2 CASAZZA AND KOVACEVIC source to a recipient. These packets are sequences of information bits of a certain length surrounded by error-control, addres

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1 Uniform Tight Frames with Erasures Peter G. Casazza Department of Mathematics, University of Missouri-Columbia Columbia, MO and Jelena Kovacevic Mathematics of Communication Research, Bell Labs, Lucent Technologies, 600 Mountain Avenue, Murray Hill, NJ USA Uniform tight frames have beenshown to be useful for robust data transmission. The losses in the network are modeled as erasures of transmitted frame coecients. We give the rst systematic study of the general class of uniform tight frames and their properties. We search for ecient constructions of such frames. We show that the only uniform tight frames with the group structure and one or two generators are the generalized harmonic frames. Finally, we give a complete classication of frames in terms of their robustness to erasures. 1. INTRODUCTION Frames are redundant sets of vectors in a Hilbert space which yield one natural representation for each vector in the space, but which may have innitely many dierent representations for a given vector [4, 5,6,7,8,9,10,11,16,18,20,23]. Frames have been used in signal processing because of their resilience to additive noise [10], resilience to quantization [14], as well as their numerical stability of reconstruction [10], and greater freedom to capture signal characteristics [2, 3]. Recently, several new applications for (uniform tight) frames have been developed. The rst, developed by Goyal, Kovacevic and Vetterli [15, 25, 24, 26], uses the redundancy of a frame to mitigate the eect of losses in packet-based communication systems. Modern communication networks transport packets of data from a Some of the results in this paper were reported at the SPIE Conf. on Wavelet Appl. in Signal and Image Proc. (San Diego, CA, July 2001). The rst author was supported by NSFDMS

2 2 CASAZZA AND KOVACEVIC source to a recipient. These packets are sequences of information bits of a certain length surrounded by error-control, addressing, and timing information that assure that the packet is delivered without errors. This is accomplished by not delivering the packet if it contains errors. Failures here are due primarily to buer overows at intermediate nodes in the network. So to most users, the behavior of a packet network is not characterized by random loss, but by unpredictable transport time. This is due to a protocol, invisible to the user, that retransmits lost packets. Retransmission of packets takes much longer than the original transmission. In many applications, retransmission of lost packets is not feasible and the potential for large delay isunacceptable. If a lost packet is independentofthe other transmitted data, then the information is truly lost to the receiver. However, if there are dependencies between transmitted packets, one could have partial or complete recovery despite losses. This leads us naturally to use frames for encoding. The question, however, is: What are the best frames for this purpose? With an additive noise model for quantization, in [25], the authors show that a uniform frame minimizes mean-squared error if and only if it is tight. So it is this class of frames - the uniform normalized tight frames - which we seek to identify and study. Another recent important application of uniform normalized tight frames is in multiple-antenna code design [17]. Much theoretical work has been done to show that communication systems which employ multiple antennas can have very high channel capacities [13, 21]. These methods rely on the assumption that the receiver knows the complex-valued Rayleigh fading coecients. To removethis assumption, in [19] new classes of unitary space-time signals are proposed. If we have N transmitting antennas and we transmit in blocks of M time samples (over which the fading coecients are approximately constant), then a constellation of K unitary space-time signals is a(weighted by p M) collection of M N complex matrices f k g for which k k = I. The nth column of any k contains the signal transmitted on antenna n as a function of time. The only structure required in general is the time-orthogonality of the signals. Originally it was believed that designing such constellations was a too cumbersome and dicult optimization problem for practice. However, in[19], it was shown that constellations arising in a \systematic" fashion can be done with relatively little eort. Systematic here means that we need to design high-rate space-time constellations with low encoding and decoding complexity. It is known that full transmitter diversity (that is, where the constellation is a set of unitary matrices whose dierences have nonzero determinant) is a desirable property for good performance. In a tour-de-force, in [17], the authors used xed-point-free groups and their representations to design high-rate constellations with full diversity. Moreover, they classied all full-diversity constellations that form a group, for all rates and numbers of transmitting antennas. For these applications, and a host of other applications in signal processing, it has become importantthatwe understand the class of uniform normalized tight frames. In this paper, we make the rst systematic study of such frames. In Section 2 we review basic notions on frames. In particular, we state the Naimark's Theorem [1] which has been rediscovered several times in recent years [16, 12] although it has been used for several decades in operator theory. We give examples of uniform

3 UNIFORM TIGHT FRAMES WITH ERASURES 3 normalized tight frames such asharmonic and Gabor frames. For harmonic frames, we dene a more general class { general harmonic frames (GHF) { and study when harmonic frames are equivalent to each other and general harmonic frames. In Proposition 2.4, we giveasimple equivalence condition on general harmonic frames which states that the inner product between any two frame vectors ' i and ' j in a GHF is equal to the inner product between ' i+1 and ' j+1. Section 3 concentrates on uniform tight frames (UTF). We start with a review in Section 3.1 and proceed with several ways of classifying UTFs including providing a correspondence between subspaces of the original space and the UTFs, obtaining UTFs as alternate dual frames for a given frame as well as nding UTFs through frames equivalenttothem. In Section 4, we shift our attention to UTFs with group structure. We show that the UNTFs generated by the family fu k ' 0 g (where U is unitary and ' 0 2 H ) are precisely the GHFs. We then extend our discussion to UNTFs generated by fu k V j ' 0 g and higher numbers of generators. Finally, Section 5 introduces erasures modeled as losses of transform coecients hf ' i i, where f is the signal to be transmitted and f' i g i2i is a set of frames vectors corresponding to erased transform coecients. We give a complete classication of frames with respect to their robustness to erasures. We study when we can obtain frames robust to a certain number of erasures as a projection from another frame with a dierent number oferasures. 2. FRAME REVIEW Asetof vectors = f' i g i2i in a Hilbert space H, is called a frame if 0 < Akxk 2 i2i jhx ' i ij 2 < Bkxk 2 < +1 x 6= 0 (1) where I is the index set and the constants A B are called frame bounds. Although many of our results hold in more general settings, in this paper, we concentrate mostly on the N-dimensional real or complex Hilbert spaces R N and C N (which we denote H N )withthe usual Euclidean inner product. When results generalize to the innite-dimensional setting, we willpoint itout. When A = B the frame is tight (TF). If A = B =1,theframe is normalized tight (NTF). A frame is uniform (UF) if all its elements havethesame norm c, 1 k' i k = c. For a uniform tight frame (UTF), the frame bound A gives the redundancy ratio. A UNTF of norm-1 vectors is an orthonormal basis (ONB). The analysis frame operator 2 F maps the Hilbert space H into `2(I) (Fx) i = h' i xi (2) 1 Often, uniform is used to mean c =1 here, however, we use the more general denition. 2 F is sometimes called just frame operator. Here, weuseanalysis frame operator for F, synthesis frame operator for F and frame operator for F F.

4 4 CASAZZA AND KOVACEVIC for i 2 I. When H = H N, the analysis frame operator is an M N matrix whose rows are the transposed frame vectors ' i : F = 0 ' 11 ::: ' 1N. :::. ' M1 ::: ' MN 1 C A : (3) We saythattwo frames f' i g i2i and f i g i2i for H are equivalent if there is an invertible operator L on H for which L' i = i for all i 2 I, and they are unitarily equivalent if L can be chosen to be a unitary operator. Adirect calculation shows that fs ;1=2 ' i g is a normalized tight framefor any framef' i g. Thus, every frame is equivalent to a normalized tight frame. If f' i g i2i is a frame with frame bounds A B, andifp is an orthogonal projection on H,thenby(1) we have that fp' i g i2i is a frame for P H with frame bounds A and B. In particular, if f' i g i2i is a normalized tight frame for H (for example, if it is an orthonormal basis for H ), then fp' i g i2i is a normalized tight framefor P H. The following theorem tells us that every normalized tight frame can be realized as a projection of an orthonormal basis from a larger space. It serves as a converse to the observation above that orthogonal projections of normalized tight frames produce normalized tight frames for their span. Theorem 2.1 (Naimark [1], Han &Larson [16]). 3 Afamily f' i g i2i in a Hilbert space H is a normalized tight frame for H if and only if there is a larger Hilbert space H K and an orthonormal basis fe i g i2i for K so that the orthogonal projection P of K onto H satises: Pe i = ' i, for all i 2 I. Let us nowgothrough certain important frame notions. Using the analysis frame operator F,(1) can be rewritten as AI F F BI: (4) We calls = F F the frame operator. It follows that S is invertible (Lemma in [10]), and furthermore B ;1 I S ;1 A ;1 I: (5) It follows that f' i g i2i is a normalized tight frameifand only if F is an isometry. Also, S is a positive self-adjoint invertible operator on H and S = AI if and only if the frame is tight. In nite dimensions, the canonical dual frame of is a frame dened as ~ =f ~' i g M = fs ;1 ' i g M, where ~' i = S ;1 ' i (6) 3 This theorem has been rediscovered by several people in recent years: The rst author rst heard it from I. Daubechies in the mid-90's. Han and Larson rediscovered it in [16] they came up with the idea that a frame could be obtained by compressing a basis in a larger space and that the process is reversible (the statement in this paper is due to Han and Larson). Finally, it was pointed out to the rst author by E.Soljanin [22] that this is, in fact, the Naimark's theorem, which has been widely known in operator theory and has been used in quantum theory.

5 UNIFORM TIGHT FRAMES WITH ERASURES 5 for k =1 ::: M. Now, f = i2i hf S ;1 ' i i' i for all f 2 H: So the canonical dual frame can be used to reconstruct the elements of H from the frame. However, there may beother sequences in H which givereconstruction. This formula points out both the strengths and weaknesses of frames. First, we see that every element f 2 H has at least one natural series representation in terms of the frame elements. Also, this element may have innitely many other representations. However, in order to nd this natural representation of f, we need to invert the frame operator, which may be dicult or even impossible in practice. The best frames then are clearly the tight frames since in this case the frame operator becomes a multiple of the identity. Noting that ~' i = ' i S;1 and stacking ~' 1, ~' 2, :::, ~' M in a matrix, the frame operator associated with ~ is ~F = FS ;1 : (7) Since F ~ F ~ = S ;1, (5) shows that B ;1 and A ;1 are frame bounds for ~. Another important concept is that of a pseudo-inverse F y. It is the analysis frame operator associated with the dual frame, F y = F ~ : (8) Note that for any matrix F with rows ' i S = F F = M ' i ' i : (9) This identity will prove to be useful in many proofs The Role of Eigenvalues The product S = F F will appear everywhere and its eigenstructure will play an important role. Denote by k 's the eigenvalues of S = F F. We nowsummarize the important eigenvalue properties. General Frame. For any frameinh N,the sum of the eigenvalues of S = F F, equals the sum of the lengths of the frame vectors: N k=1 k = M k' i k 2 : (10) Uniform Frame. For a uniform frame, that is, when k' i k = c, i =1 ::: M, N k = M k=1 k' i k 2 = M c 2 : (11)

6 6 CASAZZA AND KOVACEVIC Tight Frame. Since tightness means A = B, for a TF, we have from (1) M jhf ' i ij 2 = Akfk 2 (12) for all f 2 H N. Moreover, according to (5), a frame is a TF if and only if F F = A I N : (13) Thus, for a TF, all the eigenvalues of the frame operator S = F F are equal to A. Then, using (10), the sum of the eigenvalues of S = F F is as follows: N A = N k=1 k = M k' i k 2 : (14) Normalized Tight Frame. If a frame is an NTF, that is, A = B =1,then M jhf ' i ij 2 = kfk 2 (15) for all f 2 H N. In operator notation, a frame is an NTF if and only if S = F F = I N : (16) For an NTF, all the eigenvalues of S = F F are equal to 1. Then, using (10), the sum of the eigenvalues of S = F F is as follows: Uniform Tight Frame. N A = N = N k=1 k = M From (11) and (14), we see that N k=1 k = M k' i k 2 : (17) k' i k 2 = M c 2 : (18) Then, from (12) and (18), M jhf ' i ij 2 = M N c2 kfk 2 (19) for all f 2 H N. The redundancy ratio is then A = M N c2 : (20) Since S = F F =(M=N)I, thefollowing is obvious: M j' ik j 2 = M N : (21)

7 UNIFORM TIGHT FRAMES WITH ERASURES 7 Uniform Normalized Tight Frame. If a frame is a UNTF, that is, we alsoask for A = B =1,and if the frame vectors have norm c, then N = N k=1 k = M k' i k 2 = c 2 M: Thus, a UNTF with norm c =1is an orthonormal basis Examples of Uniform Normalized Tight Frames We start with a simple example of a frame 3 vectors in 2 dimension. The particular frame we examine is termed Mercedes-Benz (MB) frame (for obvious reasons, just draw thevectors). The UNTF version of it is given by the analysis frame operator r F 0 1 ; p 3=2 ;1=2A 3 p : (22) 3=2 ;1=2 This is obviously a UNTF since (2=3)F F = I. We couldmake this frame just a UTF with norm 1 (as in [25]) by having the analysis frame operator be simply F. We now review two general classes of uniform tight frames which are commonly used. The (general) harmonic frames and the tight Gabor frames. More general classes of uniform tightframes are the full-diversityconstellations that form a group given in [17] Harmonic Frames Harmonic tight frames (HTF) are obtained by keeping the rst N coordinates of an M M discrete Fourier transform basis as in Naimark's Theorem 2.1. They have been proven to be useful in applications [19]. An HTF is given by: ' k = (w k 1 wk 2 wk N) (23) for k = 0 ::: M ; 1, where w i are distinct Mth roots of unity. This is a TF { F F = MI. A more general denition of the harmonic frame (general harmonic frame) is as follows: Definition 2.1. Fix M N, jcj =1and fb i g N with jb i j = 1 p M. Let fc i g N be distinct Mth roots of c, andfor 0 k M ; 1, let ' k = (c k 1b 1 c k 2b 2 ::: c k Nb N ): Then f' k g is a uniform normalized tight frame for H N called a general harmonic frame (GHF). Note the dierence between the HTFs and GHFs the former ones are only tight, while the latter ones are uniform, normalized and tight. We will now examine the general harmonic frames in detail. Our goal is to determine when two such frames are unitarily equivalent. We start byshowing that the harmonic frames (not GHFs) are unique up to a permutation of the orthonormal basis (that is, if and only if their columns are permutations of each other).

8 8 CASAZZA AND KOVACEVIC Proposition 2.1. Let f' k g and f k g be harmonic frames on H N. Then f' k g is equivalent to f kg if and only if there isapermutation of f1 2 ::: Ng so that ' kj = k(j), for all 0 k M ;1 and all 1 j N. Hence, two harmonic frames are equivalent if and only if they are unitarily equivalent. Proof and and (Proposition 2.1). Let fe j g N j=1 be the natural orthonormal basis for H N ' k = (w k 1 w k 2 w k N) = k = (v k 1 vk 2 vk N) = N j=1 N j=1 where w j v j are sets of distinct Mth roots of unity. If w k j e j v k j e j fw j j1 j Ng 6= fv j j1 j Ng then, without loss of generality, we may assume that v 1 Therefore, On the other hand, v k 1 ' k = N j=1 v k 1 k (1) = (v 1 w j ) k! e j = N j=1 jv k 1 j 2 = M: =2 fw j j1 j Ng. 0 e j = 0: It follows that f' k g is not equivalenttof kg. The other direction is immediate. We now show that every general harmonic frame is unitarily equivalent to a simple variation of a harmonic frame. Proposition 2.2. Every general harmonic frame is unitarily equivalent to a frame of the form fc k k g,where jcj =1and f kg is a harmonic frame. Proof (Proposition 2.2). Let M N, jcj =1andjb k j =1= p M,forall1k N. Let fc j g N j=1 be distinct Mth roots of c and let the GHF be ' k = (c k 1 b 1 c k 2 b 2 ::: c k N b N) for all 0 k M ; 1. If c = e i, let d = e i=m. Then there exist distinct Mth roots of unity w 1 w 2 ::: w N with c j = e i=m w j = dw j. For all sets of complex numbers fa k g wehave: k a k ' k k 2 = N j=1 j a k c k j b j j 2 = 1 M N j=1 j a k c k j j 2 =

9 UNIFORM TIGHT FRAMES WITH ERASURES 9 1 M N j=1 j a k e ik=m wj k j2 = 1 M k a k d k kk 2 where k =(w k 1 wk 2 ::: wk N ). Thus, the GHF f' kg is unitarily equivalent to p1 M f k g,with f kg an HTF. Finally, we show that the frames given in Proposition 2.2 are not unitarily equivalent to each other or to harmonic frames except in the trivial case. Proposition 2.3. Let f' k g and f kg beharmonic frames and let jcj = 1. Then fc k ' k g is equivalent to f k g if and only if c is an Mth root of unity and there is a permutation of f1 2 ::: Ng so that ' kj = k(j), for all 0 k M ; 1 and all 1 j N. In particular, a general harmonic frame is equivalent to a harmonic tight frame if and only if it equals a harmonic tight frame. P N Proof (Proposition 2.3). Let ' k = j=1 wk j e j, for all 0 k M ;1. If c = w j ;1, for some j, we are P P done. Otherwise, since k =0,iffc k ' k g is equivalent to f k g then also ck ' k =0. Hence, for all 1 j N we have Hence, (cw j ) M = c M w M j = c M =1. The proposition now follows from Proposition 2.1. c k w k j = We now have immediately, (cw j ) k = 1 ; (cw j) M 1 ; cw j = 0: Corollary 2.1. Let f' k g and f k g be harmonic frames and let jcj = jdj =1. The frames fc i ' k g and fd i k g are equivalent if and only if c = d and there isapermutation of f1 2 ::: Ng so that ' kj = k(j), for all 0 k M ; 1. The next proposition gives a classication of GHFs. Proposition 2.4. A family f' i g in H N is a GHF if and only if for all 0 i j M ; 1 we have where ' M = ' 0. h' i ' j i = h' i+1 ' j+1 i Proof. If f' i g in H N is a generalized harmonic frame then there is a unitary operator U on H N so that U i ' 0 = ' i, for all 0 i M ; 1. Hence, for all 0 i j M ; 1wehave h' i+1 ' j+1 i = hu i+1 ' 0 U j+1 ' 0 i = hu i ' 0 U j ' 0 i = h' i ' j i:

10 10 CASAZZA AND KOVACEVIC Conversely, given our equality, for any sequence of scalars fa i g, k a i ' i+1 k 2 = h a i ' i+1 a i ' i+1 i = i j=0 = h a i a j h' i+1 ' j+1 i = a i ' i i j=0 a i ' i i = k a i a j h' i ' j i a i ' i k 2 : It follows that U' i = ' i+1 is a unitary operator and U i ' 0 = ' i, for all 0 i M ; 1. As we will see in the next section, UTFs with M = N +1 are all unitarily equivalent. Thus, as a direct consequence of Theorem 3.3, any UTF with M = N +1 is unitarily equivalent to the HTF with M = N + 1. This is a very useful result since we have HTFs for any N and M thus, for M = N +1, wealways have an expression for all UTFs. For example, this means that the MB frame we introduced earlier is equivalent tothehtf with M =3andN =2. Another interesting property of an HTF is that it is the only UNTF such that its elements are generated by a group of unitary operators with one generator, as we will see in Section 4. That is, = f' i g M = fu i ' 0 g M, where U is a unitary operator. Moreover, HTFs have a very convenient property when it comes to erasures. We can erase any e (M ; N) elements from the original frame and what is left is still a frame(theorem 4.2 from [25]). We provide a complete classication of frames in terms of their robustness to erasures in Section Gabor or Weyl-Heisenberg Frames For the other general class of frames, weintroduce twospecial operators on L 2 (R). Fix 0 <a band for f 2 L 2 (R) dene translation by a as and modulation by b as T a f(t) = f(t ; a) E b f(t) = e 2imbt f(t): Now, x g 2 L 2 (R). If fe mb T na gg m n2z is a frame for L 2 (R), we call it a Gabor frame (or a Weyl-Heisenberg frame). It is clear that in this class the frames are uniform. Also, since the frame operator S for a Gabor frame fe mb T na gg m n2z must commute with translation and modulation, each Gabor frame is equivalent to the (uniform) normalized tight Gabor frame fe mb T na S ;1=2 gg m n2z. For an introduction to Gabor frames we refer the reader to [7] and [18].

11 UNIFORM TIGHT FRAMES WITH ERASURES UNIFORM TIGHT FRAMES 3.1. What is Known about UTFs? As wehave mentioned in the introduction, UNTFs havebecomepopular in applications. In particular, in [25], the authors attack the problem of robust transmission over the Internet by using frames. Being redundant sets of vectors, they provide robustness to losses, which are modeled as erasures of certain frame coecients. It is further shown in [25] that, assuming a particular quantization model, a uniform frame with quantized coecients, minimizes mean-squared error if and only if it is tight. Moreover, the same is true when we consider one erasure and look at both the average- and worst-case MSE. As a result, our aim is to construct useful families of frames for such applications. To that end, it is important to classify uniform normalized tight frames in order to facilitate the search for useful families. Our notion of usefulness includes any families which would be computationally ecient for example, those that can be obtained from one or more generating vectors or those with a simple structure, for example, such that any of the frame operators could be expressed as a product of sparse matrices. Finally, we are interested in the robustness of our frames to coecient (frame vector) erasures. These and other issues will be explored in the rest of the paper Construction of UTFs There is a general well known method for getting nite UNTFs. We state this result here in its standard form and will give anew proof of the result in Section 5. In Section 5 we will also extend this result to uniform frames. Theorem 3.1. There is a unique way to get UNTFs with M elements in H N. Take any orthonormal set f k g N k=1 in H M which has the property N k=1 j ki j 2 = c for all i: Thinking of the k as row vectors, switch to the M column vectors and divide by p c. This family is a uniform tight frame for H M with M elements, and all UNTFs for H N with M elements are obtained in this way. Corollary 3.1. If f' i g M is a uniform normalized tight frame for H N then k' i k 2 = M N : There is a detailed discussion concerning the uniform normalized tight frames for R 2 in [25]. In [17], there is a deep classication of groups of unitary operators which generate uniform normalized tight frames. The simplest case of this is the harmonic frames. In Section 4 we do not assume that we have a\group" of unitaries, but instead conclude that our family of unitaries must be a group. The picture becomes much more complicated if the uniform normalized tight frame is generated by a group of unitaries with more than one generator (see [17])

12 12 CASAZZA AND KOVACEVIC or worse, if the uniform tight frame comes from a subset of the elements of such a group Classication of UNTFs Although we would like to classify all uniform tight frames, especially those which can be obtained byreasonable algorithms, this is an impossible task in general. The following theorem shows that every nite family of norm-1 vectors in a Hilbert space can be extended to become a uniform tight frame. Theorem 3.2. If f' i g M is a family of norm-1 vectors in a Hilbert space H, then there isauniform tight frame for H which contains the family f' i g M. Proof (Theorem 3.2). For each 1 i M choose an orthonormal basis fe ij g j2j for H which contains the vector ' i. Now the family fe ij g M j2j is made up of norm-1 vectors and for any f 2 H we have M jhf e ij ij 2 = M j2j kfk 2 = Mkfk 2 : In the above construction we get as a tight frame bound the number of elements M in the family f' i g M. In general, this is the best possible. For example, just let ' i = ' j be norm-1 vectors for all 1 i j M. Then M j=1 jh' i ' j ij 2 = M: Hence, any tight frame containing the family f' i g has the tight frame bound at least M. There is another general class of uniform tightframes (see [25]). The result below tells us that all UNTFs with M = N +1 are unitarily equivalent. It is a direct consequence of Theorem 2.6 from [25] where it is stated for UTFs with norm 1. Theorem 3.3 (Goyal, Kovacevic and Kelner [25]). Afamily f' i g N+1 is a uniform normalized tight frame for H N if and only if f' i g N+1 is unitarily equivalent to the frame fpe i g N+1 where fe i g N+1 is an orthonormal basis for H N+1 and P is the orthogonal projection of H N+1 onto P the orthogonal complement of the onedimensional subspace ofhn+1 spanned by e N+1 i. Since there are HTFs with N + 1-elements in H N,it follows that every UNTF for H N with N + 1-elements is unitarily equivalent to a HTF Normalized and Uniform Tight Frames and Subspaces of the Hilbert Space Here, we begin to classify NTFs and UNTFs by providing a correspondence with subspaces of the original Hilbert space. We give two results, one for NTFs and another for UNTFs with the correspondence being one-to-one. The material in this section grew out of conversations between the rst author and V. Paulsen.

13 UNIFORM TIGHT FRAMES WITH ERASURES 13 As we sawintheorem 2.1, there is a unique way togetnormalized tight frames in H N with M elements. Namely, we take anorthonormal basis fe i g M for H M and take the orthogonal projection PH N of H M onto H N. Then fph N e i g M is a normalized tight frame for H N with M elements. In particular, there is a natural one-to-one correspondence between the normalized tight frames for H N with M elements and the orthonormal bases for H M. Then, the uniform normalized tight frames for H N are the ones for which kpe i k = kpe j k, for all 1 i j M. We use this to exhibit a natural correspondence between these families and certain subspaces of H M. Here we treat two frames as the same if they are unitarily equivalent. Theorem 3.4. Let P be arank-n orthogonal projection on H M and let fe i g M be an orthonormal basis for H M. There is a natural one-to-one correspondence between the normalized tight frames for P H M with M elements and the family of all N-dimensional subspaces of H M. Proof (Theorem 3.4). If f' i g M is anynormalized tight frame for P H M,thenby Naimark's Theorem, there is an orthonormal basis fe 0 i gm for H M so that Pe 0 i = ' i. Dene a unitary operator U on H M by Ue i = e 0 i. Now, ' i = PUe i which is unitarily equivalent to' 0 i = U PUe i. So we will associate our normalized tight frame f' i g with the subspace U PUH M. Now we need to check that this correspondence is one-to-one. Let f i g be another normalized tight frame for P H M which is associated with the same subspace of H M,namely U PUH M. Then there is an orthonormal basis fe 00 i g and a unitary operator Ve i = e 00 i with V PV = U PU and PVe i = e 00 i. Hence, V PVe i = V i = U PUe i = U ' i. This implies that the normalized tight frames f' i g and f i g are unitarily equivalent (andhence the same). Finally, we need to see that this correspondence covers all subspaces. If W is any subspace of H M of dimension N, we can dene a unitary operator U on H M so that UW = P H M. Then U PU = P W while fpue i g M is a normalized tight frame for P H M (which underour association corresponds to W ). One of the consequences of the above result is that if we have one normalized tight frame for H N,then all the others can be obtained from it by thisprocess. We now turn our attention to uniform frames: Theorem 3.5. Fix an orthonormal basis fe i g M for H M so that if P is the orthogonal projection of H M onto H N then fpe i g M is a uniform normalized tight frame for H N. Then there is a natural one-to-one correspondence between the uniform normalized tight frames for H N with M elements and the subspaces W of H M for which kp W e i k 2 = M=N, for all 1 i M. Proof (Theorem 3.5). We already have our classication of the normalized tight frames in terms of all subspaces. Now we need to see which of these subspaces correspond to the uniform normalized tight frames. Suppose U is any unitary operator on H so that fpue i g is a uniform normalized tight frame for P H M. Then this frame is associated to the subspace W = U P H M and P W = U PU. Hence, PUe i = UP W e i, for all i. But, fpue i g is a uniform normalized tight frame for

14 14 CASAZZA AND KOVACEVIC P H M and so kpue i k = N=M, for all i. Hence, kpue i k = N M = kup W e i k = kp W e i k: So our association between normalized tight frames and subspaces given in Theorem 3.4 identies the subspaces W of H M for which kp W e i k = N=M, for all i Uniform Dual Frames Another method for classifying UTFs uses a back door: get UTFs as alternate dual frames for a given frame. To dothis,we need the denition of alternate dual frames: Definition 3.1. Let f' i g i2i be a frame for a Hilbert space H. Afamilyf i g i2i is called an alternate dual frame for f' i g i2i if f = i2i hf i i' i for all f 2 H : In Denition 3.1, if we letf 1 (respectively, F 2 )bethe analysis frame operators for f' i g (respectively f i g), then we seethatf 2 F 1 = I. There are many alternate dual frames for a given frame. In fact, a frame has a unique alternate dual frame (the canonical dual frame) if and only if it is a Riesz basis [16]. Moreover, no two distinct alternate dual frames for a given frame are equivalent [16]. For a normalized tight frame, its canonical dual frame is the frame itself since ~ F = FS ;1 = F I = F. Moreover, Proposition 3.1. If f' i g i2i is a normalized tight frame for H N,then the only normalized tight alternate dual frame for f' i g i2i is f' i g i2i itself. Proof (Proposition 3.1). Let F 1 be the analysis frame operator for f' i g i2i. Let f i g i2i be any normalized tight alternate dual frame for f' i g i2i with analysis frame operator F 2. It follows that F 1 F 2 are isometries. Now, F 2 (F 1 ; F 2 ) = F 2 F 1 ; F 2 F 2 = I ; I = 0: Hence, for every f g 2 H we have hf 2 g (F 1 ; F 2 )fi = hg F 2 (F 1 ; F 2 )fi = hg 0i = 0: Since f' i g i2i is an NTF, for every f 2 H kfk 2 = We can further expand this as kf 1 fk 2 = k(f 1 F 2 )fk 2 M we have jhf ' i ij 2 = kf 1 fk 2 : (24) = kf 2 fk 2 + k(f 1 ; F 2 )fk 2 + hf 2 (F 1 ; F 2 )f fi + hf 2 (F 1 ; F 2 )f fi : {z } 0 {z } 0

15 UNIFORM TIGHT FRAMES WITH ERASURES 15 However, since F 2 corresponds to an NTF as well, kf 2 fk 2 + k(f 1 ; F 2 )fk 2 = kfk 2 + k(f 1 ; F 2 )fk 2 : (25) Equating (24) and (25), we get k(f 1 ; F 2 )fk 2 = 0: Hence, F 1 = F 2 and so ' i = i, for all i 2 I. If we ask for the dual frame just to be tight, but not normalized, then: Proposition 3.2. If f' i g M is a normalized tight frame for H N and M<2N, then the only tight dual frame for f' i g M is f' ig M itself. If M 2N then there are innitely many (nonequivalent) tight alternate dual frames for f' i g M. Proof (Proposition 3.2). With the notation of the proof of Proposition 3.1, the only change is that, since the dual frame is only tight and not normalized tight, there is a frame bound A>0sothat kf 2 k 2 = Akfk 2 for all f 2 H : Hence, as in the proof of Proposition 3.1 we have kfk 2 = kf 1 fk 2 = kf 2 fk 2 + k(f 1 ; F 2 )fk 2 = Akfk 2 + k(f 1 ; F 2 k 2 : Hence, k(f 1 ; F 2 )fk 2 =(1; A)kfk 2 for all f 2 H. Hence, either F 1 ; F 2 =0and A =1which gives us the NTF as in the previous proposition, (and so ' i = i for all 1 i M) or(1= p 1 ; A)(F 2 ; F 1 )isanisometry on H N. In the latter case it follows that dim (F 1 H N )? N, and so M 2N. Thus, if M < 2N, theonly tight dual frame is the frame itself. On the other hand, if M 2N given any F : H N! `M2 which isaconstant times an isometry, andwith F 1 H N? F H N, F 1 +F denes a tight frame which isanalternate dual frame for f' i g. However, since it is really the uniform case we are interested in, Proposition 3.3. If f' i g M is a uniform normalized tight frame for H N and M = 2N, then there are innitely many uniform tight alternate dual frames for f' i g 2N. Proof (Proposition 3.3). Again we use the notation of the proof of Proposition 3.1. Choose any F 2 : H N! `M2 which isaconstant multiple of an isometry and with F 2 H N = (F 1 H N )?. Then F 1 + F 2 denes a tight alternate dual frame for f' i g M, say i = (F 1 + F 2 )e i, for all 1 i M. We just need to check that this frame is uniform. Let P : `M2! F 1 H N be the orthogonal projection, so that Pe i = F 1 ' i, for all 1 i M. It follows that kpe i k 2 = N=M and k(i ; P )e i k 2 =1; N=M, forall1im. Now, i = F 1 e i + F 2 e i = ' i + F 2 (I ; P )e i :

16 16 CASAZZA AND KOVACEVIC Also, there is an a>0sothatforevery 1 i M we have k i k 2 = k i k 2 + kf 2 (I ; P )e i k 2 = N M + ak(i ; P )e ik 2 Hence, f i g M is a uniform tight frame. = N M + a 1 ; N = a +(1; a) N M M : It can be shown that the results of this section actually classify when NTFs or UNTFs have tight (respectively uniform tight) alternate dual frames, that is, all tightalternate dual frames for f' i g are obtained bythe methods of this section. We do not know in general which frames (not tight) have tight (respectively uniform tight) alternate dual frames Frames Equivalent to UTFs Another approach to the classication of UTFs looks into families of frames obtained from a UTF by equivalence relations. For example, we know that every frame is equivalent toanormalized tight frame. That is, given any frame f' i g i2i with the frame operator S, the frame fs ;1=2 ' i g i2i is a normalized tight frame which isequivalent tof' i g i2i. Therefore, it is natural to try to nd ways to turn frames into uniform normalized tight frames. As it turns out, this is not possible in most cases: Theorem 3.6. If a frame f' i g i2i with the frame operator S is equivalent to a UTF, then fs ;1=2 ' i g i2i is a UNTF. In particular, a tight frame which is not uniform cannot be equivalent to any UNTF. Proof (Theorem 3.6). It is known that fs ;1=2 ' k g is a normalized tight frame which isequivalent tof' k g. So if f' k g is equivalent toauniform tight frame, say f k g, and k k k = c, forall k, then fs ;1=2 ' k g is equivalent tof k g. That is, there is an invertible operator T on H so that TS ;1=2 ' k = k. NowweshowthatT= p A is a unitary operator. Let A be the tightframeconstant of fts ;1=2 ' k g. Then for all f 2 H, the tightness of fts ;1=2 ' k g implies Akfk 2 = k jhf TS ;1=2 ' k ij 2 = k jht f S ;1=2 ' k ij 2 : (26) However, since fs ;1=2 ' k g is an NTF, then this leads us to Equating (26) and (27), we get that k jht f S ;1=2 ' k ij 2 = kt fk 2 : (27) Akfk 2 = kt fk 2 that is, T= p A is unitary. Hence, the frame S ;1=2 ' k = T ;1 k

17 UNIFORM TIGHT FRAMES WITH ERASURES 17 is a UNTF since kt ;1 kk = 1 p A k k k = 1 p A c for all k =1 ::: M: 4. UNIFORM TIGHT FRAMES WITH GROUP STRUCTURE In this section, we examine frames which are generated by a group of unitary operators applied to a xed vector in the Hilbert space. These classes are especially important inapplications. The traditional use of one of these classes is in the Gabor frames used in signal/image processing. These are groups with two generators. There is also a class of wavelet frames which hastwo generators [9]. Recently, several new applications havearisen. In [25], the authors proposed using the redundancy of frames to mitigate the losses in packet-based communication systems such as the Internet. In [17], frames are used for multiple antenna coding/decoding (see the discussion Section 4.3 for further explanation). In [22, 12], connections between quantum mechanics and tight frames are established. Although each ofthese applications requires a dierent class of UNTFs, they all have one important common constraint. Namely, calculations for the frame must be easily implementable on the computer. Since UNTFs generated by agroup of unitary operators satisfy this constraint, this class is one of the most important tounderstand. In this section we will showthat the UNTFs generated by the family fu k ' 0 g (where U is unitary and ' 0 2 H ) are precisely the GHFs. We then extend our discussion to UNTFs generated by fu k V j ' 0 g and higher numbers of generators. In the discussion Section 4.3 we will relate the results of this section to the literature UTFs with a Single Generator Here, we giveacomplete classication of UNTFs of the form fu k ' 0 g where U is a unitary operator on H N. We will,in fact, nd that in this case fu k g must be a group. Moreover, we willsee the the GHFs will be that special class. Since the proofs of the following results are long, the reader can nd them in the Appendix. Our rst result classies GHFs as those frames generated by powers of aspecial class of unitary operators applied to a xed vector in H. Proposition 4.1. f' k g is a general harmonic frame for H N if and only if there isavector ' 0 2 H N with k' 0 k 2 = M N,anorthonormal basis fe ig N for H N and a unitary operator U on H N with Ue i = c i e i, with fc i g N distinct Mth roots of some jcj =1so that ' k = U k ' 0, for all 0 k M ; 1. Proof. See Appendix A.1. We nowshow that the only UNTFs with group structure and a single generator are GHFs. This result is an excellentresult and a disappointment atthesame time. On the one hand, the fact that GHFs are the only ones with such agroup structure proves once more why they are so universally used. It also spares us the trouble of looking for other such UNTFs. On the other hand, we cannot nd any other useful families via this route as we were hoping for. Another important property of

18 18 CASAZZA AND KOVACEVIC GHFs (Theorem 4.2 from [25]) is that GHFs are robust to the maximum number of possible erasures, that is, a GHF with M elements in H N is robust to e erasures for any e (M ; N). Our next theorem completes the classication of GHFs as being those UNTFs of the form fu k ' 0 g where U is a unitary operator on H. As a consequence, we discover that, in this case, fu k g is a group. Theorem 4.1. Let U be a unitary operator on H N, ' 0 2 H N and assume fu k ' 0 g is a UNTF for H N. Then U M = ci for some jcj =1and fu k ' 0 g is a general harmonic frame. That is, the GHFs are theonly UNTFs generated by agroup of unitary operators with a single generator. Proof. See Appendix A UTFs with Two ormore Generators Having completely characterized UTFs whichcome from a group of unitary operators with one generator, we nowturn our attention to those with more than one. The fundamental examples of frames with two generators are the Gabor frames fe mb T ma gg m n2z (or, in our case, the nite discrete Gabor frames). Each ofthese frames is equivalent to the UNTF Gabor frame fe mb T na S ;1 gg m n2z where S is the frame operator for fe mb T ma gg m n2z. We will classify the UNTFs of the form fu i V j ' 0 g L;1 j=0 where fv j ' 0 g j=0 is a UNTF for its span. In the process of proving these results, we introduce a general method for using special families of nite-rank projections to produce frames for a space. If we associate a frame f' i g M with the family of rank-1 orthogonal projections P i taking H onto span ' i,then we canview the results of this section as a generalization of the very notation of a frame to the case where the P i have arbitrary rank. Our rst result states that given aghfgenerated by aunitary operator V, a unitary operator U and an integer M, we can always nd a corresponding UNTF generated by U V. Proposition 4.2. Let V be aunitary operator on H N, let m 2 N and 0 2 H so that fv i 0 g m;1 is a UNTF for H N. Then for every unitary operator U on H N and every M 2 N, there is a vector ' 0 2 H N such that fu k V j ' 0 g m;1 j=0 is a uniform, normalized tight frame for H N. Proof (Proposition 4.2). Let ' 0 = 0 = p M. Now, for every f 2 H N we have m;1 j=0 Since fv j 0 g m;1 j=0 jhf U k V j ' 0 ij 2 = m;1 j=0 jhu ;k f V j ' 0 ij 2 = 1 M is a UNTF, the previous sum equals 1 ku ;k fk 2 = 1 M M kfk 2 = kfk 2 : m;1 j=0 jhu ;k f V j 0ij 2 :

19 UNIFORM TIGHT FRAMES WITH ERASURES 19 It is easily seen that an invertible operator on H maps a frame for H to another frame for H. We next observe that there is a natural relationship between the frame operators of these equivalent frames. Proposition 4.3. If f' k g is aframe for H with the frame operator S, and T is an invertible operator on H, then TST is the frame operator for the frame ft' k g. In particular, if f' k g is a UNTF then every frame unitarily equivalent to f' k g is also a UNTF. Proof (Proposition 4.3). Let L be the frame operator for ft' k g. For any f 2 H we have:! Lf = hf T' k it' k = T ht f ' k i' k = T [S(T )f]: k k As a consequence of Proposition 4.3, we can apply a unitary operator to any UNTF and get a new UNTF. If this unitary operator is part of the generating set for our UNTF, then our new UNTF has a special form as given in the next proposition. Proposition 4.4. If U is unitary and fu k V j ' 0 g is a UNTF for H N, then for every M 2 Z and for every f 2 H N we have, f = hf U k+m V j ' 0 iu k+m V j ' 0 k j that is, fu k+m V j ' 0 g k j is a UNTF as well. Proof (Proposition 4.4). Since U M is unitary and fu k V j ' 0 g k j is a UNTF for H N,itfollows that fu M U k V j ' 0 g k j is also a UNTF for H N. To explain where we are going from here, let us revisit the notion of a UNTF and look at it from a slightly dierent pointofview. Suppose f' i g i2i is a UNTF for a (nite or innite-dimensional) Hilbert space H with k' i k = c, for all i 2 I. For each i 2 I, letp i be the orthogonal projection of H onto the one-dimensional subspace of H spanned by ' i. Then, for all f 2 H i2i kp i fk 2 = khf ' i i' i k 2 i2i = jhf ' i ij 2 k' i k 2 i2i = c 2 i2i jhf ' i ij 2 = c 2 kfk 2 : Conversely, letfp i g i2i be rank-1 orthogonal projections on H i2i kp i fk 2 = akfk 2 for all f 2 H : satisfying

20 20 CASAZZA AND KOVACEVIC Then f' i g i2i is a UNTF for H where ' i 2 P i H and k' i k 2 =1=a. That is, if f 2 H then i2i jhf ' i ij 2 = i2i = 1 a i2i jhp i f ' i ij 2 i2i kp i fk 2 k' i k 2 kp i fk 2 = 1 a akfk2 = kfk 2 : It follows that the condition on f' i g which guarantees that we have a UNTF is really a condition on the rank-1 orthogonal projections of H onto the span of ' i. It is this which we now generalize to nd our next general class of UNTFs. Since this material is also new for general frames, we start here. Proposition 4.5. Let fp i g i2i be a family of projections (of any rank) on a (nite or innite-dimensional) Hilbert space H and assume there are constants 0 < A B satisfying, Akfk 2 i2i kp i fk 2 Bkfk 2 for all f 2 H : Let f' ij g Mi be a frame for P j=1 ih with frame bounds A i B i, for all i 2 I. Then f' ij g Mi j=1 i2i has frame bounds, A inf i2i A i B sup i2i B i : Proof. For any f 2 H we have M i j=1 i2i jhf ' ij ij 2 = i2i M i j=1 M i jhf ' ij ij 2 = jhp i f ' ij ij 2 B i kp i fk 2 i2i i2i sup i2i B i kp i fk 2 B (sup i2i B i ) kfk 2 : i2i The lower frame bound is computed similarly. The cases of interest to us are the next two corollaries: the rst for NTFs and the second for UNTFs. Corollary 4.1. Let fp i g i2i be a family of projections on a Hilbert space H satisfying: i2i kp i fk 2 = a 2 kfk 2 for all f 2 H : Then a 1, andiff' ij g Mi j=1 is an NTF for P ih, for all i 2 I, then f 1 a ' ijg Mi j=1 i2i is an NTF for H.

21 UNIFORM TIGHT FRAMES WITH ERASURES 21 In the next corollary we assume that the projections P i all have thesame rank and the frames for P i H all have the same number of elements. This assumption is necessary to guarantee that our frame is uniform. That is, as we observed in Section 2.1, if dim P i H = N and f' i j g j2j is a UNTF for P i H then k' ij k 2 = N jjj : Corollary 4.2. Let fp i g i2i be a sequence of projections (all with the same rank) on a Hilbert space H satisfying: i2i kp i fk 2 = a 2 kfk 2 for all f 2 H : Then a 1, andif f' ij g j2j is a UNTF for P i H, for all i 2 I, then f 1 a ' ijg j2j i2i is a UNTF for H. We are now ready to consider UNTFs with two generators. Theorem 4.2. Let U be aunitary operator on H, and let fv j 0 g j=0 be autf (with tight frame bound 1=a) forits closed linear span in H. Let P 0 betheorthogonal projection of H onto this span. Assume that fu i V j ' 0 g L;1 j=0 is a UNTF for H. If P i is the orthogonal projection of H onto the span of U i P 0 H, for all 1 i L;1, then L;1 kp i fk 2 = akfk 2 for all f 2 H : Proof. Note that for all 1 i K ; 1wehave P i = U i P 0 U ;i. Now, for all f 2 H we compute, L;1 kp i fk 2 = = = a L;1 L;1 K;1 ku i P 0 U ;i fk 2 kp 0 U ;i fk 2 = j=0 L;1 a j=0 jhf U i V j ' 0 ij 2 = akfk 2 : jhu ;i f V j ' 0 ij 2 The next theorem is the converse of the above. Theorem 4.3. Let fp i g L;1 L;1 beafamily of orthogonal projections on H satisfying kp i fk 2 = a 2 kfk 2 for all f 2 H : Assume that U is a unitary operator on H so that U i P 0 H = P i H. Let ' 0 2 P 0 H and let V be aunitary operator on P 0 H such that fv j ' 0 g j=0 is a UNTF for P 0 H. Then f 1 a U i V j ' 0 g L;1 j=0 is a UNTF for H.

22 22 CASAZZA AND KOVACEVIC Proof. For all f 2 H we have, L;1 j=0 jhf 1 a U i V j ' 0 ij 2 = 1 a 2 L;1 j=0 jhu ;i f V j ' 0 ij 2 = L;1 1 kp a 2 0 U ;i fk 2 = 1 L;1 ku i P a 2 0 U ;i fk 2 = 1 L;1 kp i fk 2 = kfk 2 : a 2 Some remarks are in order: 4.3. Discussion 1. The assumptions in Theorems 4.2 and 4.3 that fv j ' 0 g j=0 is a UNTF for its span is sucient for the conclusion of the theorems but not necessary. For example, it is known to Gabor frame specialists that for ab < 1, there are Gabor frames fe mb T na gg m n2zfor which neither of fe mb T na gg m2znor fe mb T na gg n2zis a frame for its closed linear span. The equivalent UNTF Gabor frame fe mb T na S ;1=2 gg m n2z fails the hypotheses of Theorems 4.2 and 4.3 but satises the conclusion. 2. The results of Section 4.2 generalize to three or more generators. For example, for three generators, Theorem 4.3 becomes: Theorem 4.4. Let U V be unitary operators with `fu i V j ' 0 g L;1 j=0 a UNTF for its span. Let P 0 be the projection of the Hilbert space H onto this span. Let W be aunitary operator on H so that K;1 kw k P 0 W ;k fk 2 = a 2 kfk 2 for all f 2 H : Then f 1 a W k U i V j ' 0 g K;1 L;1 j=0 is a UNTF for H. 3. It would be very interesting to classify when a nite group of unitaries G generates a UNTF for H of the form fu' 0 g U2G. Since nite Abelian groups are isomorphic to a direct product of cyclic groups, the results of this section give sucient conditions for fu' 0 g U2G to generate a UNTF for H. Frames generated by Abelian groups of unitaries were studied in [4] where they are called geometrically uniform frames (GU frames). It is shown there that the canonical dual frame of a GU frame is also a GU frame and that the equivalent NTF frame is also GU. Since GU frames have strong symmetry properties, they are particularly useful in applications. In [4], it is further shown that the frame bounds resulting from removing a single vector of aguframe are the same regardless of the particular vector removed. This result for HTFs was observed in [25]. 4. Frames generated by possibly noncommutative groups of unitaries are important in multiple antenna coding and decoding [17]. Here, one needs classes of

23 UNIFORM TIGHT FRAMES WITH ERASURES 23 unitary space-time signals (that is, frames generated by classes of unitary operators) called constellations. It is also desirable in this setting to have full transmitter diversity, meaning that det (I ; U) 6= 0 for all I 6= U 2 G: Groups with this property are called xed-point free groups. In a tour-de-force, the authors in [17] classify all full-diversity constellations that form a group, for all rates and numbers of transmitting antennas. Along the way they correct some errors in the classication theory of xed point free groups. 5. CLASSIFICATION OF UNIFORM TIGHT FRAMES WITH ERASURES As mentioned in the introduction, one of the main applications that motivated us to examine uniform tight framesisthat of robust data transmission. This means that at some point, the system experiences losses. These losses are modeled as erasures of transmitted frame coecients. Since at the receiver side, this looks like the original frame was the one without vectors corresponding to erased coecients, we examine the structure of our frames after losses. The rst question is whether after erasures what we have isstill a frame? For the MB frame, if we erase any one element, the remaining two are enough for reconstruction. However, if we erase any two, we have lost one subspace and reconstruction is not possible. We can provide more robustness by adding more vectors. For example, take auniform tight frame in R 2 made up of two orthonormal bases (1 0) (0 1) (;1 0) (0 ;1). This frame is robust to any one erasure what is left contains at least one of the two bases. However, if two elements are erased, the situation is not that clear anymore. We could erase coecients corresponding to one of the bases and still have abasis able to reconstruct. If, on the other hand, we lose coecients corresponding to two collinear vectors, we are stuck we have lost one entire subspace and cannot reconstruct. We could, however, take another UNTF with M =4vectors in R 2 {the HTF with the analysis frame operator F = 0 p 1 p 0 2=2 2=2 0 1 ; p 2=2 p 2=2 1 C A : (28) This frame is also made up of two orthonormal bases. However, this frame is robust to any two erasures. It is thus more robust than the MB frame we pay for this added benet with more redundancy (more vectors). It is also more robust than the previous frame with 4 vectors. These simple examples demonstrate the types of questions we willbeaskingin this section. One might think that starting with a tight frame might provide some resilience to losses (as opposed to starting with a general frame). As we have seen in our example, this is not the case. Thus, we are searching for frames robust to a certain number oferasures. That is, frames which remain frames after erasures. There are frames on H N with M elements which are robust to M ; N erasures,