On the Degrees of Freedom Achievable Through Interference Alignment in a MIMO Interference Channel
aa r X i v : . [ c s . I T ] S e p On the Degrees of Freedom AchievableThrough Interference Alignment in a MIMOInterference Channel ∗ Meisam Razaviyayn † , Gennady Lyubeznik ‡ and Zhi-Quan Luo † Abstract
Consider a K -user flat fading MIMO interference channel where the k -th transmitter (or receiver)is equipped with M k (respectively N k ) antennas. If an exponential (in K ) number of generic channelextensions are used either across time or frequency, Cadambe and Jafar [1] showed that the total achievabledegrees of freedom (DoF) can be maximized via interference alignment, resulting in a total DoF thatgrows linearly with K even if M k and N k are bounded. In this work we consider the case where nochannel extension is allowed, and establish a general condition that must be satisfied by any degrees offreedom tuple ( d , d , ..., d K ) achievable through linear interference alignment. For a symmetric systemwith M k = M , N k = N , d k = d for all k , this condition implies that the total achievable DoF cannotgrow linearly with K , and is in fact no more than K ( M + N ) / ( K + 1) . We also show that this boundis tight when the number of antennas at each transceiver is divisible by d , the number of data streamsper user. I. I
NTRODUCTION
Consider a multiuser communication system in which a number of transmitters must share commonresources such as frequency, time, or space in order to send information to their respective receivers.The mathematical model for this communication scenario is the well-known interference channel , whichconsists of multiple transmitters simultaneously sending messages to their intended receivers while causinginterference to each other. † Department of Electrical and Computer Engineering, University of Minnesota, Minneapolis, 55455, USA. (e-mail: { meisam,luozq } @umn.edu) ‡ Department of Mathematics, University of Minnesota, Minneapolis, 55455, USA. ∗ This work was supported in part by the Army Research Office, grant number W911NF-09-1-0279, and in part by a researchgift from Huawei Technologies Inc.
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A central issue in the study of interfering multiuser systems is how to mitigate multiuser interference.In practice, there are several commonly used methods for dealing with interference. First, we can treatthe interference as noise and just focus on extracting the desired signals. This approach is widelyused in practice because of its simplicity and ease of implementation, but is known to be non-capacityachieving in general. An alternative technique is channel orthogonalization whereby transmitted signalsare chosen to be nonoverlapping either in time, frequency or space, leading to Time Division MultipleAccess, Frequency Division Multiple Access, or Space Division Multiple Access respectively. Whilechannel orthogonalization effectively eliminates multiuser interference, it can lead to inefficient use ofcommunication resources and is also generally non-capacity achieving. Another interference managementtechnique is to decode and remove interference. Specifically, when interference is strong relative to desiredsignals, a user can decode the interference first, then subtract it from the received signal, and finallydecode its own message. Unfortunately, none of the aforementioned interference management techniquescan achieve the maximum system throughput in general.Theoretically, what is the optimal transmit/receive strategy in a MIMO interference channel? Theanswer is related to the characterization of the capacity region of an interference channel, i.e., determiningthe set of rate tuples that can be achieved by the users simultaneously. In spite of intensive research onthis subject over the past three decades, the capacity region of interference channels is still unknown(even for small number of users). The lack of progress to characterize the capacity region of the MIMOinterference channel has motivated researchers to derive various approximations of the capacity region.For example, the maximum total degrees of freedom (DoF) corresponds to the first order approximationof sum-rate capacity in the high SNR regime. Specifically, in a K -user interference channel, we definethe degrees of freedom region as the following [1]: D = (cid:26) ( d , d , . . . , d K ) ∈ R K + | ∀ ( w , w , . . . , w K ) ∈ R K + , K X k =1 w k d k ≤ lim sup SNR →∞ " sup R ∈C K X k =1 w k R k , (1)where C is the capacity region and R k is the rate of user k . We can further define the total DoF in thesystem as the following: η = max ( d ,d ,...,d K ) ∈D d + d + . . . + d K . Intuitively, the total DoF is the number of independent data streams that we can communicate interference-
October 22, 2018 DRAFT free in the channel.It is well known that for a point-to-point MIMO channel with M antennas at the transmitter and N antennas at the receiver, the total DoF is η = min { M, N } . Different approaches such as SVD precoderor V-BLAST can be used to achieve this DoF bound. For a 2-user MIMO fading interference channel withuser k equipped with M k transmit antennas and N k receive antennas ( k = 1 , ), Jafar and Fakhereddin[9] proved that the maximum total DoF is η = min { M + M , N + N , max { M , N } , max { M , N }} . This result shows that for the case of M = M = N = N , the total DoF in the system is the same asthe single user case. In other words, we do not gain more DoF by increasing the number of users from oneto two. Interestingly, if generic channel extensions (drawn from a continuous probability distribution) areallowed either across time or frequency, Cadambe and Jafar [1] showed that the total DoF is η = KM / for a K -user MIMO interference channel, where M is the number of transmit/receive antennas peruser. This result implies that each user can effectively utilize half of the total system resources in aninterference-free manner by aligning the interference at all receivers . The principal assumption enablingthis surprising result is that the channel extensions are exponentially long in K and are generic (e.g.,drawn from a continuous probability distribution). If channel extensions are restricted to have a polynomiallength or are not generic, the total DoF for a MIMO interference channel is still largely unknown evenfor the Single-Input-Single-Output (SISO) interference channel. For the 3-user special case, reference[7] provided a characterization of the total achievable DoF as a function of the diversity. In the absenceof channel extensions, the computational complexity of numerically designing an interference alignmentscheme has been shown to be NP-hard [12] in the number of users.The main theoretical investigation pertaining to the current work is [2] by Yetis et al. who studiedthe maximum achievable DoF for a MIMO interference channel without channel extension. In general,linear interference alignment can be described by a set of bilinear equations which correspond to thezero-forcing conditions at each receiver. For a K -user system, there are a total of K ( K − suchcoupled quadratic matrix equations whose unknowns are the transmit/receive beamforming matrices tobe designed. Moreover, the achievability of a given tuple of DoF corresponds to these quadratic equationshaving a solution (in the form of beamforming matrices) whose individual matrix ranks are given by the The idea of interference alignment was introduced in [3]–[5] and the terminology “interference alignment” was first used in[6].
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DoFs. One can easily count the number of “independent unknowns” and the number of scalar equationsin this quadratic system defining interference alignment. It is then tempting to conjecture, as was done in[2], that the interference alignment is feasible if and only if the number of equations is no more than thenumber of unknowns in each subsystem of the quadratic equations. When the latter is true, the authorsof [2] called the corresponding system proper . However, except for some special cases involving a smallnumber of users and antennas, the investigation of [2] was largely inconclusive.In this paper, we settle the conjecture of [2] completely in one direction, and partially in the other. Inparticular, we consider the case where no channel extension is allowed, and use results from the fieldtheory to establish a general condition that must be satisfied by any DoF tuple achievable through linearinterference alignment. This condition shows that the improperness property (in the sense of [2]) indeedimplies the infeasibility of interference alignment. For the symmetric system with M k = M and N k = N for all k , this condition implies that the total achievable DoF cannot grow linearly with the number ofusers, and is in fact no more than M + N − . This is in sharp contrast to the case with independent channelextensions for which the total DoF can grow linearly with the number of users. For the converse direction,we show that if all users have the same DoF d and the number of antennas M k , N k are divisible by d for each k , then the properness of the quadratic system implies the feasibility of interference alignmentfor generic choice of channel coefficients (e.g., drawn from a continuous probability distribution). Ifin addition, M k = M and N k = N for all k and M, N are divisible by d , then our results imply thatinterference alignment is achievable if and only if ( M + N ) ≥ d ( K +1) . In the simulation section, we usethese established DoF bounds to numerically benchmark the performance of several existing algorithmsfor interference alignment and sum-rate maximization.II. S YSTEM M ODEL
Consider a MIMO interference network consisting of K transmitter - receiver pairs, with transmitter k sending d k independent data streams to receiver k . Let H kj be an M j × N k matrix that represents thechannel gain matrix from transmitter j to receiver k where M j and N k denote the number of antennasat transmitter j and receiver k , respectively. The received signal at receiver k is given by y k = K X j =1 H kj x j + n k where x j is an M j × random vector that represents the transmitted signal of user j and n k ∼ N ( , σ I ) is a zero mean additive white Gaussian noise. October 22, 2018 DRAFT
Throughout this paper, we focus on linear transmit and receive strategies that can maximize systemthroughput. In this case, transmitter k uses a beamforming matrix V k in order to send a signal vector s k to its intended receiver k . On the other side, receiver k estimates the transmitted data vector s k by usinga linear beamforming matrix U k , i.e., x k = V k s k , ˆ s k = U Hk y k where the power of the data vector s k ∈ R d k × is normalized such that E [ s k s Hk ] = I , and ˆ s k is theestimate of s k at the k -th receiver. The matrices V k ∈ C M k × d k and U k ∈ C N k × d k are the beamformingmatrices at the k -th transmitter and receiver respectively. Without channel extension, the linear interferencealignment conditions can be described by the following zero-forcing conditions [2], [12] U Hk H kj V j = , ∀ j = k, (2) rank (cid:0) U Hk H kk V k (cid:1) = d k , ∀ k. (3)The first equation guarantees that all the interfering signals at receiver k lie in the subspace orthogonalto U k , while the second one assures that the signal subspace H kk V k has dimension d k and is linearlyindependent of the interference subspace. Intuitively, as the number of users K increases, the numberof constraints on the beamformers { U k , V k } increases quadratically in K , while the number of designvariables in { U k , V k } only increases linearly. This suggests the above interference alignment can nothave a solution unless K or d k is small.The interference alignment conditions (2) and (3) imply that each transmitter k can use a lineartransmit/receive strategy to communicate d k interference-free independent data streams to receiver k (per channel use). In this case, it can be checked that d k represents the DoF achieved by the k -thtransmitter/receiver pair in the information theoretic sense of (1). In other words, the vector ( d , d , ..., d K ) in (2) and (3) represents the tuple of DoF achieved by linear interference alignment. Intuitively, the largerthe values of d , d ,..., d K , the more difficult it is to satisfy the interference alignment conditions (2) and(3). III. B OUNDING THE T OTAL D O F A
CHIEVABLE VIA L INEAR I NTERFERENCE A LIGNMENT
Our goal is to study the solvability of the interference alignment problem (2)-(3) and derive a generalcondition that must be satisfied by any DoF tuple ( d , d , ..., d K ) achievable through linear interferencealignment for generic choice of channel matrices. We will also provide some conditions under which thisupper bound is achievable. October 22, 2018 DRAFT
Let us denote the polynomial equations in (3) by the index set J , { ( k, j ) | ≤ k = j ≤ K } . The following theorem provides an upper bound on the total achievable DoF when no channel extensionis allowed.
Theorem 1
Consider a K -user flat fading MIMO interference channel where the channel matrices { H ij } Ki,j =1 are generic (e.g., drawn from a continuous probability distribution). Assume no channelextension is allowed. Then any tuple of degrees of freedom ( d , d , ..., d K ) that is achievable throughlinear interference alignment (2) and (3) must satisfy the following inequalities min { M k , N k } ≥ d k , ∀ k, (4) max { M k , N j } ≥ d k + d j , ∀ k, j, k = j, (5) X k :( k,j ) ∈I ( M k − d k ) d k + X j :( k,j ) ∈I ( N j − d j ) d j ≥ X ( k,j ) ∈I d k d j , ∀ I ⊆ J . (6)Condition (6) in Theorem 1 can be used to bound the total DoF achievable in a MIMO interferencechannel. The following corollary is immediate. Corollary 1
Assume the setting of Theorem 1. Then the following upper bounds hold true.(a) In the case of d k = d for all k , interference alignment is impossible unless d ≤ K ( K + 1) K X k =1 ( M k + N k ) . (b) In the case of M k + N k = M + N , interference alignment requires K X k =1 d k ! + K X k =1 d k ≤ ( M + N ) K X k =1 d k which further implies K X k =1 d k < ( M + N ) . Part (b) of Corollary 1 shows that the total achievable DoF in a MIMO interference channel is boundedby a constant M + N − , regardless of how many users are present in the system. While this bound isan improvement over the single user case which has a maximum DoF of min { M, N } , it is significantlyweaker than the maximum achievable total DoF for a diagonal frequency selective (or time varying) October 22, 2018 DRAFT interference channel. The latter grows linearly with the number of users in the system [1].The rest of this section is devoted to the proof of Theorem 1 and its converse. Since we will useseveral concepts and results from the field theory [11] and algebraic geometry [14], [16], we first providea brief review of the necessary algebraic background.
A. Algebraic Preliminaries
Let K , F be two fields such that K ⊆ F . In this case, we say F is an extension of K , denotedby F / K . Let us use K [ z , z , . . . , z n ] to denote the ring of polynomials with coefficients drawn from K . We say α , α , . . . , α n ∈ F are algebraically dependent over K if there exists a nonzero polyno-mial f ( z , z , . . . , z n ) ∈ K [ z , z , . . . , z n ] such that f ( α , α , . . . , α n ) = 0 . (7)Otherwise, we say that they are algebraically independent over K . The largest cardinality of an alge-braically independent set is called the transcendence degree of F over K . An element α ∈ F is said tobe algebraic over K if there exists a nonzero polynomial f ∈ K [ z ] such that f ( α ) = 0 ; else, we say α is transcendental over K . Example 1.
Let K = C be the field of complex numbers and F = C ( x , x ) be the field of rationalfunctions in variables x , x . Then, the polynomials g = x x , g = x , g = x x are algebraically dependent over C because f ( g , g , g ) = 0 identically for all ( x , x ) , where f ( z , z , z ) = z z − z . Example 2.
The two complex numbers a = √ π, b = 3 π + 2 are algebraically dependent over the fieldof rational numbers because by defining f ( z , z ) = 3 z − z + 2 , we have f ( a, b ) = 0 .Notice that the definition of algebraic independence is in many ways similar to the standard notionof linear independence from linear algebra. In fact, if the function f in (7) is required to be linear, thenalgebraic independence reduces to the usual concept of linear independence. Similar to linear algebra,we can define a basis for the field F using the notion of algebraic independence. In particular, givenany algebraically independent set S over the field K , let K ( S ) denote the field of rational functions in October 22, 2018 DRAFT S with coefficients taken from the field K . For any field extension F / K , it is always possible to find aset S in F , algebraically independent over K , such that F is an algebraic extension of K ( S ) . Such a set S is called a transcendence basis of F over K . All transcendence bases have the same cardinality, equalto the transcendence degree of the extension F / K . If every element in F is algebraic over K , then wesay F / K is an algebraic extension. In this case, the transcendence degree of F over K is zero. Example 3.
The two polynomials g and g in Example 1 are algebraically independent over C . Together,they constitute a transcendental basis for C ( x , x ) over C .The following table shows similar concepts between linear algebra and transcendental field extension(see [11], [16] for more details). Linear algebra Transcendental field extension linear independence algebraic independence A ⊆ span( B ) A algebraically dependent on B linear basis transcendence basisdimension transcendence degreeIn linear algebra, it is well known that any ( n + 1) vectors v , v , ..., v n +1 in an n -dimensional vectorspace must be linearly dependent. In other words, there exists a nonzero linear function f ( z , z , ..., z n +1 ) such that f ( v , v , ..., v n +1 ) = 0 . A similar result holds for algebraic independence. For example, any ( n +1) polynomials g , g ,..., g n +1 defined on n variables ( x , x , ..., x n ) must be algebraically dependent.Consequently, there exists a nonzero polynomial f ( z , z , ..., z n +1 ) such that f ( g , g , ..., g n +1 ) = 0 , ∀ ( x , x , ..., x n ) . Example 1 is an instance of this property with n = 2 . The following example states this property, to beused in the proof of Theorem 1, in a more formal setting. Example 4.
Let C ( z , z , . . . , z n ) denote the field of rational functions in n variables with coefficientsin C . The set { z , z , . . . , z n } is a maximal algebraically independent set in C ( z , z , . . . , z n ) . Hence thetranscendence degree of the field extension C ( z , z , . . . , z n ) / C is n . Furthermore, for any m polynomials g ( z , z , . . . , z n ) , g ( z , z , . . . , z n ) , . . . , g m ( z , z , . . . , z n ) , October 22, 2018 DRAFT where m > n , there exists a nonzero polynomial f ( · ) such that f ( g , g , . . . , g m ) = 0 , ∀ z , z , . . . , z n . Next we describe a useful local expansion of a multivariate polynomial function. Recall that for anyunivariate polynomial f and any ¯ x ∈ C , there holds f ( x ) = f (¯ x ) + ( x − ¯ x ) g ( x ) , for all x ∈ C ,where g is some polynomial dependent on ¯ x and the coefficients of f only. Similarly, for a n -variatepolynomial f defined on the variables x = ( x , x , ..., x n ) and any ¯ x ∈ C n , we have f ( x ) = f (¯ x ) + n X i =1 ( x i − ¯ x i ) g i ( x ) = f (¯ x ) + ( x − ¯ x ) T g ( x ) , ∀ x ∈ C n , where each g i is some polynomial dependent on ¯ x and the coefficients of f only. If we replace the scalarvariable x i by a matrix variable X i , then we can write f ( X ) = f ( ¯ X ) + n X i =1 Tr (cid:0) ( X i − ¯ X i ) G i ( X ) (cid:1) , ∀ X , (8)where each G i is a matrix whose entries are polynomials dependent on the entries of ¯ X and the coefficientsof f only. The local expansion (8) will be used in the proof of Theorem 1.To prove the converse of Theorem 1, we will use the concepts of Zariski topology and a Zariskiconstructible set. We briefly review these concepts next (see [14] for more details). Consider C n , the n -dimensional vector space over the field of complex numbers C . [One can replace C by any algebraicallyclosed field.] The Zariski topology for C n is defined by specifying its closed sets, and these are takensimply to be all the algebraic sets in C n . That is, the closed sets under Zariski topology are those of theform S = { x ∈ C n | f i ( x ) = 0 , i = 1 , , ..., m } where { f i } mi =1 is any set if polynomials with coefficients taken from C . For example, the entire space C n is Zariski closed (Take m = 1 and f to be the zero function, i.e., f ( x ) = 0 , ∀ x ). All otherZariski closed sets have zero measure. A nonempty Zariski open set (the complement of a Zariski closedset) always has dimension n . If a property holds over a Zariski open set, we say the property holds generically .In topology, a set is locally closed if it is the intersection of an open set with a closed set. A constructibleset is defined as a finite union of locally closed sets. Thus, a Zariski constructible set is simply a finite October 22, 2018 DRAFT0 collection of sets, each defined by the feasible set of finitely many polynomial equations and polynomialinequalities. Clearly, if a Zariski constructible set has dimension n , then it must contain a Zariski opensubset.Let φ , φ , . . . , φ n be polynomials in x , x , . . . , x n with coefficients from C . They define a map Φ : C n C n as follows: Φ( x ) = ( φ ( x ) , φ ( x ) , . . . , φ n ( x )) ∈ C n . Chevalley’s Theorem says that the imageof this map is a constructible set (see [16] for more details). Example 5.
Let
Φ : C C be defined by Φ( x ) = ( φ ( x ) , φ ( x )) where φ ( x ) = x and φ ( x ) = x x .Let L be the line { x ∈ C : x = 0 } . The image of Φ is the union of two locally closed sets, C \L (which is in fact open) and the point (0 , (which is indeed closed).Let the image of Φ be the union of locally closed subsets W , W , . . . , W p where W i = U i T V i and V i is closed and U i is open. Assume the Jacobian of φ , φ , . . . , φ n is nonsingular at some point x ∈ C n .The Implicit Function Theorem says that the image of Φ contains a small open disc around Φ( x ) , hencethe measure of the image is nonzero. This implies that for some i , V i = C n and W i = U i , i.e., the imageof the map Φ( · ) contains a Zariski open set. Thus, if a certain property is shown to hold over the imageof a polynomial map Φ : C n C n whose Jacobian is nonsingular at some point, then this property musthold generically. We will use this approach to establish the generic feasibility of interference alignmentfor certain MIMO interference channels (Theorem 2). B. Proof of Theorem 1
We now use the transcendental field extension theory to establish Theorem 1.
Proof:
The inequality (4) is obvious due to (3). To prove (5), assume M j ≤ N k . Since H kj isgeneric, rank( H kj V j ) = d j . Furthermore, due to (3), the beamformer U k must be full rank and hence d k + d j must be no more than the total dimension N k . Similar argument shows that d k + d j ≤ M j when M j ≥ N k . Thus, d k + d j ≤ max { M j , N k } .For simplicity of notations, we prove (6) for the case I = J . When I ⊂ J , the proof is the sameexcept that we need to focus on a subset of equations/variables. Now, we prove (6) for the case of I = J October 22, 2018 DRAFT1 by contradiction. Assume the contrary that K X k =1 ( M k − d k ) d k + K X j =1 ( N j − d j ) d j < K X k,j =1 ,k = j d k d j , (9)and the interference alignment conditions in (2) and (3) are satisfied. The interference alignment condition(3) implies that U k and V k must have full column rank. By applying appropriate linear transformationsto the rows of U k and V k , we can write U k = P uk I ¯ U k Q uk , V k = P vk I ¯ V k Q vk , ∀ k, (10)where ¯ U k and ¯ V k are some matrices of size ( N k − d k ) × d k and ( M k − d k ) × d k respectively. Thematrices P uk and P vk are square permutation matrices of size N k × N k and M k × M k respectively, while Q uk , Q vk are some invertible matrices of size d k × d k . Define ¯ H ij = P u − i H ij P v − j to be the permutedversion of H kj . We can partition the matrix ¯ H kj as ¯ H kj = ¯ H (1) kj ¯ H (2) kj ¯ H (3) kj ¯ H (4) kj where ¯ H (1) kj is of size d k × d j . Since the channel matrices { H kj } k = j are drawn from a continuousprobability distribution, the transformed channel matrices { ¯ H (1) kj } k = j remain generic. Rewriting the linearinterference alignment condition (2) in terms of ¯ U k and ¯ V k , we obtain h I ¯ U Hk i ¯ H (1) kj ¯ H (2) kj ¯ H (3) kj ¯ H (4) kj I ¯ V j = (11)or equivalently ¯ H (1) kj + ¯ U Hk ¯ H (3) kj + ¯ H (2) kj ¯ V j + ¯ U Hk H (4) kj ¯ V j = , ∀ j = k. (12)The above system of quadratic equations, first derived in [2], is equivalent to the interference alignmentcondition (2). The number of scalar equations in (12) is K X j,k =1 ,j = k d k d j , while the total number of scalar variables (i.e., the scalar entries of the unknown matrices { ¯ U k } ’s and October 22, 2018 DRAFT2 { ¯ V k } ’s) is K X k =1 ( M k − d k ) d k + K X k =1 ( N k − d k ) d k = K X k =1 ( M k + N k − d k ) d k . So if K X k =1 ( M k + N k − d k ) d k < K X j,k =1 ,j = k d k d j , (13)then we would have more constraints than unknowns in the interference alignment condition (12), whichwe will argue cannot hold.Let us consider the field F defined over the field of complex numbers C , consisting of all rationalfunctions in the entries of the matrices { ¯ U k } Kk =1 and { ¯ V k } Kk =1 . Note that the entries of the matri-ces { ¯ U k , ¯ V k } Kk =1 form a transcendence basis for F over C . Thus, the transcendence degree of F is P Kk =1 ( M k + N k − d k ) d k , which is equal to the number of entries in the matrices { ¯ U k , ¯ V k } Kk =1 .Now, let us consider the matrices H (2) kj , H (3) kj , H (4) kj for all k, j, k = j and define the matrix F kj : F kj ( ¯ U , ¯ V ) , − (cid:16) ¯ U Hk ¯ H (3) kj + ¯ H (2) kj ¯ V j + ¯ U Hk ¯ H (4) kj ¯ V j (cid:17) , (14)for all k, j with k = j . Note that F kj is a d k × d j matrix, with each entry being a quadratic polynomialfunction of the entries in the matrices ¯ U k and ¯ V k . As a result, the entries of F kj belong to the field F . Moreover, if (13) holds, then the number of quadratic polynomials given in the matrices { F kj } k = j isstrictly larger than the transcendence degree of F over C . Hence, as we discussed in the algebraicpreliminaries (Section III-A; see also [11, Chapter 8]), these quadratic polynomials in F must bealgebraically dependent. This implies that there exists a nonzero polynomial p which vanishes at thequadratic polynomials corresponding to the entries of the matrices { F kj } k = j , i.e., p (cid:0) F ( ¯ U , ¯ V ) , F ( ¯ U , ¯ V ) , . . . , F K ( K − ( ¯ U , ¯ V ) (cid:1) = 0 , for all { ¯ U k , ¯ V k } Kk =1 . Notice that the polynomial p is independent of the channel matrices n ¯ H (1) kj o k = j ,even though it does depend on the matrices n ¯ H (2) kj , ¯ H (3) kj , ¯ H (4) kj o k = j . When viewed as a polynomial of thematrix variable X := (cid:16) ¯ H (1)12 , ¯ H (1)13 , . . . , ¯ H (1) K ( K − (cid:17) , p ( · ) can be expanded locally at ¯ X := ( F ( ¯ U , ¯ V ) , F ( ¯ U , ¯ V ) , . . . , F K ( K − ( ¯ U , ¯ V )) October 22, 2018 DRAFT3 using (8): p (cid:16) ¯ H (1)12 , ¯ H (1)13 , . . . , ¯ H (1) K ( K − (cid:17) = p (cid:0) F ( ¯ U , ¯ V ) , F ( ¯ U , ¯ V ) , . . . , F K ( K − ( ¯ U , ¯ V ) (cid:1) + X k = j Tr (cid:16) ( ¯ H (1) kj − F kj ( ¯ U , ¯ V )) Q kj ( ¯ U , ¯ V ) (cid:17) , for all { ¯ U k , ¯ V k } Kk =1 , where Q kj is some polynomial matrix of size d j × d k . Combining the above twoidentities yields p (cid:16) ¯ H (1)12 , ¯ H (1)13 , . . . , ¯ H (1) K ( K − (cid:17) = X k = j Tr (cid:16) ( ¯ H (1) kj − F kj ( ¯ U , ¯ V )) Q kj ( ¯ U , ¯ V ) (cid:17) . (15)Notice that this equality holds for all choices of { ¯ U k , ¯ V k } Kk =1 . If the interference alignment condition(12) holds, then we have ¯ H (1) kj − F kj ( ¯ U , ¯ V ) = 0 , for all k, j with k = j, for some special choices of the matrices { ¯ U k , ¯ V k } Kk =1 . Substituting this condition into the right handside of (15), we obtain p (cid:16) ¯ H (1)12 , ¯ H (1)13 , . . . , ¯ H (1) K ( K − (cid:17) = 0 . (16)Notice that the polynomial p is independent of the channel matrices { ¯ H (1) kj } k = j . Under our channelmodel, the channel matrices { ¯ H (1) kj } k = j are drawn from a continuous probability distribution. It followsthat the condition (16) cannot hold unless p is identically zero, which contradicts the requirement p = 0 .Theorem 1 settles the conjecture of [2] in one direction, namely, the improperness of polynomial system(2) and (3) implies the infeasibility of interference alignment. From the proof of Theorem 1, it can beseen that the upper bound (6) holds for any choice of fixed channel matrices { ¯ H (2) kj , ¯ H (3) kj , ¯ H (4) kj } k = j aslong as the channel matrices { ¯ H (1) kj } k = j are generic.Also, we remark that the proof technique for Theorem 1 can be used to bound the DoF for a singleantenna parallel interference channel (e.g., the OFDM channel). In particular, consider a single inputsingle output interference channel with M channel extensions, i.e., the channel matrices are diagonal andof the size M × M . Assuming each user transmits one data stream ( d k = 1 for all k ), we can check October 22, 2018 DRAFT4 that the properness of the interference alignment condition (2)-(3) is equivalent to K + 1 ≤ M (see [2,Theorem 1]). Using a completely identical proof, we can show that the properness condition K +1 ≤ M is a necessary condition for the feasibility of interference alignment. This implies that for the single beamcase the total DoF per channel extension is upper bounded by 2, regardless of the number of channelextensions. This DoF bound has also been proposed recently in [17]. C. The Converse Direction
In the remainder of this section, we consider the converse of Theorem 1. In particular, we show thatthe upper bound in Theorem 1 is tight for a special case where all users have the same DoF d and numberof antennas is divisible by d . In this case, we have K ( K − matrix equations in (12), each giving riseto d scalar equations. For any subset of these matrix equations indexed by I , with I ⊆ J , the numberof corresponding scalar equations is equal to d |I| , whereas the number of scalar variables involved inthe equations indexed by I is X k :( k,j ) ∈I ( M k − d ) + X j :( k,j ) ∈I ( N j − d ) d. The next result shows that the bound in Theorem 1 is tight if the polynomial system (12) defininginterference alignment is proper, i.e., for each
I ⊆ J , the number of variables involved in each set ofequations indexed by I is no less than d |I| , the number of scalar equations. The proof of this result usesthe Implicit Function Theorem which involves checking the Jacobian matrix of the polynomial map (14)is nonsingular at some channel realization { ¯ H kj } k = j . Notice that the feasibility of interference alignmentcondition (12) at a given channel realization { ¯ H kj } k = j is equivalent to { ¯ H (1) kj } k = j being containedin the image of the polynomial map (14) which is defined by { ¯ H (2) kj , ¯ H (3) kj , ¯ H (4) kj } k = j . Fix a genericchoice of { ¯ H (2) kj , ¯ H (3) kj , ¯ H (4) kj } k = j for which the Jacobian of the polynomial map (14) is nonsingular. TheImplicit Function Theorem allows us to establish the existence of a locally invertible map from thespace of channel submatrices { ¯ H (1) kj } k = j to the space of beamforming matrices, and that the image ofthis polynomial map (14) is locally full-dimensional. Therefore, for all channel submatrices near thegiven channel realization { ¯ H (1) kj } k = j , the interference alignment condition (12) can be satisfied by somebeamforming matrices. By Chevalley’s Theorem from algebraic geometry [14] (see also the discussionat the end of Section III-A), the “local full-dimensionality” of the image of (14) implies that this image,which is a constructible set, must contain a nonempty Zariski open set. As a result, the whole image October 22, 2018 DRAFT5 of polynomial map (14) contains all generically generated channel sub-matrices { ¯ H (1) kj } k = j . Since thechoice of channel submatrices { ¯ H (2) kj , ¯ H (3) kj , ¯ H (4) kj } k = j is also generic, this then establishes the feasibilityof interference alignment for all generically generated channel matrices { ¯ H kj } k = j . Theorem 2
Assume that all users have the same DoF d k = d , where ≤ d ≤ min { M k , N k } , ∀ k .Furthermore, suppose that M k and N k are divisible by d for all k . Then interference alignment isachievable for generic channel coefficients if and only if for each subset I of equations in (12) , thenumber of variables involved in these equations is no less than the number of matrix equations times d ,or equivalently, |I| d ≤ X k :( k,j ) ∈I ( M k − d ) + X j :( k,j ) ∈I ( N j − d ) , ∀ I with I ⊆ J . (17) Proof:
First of all, the “only if” direction is a direct consequence of Theorem 1. We now focus onthe “if” direction. Consider the polynomial map that we get by concatenating all maps in (14) for all ( k, j ) ∈ J , i.e., F ( ¯ U , ¯ V ) = − (cid:16) ¯ U H ¯ H (3)12 + ¯ H (2)12 ¯ V + ¯ U H ¯ H (4)12 ¯ V (cid:17) , F ( ¯ U , ¯ V ) = − (cid:16) ¯ U H ¯ H (3)13 + ¯ H (2)13 ¯ V + ¯ U H ¯ H (4)13 ¯ V (cid:17) , ... F K ( K − ( ¯ U , ¯ V ) = − (cid:16) ¯ U HK ¯ H (3) K ( K − + ¯ H (2) K ( K − ¯ V K − + ¯ U HK ¯ H (4) K ( K − ¯ V K − (cid:17) , (18)which maps the variables { ¯ U k , ¯ V k } Kk =1 to the { F k,j } k = j space. We will first show that for a specificset of channel matrices, the rank of the Jacobian of this polynomial map is K ( K − d , equal to thenumber of equations. Hence, if we restrict the equations to a subset of variables of size K ( K − d , thedeterminant of the Jacobian matrix of the polynomial map (18) does not vanish identically. This step willestablish the existence of a locally invertible map from the space of beamforming matrices to the spaceof channel matrices. By Chevalley’s Theorem (see [14, Chapter 2, 6.E.]), this image is a constructiblesubset under Zariski topology. This, plus the fact that the image is locally full-dimensional, implies thatthe interference alignment condition (12) is feasible for all generically chosen channel matrices. Thisthen will show the “if” direction of Theorem 2.To show the nonsingularity of the Jacobian matrix, we need to remove some redundant variables in { ¯ U k , ¯ V j } k,j (this occurs when there are more variables than equations), and then construct a specific setof channel matrices { H (2) kj , H (3) kj , H (4) kj } k = j and a solution { ¯ U k , ¯ V j } k,j at which the Jacobian matrix of October 22, 2018 DRAFT6 (18) is nonsingular. Before providing a rigorous description for such a construction, we first consider atoy example with K = 3 users where M k = 3 , N k = 2 , d k = 1 , for k = 1 , , . For this specific example,the assumption (17) is satisfied and the equations in (18) can be rewritten as F ( ¯ U , ¯ V ) = − (cid:16) ¯ U H ¯ H (3)12 + ¯ H (2)12 ¯ V + ¯ U H ¯ H (4)12 ¯ V (cid:17) , F ( ¯ U , ¯ V ) = − (cid:16) ¯ U H ¯ H (3)13 + ¯ H (2)13 ¯ V + ¯ U H ¯ H (4)13 ¯ V (cid:17) , F ( ¯ U , ¯ V ) = − (cid:16) ¯ U H ¯ H (3)21 + ¯ H (2)21 ¯ V + ¯ U H ¯ H (4)21 ¯ V (cid:17) , F ( ¯ U , ¯ V ) = − (cid:16) ¯ U H ¯ H (3)23 + ¯ H (2)23 ¯ V + ¯ U H ¯ H (4)23 ¯ V (cid:17) , F ( ¯ U , ¯ V ) = − (cid:16) ¯ U H ¯ H (3)31 + ¯ H (2)31 ¯ V + ¯ U H ¯ H (4)31 ¯ V (cid:17) , F ( ¯ U , ¯ V ) = − (cid:16) ¯ U H ¯ H (3)32 + ¯ H (2)32 ¯ V + ¯ U H ¯ H (4)32 ¯ V (cid:17) , where ¯ V k = [ v k v k ] T ∈ C × , ¯ U k = [ u k ] ∈ C , for k = 1 , , , and ¯ H (2) kj = [¯ h (2) , kj ¯ h (2) , kj ] T ∈ C × , ¯ H (3) kj = [¯ h (3) kj ] ∈ C , for k = j . If we set ¯ H (4) kj = 0 for all channels, one can write the Jacobian of [ F F F F F F ] with respect to the variables [ u u u v v v v v v ] as − ¯ h (3)12 − ¯ h (3)13 − ¯ h (3)21 − ¯ h (3)23 − ¯ h (3)31 − ¯ h (3)32 − ¯ h (2) , − ¯ h (2) ,
00 0 − ¯ h (2) , − ¯ h (2) , − ¯ h (2) , − ¯ h (2) , − ¯ h (2) , − ¯ h (2) , − ¯ h (2) , − ¯ h (2) , − ¯ h (2) , − ¯ h (2) , . One can easily observe that by removing the variables { v , v , v } and setting ¯ h (3)12 = ¯ h (3)23 = ¯ h (3)31 = ¯ h (2) , = ¯ h (2) , = ¯ h (2) , = 1 , ¯ h (3)13 = ¯ h (3)21 = ¯ h (3)32 = ¯ h (2) , = ¯ h (2) , = ¯ h (2) , = 0 , October 22, 2018 DRAFT7 the Jacobian of the mapping (18) with respect to the remaining variables becomes − − − − − − , which is clearly nonsingular since there exists exactly one nonzero element in each column/row.Next we argue that the above construction procedure can be generalized to the case where M k and N k are divisible by d , provided that the assumption (17) is satisfied. The construction of thesechannel/beamforming matrices and the removal of redundant variables are outlined below. First, we set H (4) kj = , for all k = j . Then we choose { ¯ U k , ¯ V j } k,j arbitrarily. It remains to specify { ¯ H (3) kj , ¯ H (2) kj } k = j .We should do so to ensure that the corresponding Jacobian matrix of (18) at { ¯ U k , ¯ V j } k,j is nonsingular.Since M k and N k are divisible by d , we can partition our variables into blocks of size d × d and rewritethe mapping (18) as F kj ( ¯ U , ¯ V ) = − h ¯ U Hk ¯ U Hk . . . ¯ U Hk sk i ¯ H (3) , kj ¯ H (3) , kj ... ¯ H (3) ,s k kj − h ¯ H (2) , kj ¯ H (2) , kj . . . ¯ H (2) ,t j kj i ¯ V j ¯ V j ... ¯ V j tj , ∀ k = j, (19)where s k = M k d − , t j = N j d − , and ¯ U k i , ¯ V j ℓ , ¯ H (2) ,ikj , ¯ H (3) ,ℓkj ∈ C d × d . Consider a bipartite graph G where the vertices are partitioned into two sets X and Y . Each block of variables will correspond to anode in X , while each matrix equation in (19) will correspond to a node in Y . We draw an edge betweena node x ∈ X and a node y ∈ Y if the block of variables corresponding to node x appears in theequation corresponding to node y . When viewed on the bipartite graph G , the assumption (17) simplysays that for any given set of nodes S ⊆ Y , the cardinality of the neighbors of S in X is no smaller thanthe cardinality of S . This condition is precisely what is required to ensure the existence of a completematching in G covering all nodes in Y (Hall’s theorem, see [15, Theorem 3.1.11]). Now consider a fixedcomplete matching in G . Let A ⊆ X be the set of vertices that are not matched to a node in Y . Then,we can set to zero all the blocks of the variables corresponding to the vertices in A , i.e., we can removethem from our equations. Now we choose the rest of the channel matrices so that the determinant of the October 22, 2018 DRAFT8
Jacobian with respect to the remaining variables is nonzero. To this end, we set ¯ H (3) ,pkj = 0 if the node for ¯ U k p is not matched to the node in Y corresponding to the equation F kj . Similarly, we set ¯ H (2) ,qkj = 0 if ¯ V j q is not matched to F kj . Moreover, we set all the remaining channel sub-matrices to the d × d identitymatrix. Since this construction is based on a complete matching, it is not hard to see that the Jacobianfor the whole system is a block permutation matrix, with nonzero blocks equal to the negative d × d identity matrix. Hence the determinant of the Jacobian matrix is equal to the product of the determinantof all nonzero blocks (up to sign), which is clearly nonzero in our case. This completes the descriptionof the procedure to remove potential redundant variables, as well as the procedure to construct all thechannel matrices { H (2) kj , H (3) kj , H (4) kj } k = j and the beamforming solution { ¯ U k , ¯ V j } k,j . The Jacobian matrixof (18) is nonsingular at this constructed channel realization and beamforming solution. Figure 1 illustratesthe construction of graph G and a complete matching (in solid lines) for the aforementioned toy example.Fig. 1: The bipartite graph G and a complete matching for the toy exampleTo complete the proof, we fix a generic choice of { ¯ H (2) kj , ¯ H (3) kj , ¯ H (4) kj } k = j for which the Jacobian of (18)is nonsingular. Let n be the total number of remaining scalar variables in { ¯ U k , ¯ V j } k,j after removing theredundant variables. Notice that n is the same as the number of scalar equations, i.e., n = d K ( K − . Let R = C [ h , h , . . . , h n ] and R = C [¯ u , ¯ u , . . . , ¯ u m , ¯ v , . . . , ¯ v n − m ] be two polynomial rings where ¯ u i ’sand ¯ v j ’s are the entries of the matrices { ¯ U k } Kk =1 and { ¯ V k } Kk =1 (after removing the redundant variables), October 22, 2018 DRAFT9 and h , h , . . . , h n are the entries of the matrices { ¯ H (1) kj } k = j . Consider { f i } ni =1 (the components of F kj ’sin (18)) as the functions of ¯ u ’s and ¯ v ’s, i.e., f i ’s are polynomials in R . These polynomials define amap a φ which maps a point c = ( c , c , . . . , c n ) to ( f ( c ) , f ( c ) , . . . , f n ( c )) . According to the ChevalleyTheorem (see [14, Chapter 2, 6.E.]), the image of this map is a Zariski constructible subset of A n C , , where A n C , is the corresponding affine space of R . Since the Jacobian of the set { f , f , . . . , f n } with respectto the variables { ¯ u , ¯ u , . . . , ¯ u m , ¯ v , . . . , ¯ v n − m } is nonsingular generically for all channel realizations, itfollows from the Implicit Function Theorem that the dimension of the image of a φ is n . Note that theimage of a φ is a Zariski constructible subset of A n C , (see Chevalley Theorem [14], Ch. 2, 6.E.) andit has full dimension. Hence, the image contains a Zariski open subset of A n C , (see the discussion insection III-A). Let U be that Zariski open subset of A n C , in the image. Since U is in the image of themap a φ , there exists a solution for interference alignment equations for any choice of { ¯ H (1) kj } k = j in U ,which implies that interference alignment is feasible for generic choice of { ¯ H (1) kj } k = j . Since the choiceof channel matrices { ¯ H (2) kj , ¯ H (3) kj , ¯ H (4) kj } k = j is also generic, this completes the proof of the “if” direction.Notice that the condition (17) is equivalent to the properness of the polynomial system (12) defininginterference alignment. For symmetric systems with M k = M , N k = N for all k , this condition simplifiesto M + N ≥ d ( K + 1) (see [2, Theorem 1]). Thus, each user can achieve d degrees of freedom as long as M + N ≥ d ( K + 1) and that d divides both M and N . In a concurrent work, the authors of [8] obtaineda similar result under a different set of assumptions. More specifically, they considered the symmetriccase with M k = N k = M, d k = d for all k , and proved that the feasibility of interference alignment inthis case is equivalent to M ≥ d ( K + 1) . This result and Theorem 2 are complementary to each other.In particular, Theorem 2 is applicable to non-symmetric systems, but does require an extra conditionabout the divisibility of the number of antennas by the number of data streams. When K is odd and ( K + 1) d = 2 M , then M must be divisible by d . This case is then covered by both Theorem 2 andthe result in [8]. However, for the case where K is even and ( K + 1) d ≤ M , Theorem 2 is no longerapplicable, whereas [8] shows that the interference alignment is achievable.A few other remarks are in order.1) Reference [2] also considered the case d k = 1 and used the Bernshtein’s theorem to numericallycompute the number of solutions, and therefore prove the feasibility, for the resulting polynomialsystem (2)-(3) when the number of antennas are small. In contrast, Theorem 2 shows the feasibilityof single beam interference alignment for all values of M k , N k as long as the system is proper. October 22, 2018 DRAFT0
2) As shown in Theorem 1, the condition (5) is necessary. For example, the system K = 2 , M = N = 3 , d = 2 satisfies the inequality (6). However, the system of equations (2)-(3) is infeasible forgeneric choice of channel coefficients. This further shows that the properness property in [2] doesnot imply feasibility in general, a fact that was first pointed out in [2, example 17].3) Theorem 2 does not contradict the NP-hardness result of [12]. Given a set of channel matrices,checking the feasibility of the interference alignment conditions (2)-(3) when M k ≥ and N k ≥ ,is NP-hard. It is true that, under the setting of Theorem 2, the interference alignment fails only fora measure zero set of channels. However, for systems not satisfying the conditions of Theorem 2,checking the feasibility of interference alignment can be hard. Moreover, the results of [12] implythat, even if a given tuple of DoF is known to be achievable via interference alignment, finding theactual linear transmit/receive beamformers to achieve it is still a NP-hard problem when the numberof users is large.4) The condition (17) implies the condition (5) if the number of antennas at each transceiver is divisibleby d . In fact, by choosing I = { ( k, j ) } , condition (17) implies that d ≤ M k + N j − d and hencethe condition (5) is satisfied.5) Theorem 2 assumes that both M k and N k are divisible by d . This condition can be weakened for asymmetric system where M k = M, N k = N, d k = d , for all k . In particular, assume that only M (not N ) is divisible by d and M, N ≥ d . If the properness condition ( K + 1) d ≤ M + N holds,then we can construct a reduced MIMO interference channel with N ′ = M − d ( K + 1) receiveantennas for each user, where M + N ′ = d ( K + 1) and M, N ′ are divisible by d . By Theorem 2, theinterference alignment condition for the reduced interference channel is feasible and therefore, so isthe interference alignment condition for the original channel since the latter has more antennas. Thisshows that if M is divisible by d and M, N ≥ d , then the interference alignment system (2)–(3) isfeasible for generic choice of channel coefficients if and only if ( K + 1) d ≤ M + N . By symmetry,the same conclusion holds for the case where N is divisible by d .IV. S IMULATION R ESULTS
In this section, we use the theoretical DoF upper bounds to benchmark an existing algorithm for sum-rate maximization. We generate MIMO interference channels using the standard Rayleigh fading model.The numerical experiments are averaged over 100 Monte Carlo runs.We consider a MIMO interference channel where each transmitter/receiver is equipped with antennas.For different number of users in the system, we maximize the sum-rate using the WMMSE algorithm [13] October 22, 2018 DRAFT1 at increasingly high SNRs. We estimate the slope of the sum-rate versus SNR and use it to approximatethe achievable total DoF. We then compare it with the value of theoretical upper bound given by theconditions in Theorem 1. The maximum gap of the two curves is one, but it is not clear if the gap isdue to the weakness of the WMMSE algorithm or the DoF upper bound. T o t a l D eg r ee s o f F r eedo m Estimated DoF using WMMSE AlgorithmTheoretical Upper Bound
Fig. 2: Achievable DoF and theoretical upper bound
Acknowledgement:
We are grateful to the authors of [8] for sharing their concurrent work.R
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