Necessary Conditions for K/2 Degrees of Freedom

Stotz et al., 2016, reported a sufficient (injectivity) condition for each user in a K-user single-antenna constant interference channel to achieve 1/2 degree of freedom. The present paper proves that this condition is necessary as well and hence provides an equivalence characterization of interference channel matrices allowing full degrees of freedom.


I. INTRODUCTION
Cadambe and Jafar [1], [2] proposed a signaling scheme-known as interference alignment-that exploits time-frequency selectivity to achieve K/2 degrees of freedom (DoF) in K-user single-antenna interference channels (ICs). In [3] and [4] it was shown that K/2 DoF can also be achieved in ICs with constant channel matrix, i.e, in the absence of selectivity. Wu et al. [5] developed a general formula for the number of DoF in single-antenna ICs, extended to vector ICs in [6]. This formula can, however, be difficult to evaluate as it is expressed in terms of Rényi information dimension [7]. Building on the work by Wu et al. [5] and a recent breakthrough result in fractal geometry by Hochman [8], Stotz and Bölcskei [9] derived a DoF-formula for single-antenna ICs, which is exclusively in terms of Shannon entropy; this formula was then used to develop an explicit sufficient condition for achieving K/2 DoF. Stotz et al. [10] later identified an even more general sufficient (injectivity) condition for each user to achieve 1/2 DoF and hence K/2 DoF in total.
The main contribution of the present paper is to establish that the sufficient condition in [10] is also necessary for each user to achieve 1/2 DoF in fully connected ICs 1 . The tools used in the proof of this result are the DoF-formula developed in [9] and the entropy version of the Plünnecke-Ruzsa inequality [11].

II. SYSTEM MODEL
We consider a single-antenna K-user (additive) IC with fully connected channel matrix H = (h ij ) 1 i,j K ∈ R K×K , i.e, h ij = 0 for all i, j, and input-output relation where X i ∈ R is the input at the i-th transmitter, Y i ∈ R is the output at the i-th receiver, and Z i ∈ R is noise of 1 Note that the sufficient condition in [10] applies to all ICs, whereas here we restrict ourselves to fully connected ICs.
The work of S. Shamai (Shitz) was supported by the European Union's Horizon 2020 Research And Innovation Programme, grant agreement no. 694630. absolutely continuous distribution satisfying h(Z i ) > −∞ and H( Z i ) < ∞. The input signals are independent across transmitters, and noise is i.i.d. across users and channel uses.
The channel matrix H is known perfectly at all transmitters and receivers. We impose the average power constraint We note that [10] studies conditions for each user to achieve 1/2 DoF as opposed to K/2 DoF for all users in total. This is done to avoid trivial exceptions induced by particular network topologies reflected by zeros in the IC matrix. Here, we also study conditions for each user to achieve 1/2 DoF. We remark, however, that topology-induced exceptions cannot occur for fully connected ICs.

III. MAIN RESULT
We start by reviewing the sufficient (injectivity) condition for each user to achieve 1/2 DoF identified in [10]. Leť h ∈ R K(K−1) denote the vector containing the off-diagonal elements of H. There are ϕ(d) := K(K − 1) + d d monomials 2 in K(K − 1) variables of degree not larger than d, which we enumerate as f 1 ,..., f ϕ(d) . We will further need the following sets It was shown in [10, Th. 1] that for each user to achieve 1/2 DoF, it is sufficient that the following condition be satisfied for either H directly or at least one scaled version of H, where 2 A monomial in k variables x i is an expression of the form x n 1 1 x n 2 2 ... x n k k , where n 1 ,..., n k ∈ N, and the degree of the monomial is n 1 + ... + n k . scaling, as defined in [10], refers to a finite number of scalings of individual rows and columns by nonzero constants: For each i = 1,..., K, the map The main result of the present paper is as follows.
Theorem 1. Let W be as in (1). Then, for almost all fully connected IC matrices H, the following condition is necessary for each user to achieve 1/2 DoF: Either the channel matrix H itself or at least one scaled version thereof satisfies condition ( * * ).
Proof: See Section V.
With the matching sufficient condition in [10, Th. 1] this yields an equivalence characterization of full-DoF-achieving IC matrices.
Remark 1. Note that thanks to W containing integer linear combinations of monomials in the off-diagonal elements of H, the condition ( * * ) is exclusively in terms of channel coefficients.
Remark 2. Almost all channel matrices H are fully connected. Since an almost all subset of an almost all set is itself almost all, the set of channel matrices H not covered by Theorem 1 has measure zero.

IV. BALANCING RESULTS
We start with a simple modification of [9,Th. 3], with the proof omitted due to space constraints. Proposition 1. For almost all fully connected IC matrices H the following holds: If each user achieves 1/2 DoF, then for ε ∈ (0, 1/2), there exist independent discrete random variables V 1 ,..., V K of finite entropy such that for i = 1,..., K, with the denominator of (2) nonzero.
The following lemma, based on Proposition 1, states "balancing" results on the entropies of signal and interference contributions and will turn out instrumental in the proof of our main statement. Lemma 1. For ε ∈ (0, 1/2), let V 1 ,..., V K be the corresponding independent discrete random variables satisfying (2) for the fully connected IC matrix H. Then, Proof: Starting from (2), we have for i = 1,..., K. Rearranging terms, we get Invoking the following inequality for independent discrete random variables X, Y [12, Ex. 2.14] on the right-hand side (RHS) of (7) yields Next, we show that for fully connected ICs To this end, w.l.o.g., we assume that repeatedly for independent discrete random variables X and Y , and arbitrary α, This proves (10) for i = 2,..., K. The case i = 1 is obtained as follows. First note that for all i = 1, where the first inequality is by (11) and the second by (9). Further, again by (11), we have for all i = 1, and inserting into (13) establishes (10) for i = 1. We can now combine (9) and (10) to get for all i, which yields (3).
We conclude this section by recording, for later use, a simple variation of the entropy version of the Plünnecke-Ruzsa inequality [11].
for i = 1,..., m. Then, for finite m, Proof: In [11, Th. 2.8.2], set log K i = H(X)O(ε) for i = 1,..., m, and take the additive group G as R. This results in H(X + Y 1 + ... + Y m ) ≤ H(X) + mH(X)O(ε). Divide this inequality by H(X) and note that owing to (11) the expression on the LHS of (19) is greater than or equal to 1. The proof is concluded by noting that O(ε)m = O(ε) for finite m.

V. PROOF OF THEOREM 1
For simplicity of exposition and due to space constraints, we detail the proof for the 3-user case only. The proof for the K-user case follows by induction over the number of users with the 3-user case constituting the base case.
Consider the 3-user fully connected IC matrix The proof is effected by contradiction, with the contradiction established by induction on the degrees of polynomials in W as defined in (1). Towards this contradiction, we assume that H is in the almost all set covered by Theorem 1, while at the same time each user achieves 1/2 DoF and condition ( * * ) is violated for H and all scaled versions thereof. In particular, condition ( * * ) must also be violated for We proceed by assuming that the map in ( * * ) is not injective for user i = 1 so that condition ( * * ) is violated. This implies the existence of w 1 , w 2 , w 1 , w 2 ∈ W such that w 1 = w 1 , w 2 = w 2 , and We can hence write where d 1 and d 2 are finite and the coefficientsâ p ,b p are integers. Since w 1 = w 1 and w 2 = w 2 , the polynomials on the RHS of (21) are nonzero.
Since H is fully connected, we know from (5) that, for i = 1, We shall show that this leads to a contradiction, by proving that (2) implies for every nonzero finite-degree polynomial in h with integer coefficients a p and b p , in particularâ p andb p . The implication (2) =⇒ (23) is established by induction over d = max{d 1 , d 2 }.
Induction step. We assume that (23) holds for d = m − 1, with m ≥ 1, and show that this implies (23) for d = m. The following lemma contains the central technical result in the induction step.
Proof: To establish (32), we apply . The corresponding conditions (18) hold as dividing X + Y 1 and X + Y 2 by h yields (23), which is satisfied by the induction hypothesis.
We now use Lemma 3 to finalize the induction argument. First, let where V 2 ,V 2 and V 3 ,V 3 are independent copies of V 2 and V 3 , respectively. Applying , noting that the corresponding conditions (18) are satisfied thanks to (32) and (33), and invoking (26), we get H(p(h)) H(V 1 ) = 1 + O(ε).