Comments on Degrees of freedom region for $K$-user interference channel with $M$ antennas

For a K-user interference channel with M antenna at each transmitter and each receiver, the maximum total DoF of this channel has been previously determined to be max ∑ K k=1 dk = MK/2. However, the DoF region remains to be unknown. In this short note, through a simple time-sharing argument, we obtain the degrees of freedom (DoF) region of this channel. Index Terms Interference alignment, Degrees of freedom region,interference channel Consider a K-user interference channel with M antennas at each transmitter and each receiver, the same as in [1]. Let dk denote the degrees of freedom (DoF) of user k, k = 1, . . . , K. The maximum total DoF max ∑K k=1 has been found in [1] to be MK/2. We have the following result regarding the DoF region. Theorem 1: The degrees of freedom region of the K-user interference channel with M-antennas at both receivers and transmitters is characterized as follows : D = {(d1, d2, · · · , dK) : di + dj ≤ M, ∀1 ≤ i, j ≤ K, i 6= j} . (1) Proof: The converse argument is the same as the converse argument in [1, Theorem 1]. We show the achievability as follows. Without loss generality, suppose d∗ 1 ≥ d∗ 2 ≥ d∗k, k = 3, · · · , K, and d ∗ i+d ∗ j ≤ d ∗ 1 +d∗ 2 ≤ M , ∀i, j ∈ [1, K]. We would like to show that (d1, d2, . . . , dM) = (d∗1, d ∗ 2 , . . . , d∗M) is achievable. It is obvious that (d1, d2, . . . , dK) = (M, 0, . . . , 0) can be achieved by single user transmission. It is also known from [1] that the point (d1, d2, . . . , dK) = (M/2,M/2, . . . ,M/2)