Tight frames

Chapter
Part of the Applied and Numerical Harmonic Analysis book series (ANHA)

Abstract

Decompositions like those in our two prototypical examples, i.e., $$f = \sum _{j\in J} \langle f, f_j\rangle f_j, \qquad \forall f\in {\mathscr {H}},$$ will come from what is called a tight frame $$(f_j)_{j\in J}$$.

Decompositions like those in our two prototypical examples, i.e.,
\begin{aligned} f = \sum _{j\in J} \langle f, f_j\rangle f_j, \qquad \forall f\in {\mathscr {H}}, \end{aligned}
(2.1)
will come from what is called a tight frame $$(f_j)_{j\in J}$$. The basic ingredients are
• $${\mathscr {H}}$$ – a real or complex Hilbert space (for us usually finite dimensional)

• J – an index set (often with a group structure)

• $$(f_j)_{j\in J}$$ – a sequence (set, or multiset) of vectors in $${\mathscr {H}}$$

• $$\sum _{j\in J}$$ – a sum (for us usually finite, but sometimes continuous)

The emphasis here is on the possible redundancy (over completeness) of the vectors $$(f_j)$$ in the expansion, i.e., the case when (2.1) is not an orthogonal expansion.

In the first instance, you are encouraged to consider $${\mathscr {H}}$$ as $$\mathbb {R}^d$$ or $$\mathbb {C}^d$$, with the usual (Euclidean) inner product, and to think in familiar matrix terms.

2.1 Normalised tight frames

The polarisation identity (see Exer. 2.1) implies that (2.1) is equivalent to
$$\Vert f\Vert ^2 = \sum _{j\in J} |\langle f, f_j\rangle |^2, \qquad \forall f\in {\mathscr {H}},$$
which explains the following definition.

Definition 2.1.

A countable sequence $$(f_j)_{j\in J}$$ in a Hilbert space $${\mathscr {H}}$$ is said to be a tight frame (for $${\mathscr {H}}$$) if there exists a (frame bound) $$A>0$$ , such that
\begin{aligned} A \Vert f\Vert ^2 = \sum _{j\in J} |\langle f, f_j\rangle |^2, \qquad \forall f\in {\mathscr {H}}. \end{aligned}
(2.2)
Further, $$(f_j)_{j\in J}$$ is normalised if $$A=1$$ , and finite if J is finite.
The Bessel identity (2.2) is equivalent (see Exer. 2.2) to either of the identities
\begin{aligned} { Parseval:}&\qquad f = {1\over A}\sum _{j\in J} \langle f, f_j\rangle f_j, \quad \forall f\in {\mathscr {H}}, \end{aligned}
(2.3)
\begin{aligned} { Plancherel:}&\qquad \langle f, g\rangle = {1\over A}\sum _{j\in J} \langle f,f_j\rangle \langle f_j,g\rangle , \quad \forall f, g\in {\mathscr {H}}. \end{aligned}
(2.4)

For $$A{=}1$$ (2.2) says that $$f\mapsto (\langle f, f_j\rangle )_{j\in J}$$ is an isometry, and so normalised  tight frames for $${\mathscr {H}}$$ are equivalent to isometries $${\mathscr {H}}\rightarrow \ell _2(J)$$. The maps taking normalised tight frames to normalised tight frames are the partial isometries (Exer. 2.7).

We prefer the term normalised tight frame 1 to Parseval frame (which is also used), as it emphasizes the fact the frame bound $$A>0$$ is simply a normalising factor, i.e., if $$(f_j)$$ is a tight frame, then $$(f_j/\sqrt{A})$$ is the unique positive scalar multiple of it which is a normalised tight frame. We will soon see that this normalised version of a tight frame is convenient in many situations.

We have defined a tight frame to be a sequence, which is standard, but not universal. By contrast with a basis (which can be a set or a sequence), a tight frame can have repeated vectors. At times, e.g., when the vectors in a frame are all distinct or the indexing is unimportant, it can be convenient to think of them as a (multi)set. We will not labour this point, making statements such as the set $$\{f_j\}_{j\in J}$$ is a tight frame, without further explanation.

Example 2.1.

(Exer. 2.4) An orthonormal basis is a normalised tight frame. These are the only normalised tight frames in which all the vectors have unit length (all vectors in a normalised tight frame have length $$\le 1$$).

Example 2.2.

(Exer. 2.5) The unitary image of a normalised tight frame is again a normalised tight frame, and the only invertible linear maps which map a normalised tight frame to a normalised tight frame are the unitary maps.

Example 2.3.

(Exer. 2.6) The orthogonal projection of a normalised tight frame is again a normalised tight frame (for its span). In particular, if U is an $$n\times n$$ unitary matrix, then the columns of any $$d\times n$$ submatrix is a normalised tight frame of n vectors for $$\mathbb {C}^d$$. This is effectively the projection of the orthonormal basis for $$\mathbb {C}^n$$ given by the columns of U onto the d-dimensional subspace of vectors which are zero in some fixed $$n-d$$ coordinates.

We say that $$(f_j)_{j\in J}$$ is an equal-norm tight frame if $$\Vert f_j\Vert =\Vert f_k\Vert$$, $$\forall j, k\in J$$, and is a unit-norm tight frame if $$\Vert f_j\Vert =1$$, $$\forall j\in J$$.

Example 2.4.

(Exer. 2.8) Equal-norm tight frames of n vectors for $$\mathbb {F}^d$$ can be obtained from an $$n\times n$$ unitary matrix U with entries of constant modulus, by taking the columns of any $$d\times n$$ submatrix. Examples of such U include the Hadamard matrices (real entries) and the Fourier (transform) matrix
\begin{aligned} U=F_n ={1\over \sqrt{n}} \begin{bmatrix} 1&1&1&\cdots&1 \\ 1&\omega&\omega ^2&\cdots&\omega ^{n-1} \\ 1&\omega ^2&\omega ^4&\cdots&\omega ^{2(n-1)} \\ \vdots&\vdots&\vdots&{\ }&\vdots \\ 1&\omega ^{n-1}&\omega ^{2(n-1)}&\cdots&\omega ^{(n-1)(n-1)} \end{bmatrix}, \qquad \omega :=e^{2\pi i\over n}. \end{aligned}
(2.5)
Equal-norm tight frames which come from the Fourier matrix in this way will be known as harmonic tight frames (see Chapter ).

In §2.6, we will show that every normalised tight frame can be obtained as the orthogonal projection of an orthonormal basis (in a larger space).

Example 2.5.

(Exer. 2.9) The tight frames for $$\mathbb {R}^2$$. A sequence of vectors $$(v_j)_{j=1}^n$$, $$v_j=(x_j, y_j)\in \mathbb {R}^2$$ is a tight frame for $$\mathbb {R}^2$$ if and only if the diagram vectors which are defined by $$w_j := (x_j+i y_j)^2\in \mathbb {C}$$ , $$1\le j\le n$$, sum to zero (in $$\mathbb {C}$$).

In applications, the interest in (2.3) is usually that it gives a decomposition of the identity into a weighted sum of projections (see Chapter ), i.e.,
\begin{aligned} I = \sum _{j\in J} c_j P_j, \qquad c_j:={\Vert f_j\Vert ^2\over A}, \quad P_j f:= {\langle f,f_j\rangle \over \langle f_j, f_j\rangle } f_j, \end{aligned}
(2.6)
where the particular (unit modulus) scalar multiple of $$f_j$$ that is used to define the orthogonal projection $$P_j$$ is unimportant. The pair $$(P_j)$$,$$(c_j)$$ above is a fusion frame (see §). When taking this point of view, we will use the epithet projective.

2.2 Unitarily equivalent finite tight frames

Before giving any further concrete examples of finite tight frames, we define an equivalence, under which any set of three equally spaced vectors with the same norm in $$\mathbb {R}^2$$ would be considered equivalent.

Definition 2.2.

We say that two normalised tight frames $$(f_j)_{j\in J}$$ for $${\mathscr {H}}$$ and $$(g_j)_{j\in J}$$ for $${\mathscr {K}}$$, with the same index set J, are (unitarily) equivalent if there is a unitary transformation $$U:{\mathscr {H}}\rightarrow {\mathscr {K}}$$, such that $$g_j=Uf_j$$ , $$\forall j\in J$$.

Since unitary transformations preserve inner products, unitarily equivalent tight frames have the same inner products (angles) between their vectors. Furthermore, these inner products uniquely determine the equivalence classes (see §2.5).

This equivalence is dependent on the indexing, which is appropriate when set J has some natural (e.g., group) structure. The normalised tight frames of two vectors $$(e_1,0)$$ and $$(0,e_1)$$ for the one-dimensional space $${\mathscr {H}}=\mathop {\mathrm{span}}\nolimits \{e_1\}$$ are not equivalent, since there is no unitary map $$e_1\mapsto 0$$ (or $$0\mapsto e_1$$). For such cases, where it is useful to consider these as equivalent, we extend our definition of equivalence as follows.

Definition 2.3.

We say that two finite normalised tight frames $$(f_j)_{j\in J}$$ for $${\mathscr {H}}$$ and $$(g_j)_{j\in K}$$ for $${\mathscr {K}}$$ are (unitarily) equivalent up to reordering if there is a bijection $$\sigma :J\rightarrow K$$ for which $$(f_j)_{j\in J}$$ and $$(g_{\sigma j})_{j\in J}$$ are unitarily equivalent.

We will say that tight frames are unitarily equivalent (up to reordering) if after normalisation they are, in which case we say they are equal up to unitary equivalence (and reordering).

Example 2.6.

Let $$u_1,u_2,u_3$$ be equally spaced unit vectors in $$\mathbb {R}^2$$, and $$R_\theta$$ be rotation through an angle $$\theta$$. Then each of the sets of six vectors
$$\{u_1,u_2,u_3,R_\theta u_1,R_\theta u_2,R_\theta u_3\}, \qquad 0<\theta \le {\pi \over 3}$$
forms a tight frame. Since unitary maps preserve angles, none of these are unitarily equivalent (up to reordering).

2.3 Projective and complex conjugate equivalences

There are other equivalences which appear in the frame literature. Most notably, the normalised tight frames of Definition 2.2 are projectively (unitarily) equivalent if
$$g_j = \alpha _j U f_j, \qquad \forall j\in J,$$
where U is unitary, and $$|\alpha _j|=1$$, $$\forall j$$.

All tight frames $$(\alpha _j f_j)$$, $$|\alpha _j|=1$$, $$\forall j$$, obtained from a given tight frame $$(f_j)$$, are projectively unitarily equivalent, but are not unitarily equivalent, in general.

Example 2.7.

For tight frames of n nonzero vectors in $$\mathbb {R}^2$$ the equivalence classes for projective equivalence up to reordering are in 1–1 correspondence with convex polygons with n sides (see Exer. 2.10).

The complex conjugation map on $$\mathbb {C}^d$$ is the antilinear map
$$\mathbb {C}^d\rightarrow \mathbb {C}^d: v=(v_j)\mapsto \overline{v}:=(\overline{v_j}).$$
Since $$\langle \overline{v},\overline{w}\rangle =\overline{\langle v, w\rangle }$$, this maps a tight frame $$(f_j)$$ for $$\mathbb {C}^d$$ to a tight frame $$(\overline{f_j})$$, and these are said to be complex conjugate (or anti) equivalent. The conjugation map $$C:{\mathscr {H}}\rightarrow \overline{{\mathscr {H}}}$$ extends these ideas to $${\mathscr {H}}$$ (see Exer. 2.11).
These basic types of equivalences can be combined, in the obvious way, to obtain others, e.g., the tight frames $$(f_j)_{j\in J}$$ and $$(g_k)_{k\in K}$$ for $${\mathscr {H}}$$ and $${\mathscr {K}}$$ would be anti projectively unitarily equivalent up to reordering if
$$\overline{ g_{\sigma j}} = \alpha _j U f_j, \qquad \forall j\in J,$$
for $$\sigma :J\rightarrow K$$ a bijection, $$|\alpha _j|=1$$, $$\forall j$$, and $$U:{\mathscr {H}}\rightarrow {\mathscr {K}}$$ unitary.

2.4 The analysis, synthesis and frame operators

The Parseval identity (2.3) consists of an extraction of “coordinates”
$$c_j=\langle f, f_j\rangle , \qquad j\in J$$
for the vector f (analysis), and a reconstruction of f from these (synthesis). Many important properties of a tight frame follow from this factorisation.

For simplicity of presentation, we suppose J is finite, write $$\mathbb {F}$$ for $$\mathbb {R}$$ or $$\mathbb {C}$$, $$\ell _2(J)$$ for $$\mathbb {F}^J$$, with the usual inner product, and $$I=I_{\mathscr {H}}$$ for the identity on $${\mathscr {H}}$$.

Definition 2.4.

For a finite sequence $$(f_j)_{j\in J}$$ in $${\mathscr {H}}$$ the synthesis operator (reconstruction operator or pre-frame operator) is the linear map
$$V:=[f_j]_{j\in J}:\ell _2(J)\rightarrow {\mathscr {H}}:a\mapsto \sum _{j\in J} a_j f_j,$$
and its dual is the analysis operator (or frame transform operator)
$$V^*:{\mathscr {H}}\rightarrow \ell _2(J): f\mapsto (\langle f, f_j\rangle )_{j\in J}.$$

It is convenient to make little distinction between the sequence $$(f_j)_{j\in J}$$ and the linear map $$V=[f_j]_{j\in J}$$, which we will say has j -th column $$f_j$$.

The product $$S:=VV^*:{\mathscr {H}}\rightarrow {\mathscr {H}}$$ is known as the frame operator. A simple calculation (see Exer. 2.12) shows the trace of S and $$S^2$$ are given by
\begin{aligned} \mathop {\mathrm{trace}}\nolimits (S) = \Vert S^{1\over 2}\Vert _F^2 = \sum _{j\in J} \Vert f_j\Vert ^2, \qquad \mathop {\mathrm{trace}}\nolimits (S^2) = \Vert S\Vert _F^2 = \sum _{j\in J}\sum _{k\in J} |\langle f_j, f_k\rangle |^2. \end{aligned}
(2.7)

Proposition 2.1.

A finite sequence $$(f_j)_{f\in J}$$ in $${\mathscr {H}}$$ is a tight frame for $${\mathscr {H}}$$ (with frame bound A) if and only if
\begin{aligned} S= VV^* = A I_{\mathscr {H}}, \qquad V:=[f_j]_{f\in J}. \end{aligned}
(2.8)
In particular, a tight frame satisfies
\begin{aligned} \sum _{j\in J} \Vert f_j\Vert ^2 = dA, \qquad d:=\dim ({\mathscr {H}}), \end{aligned}
(2.9)
and
\begin{aligned} \sum _{j\in J}\sum _{k\in J} |\langle f_j, f_k\rangle |^2 = {1\over d} \Bigl (\sum _{j\in J}\langle f_j, f_j\rangle \Bigr )^2. \end{aligned}
(2.10)

Proof.

Since
$$Sf = VV^*f = \sum _j\langle f, f_j\rangle f_j, \qquad \forall f\in {\mathscr {H}},$$
the Parseval identity (2.3) implies the condition (2.2) is equivalent to (2.8). Taking the trace of (2.8) and its square gives
\begin{aligned} \sum _j\Vert f_j\Vert ^2&= \mathop {\mathrm{trace}}\nolimits (S) =\mathop {\mathrm{trace}}\nolimits (A I_{\mathscr {H}}) =dA, \\ \sum _j \sum _k |\langle f_j, f_k\rangle |^2&= \mathop {\mathrm{trace}}\nolimits (S^2) =\mathop {\mathrm{trace}}\nolimits (A^2 I_{\mathscr {H}}) = {1\over d}(Ad)^2 = {1\over d} \Bigl (\sum _j\langle f_j, f_j\rangle \Bigr )^2, \end{aligned}
which are (2.9) and (2.10).    $$\square$$

The equations (2.3) and (2.8) will be referred to as the Parseval identity, (2.9) as the trace formula, and (2.10) as the variational formula.

For $${\mathscr {H}}=\mathbb {F}^d$$ and $$|J|=n$$, V is a $$d\times n$$ matrix, and the condition (2.8) says that the columns of V are orthogonal and of length $$\sqrt{A}$$, i.e., $$V/\sqrt{A}$$ is a coisometry, equivalently, $$V^*/\sqrt{A}$$ is an isometry.

In §, we show that the variational formula characterises tight frames for finite dimensional spaces. There is no infinite dimensional counterpart for this result.

2.5 The Gramian

Unitary equivalence has the advantage (over projective unitary equivalence) that it preserves the inner product between vectors, and hence the Gramian matrix. Indeed, we will show that the Gramian characterises the equivalence class.

Definition 2.5.

For a finite sequence of n vectors $$(f_j)_{j\in J}$$ in $${\mathscr {H}}$$, the Gramian 2 or Gram matrix is the $$n\times n$$ Hermitian matrix
$$\mathop {\mathrm{Gram}}\nolimits ((f_j)_{j\in J}) := [\langle f_k,f_j\rangle ]_{j, k\in J}.$$

This is the matrix representing the linear map $$V^*V:\ell _2(J)\rightarrow \ell _2(J)$$ with respect to the standard orthonormal basis $$\{e_j\}_{j\in J}$$.

The possible Gramian matrices are precisely the orthogonal projections:

Theorem 2.1.

An $$n \times n$$ matrix $$P=[p_{jk}]_{j, k\in J}$$ is the Gramian matrix of a normalised tight frame $$(f_j)_{j\in J}$$ for the space $${\mathscr {H}}:=\mathop {\mathrm{span}}\nolimits \{f_j\}_{j\in J}$$ if and only if it is an orthogonal projection matrix, i.e., $$P = P^* = P^2$$. Moreover,
\begin{aligned} d=\dim ({\mathscr {H}})= \mathop {\mathrm{rank}}\nolimits (P)=\mathop {\mathrm{trace}}\nolimits (P)=\sum _{j\in J}\Vert f_j\Vert ^2. \end{aligned}
(2.11)

Proof.

($$\Longrightarrow$$) Let $$\varPhi =(f_j)_{j\in J}$$ be a normalised tight frame, and $$P=\mathop {\mathrm{Gram}}\nolimits (\varPhi )$$. Take $$f=f_\ell$$ in (2.3) to get $$f_\ell =\sum _{j\in J}\langle f_\ell , f_j\rangle f_j,$$ and take the inner product of this with $$f_k$$ to obtain
$$\langle f_k,f_\ell \rangle =\sum _{j\in J}\langle f_j,f_\ell \rangle \langle f_k, f_j\rangle \iff p_{\ell k} = \sum _{j\in J} p_{\ell j}p_{jk} \iff P=P^2.$$
But P is Hermitian, since $$\overline{p_{jk}}=\overline{\langle \phi _k,\phi _j\rangle } = \langle \phi _j,\phi _k\rangle = p_{kj}$$, and so is an orthogonal projection.
($$\Longleftarrow$$) Suppose that P is an $$n \times n$$ matrix, such that $$P=P^* =P^2$$. The columns of P are $$f_j:=Pe_j$$, $$j\in J$$, where $$\{e_j\}_{j\in J}$$ is the standard orthonormal basis of $$\ell _2(J)$$. Fix $$f\in {\mathscr {H}}:=\mathop {\mathrm{span}}\nolimits \{f_j\}_{j=1}^n\subset \ell _2(J)$$. Then $$f=Pf$$, so that
$$f = P \Bigl ( \sum _{j\in J} \langle Pf,e_j\rangle e_j \Bigr ) = \sum _{j\in J} \langle f,P e_j\rangle Pe_j = \sum _{j\in J} \langle f, f_j\rangle f_j,$$
i.e., $$(f_j)_{j=1}^n$$ is a normalised tight frame for $${\mathscr {H}}$$, with Gramian P.

Finally, taking the trace of P gives (2.11).   $$\square$$

The condition that $$P=\mathop {\mathrm{Gram}}\nolimits (\varPhi )$$ be an orthogonal projection is equivalent to it having exactly d nonzero eigenvalues all equal to 1 (see Exer. 2.17).

Corollary 2.1.

(Characterisation of unitary equivalence) Normalised tight frames are unitarily equivalent if and only if their Gramians are equal.

Proof.

Let $$\varPhi =(f_j)_{j\in J}$$, $$\varPsi =(g_j)_{j\in J}$$ be normalised tight frames for $${\mathscr {H}}$$ and $${\mathscr {K}}$$.

($$\Longrightarrow$$) If $$\varPhi$$ and $$\varPsi$$ are unitarily equivalent, i.e., $$g_j=Uf_j$$, $$\forall j$$, for some unitary $$U:{\mathscr {H}}\rightarrow {\mathscr {K}}$$, then their Gramians are equal since
$$\langle g_j,g_k\rangle =\langle Uf_j,Uf_k\rangle =\langle f_j, f_k\rangle .$$
($$\Longleftarrow$$) Suppose the Gramians of $$\varPhi$$ and $$\varPsi$$ are equal, i.e., $$\langle g_j,g_k\rangle =\langle f_j, f_k\rangle$$, $$\forall j, k$$. Then, by Exer. 2.19, there is a unitary $$U:{\mathscr {H}}\rightarrow {\mathscr {K}}$$ with $$g_j=Uf_j$$, $$\forall j$$. Hence $$\varPhi$$ and $$\varPsi$$ are unitarily equivalent.    $$\square$$

In other words:

The properties of a tight frame (up to unitary equivalence) are determined by its Gramian.

Example 2.8.

Equal-norm tight frames of three vectors for $$\mathbb {C}^2$$ are given by
$$\varPhi :=(\begin{bmatrix}1\\ 1\end{bmatrix},\begin{bmatrix}\omega \\ \omega ^2\end{bmatrix},\begin{bmatrix}\omega ^2\\ \omega \end{bmatrix}), \qquad \varPsi :=(\begin{bmatrix}1\\ 1\end{bmatrix},\begin{bmatrix}1\\ \omega \end{bmatrix},\begin{bmatrix}1\\ \omega ^2\end{bmatrix}), \qquad \omega :=e^{2\pi i\over 3}.$$
These harmonic frames are not unitarily equivalent since their Gramians
$$\mathop {\mathrm{Gram}}\nolimits (\varPhi ) = \begin{pmatrix}2&{}-1&{}-1\\ -1&{}2&{}-1\\ -1&{}-1&{}2\end{pmatrix}, \qquad \mathop {\mathrm{Gram}}\nolimits (\varPsi ) = \begin{pmatrix}2&{}1+\omega &{}1+\omega ^2\\ 1+\omega ^2&{}2&{}1+\omega \\ 1+\omega &{}1+\omega ^2&{}2\end{pmatrix}$$
are different. They are however projectively unitarily equivalent (see Exer. 2.21).

2.6 Tight frames as orthogonal projections

We have seen (Exer. 2.6) that the orthogonal projection of an orthonormal basis is a normalised tight frame (for its span). The converse is also true.

Theorem 2.2.

(Naĭmark) Every finite normalised tight frame $$\varPhi =(f_j)_{j\in J}$$ for $${\mathscr {H}}$$ is the orthogonal projection of an orthonormal basis for $$\ell _2(J)$$. Indeed, the orthogonal projection $$P=\mathop {\mathrm{Gram}}\nolimits (\varPhi )$$ of the standard orthonormal basis $$(e_j)_{j\in J}$$ (onto the column space of the Gramian) is unitarily equivalent to $$\varPhi$$ via $$f_j\mapsto Pe_j$$, i.e.,
$$\langle Pe_j, Pe_k\rangle _{\ell _2(J)} = \langle f_j,f_k\rangle _{\mathscr {H}}, \qquad \forall j, k\in J.$$

Proof.

By Theorem 2.1, $$P=\mathop {\mathrm{Gram}}\nolimits (\varPhi )$$ is an orthogonal projection, and so
$$\langle Pe_j,Pe_k\rangle = \langle Pe_j,e_k\rangle = (k,j)\text {-}\text {entry of }P = \langle f_j, f_k\rangle .$$

$$\square$$

When $$\varPsi$$ and $$\varPhi$$ are unitarily equivalent, then we will say that $$\varPsi$$ is a copy of $$\varPhi$$. With this terminology, Naĭmark’s theorem says:

A canonical copy of a tight frame $$\varPhi$$ is given by the columns of $$\mathop {\mathrm{Gram}}\nolimits (\varPhi )$$.

This is one of those often rediscovered theorems, which can be considered as a special case of Naĭmark’s theorem (see [AG63] and Exer. 2.26). Hadwiger [Had40] showed that $$(f_j)_{j=1}^n$$ in $$\mathbb {R}^d$$ is a coordinate star (normalised tight frame) if and only if it is a Pohlke normal star (projection of an orthonormal basis). In signal processing, this method of obtaining tight frames is called seeding.

Example 2.9.

The Gramian of the three equally spaced vectors in $$\mathbb {R}^2$$ is
(2.12)
The (particular choice of) vectors are normalised $$\Vert v_j\Vert =\sqrt{2\over 3}$$ (see Exer. 2.15), so that $$(v_j)$$ is a normalised tight frame, and hence P is an orthogonal projection.
The columns $$(Pe_j)$$ of P give a canonical copy of this normalised tight frame (up to unitary equivalence), e.g.,
$$\langle Pe_1,Pe_1\rangle = \langle \begin{pmatrix}{2\over 3}\\ -{1\over 3}\\ -{1\over 3}\end{pmatrix}, \begin{pmatrix}{2\over 3}\\ -{1\over 3}\\ -{1\over 3}\end{pmatrix}\rangle ={2\over 3}, \quad \langle Pe_1,Pe_2\rangle = \langle \begin{pmatrix}{2\over 3}\\ -{1\over 3}\\ -{1\over 3}\end{pmatrix}, \begin{pmatrix}-{1\over 3}\\ {2\over 3}\\ -{1\over 3}\end{pmatrix} \rangle =-{1\over 3}.$$

Example 2.10.

A cross in $$\mathbb {R}^n$$ is the set obtained by taking an orthonormal basis and its negatives $$\{\pm e_1,\ldots ,\pm e_n\}$$, and the orthogonal projection of a cross onto a d-dimensional subspace V is a called a eutactic star (see Coxeter [Cox73]). In view of Theorem 2.2, a eutactic star is precisely a tight frame of the form $$\{\pm a_1,\ldots ,\pm a_n\}$$ for V, i.e., the union of a tight frame $$\{a_1,\ldots , a_n\}$$ and the equivalent frame obtained by taking its negative. When the vectors $$a_i$$ all have the same length, one obtains a so-called normalised eutactic star. Since equal-norm tight frames always exist (see Chapters and ), so do normalised eutactic stars in $$\mathbb {R}^d$$ for every $$n\ge d$$.

2.7 The construction of tight frames from orthogonal projections

The Gramian of a normalised tight frame $$\varPhi =(v_j)_{j=1}^n$$ for a d-dimensional space is an orthogonal projection P. By Theorem 2.2, the columns $$(Pe_j)$$ of P give a canonical copy of the frame (up to unitary equivalence) as a d-dimensional subspace of $$\mathbb {F}^n$$. To obtain a copy of $$\varPhi$$ in $$\mathbb {F}^d$$, we consider the rows of $$P=\mathop {\mathrm{Gram}}\nolimits (\varPhi )$$.

Theorem 2.3.

(Row construction). Let $$P\in \mathbb {C}^{n\times n}$$ be an orthogonal projection matrix of rank d. The columns of $$V=[v_1,\ldots , v_n]\in \mathbb {C}^{d\times n}$$ are a normalised tight frame for $$\mathbb {C}^d$$ with Gramian P if and only if the rows of V are an orthonormal basis for the row space of P. In particular, such a V can always be obtained by applying the Gram–Schmidt process to the rows of P.

Proof.

($$\implies$$) Suppose the columns of V are a normalised tight frame for $$\mathbb {C}^d$$ with Gramian P, i.e., $$VV^*=I$$ (the rows of V are orthonormal) and $$P=V^*V$$. Then
$$\mathop {\mathrm{row}}\nolimits (P)=\mathop {\mathrm{row}}\nolimits (V^*V) \subset \mathop {\mathrm{row}}\nolimits (V) = \mathop {\mathrm{row}}\nolimits (VV^*V) \subset \mathop {\mathrm{row}}\nolimits (V^*V) =\mathop {\mathrm{row}}\nolimits (P),$$
so that $$\mathop {\mathrm{row}}\nolimits (P)=\mathop {\mathrm{row}}\nolimits (V)$$, and the rows of V are an orthonormal basis for $$\mathop {\mathrm{row}}\nolimits (P)$$.
($$\Longleftarrow$$) Suppose the rows of V are an orthonormal basis for $$\mathop {\mathrm{row}}\nolimits (P)$$. Then $$VV^*=I$$, and we have
$$(V^*V)^2= V^*(VV^*)V=V^*V,$$
so that $$V^*V$$ is an orthogonal projection matrix with the same row space (and hence column space) as P. Thus, $$V^*V=P$$.   $$\square$$

In other words:

A frame $$V=[v_1,\ldots , v_n]$$ is a copy of a normalised tight frame $$\varPhi$$ if and only if the rows of V are an orthonormal basis for the row space of $$\mathop {\mathrm{Gram}}\nolimits (\varPhi )$$.

Example 2.11.

For the three equally spaced vectors of Example 2.9, applying the Gram–Schmidt process to the first two rows of the Gramian P gives
(2.13)

Example 2.12.

For $$\varPhi$$ a tight frame with frame bound A, the matrix $$P={1\over A}\mathop {\mathrm{Gram}}\nolimits (\varPhi )$$ is an orthogonal projection, and so a copy of $$\varPhi$$ is given by an orthogonal basis for the row space of $$Q=\mathop {\mathrm{Gram}}\nolimits (\varPhi )$$ consisting of vectors of length $$\sqrt{A}$$. For the four equally spaced unit vectors of (), we have $$A=2$$, and applying the Gram–Schmidt process to the first two rows of the Gramian Q gives
$$Q = \begin{bmatrix}1&{1\over \sqrt{3}}&{1\over \sqrt{3}}&-{i\over \sqrt{3}}\\ {1\over \sqrt{3}}&1&-{i\over \sqrt{3}}&{1\over \sqrt{3}}\\ {1\over \sqrt{3}}&{i\over \sqrt{3}}&1&-{1\over \sqrt{3}}\\ {i\over \sqrt{3}}&{1\over \sqrt{3}}&-{1\over \sqrt{3}}&1 \end{bmatrix} \quad \longrightarrow \quad V=\begin{bmatrix} 1&{1\over \sqrt{3}}&{1\over \sqrt{3}}&-{i\over \sqrt{3}}\\ 0&{\sqrt{2}\over \sqrt{3}}&-{i\over \sqrt{2}}-{1\over \sqrt{2}\sqrt{3}}&{1\over \sqrt{2}}+{i\over \sqrt{2}\sqrt{3}} \end{bmatrix}.$$
Here $$\mathop {\mathrm{col}}\nolimits (Q)\ne \mathop {\mathrm{row}}\nolimits (Q)$$, and applying Gram–Schmidt to the columns of Q (instead of the rows) does not give a copy of $$\varPhi$$.

2.8 Complementary tight frames

Tight frames are determined (up to unitary equivalence) by their Gramian matrix P, which is an orthogonal projection matrix (when the frame is normalised), and all orthogonal projection matrices correspond to normalised tight frames. Thus there is normalised tight frame with Gramian given by the complementary projection $$I-P$$.

Definition 2.6.

Given a finite normalised tight frame $$\varPhi$$ with Gramian P, we call any normalised tight frame with Gramian $$I-P$$ its complement. More generally, we say that two tight frames are complements of each other, if after normalisation the sum of their Gramians is the identity I.

The complement of a finite tight frame is unique up to unitary equivalence (and normalisation), the complement of the complement is the frame itself, and a tight frame is equiangular (or equal-norm) if and only if its complement is.

In view of (2.11), the complement of a tight frame of n vectors for a space of dimension d is a tight frame of n vectors for a space of dimension $$n-d$$.

A tight frame and its complement can never be unitarily or projectively unitarily equivalent (Exer. 2.23), though they do have the same symmetries (see §, §).

Example 2.13.

The complement of an orthonormal basis for $$\mathbb {C}^d$$ is the frame for the zero vector space given by d zero vectors.

Example 2.14.

By (2.12) the Gramian of the complementary frame to the three equally spaces vectors in $$\mathbb {R}^2$$ is
$$Q=I-P = \begin{bmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{bmatrix} - \begin{bmatrix}{2\over 3}&-{1\over 3}&-{1\over 3}\\ -{1\over 3}&{2\over 3}&-{1\over 3} \\ -{1\over 3}&-{1\over 3}&{2\over 3}\end{bmatrix} = \begin{bmatrix}{1\over 3}&{1\over 3}&{1\over 3} \\{1\over 3}&{1\over 3}&{1\over 3} \\ {1\over 3}&{1\over 3}&{1\over 3}\end{bmatrix} .$$
By the row construction (Theorem 2.3), this is the Gramian of normalised tight frame $$\{{1\over \sqrt{3}},{1\over \sqrt{3}},{1\over \sqrt{3}}\}$$ of three repeated vectors for $$\mathbb {R}$$. The Gramian of an arbitrary normalised tight frame of three vectors for $$\mathbb {C}^2$$ is considered in Exer. 2.22.
We call the tight frame of $$n=d+1$$ vectors for $$\mathbb {R}^d$$ which is the complement of $$\{ {1\over \sqrt{d+1}},\ldots , {1\over \sqrt{d+1}}\}$$, the vertices of the (regular) simplex in $$\mathbb {R}^d$$. This has Gramian
\begin{aligned} P=[p_{jk}], \qquad p_{jk} := {\left\{ \begin{array}{ll} {-1\over d+1}, &{} j\ne k; \\ {d\over d+1}, &{} j= k. \end{array}\right. } \end{aligned}
(2.14)
To find a copy in $$\mathbb {R}^d$$, one can apply the method of Theorem 2.3 to the Gramian. This example can be generalised to obtain partition frames.

Definition 2.7.

Let $$\alpha =(\alpha _1,\ldots ,\alpha _k)\in \mathbb {Z}^k$$ be a partition of n, i.e.,
$$n=\alpha _1+\cdots +\alpha _k, \qquad 1\le \alpha _1\le \alpha _2\le \cdots \le \alpha _k.$$
The $$\alpha$$ -partition frame for $$\mathbb {R}^d$$ , $$d=n-k$$, is the complement of the normalised tight frame of n vectors for $$\mathbb {R}^k$$ given by
\begin{aligned} \Bigl ( \underbrace{{e_1\over \sqrt{\alpha _1}},\ldots ,{e_1\over \sqrt{\alpha _1}}}_{\alpha _1\ \mathrm{times}}, \ldots , \underbrace{{e_k\over \sqrt{\alpha _k}},\ldots ,{e_k\over \sqrt{\alpha _k}}}_{\alpha _k\ \mathrm{times}} \Bigr ). \end{aligned}
(2.15)
It is said to be proper if $$\alpha _j\ge 2$$, $$\forall j$$.
The Gramian of the $$\alpha$$-partition frame is the block diagonal $$n\times n$$ matrix
\begin{aligned} P = \begin{bmatrix} B_1&\\&\ddots&\\&B_j&\\&\ddots&\\&B_k\\\end{bmatrix}, \qquad B_j := \begin{bmatrix} {\alpha _j-1\over \alpha _j}&{-1\over \alpha _j}&{-1\over \alpha _j}&\cdots&{-1\over \alpha _j}\\ {-1\over \alpha _j}&{\alpha _j-1\over \alpha _j}&{-1\over \alpha _j}&\cdots&{-1\over \alpha _j}\\ {-1\over \alpha _j}&{-1\over \alpha _j}&{\alpha _j-1\over \alpha _j}&{-1\over \alpha _j}\\ \vdots&\vdots&\ddots&{-1\over \alpha _j}\\ {-1\over \alpha _j}&{-1\over \alpha _j}&{-1\over \alpha _j}&{-1\over \alpha _j}&{\alpha _j-1\over \alpha _j} \\\end{bmatrix} \end{aligned}
(2.16)
where the above $$B_j$$ is a $$\alpha _j\times \alpha _j$$ orthogonal projection matrix of rank $$\alpha _j-1$$. Since each normalised tight frame is unitarily equivalent to the columns of its Gramian, it follows that the vectors in a proper partition frame are distinct and nonzero. If $$\alpha _j=1$$, then the corresponding partition frame vector is zero.

Example 2.15.

(Simplex) For $$n=d+1$$, the trivial partition $$\alpha =(d+1)$$ gives the vertices of the simplex in $$\mathbb {R}^d$$.

Table 2.1:

The proper partition frames in $$\mathbb {R}^2$$ and $$\mathbb {R}^3$$. Here |G| is the order of their symmetry group $$G=\mathop {\mathrm{Sym}}\nolimits (\varPhi )$$ (see Chapter , Exer. ).

Partition

n

Description of partition frame

|G|

(3)

3

three equally spaced vectors in $$\mathbb {R}^2$$

6

(2, 2)

4

four equally spaced vectors in $$\mathbb {R}^2$$

8

(4)

4

vertices of the tetrahedron in $$\mathbb {R}^3$$

24

(2, 3)

5

vertices of the trigonal bipyramid in $$\mathbb {R}^3$$

12

(2, 2, 2)

6

vertices of the octahedron in $$\mathbb {R}^3$$

48

2.10 Real and complex tight frames

A tight frame for a real Hilbert space is a tight frame for its complexification (see Exer. 2.28). We will call frames that come in this way real tight frames.

Definition 2.8.

We say that a tight frame $$(f_j)_{j\in J}$$ is real if its Gramian is a real matrix, and otherwise it is complex.

By Theorem 2.3 (row construction), a tight frame for a space $${\mathscr {H}}$$ of dimension d is real if and only if there is a unitary matrix $$U:{\mathscr {H}}\rightarrow \mathbb {F}^d$$ for which $$Uf_j\in \mathbb {R}^d$$, $$\forall j$$. Moreover, a frame is complex if and only if its complementary frame is.

Example 2.16.

The vertices of a simplex (or partition frame) are a real frame, by definition. The second prototypical example () and the example (a) of Exer. are complex frames, since their Gramians (after normalisation) are
\begin{aligned} \begin{bmatrix} {1\over 2}&{1\over 2\sqrt{3}}&{1\over 2\sqrt{3}}&{-i\over 2\sqrt{3}} \\ {1\over 2\sqrt{3}}&{1\over 2}&{-i\over 2\sqrt{3}}&{1\over 2\sqrt{3}} \\ {1\over 2\sqrt{3}}&{i\over 2\sqrt{3}}&{1\over 2}&{-1\over 2\sqrt{3}} \\ {i\over 2\sqrt{3}}&{1\over 2\sqrt{3}}&{-1\over 2\sqrt{3}}&{1\over 2} \end{bmatrix}, \qquad \begin{bmatrix} {2\over 3}&{-\omega ^2\over 3}&-{\omega \over 3} \\ {-\omega \over 3}&{2\over 3}&-{\omega ^2\over 3} \\ {-\omega ^2\over 3}&-{\omega \over 3}&{2\over 3} \end{bmatrix}. \end{aligned}
(2.17)

There are intrinsic differences between the classes of real and complex frames, e.g., see Exer. , §, §, § and §.

The real algebraic variety of real and complex normalised finite tight frames (and unit-norm tight frames) is considered in Chapter .

Remark 2.1.

One could extend the Definition 2.8 to other fields, e.g., say that the three equally spaced unit vectors in $$\mathbb {R}^2$$ are a rational tight frame, since their Gramian has rational entries. In this case, the columns of the Gramian give a copy of this frame in a rational inner product space (Example 2.9), but the row construction (Example 2.11) does not give a copy in $$\mathbb {Q}^2$$ (with the Euclidean inner product). These ideas are explored in [CFW15].

2.11 SICs and MUBs

There are many interesting and useful examples of equal-norm tight frames $$(v_j)$$, e.g., group frames (see §). Those for which the cross-correlation $$|\langle v_j, v_k\rangle |$$, $$j\ne k$$, takes a small number of values are of particular interest (especially for applications). We briefly mention two such classes of frames: the SICs and the MUBs. These are simple to describe, and come with some intriguing conjectures, which are still unproven (despite considerable work on them). Indeed, the construction of SICs (see §) and maximal sets of MUBs (see §, §) are two central problems in the theory of finite tight frames.

Definition 2.9.

A tight frame of $$d^2$$ unit vectors $$(v_j)$$ for $$\mathbb {C}^d$$ is a SIC if
$$|\langle v_j, v_k\rangle |^2 ={1\over d+1}, \qquad j\ne k.$$

SICs can be viewed as maximal sets of complex equiangular lines. It follows from the bounds of Theorem on such lines that SICs are complex frames. Their origins as quantum measurements and the known constructions are detailed in Chapter . The conjecture that SICs exist in every dimension d is known as Zauner’s conjecture or the SIC problem.

Example 2.17.

The second prototypical example () is a SIC for $$\mathbb {C}^2$$. The case $$d=3$$ seems to be an exception for SICs. Here the SICs form a continuous family, while for $$d\ne 3$$, there is currently only a finite number of SICs for $$\mathbb {C}^d$$ known.

Definition 2.10.

A tight frame consisting of m orthonormal bases $${\mathscr {B}}_1,\ldots ,{\mathscr {B}}_m$$ is said to be a MUB (or a set of m MUBs) for $$\mathbb {C}^d$$ if the bases are mutually unbiased, i.e.,
$$|\langle v, w\rangle |^2={1\over {d}}, \qquad v\in {\mathscr {B}}_j, \ w\in {\mathscr {B}}_k, \quad j\ne k.$$
Mutually unbiased bases have similar uses in quantum-state determination as SICs do. The maximal number $${\mathscr {M}}(d)$$ of MUBs for $$\mathbb {C}^d$$ is bounded above by $$d+1$$ (Proposition ). This bound is attained for d a prime power. Beyond this not much is known [BWB10], e.g., for $$d=6$$ (the first d which is not a prime power), it is only known that
$$3\le {\mathscr {M}}(6)\le 7.$$
The MUB problem is to say anything more, e.g., to show that $${\mathscr {M}}(6)=3$$ (as is commonly believed).

Example 2.18.

Three mutually unbiased bases in $$\mathbb {C}^2$$ are given by
$${\mathscr {B}}_1=\Bigl \{\begin{bmatrix}1\\0\end{bmatrix},\begin{bmatrix}0\\1\end{bmatrix}\Bigr \},\quad {\mathscr {B}}_2=\Bigl \{ {1\over \sqrt{2}}\begin{bmatrix}1\\1\end{bmatrix},{1\over \sqrt{2}}\begin{bmatrix}1\\-1\end{bmatrix}\Bigl \},\quad {\mathscr {B}}_3=\Bigl \{{1\over \sqrt{2}}\begin{bmatrix}1\\i\end{bmatrix},{1\over \sqrt{2}}\begin{bmatrix}1\\-i\end{bmatrix}\Bigr \}.$$
The first two are real MUBs. The question on how many real MUBs there are and its connection with association schemes is of interest (see [LMO10]). Three MUBs for $$\mathbb {R}^4$$ can be obtained by choosing a subset of the vertices of the 24-cell in $$\mathbb {R}^4$$.

Footnotes

1. 1.

This term dates back to [HL00]. Just to confuse matters, the term normalised tight frame has also been used for a tight frame with $$\Vert f_j\Vert =1$$, $$\forall j\in J$$ (we call these unit-norm tight frames).

2. 2.

Note the (jk)-entry of the Gramian is $$\langle f_k, f_j\rangle =f_j^*f_k$$ (so it factors $$V^*V$$), not $$\langle f_j, f_k\rangle$$, which is sometimes used to define the Gramian.