Sparse sums of squares on finite abelian groups and improved semidefinite lifts

Abstract

Let G be a finite abelian group. This paper is concerned with nonnegative functions on G that are sparse with respect to the Fourier basis. We establish combinatorial conditions on subsets ${\mathcal {S}}$ and ${\mathcal {T}}$ of Fourier basis elements under which nonnegative functions with Fourier support ${\mathcal {S}}$ are sums of squares of functions with Fourier support ${\mathcal {T}}$. Our combinatorial condition involves constructing a chordal cover of a graph related to G and ${\mathcal {S}}$ (the Cayley graph ${{\mathrm{Cay}}}(\widehat{G},{\mathcal {S}})$) with maximal cliques related to ${\mathcal {T}}$. Our result relies on two main ingredients: the decomposition of sparse positive semidefinite matrices with a chordal sparsity pattern, as well as a simple but key observation exploiting the structure of the Fourier basis elements of G (the characters of G). We apply our general result to two examples. First, in the case where $G = {\mathbb {Z}}_2^n$, by constructing a particular chordal cover of the half-cube graph, we prove that any nonnegative quadratic form in n binary variables is a sum of squares of functions of degree at most $\left\lceil n/2 \right\rceil$, establishing a conjecture of Laurent. Second, we consider nonnegative functions of degree d on ${\mathbb {Z}}_N$ (when d divides N). By constructing a particular chordal cover of the dth power of the N-cycle, we prove that any such function is a sum of squares of functions with at most $3d\log (N/d)$ nonzero Fourier coefficients. Dually this shows that a certain cyclic polytope in ${\mathbb {R}}^{2d}$ with N vertices can be expressed as a projection of a section of the cone of positive semidefinite matrices of size $3d\log (N/d)$. Putting $N=d^2$ gives a family of polytopes in ${\mathbb {R}}^{2d}$ with linear programming extension complexity $\varOmega (d^2)$ and semidefinite programming extension complexity $O(d\log (d))$. To the best of our knowledge, this is the first explicit family of polytopes $(P_d)$ in increasing dimensions where ${{\mathrm{xc_{\text {PSD}}}}}(P_d) = o({{\mathrm{xc_{\text {LP}}}}}(P_d))$, where ${{\mathrm{xc_{\text {PSD}}}}}$ and ${{\mathrm{xc_{\text {LP}}}}}$ are respectively the SDP and LP extension complexity.

Additional proofs

1.1 Proofs from Sect. 3.2

In this section we provide detailed proofs of the main results in Sect. 3.2.

Proof of Proposition 4

Suppose $\mu$ is a probability measure supported on G. Then because $\mu$ is a probability measure, $({\mathbb {E}}_{\mu }[\chi ])_{\chi \in {\mathcal {S}}}$ certainly satisfies ${\mathbb {E}}_{\mu }[1_{\widehat{G}}] = 1$ and ${\mathbb {E}}_{\mu }[|f|^2] \ge 0$ whenever $f\in {\mathcal {F}}(G,{\mathbb {C}})$. Conversely, suppose $\ell$ is a linear functional on ${\mathcal {F}}(G,{\mathbb {C}})$ such that $\ell (1_{\widehat{G}})=1$ and $\ell (|f|^2) \ge 0$ for all $f\in {\mathcal {F}}(G,{\mathbb {C}})$. We show that $\ell (\cdot )$ coincides with ${\mathbb {E}}_{\mu }[\cdot ]$ for some probability measure $\mu$ supported on G. For any $x\in G$ define $\mu (\{x\}) = \ell (\delta _x)$. Since $\delta _x = |\delta _x|^2$ we have that $\mu (\{x\})=\ell (|\delta _x|^2)\ge 0$ for all $x\in G$. Since $\ell$ is linear

$$\begin{aligned} \sum _{x\in G}\mu (\{x\}) = \ell \left( \sum _{x\in G}\delta _x\right) = \ell (1_{\widehat{G}}) = 1. \end{aligned}$$

Hence $\mu$ defines a probability measure supported on G and $\ell (\cdot )$ is exactly the corresponding expectation ${\mathbb {E}}_\mu [\cdot ]$. The second description of ${\mathcal {M}}(G,\widehat{G})$ in the lemma follows by rewriting the condition $\ell (|f|^2)$ for all $f\in {\mathcal {F}}(G,{\mathbb {C}})$ in coordinates with respect to the character basis as

$$\begin{aligned} \ell \left( \big |\sum _{\chi \in \widehat{G}} \widehat{f}(\chi )\chi \big |^2\right) = \sum _{\chi ,\chi '\in \widehat{G}}\ell (\overline{\chi }\chi ')\overline{\widehat{f}(\chi )}\widehat{f}(\chi ')\quad \text {for all }(\widehat{f}(\chi ))_{\chi \in \widehat{G}}\in {\mathbb {C}}^{\widehat{G}}. \end{aligned}$$

This is exactly the definition of the matrix $M(\ell ) = [\ell _{\overline{\chi }\chi '}]_{\chi ,\chi '\in \widehat{G}}$ being Hermitian positive semidefinite. $\square$

Proof of Proposition 5

If $(\ell _{\chi })_{\chi \in {\mathcal {S}}}\in {\mathcal {M}}(G,{\mathcal {S}})$ then from Corollary 2 there is $(y_{\chi })_{\chi \in \widehat{G}}$ such that $y_\chi = \ell _\chi$ for all $\chi \in {\mathcal {S}}$, $y_{1_{\widehat{G}}}=1$, and $M(y) \succeq 0$. Hence we can take $Y_{\chi ,\chi '} = y_{\overline{\chi }\chi '}$ to show that $(\ell _\chi )_{\chi \in {\mathcal {S}}}$ is an element of the right hand side of (21).

Conversely, suppose there exists $Y\in \mathbf {H}_+^{\widehat{G}}$ with $Y_{\chi ,\chi '} = \ell _{\overline{\chi }\chi '}$ whenever $\overline{\chi }\chi '\in {\mathcal {S}}$ and $Y_{\chi ,\chi }=1$ for all $\chi \in \widehat{G}$. Our task is to construct, from Y some $(y_\chi )_{\chi \in \widehat{G}}$ such that $y_{\chi }=\ell _\chi$ for all $\chi \in {\mathcal {S}}$, $y_{1_{\widehat{G}}}=1$, and $M(y)\succeq 0$. Observe that $\widehat{G}$ acts on Hermitian matrices $\mathbf {H}^{\widehat{G}}$ indexed by elements of $\widehat{G}$ by simultaneously permuting the rows and columns, i.e. by $[\lambda \cdot Y]_{\chi ,\chi '}= Y_{\overline{\lambda }\chi ,\overline{\lambda }\chi '}$. We construct a new matrix Z by averaging Y over this group action:

$$\begin{aligned} Z_{\chi ,\chi '} := \frac{1}{|\widehat{G}|}\sum _{\lambda \in \widehat{G}} [\lambda \cdot Y]_{\chi ,\chi '} = \frac{1}{|\widehat{G}|} \sum _{\lambda \in \widehat{G}} Y_{\overline{\lambda }\chi ,\overline{\lambda }\chi '}. \end{aligned}$$

Since the action of $\lambda$ is by simultaneously permuting rows and columns each $\lambda \cdot Y$, and hence Z itself, is positive semidefinite with ones on the diagonal. Since we have constructed Z by averaging over a group action, Z is fixed by the action and so satisfies $Z_{\overline{\lambda }\chi ,\overline{\lambda }\chi '} = Z_{\chi ,\chi '}$ for all $\chi ,\chi '\in \widehat{G}$. Consequently there is some $(y_\chi )_{\chi \in \widehat{G}}$ such that $Z_{\chi ,\chi '} = y_{\overline{\chi }\chi '}$ for all $\chi ,\chi '\in \widehat{G}$. It remains to show that if $\overline{\chi }\chi '\in {\mathcal {S}}$ then $y_{\overline{\chi }\chi '} := Z_{\chi ,\chi '} = Y_{\chi ,\chi '} := \ell _{\overline{\chi }\chi '}$. This holds because if $\overline{\chi }\chi '\in {\mathcal {S}}$ then

$$\begin{aligned} Z_{\chi ,\chi '} = \frac{1}{|\widehat{G}|} \sum _{\lambda \in \widehat{G}} Y_{\overline{\lambda }\chi ,\overline{\lambda }\chi '} = \frac{1}{|\widehat{G}|}\sum _{\lambda \in \widehat{G}}\ell _{\overline{\chi }\chi '} = \ell _{\overline{\chi }\chi '}. \end{aligned}$$

Hence y has all the desired properties, completing the proof. $\square$

Proof of Theorem 1D

First we show that ${\mathcal {M}}(G,{\mathcal {S}})$ is a subset of the right-hand-side of (22). To see this observe that the right-hand-side of (21) is certainly contained in the right-hand-size of (22).

We now establish the reverse inclusion. Suppose $(\ell _\chi )_{\chi \in {\mathcal {S}}}$ is such that there exists $(y_\chi )_{\chi \in {\mathcal {T}}^{-1}{\mathcal {T}}}$ with $y_\chi = \ell _\chi$ for all $\chi \in {\mathcal {S}}$, $y_{1_{\widehat{G}}}=1$ and $M_{{\mathcal {T}}}(y)\succeq 0$. Let $\varGamma$ be a chordal cover of ${{\mathrm{Cay}}}(\widehat{G},{\mathcal {S}})$ with Fourier support ${\mathcal {T}}$. Specifically $\varGamma$ has the property that for every maximal clique ${\mathcal {C}}$ of $\varGamma$ there is $\chi _{{\mathcal {C}}}\in \widehat{G}$ such that $\chi _{{\mathcal {C}}}{\mathcal {C}}\subseteq {\mathcal {T}}$.

Define the $\varGamma$-partial matrix $Y_{\eta ,\eta '} = y_{\overline{\eta }\eta '}$ whenever $(\eta ,\eta ')$ form an edge of $\varGamma$ and $Y_{\chi ,\chi }=1$ for all $\chi \in \widehat{G}$. This is well defined because if $(\eta ,\eta ')$ is an edge of $\varGamma$, then $\overline{\eta }\eta '\in {\mathcal {T}}^{-1}{\mathcal {T}}$. To see this observe that any edge of $\varGamma$ is contained in a maximal clique ${\mathcal {C}}$, and so there is some $\chi _{{\mathcal {C}}}\in \widehat{G}$ such that $\chi _{{\mathcal {C}}}\eta \in {\mathcal {T}}$ and $\chi _{{\mathcal {C}}}\eta '\in {\mathcal {T}}$. Consequently $\overline{\eta }\eta ' = \overline{\chi _{{\mathcal {C}}}\eta }\chi _{{\mathcal {C}}}\eta ' \in {\mathcal {T}}^{-1}{\mathcal {T}}$.

We show that $Y[{\mathcal {C}},{\mathcal {C}}] \succeq 0$ for all maximal cliques ${\mathcal {C}}$ of the chordal graph $\varGamma$. This holds because

$$\begin{aligned} Y[{\mathcal {C}},{\mathcal {C}}] = [y_{\overline{\eta }\eta '}]_{\eta ,\eta '\in {\mathcal {C}}} = [y_{\overline{\chi _{{\mathcal {C}}}\eta }\chi _{{\mathcal {C}}}\eta '}]_{\eta ,\eta '\in {\mathcal {C}}} = [y_{\overline{\chi }\chi '}]_{\chi ,\chi '\in \chi _{{\mathcal {C}}}{\mathcal {C}}} \end{aligned}$$

which, since $\chi _{{\mathcal {C}}}{\mathcal {C}}\subseteq {\mathcal {T}}$, is a principal submatrix of the positive semidefinite (by assumption) matrix $M_{{\mathcal {T}}}(y) = [y_{\overline{\chi }\chi '}]_{\chi ,\chi '\in {\mathcal {T}}}$. By the chordal completion theorem (Theorem 6) we can complete Y to a positive semidefinite matrix $Y\in \mathbf {S}_+^{\widehat{G}}$. The completed matrix has unit diagonal and, whenever $\overline{\chi }\chi '\in {\mathcal {S}}$,

$$\begin{aligned} Y_{\chi ,\chi '} = y_{\overline{\chi }\chi '} = \ell _{\overline{\chi }\chi '} \end{aligned}$$

where the first equality is because the edge set of $\varGamma$ contains the edge set of ${{\mathrm{Cay}}}(\widehat{G},{\mathcal {S}})$ and the second is from the definition of y. Hence, by Proposition 5, $(\ell _{\chi })_{\chi \in {\mathcal {S}}}\in {\mathcal {M}}(G,{\mathcal {S}})$, as we require. $\square$

1.2 Proof of Lemma 1

Proof of Lemma 1

First note that $L \in \mathbf {H}_+^d$ if and only if $\overline{L}\in \mathbf {H}_+^d$ which holds if and only if the block diagonal matrix $\left[ {\begin{matrix} L &{} 0\\ 0 &{} \overline{L}\end{matrix}}\right] \in \mathbf {H}^{2d}_+$. Conjugating by a unitary matrix we obtain

$$\begin{aligned} \begin{bmatrix} \frac{1}{\sqrt{2}}I&\quad \frac{1}{\sqrt{2}}I\\ \frac{i}{\sqrt{2}}I&\quad -\frac{i}{\sqrt{2}}I\end{bmatrix} \begin{bmatrix} L&\quad 0\\0&\quad \overline{L}\end{bmatrix}\begin{bmatrix}\frac{1}{\sqrt{2}}I&\quad \frac{1}{\sqrt{2}}I\\ \frac{i}{\sqrt{2}}I&\quad -\frac{i}{\sqrt{2}}I \end{bmatrix}^* = \begin{bmatrix} \quad {{\mathrm{Re}}}[L]&\quad {{\mathrm{Im}}}[L]\\ -{{\mathrm{Im}}}[L]&\quad {{\mathrm{Re}}}[L] \end{bmatrix}. \end{aligned}$$

(31)

We have simply recovered the familiar realization of $\mathbf {H}_+^d$ as a section of $\mathbf {S}_+^{2d}$, and have not yet used any special properties of ${\mathcal {L}}$. To complete the proof it remains to carefully choose a $2d\times 2d$ orthogonal matrix Q (depending on J) such that

$$\begin{aligned} Q \begin{bmatrix} {{\mathrm{Re}}}[L]&{{\mathrm{Im}}}[L]\\ -{{\mathrm{Im}}}[L]&\quad {{\mathrm{Re}}}[L] \end{bmatrix} Q^T = \begin{bmatrix} {{\mathrm{Re}}}[L]-J{{\mathrm{Im}}}[L]&\quad 0 \\ 0&\quad {{\mathrm{Re}}}[L] - J{{\mathrm{Im}}}[L]\end{bmatrix}\quad \text {for all }L\in {\mathcal {L}}. \end{aligned}$$

Observe that $J^2 = I$ and $J^TJ=I$ imply that $J= J^T$. Since $JLJ^T = \overline{L}$ we have that for all $L\in {\mathcal {L}}$,

$$\begin{aligned} {{\mathrm{Re}}}[L] = \frac{L+JLJ}{2}\quad \text {and}\quad {{\mathrm{Im}}}[L] = \frac{L - JLJ}{2i}. \end{aligned}$$

(32)

It follows that for all $L\in {\mathcal {L}}$, ${{\mathrm{Re}}}[L]$ and ${{\mathrm{Im}}}[L]$ commute and anti-commute respectively with J, i.e.,

$$\begin{aligned} J{{\mathrm{Re}}}[L] = {{\mathrm{Re}}}[L]J\quad \text {and}\quad J{{\mathrm{Im}}}[L] = -{{\mathrm{Im}}}[L]J. \end{aligned}$$

(33)

Choosing Q to be the orthogonal matrix $Q = \frac{1}{\sqrt{2}}\left[ {\begin{matrix} I&JJ&I\end{matrix}}\right]$ we obtain

$$\begin{aligned}&\begin{bmatrix} \frac{1}{\sqrt{2}}I&\quad \frac{1}{\sqrt{2}}J\\ -\frac{1}{\sqrt{2}}J&\quad \frac{1}{\sqrt{2}}I\end{bmatrix} \begin{bmatrix} {{\mathrm{Re}}}[L]&\quad {{\mathrm{Im}}}[L]\\ -{{\mathrm{Im}}}[L]&\quad {{\mathrm{Re}}}[L] \end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt{2}}I&\quad \frac{1}{\sqrt{2}}J\\ -\frac{1}{\sqrt{2}}J&\quad \frac{1}{\sqrt{2}}I \end{bmatrix}^T\\&\quad = \begin{bmatrix} {{\mathrm{Re}}}[L] - J{{\mathrm{Im}}}[L]&\quad 0\\0&\quad {{\mathrm{Re}}}[L]-J{{\mathrm{Im}}}[L]\end{bmatrix}. \end{aligned}$$

Clearly this last matrix is positive semidefinite if and only if the real symmetric matrix ${{\mathrm{Re}}}[L]-J{{\mathrm{Im}}}[L]$ is positive semidefinite, completing the proof. $\square$

1.3 Proof of Proposition 6

Proof of Proposition 6

The proof proceeds as follows. First we define a graph $\varGamma$ and prove that it is a chordal cover of ${{\mathrm{Cay}}}(\widehat{G},{\mathcal {S}})$. We then characterize the maximal cliques of $\varGamma$. Finally we show that for any maximal clique ${\mathcal {C}}$ of $\varGamma$ there is some $S\in 2^{[n]}$ such that $S\triangle {\mathcal {C}}\subseteq {\mathcal {T}}$, establishing the stated result. We consider the two cases $\lceil n/2\rceil$ even and $\lceil n/2\rceil$ odd separately. We describe the argument in detail in the case where $\lceil n/2\rceil$ is even, and just sketch the required modifications in the case where $\lceil n/2\rceil$ is odd.

Assume that $\lceil n/2\rceil$ is even. Let $\varGamma$ be the graph with vertex set $2^{[n]}$ such that two vertices S, T are adjacent in $\varGamma$ if and only if either

• |S| and |T| are both even and $||S|-|T|| \le 2$ or

• |S| and |T| are both odd and $||\phi (S)| - |\phi (T)|| \le 2$.

Note that just like ${{\mathrm{Cay}}}(\widehat{G},{\mathcal {S}})$, the graph $\varGamma$ also has two connected components with vertex sets ${\mathcal {T}}_{\text {even}}$ and ${\mathcal {T}}_{\text {odd}}$. Furthermore, $\phi$ (defined in Sect. 4.2) is also an automorphism of $\varGamma$ that exchanges these two connected components. Observe that if $|S\triangle T| =2$ (i.e. S and T are adjacent in ${{\mathrm{Cay}}}(\widehat{G},{\mathcal {S}})$) then both $||S|-|T|| \le 2$ and $||\phi (S)| - |\phi (T)|| \le 2$ hold. Hence if S and T are adjacent in ${{\mathrm{Cay}}}(\widehat{G},{\mathcal {S}})$ they are also adjacent in $\varGamma$.

We now show that $\varGamma$ is a chordal graph. Let the vertices $S_1,S_2,S_3,\ldots ,S_k$ form a k-cycle (with $k\ge 4$) in $\varGamma$ such that each of the $S_i\in {\mathcal {T}}_{\text {even}}$. Without loss of generality assume that $|S_1| \le |S_i|$ for $1\le i \le k$. We show that the cycle $S_1,S_2,S_3,\ldots ,S_k$ has a chord. If $|S_2|=|S_1|$ then $||S_1|-|S_3|| = ||S_2|-|S_3||\le 2$ (since $S_2$ and $S_3$ are adjacent) and so there is a chord between $S_1$ and $S_3$. Otherwise suppose $|S_2| = |S_1|+2$. Because $S_1$ and $S_k$ are adjacent we see that either $|S_k|=|S_1| = |S_2|-2$ or $|S_k| = |S_1|+2 = |S_2|$ and so there is a chord between $S_2$ and $S_k$. Now suppose $S_1,S_2,S_3,\ldots ,S_k$ form a k-cycle (with $k\ge 4$) in $\varGamma$ such that each of the $S_i\in {\mathcal {T}}_{\text {odd}}$. Then the image of the cycle under $\phi$ is a k-cycle in $\varGamma$ with vertices in ${\mathcal {T}}_{\text {even}}$ and so it has a chord. Since $\phi$ is an automorphism of $\varGamma$ it follows that $S_1,S_2,S_3,\ldots ,S_k$ also has a chord. So $\varGamma$ is a chordal cover of ${{\mathrm{Cay}}}(\widehat{G},{\mathcal {S}})$.

The subgraphs of $\varGamma$ induced by the vertex sets ${\mathcal {C}}_k:= {\mathcal {T}}_{k} \cup {\mathcal {T}}_{k+2}$ (for $k=0,2,\ldots ,2\lfloor n/2\rfloor -2$) and the vertex sets $\phi ({\mathcal {C}}_k)$ (for $k=0,2,\ldots ,2\lfloor n/2\rfloor -2$) are cliques in $\varGamma$. In fact, these are maximal cliques in $\varGamma$. To show that each ${\mathcal {C}}_k$ is a maximal clique, suppose S is a vertex that is not in ${\mathcal {C}}_k$. Then either |S| is odd (in which case S is not adjacent to any element of ${\mathcal {C}}_k$) or $|S| \le k-2$ (in which case S is not adjacent to any $T\in {\mathcal {T}}_{k+2}$) or $|S|\ge k+4$ (in which case S is not adjacent to any $T\in {\mathcal {T}}_{k}$). Hence there is no inclusion-wise larger clique of $\varGamma$ containing ${\mathcal {C}}_k$. Since $\phi$ is an automorphism of $\varGamma$ it follows that the $\phi ({\mathcal {C}}_k)$ are also maximal cliques of $\varGamma$. Finally, there are no other maximal cliques in $\varGamma$ because every edge of $\varGamma$ is contained either in ${\mathcal {C}}_k$ or $\phi ({\mathcal {C}}_k)$ for some $k=0,2,\ldots ,2\lfloor n/2\rfloor -2$.

It remains to show that for any maximal clique ${\mathcal {C}}_k$ (for $k=0,2,\ldots ,2\lfloor n/2\rfloor -2$) of $\varGamma$ there is $S_k\in 2^{[n]}$ such that $S_k\triangle {\mathcal {C}}_k \subseteq {\mathcal {T}}$. This is sufficient to establish that ${{\mathrm{Cay}}}(\widehat{G},{\mathcal {S}})$ has a chordal cover with Fourier support ${\mathcal {T}}$ because for the cliques $\phi ({\mathcal {C}}_k)$ we have that $\phi (S_k)\triangle \phi ({\mathcal {C}}_k) = S_k\triangle {\mathcal {C}}_k \subseteq {\mathcal {T}}$. The following gives valid choices of $S_k$ (for $k=0,2,\ldots ,2\lfloor n/2\rfloor -2$).

• If $k\le \lceil n/2\rceil -2$ then ${\mathcal {C}}_k\subseteq {\mathcal {T}}$ so we can take $S_k=\emptyset$.

• If $k \ge \lceil n/2\rceil$ and n is even then $n=2\lceil n/2\rceil$ and so $n-k-2 \le \lceil n/2\rceil -2$. Hence $[n]\triangle {\mathcal {C}}_k = {\mathcal {C}}_{n-k-2} \subseteq {\mathcal {T}}$ so we can take $S_k = [n]$.

• If $k \ge \lceil n/2\rceil$ and n is odd then $n=2\lceil n/2\rceil -1$ and so $n-k+1 \le \lceil n/2\rceil$. Hence

  $$\begin{aligned} \phi ([n])\triangle {\mathcal {C}}_k= & {} [n]\triangle \phi ({\mathcal {C}}_k) \subseteq [n]\triangle ({\mathcal {T}}_{k-1} \cup {\mathcal {T}}_{k+1} \cup {\mathcal {T}}_{k+3})\\&\subseteq {\mathcal {T}}_{n-k-3}\cup {\mathcal {T}}_{n-k-1} \cup {\mathcal {T}}_{n-k+1} \subseteq {\mathcal {T}}\end{aligned}$$

  so we can take $S_k = \phi ([n])$.

This completes the argument in the case where $\lceil n/2\rceil$ is even.

In the case where $\lceil n/2\rceil$ is odd we exchange the roles of the odd and even components in the definition of $\varGamma$ and throughout the argument. More precisely, two vertices S, T are adjacent in $\varGamma$ if and only if either

• |S| and |T| are both odd and $||S|-|T|| \le 2$ or

• |S| and |T| are both even and $||\phi (S)| - |\phi (T)|| \le 2$.

It is still the case that $\varGamma$ is a chordal cover of ${{\mathrm{Cay}}}(\widehat{G},{\mathcal {S}})$. Its maximal cliques are now ${\mathcal {C}}_k:= {\mathcal {T}}_{k} \cup {\mathcal {T}}_{k+2}$ for $k=1,3,\ldots ,2\lceil n/2\rceil -3$ together with the $\phi ({\mathcal {C}}_k)$. Note that the cliques are now indexed by odd integers. As before, we can choose the $S_k$ (for $k=1,3,\ldots ,2\lceil n/2\rceil -3$) to be $S_k = \emptyset$ if $k\le \lceil n/2\rceil -2$, $S_k = [n]$ if $k\ge \lceil n/2\rceil$ and n is even, and $S_k = \phi ([n])$ if $k\ge \lceil n/2\rceil$ and n is odd.

This completes the argument in the case where $\lceil n/2\rceil$ is odd. $\square$

1.4 Proof of Theorem 9: chordal cover of the cycle graph

In this appendix we prove Theorem 9 where we construct a chordal cover of the cycle graph $C_N$. Theorem 10 below shows how to construct a chordal cover of the cycle graph $C_{N+1}$ on $N+1$ nodes, by induction. The chordal cover of $C_N$ used to obtain Theorem 9 will then be obtained simply by contracting a certain edge of the chordal cover of $C_{N+1}$ (more details below). We thus start by describing a chordal cover of the $N+1$-cycle.

Theorem 10

(Chordal cover of the cycle graph on $N+1$ vertices). Let N be an integer greater than or equal 2. Let $k_1 < \dots < k_l$ be the position of the nonzero digits in the binary expansion of N, i.e., $N = \sum _{i=1}^l 2^{k_i}$. Let k be the largest integer such that $2^k < N$ (i.e., $k = k_l - 1$ if N is a power of two and $k=k_l$ otherwise). Then there exists a chordal cover of the cycle graph $C_{N+1}$ on $N+1$ nodes with Fourier support

$$\begin{aligned} {\mathcal {T}}= \{0\} \cup \{ \pm 2^i,i=0,\dots ,k\} \cup \left\{ \sum _{j=1}^i 2^{k_j},i=1,\dots ,l-1\right\} . \end{aligned}$$

(34)

Proof

The proof of the theorem is by induction on N. Consider the cycle graph on $N+1$ nodes where nodes are labeled $0,1,\dots ,N$. To construct a chordal cover of the graph, we first put an edge between nodes 0 and $2^k$ and another edge between nodes $2^k$ and N, where $2^k$ is the largest power of two that is strictly smaller than N. This is depicted in Fig. 7.

Fig. 7

ecursive construction of a chordal cover of the cycle $0\dots N$ on $N+1$ vertices

Note that the triangle $\{0,2^k,N\}$ is equivalent, by translation, to $\{-2^k,0,N-2^k\}$. We now use induction to construct a chordal cover of the two remaining parts of the cycle (denoted (a) and (b) in Fig. 7).

• For part (a), which is a cycle graph labeled $0\dots N'$ with $N'=2^k$, the induction hypothesis gives us a chordal cover with Fourier support

  $$\begin{aligned} {\mathcal {T}}_a = \{0\}\cup \{ \pm 2^i,i=0,\dots ,k-1\}. \end{aligned}$$

  (35)

• For part (b) of the graph, we use induction on the cycle $2^k\dots N$ which is, by translation, equivalent to the cycle with labels $0\dots N''$ where $N''=N-2^k$. We distinguish two cases.

  • If $N = 2^{k+1}$, then we have $N'' = 2^k$ and induction gives a chordal cover of (b) with the same Fourier support as for part (a). Thus in this case we get a chordal cover of the full $(N+1)$-cycle with Fourier support

    $$\begin{aligned} {\mathcal {T}}_a \cup \{-2^k,0,2^k\} = \{0\}\cup \{ \pm 2^i,i=0,\dots ,k\} \end{aligned}$$

    which is what we want.

  • Now assume that $N < 2^{k+1}$, which means that the most significant bit of N is at position $k=k_l$. Thus the binary expansion of $N''=N-2^k$ is the same as that of N except that the bit at position $k=k_l$ is replaced with a 0. Let $k''$ be the largest integer such that $2^{k''} < N''$. Using induction we get a chordal cover of the cycle $0\dots N''$ with Fourier support

    $$\begin{aligned} {\mathcal {T}}_b = \{0\} \cup \{\pm 2^i,i=0,\dots ,k''\} \cup \left\{ \sum _{j=1}^i 2^{k_j}, j=1,\dots ,l-2\right\} . \end{aligned}$$

    (36)

    Combining the chordal cover of parts (a) and part (b) we get a chordal cover of the $(N+1)$-cycle with Fourier support

    $$\begin{aligned} \underbrace{\{-2^k,0,N-2^k\}}_{\text {triangle }\{0,2^k,N\}} \cup {\mathcal {T}}_a \cup {\mathcal {T}}_b. \end{aligned}$$

    Given the expressions (35) and (36) for ${\mathcal {K}}_a$ and ${\mathcal {K}}_b$, and noting that $k'' \le k-1$ and that $N - 2^k = \sum _{j=1}^{l-1} 2^{k_j}$, one can check that the chordal cover has Fourier support

    $$\begin{aligned} {\mathcal {T}}= \{0\} \cup \{\pm 2^i,i=0,\dots ,k\} \cup \left\{ \sum _{j=1}^i 2^{k_j},i=1,\dots ,l-1\right\} . \end{aligned}$$

    which is exactly what we want.

To complete the proof, it remains to show the base case of the induction. We will show the base cases $N=2$ and $N=3$. For $N=2$, note that the $(N+1)$-cycle is simply a triangle which is already chordal has Fourier support $\{-1,0,1\}$. If we evaluate expression (34) for $N=2$ (note that here $k=0$) we get ${\mathcal {T}}= \{-1,0,1\}$, as needed.

For $N=3$ (the 4-cycle), we have $k=1$ and $l=2$ with $k_1 = 0$ and $k_2 = 1$. Thus expression (34) evaluates to ${\mathcal {T}}= \{0\} \cup \{\pm 1, \pm 2\} \cup \{1\} = \{-2,-1,0,1,2\}$. It is easy to construct a chordal cover of the 4-cycle with such Fourier support (one can even construct one where ${\mathcal {T}}= {\mathcal {K}}\cup (-{\mathcal {K}}) = \{-1,0,1\}$). $\square$

Example 8

Figure 8 shows the recursive construction for the case $N=8$. We have indicated in each triangle (3-clique) the associated Fourier support.

Fig. 8

Illustration of the recursive chordal cover of the $(N+1)$-cycle for $N=8$

Proof of Theorem 9

To prove Theorem 9 for the N-cycle, we use the chordal cover of the $(N+1)$-cycle of Theorem 10 except that we regard nodes 0 and N as the same nodes (they collapse into a single one). Thus this means that the triangle in Fig. 7 labeled $\{-2^k,0,N-2^k\}$ also collapses and we only have to look at the Fourier support for parts (a) and (b). It is not hard to show that the Fourier support we get is the same as Eq. (34) except that in the middle term the index i goes from 0 to $k-1$ (instead of from 0 to k), and in the last term the index i goes from 1 to $l-2$ (instead of from 1 to $l-1$). This modification gives exactly the set ${\mathcal {T}}$ of Eq. (26). $\square$

Fig. 9

Chordal cover of the 9-cycle with Fourier support ${\mathcal {T}}= \{0,\pm 1,\pm 3\}$ and of the 27-cycle with Fourier support ${\mathcal {T}}=\{0,\pm 1,\pm 3,\pm 9\}$

Note that there are actually many different ways of constructing chordal covers for the cycle graph, and different constructions will lead to different valid Fourier supports. For instance, for the cycle graph $C_{N}$ one can actually construct a chordal cover where the size of the Fourier support is related to the logarithm of N base 3. When N is a power of three the Fourier support consists precisely of the powers of 3 that are smaller than N. We omit the precise description of this construction, but Fig. 9 shows the chordal cover for the 9-cycle and 27-cycle.