_\mu )_\nu ^-)_\nu ^\sim )\\= & {} \sum _{d_\nu (v)=\ell } \sum _{d_\mu (v)=\ell } f((({\tilde{v}}_\nu )_\nu ^-)_\mu ^\sim )\\= & {} \sum _{d_\nu (v)=\ell }\left[ f((({\tilde{v}}_\nu )_\nu ^-)_\nu ^\sim )+ \sum _{d_\mu (({\tilde{v}}_\nu )_\nu ^-)=\ell } f((({\tilde{v}}_\nu )_\nu ^-)_\mu ^\sim )\right] \\= & {} \sum _{d_\nu (v)=\ell }f(v_\nu ^-)+ (R_\ell f)(({\tilde{v}}_\nu )_\nu ^-) \\= & {} (A_{+,\ell }f)(v)+([R_\ell ,A_{+,\ell }] R_\ell f)(v)\, . \end{aligned}$$

Subtracting $([R_\ell ,A_{+,\ell }] R_\ell f)(v)$ from both sides proves Lemma 20.iii. $\square$

Proof of Lemma 21

The proof is by induction on k. If $k=1$ then by Lemma 19, we can write $p_1({\varvec{R}})A_{+,\ell +1}=c_1 A_{+,\ell } p_{1,+}({\varvec{R}}) +c_2 R_\ell A_{+,\ell } q_{1,+}({\varvec{R}})$ for certain constants $c_1,c_2$ depending on $q_0,\dots , q_M$. If g is a common eigenvector of ${\varvec{R}}$ then we can write $p_1({\varvec{R}})A_{+,\ell +1} g=(c_1 A_{+,\ell } p_{1,+}({\varvec{\lambda }}) g +c_2 R_\ell A_{+,\ell } q_{1,+}({\varvec{\lambda }}) g) =(c_1' A_{+,\ell } g +c_2' R_\ell A_{+,\ell } g)$ since $p_{1,+}({\varvec{\lambda }})$ and $q_{1,+}({\varvec{\lambda }})$ are in ${\mathbb {C}}$. Suppose that for fixed k, any $p_k({\varvec{R}})A_{+,\ell _k}\cdots p_1({\varvec{R}})A_{+,1} g$ can be written as a linear combination of at most $2^{k-1}$ terms of the form $B_{+,\ell _k}\cdots B_{+,\ell _1}g$ with $B_{+,\ell }$ as in the lemma, for the same $g\in {\mathcal {W}}$. Let f have this form and set $h=p_{k+1}({\varvec{R}}) A_{+,\ell _{k+1}} f$. Set $\ell =\ell _{k+1}$. Again by Lemma 19, we can write

$$\begin{aligned} p_{k+1}({\varvec{R}}) A_{+,\ell } =c_1 A_{+,\ell } p_{+,k+1}({\varvec{R}}) +c_2 R_{\ell } A_{+,\ell } q_{+,k+1}({\varvec{R}})\, . \end{aligned}$$

(14)

By the induction hypothesis $p_{+,k+1}({\varvec{R}}) f$ is a linear combination of at most $2^{k-1}$ terms of the form $B_{+,\ell _k}^{(1)}\cdots B_{+,\ell _1}^{(1)} g$ and likewise $q_{+,k+1}({\varvec{R}})f$ with terms $B_{+,\ell _k}^{(2)}\cdots B_{+,\ell _1}^{(2)} g$. Applying (14) we conclude then that $p_{k+1}({\varvec{R}}) A_{+,\ell } f$ can be expressed as a sum of at most $2^{k}$ terms of the form $B_{+,\ell _k}\cdots B_{+,\ell _1}g$ for the same $g\in {\mathcal {W}}$. $\square$

Proof of Proposition 9

The main idea is to associate with each r-element set $S\subset \{1,\dots , N\}$ a Hadamard-type matrix whose columns form an orthonormal basis for the $2^r$-dimensional space $\ell ^2(\Sigma _{r,S})$ of vectors f such that $f(v)\ne 0$ and $d_\nu (v)>0$ implies $\nu \in S$. The standard Hadamard matrix of size $2^r\times 2^r$ is (up to normalization) the orthogonal matrix $\otimes ^r H$ where $H=\frac{1}{\sqrt{2}}\left( {\begin{matrix}1 &{} 1\\ 1&{} -1\end{matrix}}\right)$. Each column of $\otimes ^r H$ has the form $\alpha _{\epsilon _1}\otimes \cdots \otimes \alpha _{\epsilon _{r}}$ where $\sqrt{2}\, \alpha _0=\left( {\begin{matrix}1 \\ 1\end{matrix}}\right)$ and $\sqrt{2}\, \alpha _1=\left( {\begin{matrix}1 \\ -1\end{matrix}}\right)$. Specifically, the kth column of $\otimes ^r H$ has $\alpha _{\epsilon _i}$ in the ith tensor slot, according to the binary expansion $k=1+\sum _{i=1}^{r} \epsilon _i 2^{i-1}$, $k=1,\dots , 2^r$. To view these vectors as vertex functions, first, for simplicity, take $S=\{1,\dots , r\}$. Any vertex $v\in \Sigma _{r,S}$ has the form $v=\sum _{\nu =1}^r (-1)^{\epsilon _\nu } e_\nu$ where $\epsilon _\nu \in \{0,1\}$. Thus v is uniquely associated among vertices in $\Sigma _{r,S}$ with $\epsilon (v)=(\epsilon _1,\dots ,\epsilon _r)\in \{0,1\}^r$. Suppose now that $\gamma =(\gamma _1,\dots ,\gamma _r)\in \{0,1\}^r$. Associate with $\gamma$ the Hadamard vector $h_\gamma (v)=2^{-r/2} (-1)^{\langle \epsilon (v),\, \gamma \rangle }$. When $S=\{1,\dots , r\}$, and $(\gamma _1,\dots ,\gamma _r)$ is associated with the integer $n(\gamma )=1+\sum _{p=1}^r \gamma _p 2^{p-1}$, $h_\gamma$ is associated with the $n(\gamma )$th column of $\otimes ^r H$ and can also be expressed as $h_\gamma =\alpha _{\gamma _1}\otimes \cdots \otimes \alpha _{\gamma _r}$.

We claim that $\rho _k h_\gamma =h_\gamma$ if $\gamma _k=0$ and $\rho _k h_\gamma =-h_\gamma$ if $\gamma _k=1$. To see this, observe that $\epsilon _k({\tilde{v}}_k)=1-\epsilon _k(v)$, $k=1,\dots , r$. Therefore

$$\begin{aligned} 2^{-r/2} h_\gamma ({\tilde{v}}_k)= & {} (-1)^{\langle \epsilon ({\tilde{v}}_k),\, \gamma \rangle } = (-1)^{(1-\epsilon _k(v))\gamma _k+ \sum _{\nu \ne k} \epsilon _\nu (v)\gamma _\nu }\\= & {} (-1)^{\gamma _k}(-1)^{\epsilon _k(v)\gamma _k}(-1)^{\sum _{\nu \ne k} \epsilon _\nu (v)\gamma _\nu } = (-1)^{\gamma _k}(-1)^{\langle \epsilon (v),\, \gamma \rangle } =(-1)^{\gamma _k} h_\gamma (v) \end{aligned}$$

since $(-1)^{-\epsilon }=(-1)^\epsilon$ if $\epsilon \in \{0,1\}$. This verifies the claim. It is well known that the matrix $\otimes ^r H$ is orthogonal, so when $S=\{1,\dots , r\}$, the vectors $h_{\gamma ,S}$ form an orthonormal basis for $\ell ^2(\Sigma _{r,S})$ as $\gamma$ ranges over $\{0,1\}^r$.

To address the general case of $S=\{\beta _1,\dots ,\beta _r\}\subset \{1,\dots , N\}$ we assume that the $\beta _i$ are listed in increasing order. Again, each $v\in \Sigma _{r,S}$ is uniquely associated with an $\epsilon =(\epsilon _{\beta _1},\dots ,\epsilon _{\beta _r})$ by $v=\sum _{\nu =1}^r (-1)^{\epsilon _{\beta _\nu }} e_{\beta _\nu }$. We can then assign accordingly $h_{\gamma ,S}(v)=2^{-r/2} (-1)^{\langle \gamma ,\epsilon \rangle }$. To define the vectors $h_{\gamma ,S}$ globally (on $\Sigma _{r}$) we define a block matrix ${\varvec{H}}_r$ of size $2^r\left( {\begin{array}{c}N\\ r\end{array}}\right) \times 2^r\left( {\begin{array}{c}N\\ r\end{array}}\right)$ having $2^r\times 2^r$ diagonal blocks each containing an isomorphic copy of $\otimes ^r H$ whose columns extend $h_{\gamma ,S}$ when the subsets $S\subset \{1,\dots , N\}$ are ordered lexicographically, say. In this way we can associate each $h_{\gamma ,S}$ with a unique column of ${\varvec{H}}_r$. We refer to each $h_{\gamma ,S}$ as a Hadamard-type vector.

Lemma 23

Fix $r\in \{1,\dots , N\}$ and for each $\gamma \in \{0,1\}^r$ and $S=\{\beta _1,\dots ,\beta _r\}\subset \{1,\dots , N\}$ define $h_{\gamma ,S}$ as above. Then

• (i) the vectors $h_{\gamma ,S}$ form an orthonormal basis for $\ell ^2(\Sigma _r)$

• (ii) if $\# \{k: \gamma _k=0\}=s$ then $A_0 h_{\gamma ,S} =(2s-r)\, h_{\gamma ,S}$

• (iii) $\mathrm{span}\{h_{\gamma ,S}:\, A_0 h_{\gamma ,S}=(2s-r) h_{\gamma ,S}\}$ has dimension $\left( {\begin{array}{c}N\\ r\end{array}}\right) \left( {\begin{array}{c}r\\ s\end{array}}\right)$

Proof

(i) follows since the $\{h_{\gamma ,S}\}$ are orthonormal for each fixed S and have disjoint supports for different S. (ii) follows from Lemma 5 and the observations above. For (iii), there are $\left( {\begin{array}{c}N\\ r\end{array}}\right)$ choices of r-element disjoint supports S and the number of s-element subsets of an r-element set is $\left( {\begin{array}{c}r\\ s\end{array}}\right)$ ($0\le s\le r$). $\square$

By the injectivity of $A_+$ on $\ell ^2(\Sigma _r)$ ($r<N/2$) and since $A_+$ is the adjoint of $A_-$, membership in ${\mathcal {W}}_r$ is the same as being in the orthogonal complement of $A_+ (\ell ^2(\Sigma _{r-1}))$ in $\ell ^2(\Sigma _r)$. Since the Hadamard-type vectors that are symmetric in s coordinates span the subspace of $\ell ^2(\Sigma _r)$ of $(2s-r)$-eigenvectors of $A_0$, and since an element $A_+h$ of $A_+ (\ell ^2(\Sigma _{r-1}))$ is a $(2s-r)$-eigenvector of $A_0$ precisely when its preimage h is a $(2s-1-r)$-eigenvector of $A_0$ on $\ell ^2(\Sigma _{r-1})$, it follows that ${\mathcal {W}}_{r;\, 2s-r}$ is equal to the orthogonal complement of the image of the $(2s-1-r)$-eigenspace of $A_0$ under $A_+$ inside the $(2s-r)$-eigenspace of $A_0$. These conditions can be described in terms of Hadamard-type vectors. Specifically, for a Hadamard-type vector $h_{\gamma ',S'}$ defined on $\Sigma _{r-1,S'}$, and S and r-element subset of $\{1,\dots , N\}$ that contains $S'$, define $h_{\gamma ',S',S}$ to be the Hadamard-type vector supported in $\Sigma _{r,S}$ such that $h_{\gamma ',S',S}$ is the truncation on $\Sigma _{r,S}$ of $A_+ h_{\gamma ',S'}$. There is a unique index $\beta \in S\setminus S'$ so $h_{\gamma ',S',S}$ can be identified with the vector $\alpha _{\gamma '_1}\otimes \cdots \alpha _{\gamma '_{i-1}}\otimes \alpha _0\otimes \alpha _{\gamma '_{i}}\otimes \cdots \otimes \alpha _{\gamma '_{r-1}}$ when $\beta$ is the ith element of S when listed in increasing order. Given a fixed $h_{\gamma ',S'}$, its image under $A_+$ can be expressed as a sum $\sum _{\# (S\setminus S')=1} h_{\gamma ',S',S}$, of one-coordinate symmetric extensions of $h_{\gamma ', S'}$.

Since $A_+$ is injective from $\ell ^2(\Sigma _{r-1})$ to $\ell ^2(\Sigma _{r})$, when $h_{\gamma ', S'}$ are $(2s-r-1)$-eigenvectors of $A_0$ in $\ell ^2(\Sigma _{r-1})$, their extensions $\sum _{\# (S\setminus S')=1} h_{\gamma ',S',S}$ are linearly independent $(2s-r)$-eigenvectors of $A_0$ in $\ell ^2(\Sigma _r)$. The latter eigenspace has dimension $\left( {\begin{array}{c}N\\ r\end{array}}\right) \left( {\begin{array}{c}r\\ s\end{array}}\right)$ and contains the image of the $(2s-r-1)$-eigenvectors of $A_0$ as a subspace of dimension $\left( {\begin{array}{c}N\\ r-1\end{array}}\right) \left( {\begin{array}{c}r-1\\ s-1\end{array}}\right)$, accounted for by the Hadamard-type elements $h_{\gamma ',S'}\in \ell ^2(\Sigma _{r-1})$ that are symmetric in $s-1$ coordinates. Thus the orthogonal complement of the span of the images $\sum _{\# (S\setminus S')=1} h_{\gamma ',S',S}$ inside the $(2s-r)$-eigenspace of $A_0$, which is equal to ${\mathcal {W}}_{r;\, 2s-r}$, has dimension $\left( {\begin{array}{c}N\\ r\end{array}}\right) \left( {\begin{array}{c}r\\ s\end{array}}\right) -\left( {\begin{array}{c}N\\ r-1\end{array}}\right) \left( {\begin{array}{c}r-1\\ s-1\end{array}}\right)$. This proves the first statement of Proposition 9.

As for the second statement, orthogonality of the spaces ${\mathcal {W}}_{r;\, 2s-r}$ for different s follows from their being spanned by Hadamard-type vectors having different numbers of symmetric coordinates. Completeness follows from the fact that the Hadamard-type vectors $h_{\gamma ,S'}$ span $\ell ^2(\Sigma _{r-1})$ as $\gamma '$ ranges over $\{0,1\}^{r-1}$ and $S'$ ranges over all $(r-1)$-elements subsets of $\{1,\dots ,N\}$. The space ${\mathcal {W}}_{r,-r}$ is spanned by those Hadamard-type vectors in $\ell ^2(\Sigma _r)$ that are antisymmetric in each coordinate. Since the image of $h_{\gamma ',S'}$ under $A_+$ is symmetric in at least one coordinate on each S, Hadamard vectors $h_{\gamma ,S}\in \ell ^2(\Sigma _{r,S})$, $\gamma =(1,\dots , 1)$) are automatically in the kernel of $A_-$. The subspace ${\mathcal {W}}_{r,-r}$ of ${\mathcal {W}}_r$ has dimension $\left( {\begin{array}{c}N\\ r\end{array}}\right)$. This proves Proposition 9.

We now provide a proof of the first part of Lemma 7—that if f is a $\lambda$-eigenvector of $A_0$ in $\ell ^2(\Sigma _r)$ on ${\mathcal {C}}_3^N$ and in the kernel of $A_-$ then $A_+^{k+1}f=0$ whenever $2N-2k-3r-\lambda \le 0$. Fix $s\in \{1,\dots , r\}$ and $\lambda =2s-r$. Then the last condition is that $k\ge N-r-s$. We will use the following fact proved in [28].

Lemma 24

Let $f\in \ell ^2(\Sigma _r({\mathcal {C}}_2^N))$ ($r<N/2$). If $A_-({\mathcal {C}}_2^N)f=0$ and $k>N-2r$, then $A_+^kf=0$.

Proof of (i) of Lemma 7

In view of the Hadamard-type analysis above, specifically that $A_-f=0$ precisely when $\langle f,\, \sum _{\beta \notin S'} h_{\gamma ',S',S'\cup \{\beta \}}\rangle =0$ for each $\gamma '\in \{0,1\}^{r-1}$ and $S'$ an $(r-1)$-element subset of $\{1,\dots ,N\}$, it suffices to assume that for each r-element set $S\subset \{1,\dots , N\}$, f is symmetric in s of the coordinates in S and antisymmetric in the others. First, suppose that f is symmetric in every coordinate ($\lambda =r$ in Lemma 7). That is, for each $v\in \Sigma _r$ and each $\nu$ such that $d_\nu (v)=1$, $f({\tilde{v}}_\nu )=f(v)$. Then f is constant on $\Sigma _{r,S}$ and can regarded as a function $f_2$ defined on the r-element subsets of $\{1,\dots , N\}$. Subsets of $\{1,\dots , N\}$ can be used to index the vertices of ${\mathcal {C}}_2^N$. Thus, f can be identified with a function $f_2$ on $V({\mathcal {C}}_2^N)$ supported in $\Sigma _r({\mathcal {C}}_2^N)$ and in the kernel of $A_-({\mathcal {C}}_2^N)$. Moreover, $(A_+^kf)(v)$ is also constant in this case on $\Sigma _{r+k,{\hat{S}}}$ where ${\hat{S}}$ is any fixed $(r+k)$-element subset of $\{1,\dots , N\}$. In fact, $A_+^kf$ can be identified in the same way with the function $A_+^k({\mathcal {C}}_2^N) f_2$. In [28] we proved that for such $f_2$, $(A_+({\mathcal {C}}_2^N))^kf_2=0$ whenever $k>N-2r$. Consequently we also have $A_+^kf=0$ on $V({\mathcal {C}}_3^N)$ when $k>N-2r$.

For the general case, if for each $v\in \Sigma _r$ and each k such that $d_\nu (v)=1$, either $(\rho _\nu f)(v)=f(v)$ or $(\rho _\nu f)(v)=-f(v)$, then we have

$$\begin{aligned} (A_-f)(v)=\sum _{\nu :d_\nu (v) =0} f(v+e_\nu )+f(v-e_\nu )=2\sum _{\nu : f(v+e_\nu )=f(v-e_\nu )} f(v+e_\nu )\nonumber \\ \end{aligned}$$

(15)

since $\sum _{\nu : f(v+e_\nu )=-f(v-e_\nu )} f(v+e_\nu )+f(v-e_\nu )=0$ automatically. Fix an $(r-s)$-element subset $S'\subset \{1,\dots , N\}$ (to be concrete, without loss of generality take $S'=\{N-(r-s)+1,\dots , N\}$) and let $V(r,S',f)$ be the set of all vertices in $\Sigma _r$ such that $f({\tilde{v}}_\nu )=-f(v)$ for each $\nu \in S'$. Fix a choice $(\ell _{N-(r-s)+1},\dots ,\ell _N)$ of the last $(r-s)$ (i.e., $S'$) coordinates and for $v\in V(r,S',f)$ let ${\bar{v}}\in V(r,S',f)$ be the vertex whose first s level-one coordinates are those of v and whose last $(r-s)$ ones are as fixed. We call $v\in V(r,S',f)\mapsto f({\bar{v}})$ the clamping of f at $(\ell _{N-(r-s)+1},\dots ,\ell _N)$. Then $f({\bar{v}})$ is constant on $\Sigma _{r,{\bar{S}}\cup S'}$ whenever ${\bar{S}}\subset \{1,\dots , N-(r-s)\}$ has s elements. We can thus identify with this restriction of f a function $f_{S'}$ defined on s-element sets ${\bar{S}}\subset \{1,\dots , N-(r-s)\}$ by $f_{S'}({\bar{S}})=f({\bar{v}})$ whenever ${\bar{v}}\in \Sigma _{r,S'\cup {\bar{S}}}$. By (15), $f_{S'}$ satisfies $\sum _{\beta \notin U} f_{S'}(U\cup \{\beta \})=0$ whenever U is an $(s-1)$-element subset of $\{1,\dots , N-(r-s)\}$. This last condition is equivalent to $f_{S'}$, thought of as a function in $\ell ^2(\Sigma _s({\mathcal {C}}_2^N))$, being in the kernel of $A_-({\mathcal {C}}_2^N)$. By the results in [28], then $A_+({\mathcal {C}}_2^N)^k f_{S'}=0$ whenever $k>(N-(r-s))-2s=N-r-s$. Returning to ${\mathcal {C}}_3^N$, the value of $(A_+^k f)({\bar{v}})$ at any vertex in $\Sigma _{r+k}$ whose last $(r-s)$ coordinates are fixed as above, is equal to a value of $A_+({\mathcal {C}}_2^N)^k f_{S'}$ on a $(k+r)$-element set. In particular, the value is equal to zero if $k>N-r-s$. This same conclusion holds if we replace this clamping procedure done for $S'=\{N-(r-s)+1,\dots , N\}$ by a parallel procedure for f restricted to vertices determined by any fixed choice $S'$ of $(r-s)$ anti-symmetry coordinates. Since f is a linear sum of such restrictions, the conclusion that $A_+^{k}f=0$ when $k>N-s-r$ follows and Lemma 7 is proved. $\square$