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The Fourier Transform of Restrictions of Functions on the Slice

Shravas Rao Northwestern University
Abstract

This paper considers the Fourier transform over the slice of the Boolean hypercube. We prove a relationship between the Fourier coefficients of a function over the slice, and the Fourier coefficients of its restrictions. As an application, we prove a Goldreich-Levin theorem for functions on the slice based on the Kushilevitz-Mansour algorithm for the Boolean hypercube.

1 Introduction

Boolean functions, or functions over {0,1}n\{0,1\}^{n}, are central to computer science. One tool that has found many uses is Fourier analysis, which involves defining an orthonormal basis separate from the standard basis, but with many useful properties. One such property is that this basis consists of eigenvectors of the Boolean hypercube, the graph whose vertices are {0,1}n\{0,1\}^{n} with an edge between two vectors uu and vv if they differ in exactly one coordinate. Fourier analysis over the Boolean hypercube has led to many applications, including in learning theory, property testing, pseudorandomness, voting theory, and more. For a more complete treatment, see [O’D14].

In recent years, much attention has been drawn to functions on the slice, ([n]k)={vv{0,1}n,i=1nvi=k}\binom{[n]}{k}=\{v\mid v\in\{0,1\}^{n},\sum_{i=1}^{n}v_{i}=k\} or the subset of {0,1}n\{0,1\}^{n} that consists of vectors with exactly kk 11’s for some fixed kk. Like in the case of the Boolean hypercube, one can also define an orthonormal basis separate from the standard basis, over functions on the slice. In particular, one can consider the eigenvectors of the Johnson graph, the graph whose vertices are elements of the slice, with an edge between two vectors uu and vv if they differ in exactly two coordinates. Thus, this basis can be seen as an analogue of the Fourier transform, but for the slice.

Such a basis was given explicitly in [Fil16] and [Sri11], and this basis has subsequently found many applications in analogues of major theorems for the Boolean hypercube, but now for the slice. These include Friedgut’s theorem [Fil16], various versions of the invariance principle [FM19, FKMW18], the Kindler-Safra theorem [KK20], and a theorem due to Nisan and Szegedy regarding juntas [FI19]. In this paper, we continue this line of work of proving analogues for the slice.

Fourier analysis on the hypercube has been particularly useful when considering restrictions of functions. In particular, if f:{0,1}nf:\{0,1\}^{n}\rightarrow\mathbb{R} is a function, and z{0,1}tz\in\{0,1\}^{t} for tnt\leqslant n, we let its restriction, fz:{0,1}ntf_{z}:\{0,1\}^{n-t}\rightarrow\mathbb{R}, be the function defined by fz(x)=f(xz)f_{z}(x)=f(x\circ z) where \circ denotes concatenation. In particular, one can relate the Fourier coefficients of ff with the Fourier coefficients of fzf_{z}.

Recall that the orthonormal basis for the Fourier transform can be indexed by subsets of the set [n][n]. Then for any tt and any S[nt]S\subseteq[n-t],

T[n]\[nt]f^(ST)2=z{0,1}tfz^(S)2\sum_{T\in[n]\backslash[n-t]}\widehat{f}(S\cup T)^{2}=\sum_{z\in\{0,1\}^{t}}\widehat{f_{z}}(S)^{2} (1)

where f^(S)\widehat{f}(S) denotes the projection onto the vector indexed by SS (see Corollary 3.22 in [O’D14]).

In the case of the slice, one can also define a restriction, where given f:([n]k)f:\binom{[n]}{k}\rightarrow\mathbb{R} and z{0,1}tz\in\{0,1\}^{t} for tnt\leqslant n, we let fz:(ntk|z|)f_{z}:\binom{n-t}{k-|z|}\rightarrow\mathbb{R} be defined by fz(x)=f(xz)f_{z}(x)=f(x\circ z), where |z|=|{i:zi=1}||z|=|\{i:\,z_{i}=1\}| denotes the Hamming weight of zz. If zz contains more than kk 11’s, then fzf_{z} is an empty function. In this work, we show that an identity similar to Eq. (1) holds for the slice.

In the slice, the orthonormal basis for the Fourier transform is now indexed by subsets of [n][n] with a certain property. Let n,min{k,nk}\mathcal{H}_{n,\min\{k,n-k\}} be the collection of subsets S={s1,s2,,si}S=\{s_{1},s_{2},\ldots,s_{i}\} of [n][n] of size at most kk such that sj2js_{j}\geqslant 2j for all jj, when the elements of SS are listed in sorted order. Then the vectors in the orthogonal basis from [Fil16] and [Sri11] can be indexed by elements of n,min{k,nk}\mathcal{H}_{n,\min\{k,n-k\}}. We note that a vector indexed by the set SS in the orthogonal basis for the slice differs significantly from the vector indexed by the same set in the orthogonal basis for the Boolean hypercube.

The main result of this paper is the following theorem, which generalizes Eq. (1) to the slice.

Theorem 1.1.

Let f:([n]k)f:\binom{[n]}{k}\rightarrow\mathbb{R}, and let iki\leqslant k.

Then for any tt and any S[nt]S\subseteq[n-t],

z{0,1}tfz^(S)2=T[n]\[nt]f^(ST)2\sum_{z\in\{0,1\}^{t}}\widehat{f_{z}}(S)^{2}=\sum_{T\subseteq[n]\backslash[n-t]}\widehat{f}(S\cup T)^{2}

Again, f^(S)\widehat{f}(S) denotes the projection onto the vector indexed by SS. For convenience, we will let f^(S)=0\widehat{f}(S)=0 if SS is not a subset in n,min{k,nk}\mathcal{H}_{n,\min\{k,n-k\}} or ff is an empty function.

Eq. (1) has many applications, one of which includes the Goldreich-Levin theorem [GL89, KM93]. Informally, this theorem says that if the Fourier coefficients of a function are concentrated on just a few vectors, then the function can be learned efficiently. As an application of Theorem 1.1, we show that a similar theorem applies in the case of the slice, following the Kushilevitz-Mansour algorithm.

Theorem 1.2.

Given query access to a function f([n]k){1,1}f\in{\binom{[n]}{k}}\rightarrow\{-1,1\} and 0<τ10<\tau\leqslant 1, there is a 𝑝𝑜𝑙𝑦(n,1/τ)\mathit{poly}(n,1/\tau)-time algorithm that with high probability outputs a list L={U1,,U}L=\{U_{1},\ldots,U_{\ell}\} of subsets of [n][n] such that

  • |f^(U)|τ(nk)|\widehat{f}(U)|\geqslant\tau\cdot\sqrt{\binom{n}{k}} implies that ULU\in L

  • ULU\in L implies that |f^(U)|τ/2(nk)|\widehat{f}(U)|\geqslant\tau/2\cdot\sqrt{\binom{n}{k}}

1.1 Future Work

It would be interesting to see if Theorem 1.1 leads to other analogues of theorems for the Boolean hypercube to the slice.

It would also be interesting to see to what extent the orthonormal basis for slice can be extended to high-dimensional expanders, and additionally, if there exist analogues of theorems for the Boolean hypercube on high dimensional expanders.

Similarly to how expander graphs can be thought of as sparse approximations of the complete graph, high dimensional expanders can be thought of as sparse approximations of the Johnson graph. In recent years, many constructions of high dimensional expanders have been developed [LSV05a, LSV05b, KO18, CTZ20]. The theory of constructions of high dimensional expanders have already led to many applications in theoretical computer science, including mixing times of Markov chains [ALGV19, AL20, CGM21] and coding theory [AJQ+20, DHK+21].

Unlike regular expander graphs, one has control beyond just the second eigenvalue, as shown in work by Kaufman and Oppenheim [KO20]. In particular, the spectrum of high dimensional expanders approximates the spectrum of the Johnson graph. Both Kaufman and Oppenheim, and work by Dikstein, Dinur, Filmus, and Harsha [DDFH18] give different decompositions into approximate eigenspaces. The latter allows one to obtain an analogue of the Freidgut-Kalai-Naor theorem for high dimensional expanders. It would be interesting to obtain more explicit descriptions of approximate eigenvectors in these approximate eigenspaces, that might allow one to prove more analogues for high dimensional expanders. In particular, one hope is that the structure of the orthonormal basis for the slice can used as inspiration to construct an approximately orthonormal basis for high dimensional expanders, which can then be used to prove analogues of theorems for the Boolean hypercube, but now for high dimensional expanders.

2 Preliminaries

Denote by [n][n] the set {1,2,,n}\{1,2,\ldots,n\} and by ([n]k)\binom{[n]}{k} the set {vv{0,1}n,i=1nvi=k}\{v\mid v\in\{0,1\}^{n},\sum_{i=1}^{n}v_{i}=k\} or the subset of {0,1}n\{0,1\}^{n} that consists of vertices with exactly kk 11’s for some fixed kk.

Given a string x{0,1}nx\in\{0,1\}^{n}, denote by |x||x| the Hamming weight of xx, that is, |x|=i=1nxi|x|=\sum_{i=1}^{n}x_{i}.

As stated in the introduction, we denote by n,k\mathcal{H}_{n,k} the set of subsets S={s1,s2,,si}S=\{s_{1},s_{2},\ldots,s_{i}\} of [n][n] of size at most kk such that sj2js_{j}\geqslant 2j for all jj, when the elements of SS are listed in sorted order. Note that by Bertrand’s ballot problem, there are (n|S|)(n|S|1)\binom{n}{|S|}-\binom{n}{|S|-1} sets in n,k\mathcal{H}_{n,k} of size |S||S|, and thus |n,k|=(nk)|\mathcal{H}_{n,k}|=\binom{n}{k}.

Given a vector vNv\in\mathbb{R}^{N} we define the p\ell_{p}-norm by

vpp=i=1N|vi|p.\|v\|_{p}^{p}=\sum_{i=1}^{N}|v_{i}|^{p}.

We define the inner product for two vectors u,vNu,v\in\mathbb{R}^{N} to be

u,v=i=1Nuivi.\langle u,v\rangle=\sum_{i=1}^{N}u_{i}v_{i}.

It will often be convenient to index vectors by vectors rather than integers, that is, to let vv be a vector in ([n]k)\mathbb{R}^{\binom{[n]}{k}} for some nn and kk. It will often be convenient to identify vectors as functions, that is, to represent vnv\in\mathbb{R}^{n} as fv:[n]f_{v}:[n]\rightarrow\mathbb{R}, or v([n]k)v\in\mathbb{R}^{\binom{[n]}{k}} as fv:([n]k)f_{v}:{\binom{[n]}{k}}\rightarrow\mathbb{R} where in both cases, one obtains fvf_{v} from vv via fv(x)=vxf_{v}(x)=v_{x}.

By the Cauchy-Schwarz inequality, one can relate the 1\ell_{1}-norm of a vector with its 2\ell_{2}-norm as follows.

Claim 2.1.

For all vNv\in\mathbb{R}^{N},

v1Nv2\|v\|_{1}\leqslant\sqrt{N}\|v\|_{2}

We denote by INI_{N} the N×N\mathbb{R}^{N\times N} identity matrix.

3 An orthonormal basis of eigenvectors for the Johnson graph

Recall that the Johnson graph is defined by G=(V,E)G=(V,E) where V=([n]k)V=\binom{[n]}{k} and E={(v,u)i=1n|viui|=2}E=\left\{(v,u)\mid\sum_{i=1}^{n}|v_{i}-u_{i}|=2\right\}. Let An,kA_{n,k} be the corresponding adjacency matrix.

In this section we give with proof an orthonormal basis of eigenvectors for the Johnson graph. The claims and theorems in this section will be used in Section 4 to prove Theorem 1.1. We stress that all of the results in this section are known, and can be found in the works of Stanley [Sta88, Sta90], Filmus [Fil16], Srinivasan [Sri11], and Filmus and Lindzey [FL20]. When illustrative, however, we give the proofs of the various statements.

We start by defining a collection of matrices that will help yield the eigenvectors of An,kA_{n,k}.

Let Pn,k([n]k+1)×([n]k)P^{\uparrow}_{n,k}\in\mathbb{R}^{\binom{[n]}{k+1}\times\binom{[n]}{k}} be the matrix defined by

Pn,k(x,y)={1if yi=1 implies xi=10otherwiseP^{\uparrow}_{n,k}(x,y)=\begin{cases}1&\text{if }y_{i}=1\text{ implies }x_{i}=1\\ 0&\text{otherwise}\end{cases} (2)

Let Pn,k+1=(Pn,k)P^{\downarrow}_{n,k+1}=(P^{\uparrow}_{n,k})^{*}.

Let

Pn,i,k=Pn,k1Pn,k1Pn,iP^{\uparrow}_{n,i,k}=P^{\uparrow}_{n,k-1}P^{\uparrow}_{n,k-1}\cdots P^{\uparrow}_{n,i}

and

Pn,k,i=Pn,i+1Pn,i+2Pn,k.P^{\downarrow}_{n,k,i}=P^{\downarrow}_{n,i+1}P^{\downarrow}_{n,i+2}\cdots P^{\downarrow}_{n,k}.

For simplicity, we let Pn,k,k=Pn,k,k=I(nk)P^{\uparrow}_{n,k,k}=P^{\downarrow}_{n,k,k}=I_{\binom{n}{k}}.

Let Pn,k=Pn,k1Pn,kP^{\vee}_{n,k}=P^{\uparrow}_{n,k-1}P^{\downarrow}_{n,k} and P=Pn,k+1Pn,kP^{\wedge}=P^{\downarrow}_{n,k+1}P^{\uparrow}_{n,k}. The following fact shows that Pn,kP^{\vee}_{n,k} and Pn,kP^{\wedge}_{n,k} differ only by a multiple of the identity matrix.

Claim 3.1.

Pn,k=Pn,k(n2k)I([n]k)P^{\vee}_{n,k}=P^{\wedge}_{n,k}-(n-2k)I_{\binom{[n]}{k}}

Similarly, one can show that An,kA_{n,k} differs from Pn,kP^{\vee}_{n,k} and Pn,kP^{\wedge}_{n,k} by a multiple of the identity matrix. Thus, for the rest of this paper, we focus on determining an orthonormal basis of eigenvectors for Pn,kP^{\vee}_{n,k} and Pn,kP^{\wedge}_{n,k}.

The following claim shows that to determine the eigenvectors of Pn,kP^{\vee}_{n,k} and Pn,kP^{\wedge}_{n,k}, it is enough to determine the nullspace of Pn,iP^{\downarrow}_{n,i} for all iki\leqslant k. This will be under the assumption that Pn,iP^{\downarrow}_{n,i} has full rank when in/2i\leqslant n/2, which we will prove in Claims 3.5 and 3.6.

Claim 3.2.

Let v([n]i)v\in\mathbb{R}^{\binom{[n]}{i}} be a vector such that Pn,iv=0P^{\downarrow}_{n,i}v=0. Then,

Pn,kPn,i,kv=(nki+1)(ki)Pn,i,kv.P^{\vee}_{n,k}P^{\uparrow}_{n,i,k}v=(n-k-i+1)(k-i)P^{\uparrow}_{n,i,k}v.

and

Pn,kPn,i,kv=(nki)(ki+1)Pn,i,kv.P^{\wedge}_{n,k}P^{\uparrow}_{n,i,k}v=(n-k-i)(k-i+1)P^{\uparrow}_{n,i,k}v.

The following claim shows that if Pn,iv=0P^{\downarrow}_{n,i}v=0, then Pn,i,kv=0P^{\uparrow}_{n,i,k}v=0 if k>nik>n-i. Thus, in the case that k>n/2k>n/2, one only needs to consider the nullspace of Pn,iP^{\downarrow}_{n,i} for inki\leqslant n-k. In fact, we will compute Pn,|S|,kv22v22\displaystyle\frac{\left\|P^{\uparrow}_{n,|S|,k}v\right\|_{2}^{2}}{\left\|v\right\|_{2}^{2}} when Pn,iv=0P^{\downarrow}_{n,i}v=0, which will be useful in Section 3.2

Claim 3.3.

Let v([n]i)v\in\mathbb{R}^{\binom{[n]}{i}} be a vector such that Pn,iv=0P^{\downarrow}_{n,i}v=0 for in/2i\leqslant n/2. Then,

Pn,i,kv22v22=(n2i)!(ki)!(nki)!,\frac{\left\|P^{\uparrow}_{n,i,k}v\right\|_{2}^{2}}{\|v\|_{2}^{2}}=\frac{(n-2i)!(k-i)!}{(n-k-i)!},

if knik\leqslant n-i, and is otherwise 0.

Proof.

We prove this using induction on ii. In the base case when k=ik=i, the Claim follows as Pn,i,kv=vP^{\uparrow}_{n,i,k}v=v.

Now, assume that the statement is true for Pn,i,k1vP^{\uparrow}_{n,i,k-1}v. Then,

Pn,i,kv22\displaystyle\left\|P^{\uparrow}_{n,i,k}v\right\|_{2}^{2} =Pn,k1Pn,i,k1v,Pn,k1Pn,ik1v\displaystyle=\left\langle P^{\uparrow}_{n,k-1}P^{\uparrow}_{n,i,k-1}v,P^{\uparrow}_{n,k-1}P^{\uparrow}_{n,ik-1}v\right\rangle
=Pn,i,k1v,Pn,k1Pn,i,k1v\displaystyle=\left\langle P^{\uparrow}_{n,i,k-1}v,P^{\wedge}_{n,k-1}P^{\uparrow}_{n,i,k-1}v\right\rangle
=(nki+1)(ki)Pn,i,k1v,Pn,i,k1v\displaystyle=(n-k-i+1)(k-i)\left\langle P^{\uparrow}_{n,i,k-1}v,P^{\uparrow}_{n,i,k-1}v\right\rangle (3)
=(nki+1)(ki)(n2i)!(ki1)!(nki+1)!\displaystyle=(n-k-i+1)(k-i)\frac{(n-2i)!(k-i-1)!}{(n-k-i+1)!}
(n2i)!(ki)!(nki)!\displaystyle\frac{(n-2i)!(k-i)!}{(n-k-i)!} (4)

as desired, where the third equality follows from Claim 3.2 and the fourth equality follows from the inductive hypothesis.

If k=ni+1k=n-i+1, then Eq. (3) is 0, and it follows that Pn,i,kv2=0\left\|P^{\uparrow}_{n,i,k}v\right\|_{2}=0 whenever k>ni+1k>n-i+1 by the definition of Pn,i,kP^{\uparrow}_{n,i,k}. ∎

Note that the assumption that Pn,iP^{\downarrow}_{n,i} has full-rank when in/2i\leqslant n/2 combined with Claims 3.2 and 3.3 yields a subspace of eigenvectors with equal eigenvalue of dimension (ni)(ni1)\binom{n}{i}-\binom{n}{i-1}. For a fixed kk and summing over all ii from 0 to min{k,nk}\min\{k,n-k\}, this is (nmin{k,nk})\binom{n}{\min\{k,n-k\}} eigenvectors, and thus, this gives a complete basis of eigenvectors.

The following claim shows that to determine an orthogonal basis of eigenvectors of Pn,kP^{\vee}_{n,k} and Pn,kP^{\wedge}_{n,k} it is enough to determine an orthogonal basis for the nullspace of Pn,iP^{\downarrow}_{n,i} for all imin{k,nk}i\leqslant\min\{k,n-k\}. That is, applying Pn,iP^{\uparrow}_{n,i^{\prime}} to two orthogonal vectors preserves orthogonality. Recall that the eigenvectors of a symmetric matrix with different eigenvalues are orthogonal, and thus it is sufficient to determine an orthogonal basis for the nullspace of Pn,iP^{\downarrow}_{n,i} for each ii.

Claim 3.4.

Let v1v_{1} and v2v_{2} be such that v1,v2=0\langle v_{1},v_{2}\rangle=0 and Pn,ivi=0P^{\downarrow}_{n,i}v_{i}=0 for all ii. Then

Pn,i,kv1,Pn,i,kv2=0\left\langle P^{\uparrow}_{n,i,k}v_{1},P^{\uparrow}_{n,i,k}v_{2}\right\rangle=0

.

Proof.

We prove this on induction on kk.

When k=ik=i, the statement is obvious.

Assume that Pn,i,k1v1,Pn,i,k1v2=0\left\langle P^{\uparrow}_{n,i,k-1}v_{1},P^{\uparrow}_{n,i,k-1}v_{2}\right\rangle=0. Then through a similar sequence of equalities as Eq. (4), one sees that

Pn,i,kv1,Pn,i,kv2\displaystyle\left\langle P^{\uparrow}_{n,i,k}v_{1},P^{\uparrow}_{n,i,k}v_{2}\right\rangle =(nki+1)(ki)Pn,i,k1v1,Pn,i,k1v2\displaystyle=(n-k-i+1)(k-i)\left\langle P^{\uparrow}_{n,i,k-1}v_{1},P^{\uparrow}_{n,i,k-1}v_{2}\right\rangle
=0\displaystyle=0

as desired. ∎

We note that Claims 3.13.2 3.3, and 3.4 can be generalized to sequentially differential posets (of which, the collection of subsets of [n][n] is an example), as is done in the work of Stanley [Sta88, Sta90].

3.1 An orthogonal basis for the nullspace of Pn,kP^{\downarrow}_{n,k}

In this section, we give an orthogonal basis for the nullspace of Pn,kP^{\downarrow}_{n,k} when kn/2k\leqslant n/2. This basis was found by Srinivasan [Sri11] and independently by Filmus [Fil16]. For completeness, we include the proof that this is an orthogonal basis due to Srinivasan.

Note that

Pn,k=[Pn1,kI(n1k)0Pn1,k1]P^{\downarrow}_{n,k}=\begin{bmatrix}P^{\downarrow}_{n-1,k}&I_{\binom{n-1}{k}}\\ 0&P^{\downarrow}_{n-1,k-1}\end{bmatrix}

if kn/2k\leqslant n/2, where if k=1k=1, then we remove the bottom two blocks. In particular, the fact that Pn,kP^{\downarrow}_{n,k} can be written recursively in terms of Pn1,kP^{\downarrow}_{n-1,k} and Pn1,k1P^{\downarrow}_{n-1,k-1} suggests that an orthogonal basis for the nullspace of the former matrix can be obtained from orthogonal bases for the nullspace of the latter matrices.

Let SS be a subset in n,k\mathcal{H}_{n,k}. Recall that this means that when the elements SS are listed in sorted order as {s1,s2,,s|S|}\{s_{1},s_{2},\ldots,s_{|S|}\}, that si2is_{i}\geqslant 2i for all ii. We associate with each SS a vector χn,S\chi_{n,S} such that Pn,|S|χn,S=0P^{\downarrow}_{n,|S|}\chi_{n,S}=0. We define χn,S\chi_{n,S} recursively as follows.

χn,S={[1]if |S|=0[χn1,S0]if nS[Pn1,|S|1χn1,S\{n}n2|S|+1χn1,S\{n}]otherwise\chi_{n,S}=\begin{cases}[1]&\text{if }|S|=0\\ \begin{bmatrix}\chi_{n-1,S}\\ 0\end{bmatrix}&\text{if }n\not\in S\\ \begin{bmatrix}-\displaystyle\frac{P^{\uparrow}_{n-1,|S|-1}\chi_{n-1,S\backslash\{n\}}}{n-2|S|+1}\\ \chi_{n-1,S\backslash\{n\}}\end{bmatrix}&\text{otherwise}\end{cases} (5)

We briefly give some intuition behind restricting SS to only those subsets in n,k\mathcal{H}_{n,k}. Consider a set SS of size at most kk that is not in n,k\mathcal{H}_{n,k}. Let S={s1,,s|S|}S=\{s_{1},\ldots,s_{|S|}\} be in sorted order, and let ii be the smallest integer such that si<2is_{i}<2i. Then either i=1i=1 and si=1s_{i}=1, or si12i2s_{i-1}\geqslant 2i-2 and si=2i1s_{i}=2i-1. Thus, si2i+1=0s_{i}-2i+1=0, and χn,S\chi_{n,S} is not well-defined.

By Bertrand’s ballot problem, there are (n|S|)(n|S|1)\binom{n}{|S|}-\binom{n}{|S|-1} sets in n,k\mathcal{H}_{n,k} of size |S||S|. Thus, proving that χn,S\chi_{n,S} is in the nullspace of Pn,|S|P^{\downarrow}_{n,|S|} and that if |S1|=|S2||S_{1}|=|S_{2}| and S1S2S_{1}\neq S_{2}, then χn,S1,χn,S2=0\left\langle\chi_{n,S_{1}},\chi_{n,S_{2}}\right\rangle=0 are both enough to obtain an orthogonal basis for the nullspace of Pn,kP^{\downarrow}_{n,k} and also show it has full rank.

We first show that χn,S\chi_{n,S} is in fact in the nullspace of Pn,|S|P^{\downarrow}_{n,|S|}.

Claim 3.5.

For all Sn,kS\in\mathcal{H}_{n,k}, Pn,|S|χn,S=0P^{\downarrow}_{n,|S|}\chi_{n,S}=0.

Proof.

We prove this using induction on nn and |S||S|.

If S={n}S=\{n^{\prime}\}, then

χn,S=[1n11n11n1n1 times100].\chi_{n,S}=\begin{bmatrix}\smash[b]{\underbrace{\begin{matrix}-\frac{1}{n^{\prime}-1}&-\frac{1}{n^{\prime}-1}&\cdots&-\frac{1}{n^{\prime}-1}\end{matrix}}_{n^{\prime}-1\text{ times}}}&1&0&0&\cdots\end{bmatrix}^{*}.

Thus, Pn,1χn,S=0P^{\downarrow}_{n,1}\chi_{n,S}=0.

If nSn\not\in S, then Pn,|S|χn,S=Pn1,|S|χn,S=0P^{\downarrow}_{n,|S|}\chi_{n,S}=P^{\downarrow}_{n-1,|S|}\chi_{n,S}=0 by the inductive hypothesis.

If nSn\in S, then

Pn,|S|χn,S\displaystyle P^{\downarrow}_{n,|S|}\chi_{n,S} =[Pn1,|S|1χn1,S\{n}n2|S|+1+χn1,S\{n}Pn1,|S|1χn1,S\{n}]\displaystyle=\begin{bmatrix}-\displaystyle\frac{P^{\wedge}_{n-1,|S|-1}\chi_{n-1,S\backslash\{n\}}}{n-2|S|+1}+\chi_{n-1,S\backslash\{n\}}\\ P^{\downarrow}_{n-1,|S|-1}\chi_{n-1,S\backslash\{n\}}\end{bmatrix}
=[Pn1,|S|1χn1,S\{n}n2|S|+1χn1,S\{n}+χn1,S\{n}0]\displaystyle=\begin{bmatrix}-\displaystyle\frac{P^{\vee}_{n-1,|S|-1}\chi_{n-1,S\backslash\{n\}}}{n-2|S|+1}-\chi_{n-1,S\backslash\{n\}}+\chi_{n-1,S\backslash\{n\}}\\ 0\end{bmatrix}
=0.\displaystyle=0.

where the second inequality follows from the relationship between Pn1,|S|1P^{\wedge}_{n-1,|S|-1} and Pn1,|S|1P^{\vee}_{n-1,|S|-1} and the inductive hypothesis, and the third inequality also follows from the inductive hypothesis. ∎

The following claim shows that if |S1|=|S2||S_{1}|=|S_{2}| and S1S2S_{1}\neq S_{2}, then χn,S1\chi_{n,S_{1}} and χn,S2\chi_{n,S_{2}} are orthogonal. Recall that eigenvectors of a symmetric matrix with different eigenvalues are orthogonal. Thus, this claim along with Claim 3.6 are enough to prove that the given basis is orthogonal.

Claim 3.6.

For S1,S2n,kS_{1},S_{2}\in\mathcal{H}_{n,k}, if |S1|=|S2||S_{1}|=|S_{2}| and S1S2S_{1}\neq S_{2}, then χn,S1,χn,S2=0\left\langle\chi_{n,S_{1}},\chi_{n,S_{2}}\right\rangle=0.

Proof.

We prove this using induction on nn.

Consider the base case of |S1|=|S2|=1|S_{1}|=|S_{2}|=1. In particular, let S1={n1}S_{1}=\{n_{1}\} and let S2={n2}S_{2}=\{n_{2}\}, and without loss of generality, assume that n2>n1n_{2}>n_{1}. Then

χn(Si)=[1ni11ni11ni1ni1 times100].\chi_{n}(S_{i})=\begin{bmatrix}\smash[b]{\underbrace{\begin{matrix}-\frac{1}{n_{i}-1}&-\frac{1}{n_{i}-1}&\cdots&-\frac{1}{n_{i}-1}\end{matrix}}_{n_{i}-1\text{ times}}}&1&0&0&\cdots\end{bmatrix}^{*}.

and the claim clearly holds.

For the inductive step, assume that the statement is true for all pairs χn(S1)\chi_{n^{\prime}}(S_{1}^{\prime}) and χn(S2)\chi_{n^{\prime}}(S_{2}^{\prime}) where n<nn^{\prime}<n and |S1|=|S2||S_{1}^{\prime}|=|S_{2}^{\prime}|. Then there are three cases.

Case 1: If neither S1S_{1} nor S2S_{2} contain nn, then χn,S1,χn,S2=0\left\langle\chi_{n,S_{1}},\chi_{n,S_{2}}\right\rangle=0 by the inductive hypothesis.

Case 2: If both S1S_{1} and S2S_{2} contain nn, then

χn,S1,χn,S2\displaystyle\left\langle\chi_{n,S_{1}},\chi_{n,S_{2}}\right\rangle =1(n2|S2|+1)2\displaystyle=\frac{1}{(n-2|S_{2}|+1)^{2}} χn1,S1\{n},Pn1,|S2|1χn1,S2\{n}\displaystyle\left\langle\chi_{n-1,S_{1}\backslash\{n\}},P^{\wedge}_{n-1,|S_{2}|-1}\chi_{n-1,S_{2}\backslash\{n\}}\right\rangle
+χn1,S1\{n},χn1,S2\{n}\displaystyle+\left\langle\chi_{n-1,S_{1}\backslash\{n\}},\chi_{n-1,S_{2}\backslash\{n\}}\right\rangle
=1(n2|S2|+1)2\displaystyle=\frac{1}{(n-2|S_{2}|+1)^{2}} χn1,S1\{n},Pn1,|S2|1χn1,S2\{n}\displaystyle\left\langle\chi_{n-1,S_{1}\backslash\{n\}},P^{\vee}_{n-1,|S_{2}|-1}\chi_{n-1,S_{2}\backslash\{n\}}\right\rangle
+n2|S2|+2(n2|S2|+1)χn1,S1\{n},χn1,S2\{n}\displaystyle+\frac{n-2|S_{2}|+2}{(n-2|S_{2}|+1)}\left\langle\chi_{n-1,S_{1}\backslash\{n\}},\chi_{n-1,S_{2}\backslash\{n\}}\right\rangle
=0\displaystyle=0

where the second equality follows from Fact 3.1 and the third inequality follows from the inductive hypothesis for S1\{n}S_{1}\backslash\{n\} and S2\{n}S_{2}\backslash\{n\}, and the fact that Pn1,|S2|1χn1(S2\{n})=0P^{\downarrow}_{n-1,|S_{2}|-1}\chi_{n-1}(S_{2}\backslash\{n\})=0.

Case 3: If only S2S_{2} contains nn, then

χn,S1,χn,S2\displaystyle\left\langle\chi_{n,S_{1}},\chi_{n,S_{2}}\right\rangle =χn1(S1),Pn1,|S2|1χn1,S2\{n}n2|S2|+1\displaystyle=\left\langle\chi_{n-1}(S_{1}),\frac{P^{\uparrow}_{n-1,|S_{2}|-1}\chi_{n-1,S_{2}\backslash\{n\}}}{n-2|S_{2}|+1}\right\rangle
=Pn1,kχn1(S1),χn1,S2\{n}n2|S2|+1\displaystyle=\left\langle P^{\downarrow}_{n-1,k}\chi_{n-1}(S_{1}),\frac{\chi_{n-1,S_{2}\backslash\{n\}}}{n-2|S_{2}|+1}\right\rangle
=0\displaystyle=0

where the last inequality follows from the fact that Pn1,|S2|χn1,S1=0P^{\downarrow}_{n-1,|S_{2}|}\chi_{n-1,S_{1}}=0. ∎

We summarize below.

Theorem 3.7.

The set of vectors

{Pn,|S|χn,SSn,min{k,nk}}\left\{P^{\uparrow}_{n,|S|}\chi_{n,S}\mid S\in\mathcal{H}_{n,\min\{k,n-k\}}\right\}

form an orthogonal basis of eigenvectors for An,kA_{n,k}, the adjacency matrix of the Johnson graph.

Proof.

The theorem follows by applying Claims 3.23.33.43.5, and 3.6, and noticing that |n,min{k,nk}|=(nk)|\mathcal{H}_{n,\min\{k,n-k\}}|=\binom{n}{k} and that eigenvectors with distinct eigenvalues are orthogonal. ∎

3.2 A computation of the norm

In this section, we compute the norm of Pn,|S|,kχn,SP^{\uparrow}_{n,|S|,k}\chi_{n,S}. This combined with Theorem 3.7 will yield a set of orthonormal eigenvectors that can be used to construct a Fourier transform over the slice. The results in this section can be found in [Fil16]. We give an inductive proof that follows the framework of Srinivasan [Sri11] and Filmus and Lindzey [FL20].

We start by computing the norm of χn,S\chi_{n,S}.

Claim 3.8.

Let S={s1,,s|S|}S=\{s_{1},\ldots,s_{|S|}\} be such that s1s2s|S|s_{1}\leqslant s_{2}\leqslant\cdots\leqslant s_{|S|}. Then,

χn,S22=i=1|S|si2i+2si2i+1\left\|\chi_{n,S}\right\|_{2}^{2}=\prod_{i=1}^{|S|}\frac{s_{i}-2i+2}{s_{i}-2i+1}
Proof.

We prove this using induction on nn. For the base case, if S={n}S=\{n^{\prime}\}, then

χn,S=[1n11n11n1n1 times100].\chi_{n,S}=\begin{bmatrix}\smash[b]{\underbrace{\begin{matrix}-\frac{1}{n^{\prime}-1}&-\frac{1}{n^{\prime}-1}&\cdots&-\frac{1}{n^{\prime}-1}\end{matrix}}_{n^{\prime}-1\text{ times}}}&1&0&0&\cdots\end{bmatrix}^{*}.

and thus,

χn,S22=1+1n1=nn1\left\|\chi_{n,S}\right\|_{2}^{2}=1+\frac{1}{n^{\prime}-1}=\frac{n^{\prime}}{n^{\prime}-1}

as desired.

If nSn\not\in S, then

χn,S22=χn1,S22=i=1|S|si2i+2si2i+1\left\|\chi_{n,S}\right\|_{2}^{2}=\left\|\chi_{n-1,S}\right\|_{2}^{2}=\prod_{i=1}^{|S|}\frac{s_{i}-2i+2}{s_{i}-2i+1}

where the second inequality follows from the inductive hypothesis.

If nSn\in S, then

χn,S22\displaystyle\left\|\chi_{n,S}\right\|_{2}^{2} =χn1,S\{n}22+1(n2|S|+1)2Pn1,|S|1χn1,S\{n}22\displaystyle=\left\|\chi_{n-1,S\backslash\{n\}}\right\|_{2}^{2}+\frac{1}{(n-2|S|+1)^{2}}\left\|P^{\uparrow}_{n-1,|S|-1}\chi_{n-1,S\backslash\{n\}}\right\|_{2}^{2}
=χn1,S\{n}22+1(n2|S|+1)2χn1,S\{n},Pn,|S|1χn1,S\{n}\displaystyle=\left\|\chi_{n-1,S\backslash\{n\}}\right\|_{2}^{2}+\frac{1}{(n-2|S|+1)^{2}}\left\langle\chi_{n-1,S\backslash\{n\}},P^{\wedge}_{n,|S|-1}\chi_{n-1,S\backslash\{n\}}\right\rangle
=(1+1n2|S|+1)χn1,S\{n}22\displaystyle=\left(1+\frac{1}{n-2|S|+1}\right)\left\|\chi_{n-1,S\backslash\{n\}}\right\|_{2}^{2}
=n2|S|+2n2|S|+1i=1k1si2i+2si2i+1\displaystyle=\frac{n-2|S|+2}{n-2|S|+1}\prod_{i=1}^{k-1}\frac{s_{i}-2i+2}{s_{i}-2i+1}
=i=1|S|si2i+2si2i+1\displaystyle=\prod_{i=1}^{|S|}\frac{s_{i}-2i+2}{s_{i}-2i+1}

where the third equality follows from the fact that χn1,S\{n}\chi_{n-1,S\backslash\{n\}} is an eigenvector of Pn,k1P^{\wedge}_{n,k-1}, and the fourth equality follows from the inductive hypothesis. ∎

Finally, we can compute Pn,|S|,kχn,S22\left\|P^{\uparrow}_{n,|S|,k}\chi_{n,S}\right\|_{2}^{2}.

Claim 3.9.
Pn,|S|,kχn,S22=(n2|S|)!(k|S|)!(nk|S|)!i=1ksi2i+2si2i+1\left\|P^{\uparrow}_{n,|S|,k}\chi_{n,S}\right\|_{2}^{2}=\frac{(n-2|S|)!(k-|S|)!}{(n-k-|S|)!}\prod_{i=1}^{k}\frac{s_{i}-2i+2}{s_{i}-2i+1}
Proof.

The claim follows by applying Claims 3.3 and 3.8. ∎

Finally, we give a formula definition for the Fourier coefficients of a function on the slice.

Definition 3.10.

For f([n]k)f\in\mathbb{R}^{\binom{[n]}{k}} and Sn,min{k,nk}S\in\mathcal{H}_{n,\min\{k,n-k\}}, define

f^(S)=1Pn,|S|,kχn,S2f,Pn,|S|,kχn,S.\widehat{f}(S)=\frac{1}{\left\|P^{\uparrow}_{n,|S|,k}\chi_{n,S}\right\|_{2}}\left\langle f,P^{\uparrow}_{n,|S|,k}\chi_{n,S}\right\rangle.

Here, PP^{\uparrow} is defined in Eq. (2) and χn,S\chi_{n,S} is defined in Eq. (5). It will be convenient to define f^(S)\widehat{f}(S) even when SS is not in n,min{k,nk}\mathcal{H}_{n,\min\{k,n-k\}}, in which case we let f^(S)=0\widehat{f}(S)=0.

4 Restrictions of functions on the slice

In this section, we state and prove our main theorem regarding a relationship between the Fourier coefficients of a function on the slice, and the Fourier coefficients of its restrictions.

Definition 4.1 (Restriction).

Let f:([n]k)f:\binom{[n]}{k}\rightarrow\mathbb{R}. For m<nm<n, and z{0,1}mz\in\{0,1\}^{m}, let fz:([nm]k|z|)f_{z}:{\binom{[n-m]}{k-|z|}}\rightarrow\mathbb{R} (where |z|=i=1mzi|z|=\sum_{i=1}^{m}z_{i} is the Hamming weight of zz) be the function defined by fz(x)=f(xz)f_{z}(x)=f(x\circ z) where \circ denotes concatenation.

We start with the following theorem.

Theorem 4.2.

Let f:([n]k)f:\binom{[n]}{k}\rightarrow\mathbb{R}. Then, for all Sn1,min{k,nk}S\in\mathcal{H}_{n-1,\min\{k,n-k\}},

f^(S)=1n2|S|(nk|S|f0^(S)+k|S|f1^(S)).\widehat{f}(S)=\frac{1}{\sqrt{n-2|S|}}\left(\sqrt{n-k-|S|}\widehat{f_{0}}(S)+\sqrt{k-|S|}\widehat{f_{1}}(S)\right).

Additionally, if |S|<min{k,nk}|S|<\min\{k,n-k\}, then

f^(S{n})=1n2|S|(k|S|f0^(S)+nk|S|f1^(S))\widehat{f}(S\cup\{n\})=\frac{1}{\sqrt{n-2|S|}}\left(-\sqrt{k-|S|}\widehat{f_{0}}(S)+\sqrt{n-k-|S|}\widehat{f_{1}}(S)\right)

Thus, if |S|<min{k,nk}|S|<\min\{k,n-k\}, then

f0^(S)=1n2|S|(nk|S|f^(S)k|S|f^(S{n}))\widehat{f_{0}}(S)=\frac{1}{\sqrt{n-2|S|}}\left(\sqrt{n-k-|S|}\widehat{f}(S)-\sqrt{k-|S|}\widehat{f}(S\cup\{n\})\right)

and

f1^(S)=1n2|S|(nk|S|f^(S{n})+k|S|f^(S))\widehat{f_{1}}(S)=\frac{1}{\sqrt{n-2|S|}}\left(\sqrt{n-k-|S|}\widehat{f}(S\cup\{n\})+\sqrt{k-|S|}\widehat{f}(S)\right)
Proof.

If k=0k=0, then f=f0f=f_{0}, and |S|=0|S|=0, and the theorem holds. Similarly, if k=nk=n, then f=f1f=f_{1} and |S|=0|S|=0 and the theorem also holds.

If |S|=k|S|=k, then f^(S)=f0^(S)\widehat{f}(S)=\widehat{f_{0}}(S) as χn,S(x)=0\chi_{n,S}(x)=0 if xn=1x_{n}=1, and again, the theorem holds. If |S|=nk|S|=n-k, then recall that Pn1,|S|,kχn1,S=0P^{\uparrow}_{n-1,|S|,k}\chi_{n-1,S}=0 by Claim 3.3, and thus f^(S)=f1^(S)\widehat{f}(S)=\widehat{f_{1}}(S), and again, the theorem holds.

Thus, we can assume that f0f_{0} and f1f_{1} are not empty functions, and that Sn1,min{k1,nk1}S\in\mathcal{H}_{n-1,\min\{k-1,n-k-1\}}, and in particular, has a corresponding vector in the orthogonal basis for both An1,kA_{n-1,k} and An1,k1A_{n-1,k-1}.

Let f0:([n]k)f_{0}^{\prime}:{\binom{[n]}{k}}\rightarrow\mathbb{R} be defined by f0(x)=f(x)f_{0}^{\prime}(x)=f(x) if xn=0x_{n}=0 and f0(x)=0f_{0}^{\prime}(x)=0 otherwise. Additionally, let f1:([n]k)f_{1}^{\prime}:{\binom{[n]}{k}}\rightarrow\mathbb{R} be defined by f1(x)=f(x)f_{1}^{\prime}(x)=f(x) if xn=1x_{n}=1, and f1(x)=0f_{1}^{\prime}(x)=0 otherwise.

Let i=|S|i=|S|. Note that because f=f0+f1f=f^{\prime}_{0}+f^{\prime}_{1}, for all SS,

f^(S)\displaystyle\widehat{f}(S) =1Pn,i,kχn,S2f0,Pn,i,kχn,S+1Pn,i,kχn,S2f1,Pn,i,kχn,S\displaystyle=\frac{1}{\left\|P^{\uparrow}_{n,i,k}\chi_{n,S}\right\|_{2}}\left\langle f_{0}^{\prime},P^{\uparrow}_{n,i,k}\chi_{n,S}\right\rangle+\frac{1}{\left\|P^{\uparrow}_{n,i,k}\chi_{n,S}\right\|_{2}}\left\langle f_{1}^{\prime},P^{\uparrow}_{n,i,k}\chi_{n,S}\right\rangle (6)

We start with the first inequality. We have that f0,Pn,i,kχn,S=f0,Pn1,i,kχn1,S\left\langle f_{0}^{\prime},P^{\uparrow}_{n,i,k}\chi_{n,S}\right\rangle=\left\langle f_{0},P^{\uparrow}_{n-1,i,k}\chi_{n-1,S}\right\rangle, as f0(x)=0f_{0}^{\prime}(x)=0 if xn=0x_{n}=0. Thus, the right-hand side of Eq. (6) is equal to

1Pn,i,kχn,S2f0,Pn1,i,kχn1,S\displaystyle\ \frac{1}{\left\|P^{\uparrow}_{n,i,k}\chi_{n,S}\right\|_{2}}\left\langle f_{0},P^{\uparrow}_{n-1,i,k}\chi_{n-1,S}\right\rangle +1Pn,i,kχn,S2Pn,k,if1,χn,S\displaystyle+\frac{1}{\left\|P^{\uparrow}_{n,i,k}\chi_{n,S}\right\|_{2}}\left\langle P^{\downarrow}_{n,k,i}f_{1}^{\prime},\chi_{n,S}\right\rangle
=Pn1,i,kχn1,S2f0^(S)Pn,i,kχn,S2+kiPn,i,kχn,S2Pn1,k1,if1,χn1,S\displaystyle=\frac{\left\|P^{\uparrow}_{n-1,i,k}\chi_{n-1,S}\right\|_{2}\widehat{f_{0}}(S)}{\left\|P^{\uparrow}_{n,i,k}\chi_{n,S}\right\|_{2}}+\frac{k-i}{\left\|P^{\uparrow}_{n,i,k}\chi_{n,S}\right\|_{2}}\left\langle P^{\downarrow}_{n-1,k-1,i}f_{1},\chi_{n-1,S}\right\rangle
=Pn1,i,kχn1,S2f0^(S)Pn,i,kχn,S2+kiPn,i,kχn,S2f1,Pn1,i,k1χn1,S\displaystyle=\frac{\left\|P^{\uparrow}_{n-1,i,k}\chi_{n-1,S}\right\|_{2}\widehat{f_{0}}(S)}{\left\|P^{\uparrow}_{n,i,k}\chi_{n,S}\right\|_{2}}+\frac{k-i}{\left\|P^{\uparrow}_{n,i,k}\chi_{n,S}\right\|_{2}}\left\langle f_{1},P^{\uparrow}_{n-1,i,k-1}\chi_{n-1,S}\right\rangle
=Pn1,i,kχn1,S2f0^(S)Pn,i,kχn,S2+(ki)f1^(S)Pn1,i,k1χn1,S2Pn,i,kχn,S2\displaystyle=\frac{\left\|P^{\uparrow}_{n-1,i,k}\chi_{n-1,S}\right\|_{2}\widehat{f_{0}}(S)}{\left\|P^{\uparrow}_{n,i,k}\chi_{n,S}\right\|_{2}}+\frac{(k-i)\widehat{f_{1}}(S)\left\|P^{\uparrow}_{n-1,i,k-1}\chi_{n-1,S}\right\|_{2}}{\left\|P^{\uparrow}_{n,i,k}\chi_{n,S}\right\|_{2}} (7)

where the sequence of inequalities uses the fact that χn,S(x)=χn1,S(x)\chi_{n,S}(x)=\chi_{n-1,S}(x) unless xn=1x_{n}=1, in which case χn,S(x)=0\chi_{n,S}(x)=0. The first equality also uses the fact that (Pn,k,if1)(x)=(ki)(Pn1,k1,if1)(x)\left(P^{\downarrow}_{n,k,i}f_{1}^{\prime}\right)(x)=(k-i)\left(P^{\downarrow}_{n-1,k-1,i}f_{1}\right)(x) if xn=0x_{n}=0. By Claim 3.9, Eq. (7) is equal to

nkin2if0^(S)+(ki)(n2i)(ki)f1^(S)=1n2i(nkif0^(S)+kif1^(S))\displaystyle\sqrt{\frac{n-k-i}{n-2i}}\widehat{f_{0}}(S)+\frac{(k-i)}{\sqrt{(n-2i)(k-i)}}\widehat{f_{1}}(S)=\frac{1}{\sqrt{n-2i}}\left(\sqrt{n-k-i}\widehat{f_{0}}(S)+\sqrt{k-i}\widehat{f_{1}}(S)\right)

as desired.

Now, let i=|S|+1i=|S|+1, and let S=S{n}S^{\prime}=S\cup\{n\}. To prove the second inequality, note that the right-hand side of Eq. (6) is equal to

1Pn,i,kχn,S2\displaystyle\frac{1}{\left\|P^{\uparrow}_{n,i,k}\chi_{n,S^{\prime}}\right\|_{2}} Pn,k,if0,χn,S+1Pn,i,kχn,S2Pn,k,if1,χn,S\displaystyle\left\langle P^{\downarrow}_{n,k,i}f_{0}^{\prime},\chi_{n,S^{\prime}}\right\rangle+\frac{1}{\left\|P^{\uparrow}_{n,i,k}\chi_{n,S^{\prime}}\right\|_{2}}\left\langle P^{\downarrow}_{n,k,i}f_{1}^{\prime},\chi_{n,S^{\prime}}\right\rangle
=Pn1,k,if0,Pn1,i1χn1,S\{n}(n2i+1)Pn,i,kχn,S2\displaystyle=\frac{-\left\langle P^{\downarrow}_{n-1,k,i}f_{0},P^{\uparrow}_{n-1,i-1}\chi_{n-1,S^{\prime}\backslash\{n\}}\right\rangle}{(n-2i+1)\left\|P^{\uparrow}_{n,i,k}\chi_{n,S^{\prime}}\right\|_{2}}
+(ki)n2i+1Pn1,k1,if1,Pn1,i1χn1,S\{n}+Pn1,k1,i1f1,χn1,S\{n}Pn,i,kχn,S2\displaystyle+\frac{\frac{-(k-i)}{n-2i+1}\left\langle P^{\downarrow}_{n-1,k-1,i}f_{1},P^{\uparrow}_{n-1,i-1}\chi_{n-1,S^{\prime}\backslash\{n\}}\right\rangle+\left\langle P^{\downarrow}_{n-1,k-1,i-1}f_{1},\chi_{n-1,S^{\prime}\backslash\{n\}}\right\rangle}{\left\|P^{\uparrow}_{n,i,k}\chi_{n,S^{\prime}}\right\|_{2}}
=Pn1,k,i1f0,χn1,S\{n}(n2i+1)Pn,i,kχn,S2+nk1+1n2i+1Pn1,k1,i1f1,χn1,S\{n}\displaystyle=\frac{-\left\langle P^{\downarrow}_{n-1,k,i-1}f_{0},\chi_{n-1,S^{\prime}\backslash\{n\}}\right\rangle}{(n-2i+1)\left\|P^{\uparrow}_{n,i,k}\chi_{n,S^{\prime}}\right\|_{2}}+\frac{n-k-1+1}{n-2i+1}\left\langle P^{\downarrow}_{n-1,k-1,i-1}f_{1},\chi_{n-1,S^{\prime}\backslash\{n\}}\right\rangle
=Pn1,i1,kχn,S\{n}2(n2i+1)Pn,i,kχn,S2f0^(S\{n})+(nki+1)Pn1,i1,k1χn,S\{n}2(n2i+1)Pn,i,kχn,S2f1^(S\{n})\displaystyle=\frac{-\left\|P^{\uparrow}_{n-1,i-1,k}\chi_{n,S^{\prime}\backslash\{n\}}\right\|_{2}}{(n-2i+1)\left\|P^{\uparrow}_{n,i,k}\chi_{n,S^{\prime}}\right\|_{2}}\widehat{f_{0}}(S^{\prime}\backslash\{n\})+\frac{(n-k-i+1)\left\|P^{\uparrow}_{n-1,i-1,k-1}\chi_{n,S^{\prime}\backslash\{n\}}\right\|_{2}}{(n-2i+1)\left\|P^{\uparrow}_{n,i,k}\chi_{n,S^{\prime}}\right\|_{2}}\widehat{f_{1}}(S^{\prime}\backslash\{n\}) (8)

where the first equality follows from the definition of χn,S\chi_{n,S} along with the fact that (Pn,k,if1)(x)=(ki)(Pn1,k1,if1)(x)\left(P^{\downarrow}_{n,k,i}f_{1}^{\prime}\right)(x)=(k-i)\left(P^{\downarrow}_{n-1,k-1,i}f_{1}\right)(x) if xn=0x_{n}=0. Again, by Claim 3.9, Eq (8) is equal to

1n2i+1(n2i+1)(n2i+1)(ki+1)n2i+2f0^(S\{n})+nki+1n2i+1(n2i+1)(n2i+1)(n2i+2)(nki+1)f1^(S\{n})=1n2i+2(ki+1f0^(S\{n})+nki+1f1^(S\{n}))\frac{-1}{n-2i+1}\sqrt{\frac{(n-2i+1)(n-2i+1)(k-i+1)}{n-2i+2}}\widehat{f_{0}}(S^{\prime}\backslash\{n\})+\frac{n-k-i+1}{n-2i+1}\sqrt{\frac{(n-2i+1)(n-2i+1)}{(n-2i+2)(n-k-i+1)}}\widehat{f_{1}}(S^{\prime}\backslash\{n\})\\ =\frac{1}{\sqrt{n-2i+2}}\left(-\sqrt{k-i+1}\widehat{f_{0}}(S^{\prime}\backslash\{n\})+\sqrt{n-k-i+1}\widehat{f_{1}}(S^{\prime}\backslash\{n\})\right)

as desired. ∎

Finally, we prove the following relationship between the Fourier coefficients of ff and the Fourier coefficients of restrictions of ff.

Corollary 4.3.

Let f:([n]k)f:\binom{[n]}{k}\rightarrow\mathbb{R}, and let iki\leqslant k. Then for all Smin{k,nk}S\in\mathcal{H}_{\min\{k,n-k\}},

f0^(S)2+f1^(S)2=f^(S)2+f^(S{n})2.\widehat{f_{0}}(S)^{2}+\widehat{f_{1}}(S)^{2}=\widehat{f}(S)^{2}+\widehat{f}(S\cup\{n\})^{2}.

Additionally,

z{0,1}tfz^(S)2=T[n]\[nt]f^(ST)2\sum_{z\in\{0,1\}^{t}}\widehat{f_{z}}(S)^{2}=\sum_{T\subseteq[n]\backslash[n-t]}\widehat{f}(S\cup T)^{2}
Proof.

The first statement follows from Theorem 4.2, and the second follows by induction. ∎

5 A Goldreich-Levin Theorem for the slice

In this section, we prove Theorem 1.2, a Goldreich-Levin theorem for the slice. Informally, this theorem states that the large Fourier coefficients of a function can be efficiently detected. With the machinery of Corollary 4.3, the algorithm and proof of correctness are essentially the same as for the Boolean hypercube. However, some care is required in estimating quantities of the form

S[n]\[i]f^(US)2\sum_{S\subseteq[n]\backslash[i]}\widehat{f}(U\cup S)^{2}

for a given function ff and a given set U[i]U\subseteq[i]. For completeness, we give the algorithm.

The algorithm uses a divide and conquer strategy due to Kushilevitz and Mansour [KM93].

Let L{}L\leftarrow\{\emptyset\}
for i0i\leftarrow 0 to n1n-1 do
       for Each ULU\in L do
             Estimate S[n]\[i]f^(US)2\sum_{S\subseteq[n]\backslash[i]}\widehat{f}(U\cup S)^{2} by w1w_{1}
            Estimate S[n]\[i]f^(U{i+1}S)2\sum_{S\subseteq[n]\backslash[i]}\widehat{f}(U\cup\{i+1\}\cup S)^{2} by w2w_{2}
            Remove UU from LL.
            if w1τ2/2w_{1}\geqslant\tau^{2}/2 then
                   Add UU to LL
             end if
            if w2τ2/2w_{2}\geqslant\tau^{2}/2 then
                   Add U{i+1}U\cup\{i+1\} to LL
             end if
            
       end for
      
end for
Algorithm 1 Goldreich-Levin for the slice

Much of the analysis is identical to the algorithm for the Boolean hypercube. As in the case of the Boolean hypercube, the algorithm identifies a set of Fourier coefficients with large total weight, starting with the set of all Fourier coefficients. At each step, it splits each set into two, estimates the total weights of all of the subsets, and removes those subsets with small total weight. If the algorithm is able to estimate the total weight accurately, then at any given point, it will only identify 𝑝𝑜𝑙𝑦(τ)\mathit{poly}(\tau) different sets. (For a complete analysis, one can also refer to Section 3.5 of [O’D14]).

Thus, it is enough to show that one can estimate S[n]\[i]f^(US)2\sum_{S\subseteq[n]\backslash[i]}\widehat{f}(U\cup S)^{2} and S[n]\[i]f^(U{i}S)2\sum_{S\subseteq[n]\backslash[i]}\widehat{f}(U\cup\{i\}\cup S)^{2} in 𝑝𝑜𝑙𝑦(n,1/τ)\mathit{poly}(n,1/\tau)-time to error (τ2/2)(nk)(\tau^{2}/2)\binom{n}{k} with probability at least 11/𝑝𝑜𝑙𝑦(n,1/τ)1-1/\mathit{poly}(n,1/\tau).

Theorem 5.1.

Given query access to a function f:([n]k){1,1}f:{\binom{[n]}{k}}\rightarrow\{-1,1\} and a set U[i]U\subseteq[i] for some integer ii, then 𝑝𝑜𝑙𝑦(n,1/ε,1/τ)\mathit{poly}(n,1/\varepsilon,1/\tau) samples are sufficient to compute S[n]\[i]f^(US)2\sum_{S\in[n]\backslash[i]}\widehat{f}(U\cup S)^{2} up to error ε([n]k)\varepsilon\binom{[n]}{k} with probability at least 11/𝑝𝑜𝑙𝑦(n,1/τ)1-1/\mathit{poly}(n,1/\tau).

Proof.

By Corollary 4.3, we have

1(nk)S[n]\[i]f^(US)2=z{0,1}ni(ik|z|)(nk)1(ik|z|)fz^(U)2.\frac{1}{\binom{n}{k}}\sum_{S\subseteq[n]\backslash[i]}\widehat{f}(U\cup S)^{2}=\sum_{z\in\{0,1\}^{n-i}}\frac{\binom{i}{k-|z|}}{\binom{n}{k}}\frac{1}{\binom{i}{k-|z|}}\widehat{f_{z}}(U)^{2}. (9)

Let xx be a random variable over {0,1}ni\{0,1\}^{n-i} where x=zx=z with probability (ik|z|)(nk)\frac{\binom{i}{k-|z|}}{\binom{n}{k}}. Note that such a probability distribution is well-defined, as z{0,1}ni(ik|z|)=(nk)\sum_{z\in\{0,1\}^{n-i}}\binom{i}{k-|z|}=\binom{n}{k}. Then the right-hand side of Eq. (9) can be rewritten as

𝔼x[fx^(S)2(ik|x|)]=𝔼x[1Pi,|S|,k|x|χi,S22fx,Pi,|S|,k|x|χi,S2(ik|x|)].\mathbb{E}_{x}\left[\frac{\widehat{f_{x}}(S)^{2}}{\binom{i}{k-|x|}}\right]=\mathbb{E}_{x}\left[\frac{1}{\left\|P^{\uparrow}_{i,|S|,k-|x|}\chi_{i,S}\right\|_{2}^{2}}\frac{\langle f_{x},P^{\uparrow}_{i,|S|,k-|x|}\chi_{i,S}\rangle^{2}}{\binom{i}{k-|x|}}\right].

For each xx in {0,1}ni\{0,1\}^{n-i}, let y1y_{1} and y2y_{2} be independent random variables over ([i]k|x|)\binom{[i]}{k-|x|} distributed according to the absolute value of the coordinates Pi,|S|,k|x|χi,SP^{\uparrow}_{i,|S|,k-|x|}\chi_{i,S}. In particular, we let the probability that y1=Ty_{1}=T be |Pi,|S|,k|x|χi,S|Pi,|S|,k|x|χi,S1\frac{\left|P^{\uparrow}_{i,|S|,k-|x|}\chi_{i,S}\right|}{\left\|P^{\uparrow}_{i,|S|,k-|x|}\chi_{i,S}\right\|_{1}} and thus we obtain

𝔼x[fx^(S)2(ik|x|)]=𝔼x,y1,y2[Pi,|S|,k|x|χi,S12Pi,|S|,k|x|χi,S22fz(y1)fz(y2)sign((Pi,|S|,k|x|χi,S)y1)sign((Pi,|S|,k|x|χi,S)y2)(ik|x|)]\mathbb{E}_{x}\left[\frac{\widehat{f_{x}}(S)^{2}}{\binom{i}{k-|x|}}\right]=\mathbb{E}_{x,y_{1},y_{2}}\left[\frac{\left\|P^{\uparrow}_{i,|S|,k-|x|}\chi_{i,S}\right\|_{1}^{2}}{\left\|P^{\uparrow}_{i,|S|,k-|x|}\chi_{i,S}\right\|_{2}^{2}}\frac{f_{z}(y_{1})f_{z}(y_{2})\operatorname{sign}\left(\left(P^{\uparrow}_{i,|S|,k-|x|}\chi_{i,S}\right)_{y_{1}}\right)\operatorname{sign}\left(\left(P^{\uparrow}_{i,|S|,k-|x|}\chi_{i,S}\right)_{y_{2}}\right)}{\binom{i}{k-|x|}}\right]

By Claim 2.1, and the fact that f(S){1,1}f(S)\in\{-1,1\} for all SS, it follows that the random variable inside the expectation is bounded above by 11. Thus, by a Chernoff bound, O(log(n)/ε2)O(\log(n)/\varepsilon^{2}) samples are sufficient to compute S[n]\[i]f^(US)2\sum_{S\subseteq[n]\backslash[i]}\widehat{f}(U\cup S)^{2} up to error ε(nk)\varepsilon\binom{n}{k} with probability at least 11/𝑝𝑜𝑙𝑦(n,1/τ)1-1/\mathit{poly}(n,1/\tau). ∎

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