WhenU is chosen to be abelian, so that all theUicommute with each other, then the matrices can be simultaneously diagonalized by a unitary change of basis matrixBso that we may writeB∗UiB=Di, whereDi is diagonal. In this case the inner product corresponding toUi will take the form
v∗Uiv=v∗B∗DiBv,
so by replacingvwithB∗vwithout loss of generality, we may assume that theUiare already diagonal.
Furthermore, since eachUi must have a multiplicative order dividing the size ofU, we may take the diagonal entries ofUi to be powers of the nth-root of unity ω :=e2πin . The matrices will then take the form
Uj= diag(ωa1,j, ..., ωam,j)∈Cm×m,
where theai,j are integers between 0 and n−1. In the language of representation theory we have decomposedU into its irreducible representations, all of which are degree-1 sinceU is abelian.
If we write the coordinates of our rotated vector as v = (v1, ..., vm)T ∈ Cm×1, then our inner
products will now take the form
v∗Ujv=
m
X
i=1
ωai,j|vi|2, (3.13)
so we see that the inner products depend only on the magnitudes of thevi, which weight the diagonal entries of theUj. In particular, for the sake of minimizing coherence, we may take the entries of v to be real. Furthermore, it turns out that in order for our abelian group frame to be tight, all the entriesvimust be of equal norm. This follows from Theorem 5.4 in [17], and we will touch on this in Section 4.1. On this note, we will consider the case wherev is a scaled vector of all 1’s,
v= 1
√m1m= 1
√m[1, ...,1]T ∈Cm×1, (3.14) where we have again chosenv, and hence all the vectorsUiv, to be unit-norm. Now the inner product norm corresponding to the elementUj becomes simply
|v∗Ujv|
||v||22
= 1 m
m
X
i=1
ωai,j
. (3.15)
Notice that from Equation (3.15), we can see that the coherence of our final matrix would remain unchanged if we choseω to be any other primitive nth root of unity. Indeed, if we replaceω with another primitive root of unity, which we may write asωb where b is relatively prime ton, then the inner product associated with Uj will become m1
Pm
i=1ωb·ai,j. But this is just the original inner product associated withUbj, which in turn generates the entire cyclic grouphUji. Hence, the inner products which arise using any two primitiventhroots of unity are the same.
When we form a frame from an abelian group in this manner, and in addition require the sets of diagonal components {ωai,j}mi=1 to form distinct representations ofU, we obtain what is called a harmonic frame, which we will define concretely as follows:
Definition 2. Let m and n be integers, ω = e2πin , and Uj = diag(ωa1,j, ..., ωam,j) ∈ Cm×m for j = 1, ..., n, where the ai,j are integers between 0 and n−1. If we set v = √1m[1, ...,1]T ∈ Cm×1, andM= [U1v, ...,Unv], then if the rows of M are distinct, we call the set of columns {Ujv}nj=1 a harmonic frame.
Remark: Note that we have sidestepped the discussion of whether a “harmonic frame” is actually a frame in the sense of Definition 1. But indeed it is, as we will discuss in the proof of Lemma 8.
Harmonic frames are one of the most thoroughly-studied types of structured frames [25, 55]. An important example of a harmonic frame arises when we choose the groupU to be cyclic, meaning that
eachUj is a power of a single matrixU, which we will explicitly write as
U= diag(ωa1, ..., ωam), (3.16)
so we may write Uj := Uj. For cyclic groups, the inner product between the columns U`1v and U`2v, after normalizing the columns, will take the form |v∗U||v||`2−`21v|
2 , which is the value of the inner product determined byU`2−`1 in (3.15).
In this case, if we again take vto be the normalized vector of all 1s, our frame matrix takes the form
M=h
v Uv . . . Un−1v
i (3.17)
= 1
√m
1 ωa1 ωa1·2 . . . ωa1·(n−1) 1 ωa2 ωa2·2 . . . ωa2·(n−1)
... ... ... ... ...
1 ωam ωam·2 . . . ωam·(n−1)
, (3.18)
where the columns form a harmonic frame precisely when the ai are distinct. In this form, we see thatMis a subset of rows of then×ndiscrete Fourier matrix, so it becomes clear that the columns ofMform a tight frame sinceMM∗= mnI∈Cm×m. In fact, this is true of all harmonic frames:
Lemma 8. A harmonic frame is a tight, unit-norm frame.
Proof. The fact that harmonic frames are unit-norm follows straight from the definition. We note that the rest of this lemma is proven in [17], and we will explain the tightness of harmonic frames in Section 4.1 when we discuss tight group frames in greater generality. For now, however, we will provide a simple, self-contained proof.
A general abelian group U can be represented as follows: first express G as a direct product of, say, L cyclic groups of orders n1, ..., nL, so that U ∼= nZ1Z ×...× nZLZ. Then let ω1, ..., ωL be the corresponding primitive roots of unity: ωj = e2πi/nj. Then we set Uj = diag(ωja1j, ..., ωajmj), where we will assume that the aij are distinct integers modulo nj. The abelian group generated by the diagonal matrices {U1, ...,UL} is isomorphic to U, and an arbitrary element will take the form Ub11Ub22. . .UbLL, where bj ∈ {0, ..., nj −1}. Our frame matrix M will then take the form M= [. . .
Ub11Ub22. . .UbLLv
. . .]0≤bj≤nj−1.
In this form, our previous cyclic frames clearly arise as subsets of the columns of M. It is not too difficult to see that the frame matrixM is a subset of rows of the Kronecker productAKron :=
A1⊗...⊗AL, where Aj=h
v,Ujv, ...,Unjj−1v
i. By the properties of the Kronecker product,
AKronA∗Kron = (A1⊗...⊗AL)(A1⊗...⊗AL)∗ (3.19)
= (A1⊗...⊗AL)(A∗1⊗...⊗A∗L) (3.20)
=A1A∗1⊗...⊗ALA∗L, (3.21)
and since eachAjA∗j is a multiple of the identity matrix, so is their Kronecker product. It follows that the columns ofMare indeed a tight frame.
As we will see in the next few sections, there is a lot we can do in optimizing frame coherence even if we restrict our attention to harmonic frames.