# A note on tight projective $2$-designs

Joseph W. Iverson111Department of Mathematics, Iowa State University, Ames, IA
Emily J. King222Department of Mathematics, Colorado State University, Fort
Collins, CO Dustin G. Mixon333Department of Mathematics, The Ohio State
University, Columbus, OH 444Translational Data Analytics Institute, The Ohio
State University, Columbus, OH

###### Abstract

We study tight projective $2$-designs in three different settings. In the
complex setting, Zauner’s conjecture predicts the existence of a tight
projective $2$-design in every dimension. Pandey, Paulsen, Prakash, and
Rahaman recently proposed an approach to make quantitative progress on this
conjecture in terms of the entanglement breaking rank of a certain quantum
channel. We show that this quantity is equal to the size of the smallest
weighted projective $2$-design. Next, in the finite field setting, we
introduce a notion of projective $2$-designs, we characterize when such
projective $2$-designs are tight, and we provide a construction of such
objects. Finally, in the quaternionic setting, we show that every tight
projective $2$-design for $\mathbb{H}^{d}$ determines an equi-isoclinic tight
fusion frame of $d(2d-1)$ subspaces of $\mathbb{R}^{d(2d+1)}$ of dimension
$3$.

##  1 Introduction

Let $S(\mathbb{C}^{d})$ denote the sphere of $x\in\mathbb{C}^{d}$ with
$\|x\|_{2}^{2}=1$, and consider its uniform probability measure $\sigma$. We
let $\operatorname{Hom}_{d}(t)$ denote the complex vector space spanned by all
monomial functions $\mathbb{C}^{d}\to\mathbb{C}$ that map
$z=(z_{1},\ldots,z_{d})$ to $z_{1}^{\alpha_{1}}\cdots
z_{d}^{\alpha_{d}}\overline{z}_{1}^{\beta_{1}}\cdots\overline{z}_{d}^{\beta_{d}}$
with $t=\sum_{j}\alpha_{j}=\sum_{j}\beta_{j}$. Next, we take $\Pi_{d}^{(t)}$
to denote orthogonal projection onto the symmetric subspace
$(\mathbb{C}^{d})_{\operatorname{sym}}^{\otimes t}$ of
$(\mathbb{C}^{d})^{\otimes t}$. Finally, put $[n]:=\\{1,\ldots,n\\}$. Having
established the neccesssary notation, a projective $t$-design for
$\mathbb{C}^{d}$ is defined to be any $\\{x_{k}\\}_{k\in[n]}$ in
$S(\mathbb{C}^{d})$ that satisfies the following equivalent properties:

######  Proposition 1 (see [45]).

Given $\\{x_{k}\\}_{k\in[n]}$ in $S(\mathbb{C}^{d})$ and $t\in\mathbb{N}$, the
following are equivalent:

  * (a)

$\frac{1}{n}\sum_{k\in[n]}p(x_{k})=\int_{S(\mathbb{C}^{d})}p(x)d\sigma(x)$ for
every $p\in\operatorname{Hom}_{d}(t)$.

  * (b)

$\frac{1}{n}\sum_{k\in[n]}(x_{k}^{\otimes t})(x_{k}^{\otimes
t})^{*}=\binom{d+t-1}{t}^{-1}\cdot\Pi_{d}^{(t)}$.

  * (c)

$\frac{1}{n^{2}}\sum_{k\in[n]}\sum_{\ell\in[n]}|\langle
x_{k},x_{\ell}\rangle|^{2t}=\binom{d+t-1}{t}^{-1}$.

In words, the cubature rule Proposition 1(a) says that, for the purposes of
integration over the sphere $S(\mathbb{C}^{d})$, a projective $2$-design
“fools” every $p\in\operatorname{Hom}_{d}(t)$ by mimicking the entire sphere.
Note that one may “trace out” one of the $t$ subsystems in Proposition 1(b) to
show that every projective $t$-design is also a projective $(t-1)$-design.
Proposition 1(b) says that $\\{x_{k}^{\otimes t}\\}_{k\in[n]}$ forms what
frame theorists would call a unit norm tight frame [9] for the
$\binom{d+t-1}{t}$-dimensional complex Hilbert space
$(\mathbb{C}^{d})^{\otimes t}_{\operatorname{sym}}$. With this perspective,
Proposition 1(c) corresponds to the frame potential [5] of $\\{x_{k}^{\otimes
t}\\}_{k\in[n]}$. One may generalize Proposition 1(c) to obtain notions of
projective $t$-designs for real, quaternionic, and octonionic spaces [36]. The
complex case with $t=2$ is particularly relevant in quantum state tomography
[41], where it is desirable to take $n$ as small as possible. This motivates
the following result, which refers to $\\{x_{k}\\}_{k\in[n]}$ in
$S(\mathbb{C}^{d})$ as equiangular if

$|\\{|\langle x_{k},x_{\ell}\rangle|^{2}:k,\ell\in[n],k\neq\ell\\}|=1.$

######  Proposition 2 (special case of Proposition 1.1 in [3]).

Consider $X=\\{x_{k}\\}_{k\in[n]}$ in $S(\mathbb{C}^{d})$.

  * (a)

If $X$ is a projective $2$-design, then $n\geq d^{2}$ with equality precisely
when $X$ is equiangular.

  * (b)

If $X$ is equiangular, then $n\leq d^{2}$ with equality precisely when $X$ is
a projective $2$-design.

In the case of equality $n=d^{2}$, $\\{x_{k}\\}_{k\in[n]}$ is known as a tight
projective $2$-design for $\mathbb{C}^{d}$, which corresponds to an object in
quantum physics known as a symmetric, informationally complete positive
operator–valued measure [10]. In his Ph.D. thesis [50], Zauner conjectured
that for every $d>1$, there exists a tight projective $2$-design for
$\mathbb{C}^{d}$ of a particular form. To date, there are only finitely many
$d\in\mathbb{N}$ for which a tight projective $2$-design is known to exist
[20, 22, 19], and the conjecture is apparently related to the Stark
conjectures in algebraic number theory [32]. A solution to Zauner’s conjecture
will be rewarded with a $2021$ EUR prize from the National Quantum Information
Centre in Poland [31].

Pandey, Paulsen, Prakash, and Rahaman [38] recently proposed a new approach to
Zauner’s conjecture. They identify an explicit quantum channel
$\mathfrak{Z}_{d}\colon\mathbb{C}^{d\times d}\to\mathbb{C}^{d\times d}$ whose
so-called entanglement breaking rank is
$\operatorname{ebr}(\mathfrak{Z}_{d})\geq d^{2}$, and furthermore, equality
holds if and only if there exists a tight projective $2$-design for
$\mathbb{C}^{d}$. As such, any new upper bound on this entanglement breaking
rank represents quantitative progress towards Zauner’s conjecture. In
addition, Pandey et al. consider various analytic approaches to obtain such
bounds in small dimenions.

As a consequence of Proposition 2, every tight projective $2$-design for
$\mathbb{C}^{d}$ is necessarily an equiangular tight frame (ETF), that is, a
unit norm tight frame that is also equiangular. ETFs correspond to optimal
codes in projective space that achieve equality in the Welch bound [48] (also
known as the simplex bound [12]). By virtue of this optimality, ETFs find
applications in wireless communication [43], compressed sensing [2], and
digital fingerprinting [35]. Motivated by these applications, many ETFs were
recently constructed using various mixtures of algebra and combinatorics [43,
49, 14, 18, 16, 7, 27, 26, 28, 29]; see [17] for a survey. Despite this flurry
of work, several problems involving ETFs (such as Zauner’s conjecture) remain
open, and a finite field model was recently proposed to help study these
remaining problems [23, 24].

Notice that if $\\{x_{k}\\}_{k\in[n]}$ is a tight projective $2$-design for
$\mathbb{C}^{d}$, then Propositions 1 and 2 together imply that
$\\{x_{k}^{\otimes 2}\\}_{k\in[n]}$ is an ETF for $(\mathbb{C}^{d})^{\otimes
2}_{\operatorname{sym}}$. This suggests another approach to Zauner’s
conjecture [1] in which one seeks ETFs of $d^{2}$ vectors in the
$\binom{d+1}{2}$-dimensional complex Hilbert space $(\mathbb{C}^{d})^{\otimes
2}_{\operatorname{sym}}$. There are several known constructions of such ETFs
[17, 7, 27, 26], but in order to correspond to a tight projective $2$-design,
the ETF must consist of rank-$1$ symmetric tensors.

We note that one may leverage linear programming bounds to obtain analogous
results to Proposition 2 that relate projective $t$-designs over different
spaces to different sized angle sets [3]. In the real case, tight projective
$2$-designs have size $\binom{d+1}{2}$ and are only known to exist for
$d\in\\{2,3,7,23\\}$, with $d=119$ being the smallest dimension for which
existence is currently unknown; see [33, 34, 4, 37, 21]. In the quaternion
case, tight projective $2$-designs have size $2d(d-1)$; they are only known to
exist for $d\in\\{2,3\\}$ [11, 15], and there is numerical evidence that they
do not exist for $d\in\\{4,5\\}$ [11]. The octonions are only capable of
supporting a projective space for $d\in\\{2,3\\}$, and tight projective
$2$-designs exist in both cases [11].

In this paper, we study tight projective $2$-designs in three different
settings. In Section 2, we consider the complex setting, specifically, the new
quantitative approach of Pandey et al. [38]. Here, we show that
$\operatorname{ebr}(\mathfrak{Z}_{d})$ is precisely the size of the smallest
weighted projective $2$-design for $\mathbb{C}^{d}$. This identification
allows us to find new upper bounds on $\operatorname{ebr}(\mathfrak{Z}_{d})$.
Next, in Section 3, we use Proposition 1(b) to find an analog of projective
$2$-designs in a finite field setting. This continues the line of inquiry from
[23, 24] of tackling hard problems from frame theory in a finite field model.
In this setting, we obtain an analog of Proposition 2, and then we construct a
family of tight projective $2$-designs. Finally in Section 4, we consider the
quaternionic setting, where we take inspiration from the fact that a tight
projective $2$-design for $\mathbb{C}^{d}$ can be used to produce an ETF of
$d^{2}$ vectors in $(\mathbb{C}^{d})^{\otimes 2}_{\operatorname{sym}}$. In
particular, we show how a tight projective $2$-design for $\mathbb{H}^{d}$ can
be used to produce an equi-isoclinic tight fusion frame of $2d(d-1)$ different
$3$-dimensional subspaces of the $d(2d+1)$-dimensional real Hilbert space of
$d\times d$ quaternionic anti-Hermitian matrices.

##  2 The complex setting

A linear map $\Phi\colon\mathbb{C}^{d\times d}\to\mathbb{C}^{m\times m}$ is
said to be entanglement breaking if it admits an entanglement breaking
decomposition:

$\Phi(X)=\sum_{k\in[n]}R_{k}XR_{k}^{*},\qquad\sum_{k\in[n]}R_{k}^{*}R_{k}=I_{d},\qquad\operatorname{rank}R_{k}=1~{}~{}\text{for
every}~{}~{}k\in[n].$ (1)

The entanglement breaking rank of $\Phi$, denoted by
$\operatorname{ebr}(\Phi)$, is the smallest $n$ for which there exists
$\\{R_{k}\\}_{k\in[n]}$ in $\mathbb{C}^{m\times d}$ satisfying (1). Let
$\\{e_{i}\\}_{i\in[d]}$ denote the standard basis in $\mathbb{C}^{d}$. The
Choi matrix of $\Phi$ is given by

$C_{\Phi}:=\sum_{i\in[d]}\sum_{j\in[d]}e_{i}e_{j}^{*}\otimes\Phi(e_{i}e_{j}^{*})\in\mathbb{C}^{d\times
d}\otimes\mathbb{C}^{m\times m}.$

In words, $C_{\Phi}$ is a $d\times d$ block array whose $(i,j)$th block is
$\Phi(e_{i}e_{j}^{*})\in\mathbb{C}^{m\times m}$. One may use $C_{\Phi}$ to
discern useful properties about $\Phi$. For example, $\Phi$ is completely
positive when $C_{\Phi}$ is positive semidefinite; see Theorem 2.22 in [47].

We are interested in the quantum depolarizing channel
$\mathfrak{Z}_{d}\colon\mathbb{C}^{d\times d}\to\mathbb{C}^{d\times d}$
defined by

$\mathfrak{Z}_{d}(X):=\frac{1}{d+1}\Big{(}X+\operatorname{tr}X\cdot
I_{d}\Big{)}.$

One may verify that $\mathfrak{Z}_{d}$ is entanglement breaking as a
consequence of its scaled Choi matrix $\frac{1}{d}C_{\mathfrak{Z}_{d}}$ being
a separable bipartite state (specifically, the isotropic state with
$\lambda=1/d$ from Example 7.25 in [47]). Pandey, Paulsen, Prakash, and
Rahaman [38] pointed to this quantum channel as an opportunity for
quantitative progress on Zauner’s conjecture:

######  Proposition 3 (cf. Corollary III.3 and Theorem V.3 in [38]).

  * (a)

$\operatorname{ebr}(\mathfrak{Z}_{d})\geq d^{2}$, with equality if and only if
there exists a tight projective $2$-design for $\mathbb{C}^{d}$.

  * (b)

$\operatorname{ebr}(\mathfrak{Z}_{d})\leq d^{2}+d$ whenever $d$ is a prime
power.

We say unit vectors $\\{x_{k}\\}_{k\in[n]}$ in $\mathbb{C}^{d}$ form a
weighted projective $t$-design if there exist weights $\\{w_{k}\\}_{k\in[n]}$
such that

$\sum_{k\in[n]}w_{k}(x_{k}^{\otimes t})(x_{k}^{\otimes
t})^{*}=\tbinom{d+t-1}{t}^{-1}\cdot\Pi_{d}^{(t)},\qquad\sum_{k\in[n]}w_{k}=1,\qquad
w_{k}\geq 0,\qquad k\in[n].$

Notice that every projective $t$-design is a weighted projective $t$-design
with weights $w_{k}=1/n$. In addition, it is known that every weighted
projective $2$-design for $\mathbb{C}^{d}$ has size at least $d^{2}$, and if
equality holds, then the weights are all $1/n$; see Theorem 4 in [41]. What
follows is the main result of this section, of which Proposition 3 is a
corollary:

######  Theorem 4.

The smallest weighted projective $2$-design for $\mathbb{C}^{d}$ has size
$\operatorname{ebr}(\mathfrak{Z}_{d})$.

In fact, one may use Theorem 4 to improve upon Proposition 3(b) by collecting
various weighted $2$-designs from the literature. Specifically, Theorem 4.1,
Proposition 4.2, and Corollary 4.2 in [40], and Corollaries 4.4 and 4.6 in [6]
give the following:

######  Corollary 5.

  * (a)

$\operatorname{ebr}(\mathfrak{Z}_{d})\leq kd^{2}+2d$ whenever $kd+1$ is a
prime power with $k\in\mathbb{N}$.

  * (b)

$\operatorname{ebr}(\mathfrak{Z}_{d})\leq d^{2}+(p+1)d$ whenever $d+1=p^{k}$
with $p$ prime and $k\in\mathbb{N}$.

  * (c)

$\operatorname{ebr}(\mathfrak{Z}_{d})\leq d^{2}+1$ whenever $d-1$ is a prime
power.

  * (d)

$\operatorname{ebr}(\mathfrak{Z}_{d})\leq d^{2}+d-1$ whenever $d$ is a prime
power.

Since $\mathfrak{Z}_{d}$ is entanglement breaking, Theorem 4 also implies the
existence of weighted projective $2$-designs; we note that this also follows
from the main result in [42].

######  Corollary 6.

For each $d\in\mathbb{N}$, there is a weighted projective $2$-design for
$\mathbb{C}^{d}$ of size $\binom{d+1}{2}^{2}$.

###### Proof.

Since $\mathfrak{Z}_{d}$ is entanglement breaking, Theorem 4 promises a
weighted projective $2$-design $\\{x_{k}\\}_{k\in[n]}$ for $\mathbb{C}^{d}$.
Then $\Pi_{d}^{(2)}$ resides in the conic hull of $\\{(x_{k}^{\otimes
2})(x_{k}^{\otimes 2})^{*}\\}_{k\in[n]}$, which in turn is contained in the
$\binom{d+1}{2}^{2}$-dimensional real vector space of Hermitian operators over
$(\mathbb{C}^{d})^{\otimes 2}_{\operatorname{sym}}$. By Carathéodory’s
theorem, there exists $S\subseteq[n]$ with $|S|\leq\binom{d+1}{2}^{2}$ and
weights $w_{k}\geq 0$ for $k\in S$ such that

$\sum_{k\in S}w_{k}(x_{k}^{\otimes 2})(x_{k}^{\otimes
2})^{*}=\tbinom{d+1}{2}^{-1}\cdot\Pi_{d}^{(2)}.$

Furthermore, taking the trace of both sides reveals that $\sum_{k\in
S}w_{k}=1$. As such, $\\{x_{k}\\}_{k\in S}$ is a weighted projective
$2$-design. ∎

The remainder of this section proves Theorem 4. We first collect a few helpful
lemmas. Let $T\colon\mathbb{C}^{d\times d}\to\mathbb{C}^{d\times d}$ denote
the tranposition operator defined by $T(X):=X^{\top}$.

######  Lemma 7 (cf. Proposition III.5 in [38]).

It holds that $\operatorname{ebr}(\Phi)=\operatorname{ebr}(T\circ\Phi)$ with

$\Phi(X)=\sum_{k\in[n]}x_{k}y_{k}^{\top}X(x_{k}y_{k}^{\top})^{*}\qquad\Longleftrightarrow\qquad(T\circ\Phi)(X)=\sum_{k\in[n]}\overline{x}_{k}y_{k}^{\top}X(\overline{x}_{k}y_{k}^{\top})^{*}.$

###### Proof.

The claim follows from the following manipulation:

$\bigg{(}\sum_{k\in[n]}x_{k}y_{k}^{\top}X\overline{y}_{k}\overline{x}_{k}^{\top}\bigg{)}^{\top}=\sum_{k\in[n]}\overline{x}_{k}(\overline{y}_{k}^{\top}X^{\top}y_{k})x_{k}^{\top}=\sum_{k\in[n]}\overline{x}_{k}(\overline{y}_{k}^{\top}X^{\top}y_{k})^{\top}x_{k}^{\top}=\sum_{k\in[n]}\overline{x}_{k}y_{k}^{\top}X\overline{y}_{k}x_{k}^{\top},$

where the second step takes the transpose of a scalar. Indeed, we have both
$(x_{k}y_{k}^{\top})^{*}=\overline{y}_{k}\overline{x}_{k}^{\top}$ and
$(\overline{x}_{k}y_{k}^{\top})^{*}=\overline{y}_{k}x_{k}^{\top}$. ∎

Both of the following lemmas were implicitly used in the proof of Corollary
III.7 in [38]. To prove them, we will repeatedly use the following identity,
which is valid for any linear $\Phi\colon\mathbb{C}^{d\times
d}\to\mathbb{C}^{m\times m}$, and any $w,y\in\mathbb{C}^{d}$ and
$x,z\in\mathbb{C}^{m}$:

$\displaystyle(w\otimes x)^{\top}C_{\Phi}(y\otimes z)$
$\displaystyle=\sum_{i\in[d]}\sum_{j\in[d]}w_{i}y_{j}\cdot
x^{\top}\Phi(e_{i}e_{j}^{*})z$
$\displaystyle=x^{\top}\Phi\bigg{(}\sum_{i\in[d]}\sum_{j\in[d]}w_{i}y_{j}\cdot
e_{i}e_{j}^{*}\bigg{)}z=x^{\top}\Phi(wy^{\top})z.$ (2)

######  Lemma 8.

An entanglement breaking map $\Phi$ has entanglement breaking decomposition

$\Phi(X)=\sum_{k\in[n]}a_{k}b_{k}^{\top}X(a_{k}b_{k}^{\top})^{*}.$

if and only if $\Phi$ has Choi matrix

$C_{\Phi}=\sum_{k\in[n]}(b_{k}\otimes a_{k})(b_{k}\otimes a_{k})^{*}.$

###### Proof.

($\Rightarrow$) For any $w,x,y,z$, we may apply (2) to get

$\displaystyle(w\otimes x)^{\top}C_{\Phi}(y\otimes z)$
$\displaystyle=x^{\top}\bigg{(}\sum_{k\in[n]}a_{k}b_{k}^{\top}(wy^{\top})\overline{b}_{k}\overline{a}_{k}^{\top}\bigg{)}z$
$\displaystyle=\sum_{k\in[n]}(x^{\top}a_{k}b_{k}^{\top}w)(y^{\top}\overline{b}_{k}\overline{a}_{k}^{\top}z)$
$\displaystyle=\sum_{k\in[n]}(w\otimes x)^{\top}(b_{k}\otimes
a_{k})(\overline{b}_{k}\otimes\overline{a}_{k})^{\top}(y\otimes z)$
$\displaystyle=(w\otimes x)^{\top}\bigg{(}\sum_{k\in[n]}(b_{k}\otimes
a_{k})(b_{k}\otimes a_{k})^{*}\bigg{)}(y\otimes z).$

Since $w,x,y,z$ are arbitrary, the result follows.

($\Leftarrow$) For any $w,x,y,z$, we may similarly apply (2) to get

$\displaystyle x^{\top}\Phi(wy^{\top})z$ $\displaystyle=(w\otimes
x)^{\top}C_{\Phi}(y\otimes z)$ $\displaystyle=(w\otimes
x)^{\top}\bigg{(}\sum_{k\in[n]}(b_{k}\otimes a_{k})(b_{k}\otimes
a_{k})^{*}\bigg{)}(y\otimes z)$
$\displaystyle=x^{\top}\bigg{(}\sum_{k\in[n]}a_{k}b_{k}^{\top}(wy^{\top})\overline{b}_{k}\overline{a}_{k}^{\top}\bigg{)}z.$

Since $w,x,y,z$ are arbitrary, the result follows. ∎

######  Lemma 9.

$C_{T\circ\mathfrak{Z}_{d}}=\frac{2}{d+1}\Pi_{d}^{(2)}$.

###### Proof.

For any $w,x,y,z$, we may apply (2) to get

$\displaystyle(w\otimes x)^{\top}C_{T\circ\mathfrak{Z}_{d}}(y\otimes z)$
$\displaystyle=x^{\top}\Big{[}(T\circ\mathfrak{Z}_{d})(wy^{\top})\Big{]}z$
$\displaystyle=x^{\top}\frac{1}{d+1}\Big{(}yw^{\top}+\operatorname{tr}wy^{\top}\cdot
I_{d}\Big{)}z$ $\displaystyle=\frac{1}{d+1}\Big{(}w^{\top}z\cdot
x^{\top}y+w^{\top}y\cdot x^{\top}z\Big{)}=(w\otimes
x)^{\top}\frac{1}{d+1}\Big{(}z\otimes y+y\otimes z\Big{)}.$

Since $w,x$ are arbitrary, it follows that

$C_{T\circ\mathfrak{Z}_{d}}(y\otimes z)=\frac{2}{d+1}\cdot\frac{1}{2}(z\otimes
y+y\otimes z)=\frac{2}{d+1}\Pi_{d}^{(2)}(y\otimes z).$

Since $y,z$ are arbitrary, the result follows. ∎

We are now ready to prove the main result of this section.

###### Proof of Theorem 4.

Suppose $\\{x_{k}\\}_{k\in[n]}$ is a weighted projective $2$-design for
$\mathbb{C}^{d}$ with weights $\\{w_{k}\\}_{k\in[n]}$. We claim that
$\operatorname{ebr}(\mathfrak{Z}_{d})\leq n$. To see this, first recall that

$\sum_{k\in[n]}w_{k}(x_{k}\otimes x_{k})(x_{k}\otimes
x_{k})^{*}=\frac{2}{d(d+1)}\cdot\Pi_{d}^{(2)}.$

Taking $a_{k}:=x_{k}$ and $b_{k}:=\sqrt{dw_{k}}x_{k}$ then gives

$\sum_{k\in[n]}(b_{k}\otimes a_{k})(b_{k}\otimes
a_{k})^{*}=d\sum_{k\in[n]}w_{k}(x_{k}\otimes x_{k})(x_{k}\otimes
x_{k})^{*}=\frac{2}{d+1}\cdot\Pi_{d}^{(2)}=C_{T\circ\mathfrak{Z}_{d}},$

where the last step applies Lemma 9. Lemmas 7 and 8 then imply that

$\operatorname{ebr}(\mathfrak{Z}_{d})=\operatorname{ebr}(T\circ\mathfrak{Z}_{d})\leq
n.$

Next, suppose
$n=\operatorname{ebr}(\mathfrak{Z}_{d})=\operatorname{ebr}(T\circ\mathfrak{Z}_{d})$
and consider entanglement breaking decomposition

$(T\circ\Phi)(X)=\sum_{k\in[n]}a_{k}b_{k}^{\top}X(a_{k}b_{k}^{\top})^{*}.$

Then Lemmas 8 and 9 together imply

$\sum_{k\in[n]}(b_{k}\otimes a_{k})(b_{k}\otimes
a_{k})^{*}=\frac{2}{d+1}\cdot\Pi_{d}^{(2)}.$

Notice that for every antisymmetric $x\in(\mathbb{C}^{d})^{\otimes 2}$, it
holds that

$\sum_{k\in[n]}|\langle b_{k}\otimes
a_{k},x\rangle|^{2}=x^{*}\bigg{(}\sum_{k\in[n]}(b_{k}\otimes
a_{k})(b_{k}\otimes a_{k})^{*}\bigg{)}x=x^{*}\frac{2}{d+1}\Pi_{d}^{(2)}x=0.$

It follows that each $b_{k}\otimes a_{k}$ is necessarily symmetric, and
therefore takes the form $\sqrt{dw_{k}}x_{k}\otimes x_{k}$ for some unit
vector $x_{k}$ and scalar $w_{k}\geq 0$. Then

$\sum_{k\in[n]}w_{k}(x_{k}\otimes x_{k})(x_{k}\otimes
x_{k})^{*}=\frac{2}{d(d+1)}\cdot\Pi_{d}^{(2)}.$

The fact that $\sum_{k\in[n]}w_{k}=1$ follows from taking the trace of both
sides. Overall, there exists a weighted projective $2$-design for
$\mathbb{C}^{d}$ of size $n=\operatorname{ebr}(\mathfrak{Z}_{d})$, as claimed.
∎

##  3 The finite field setting

In this section, we introduce a notion of projective $2$-designs in a finite
field setting. Here, we will find an analog to Proposition 2 in which tight
projective $2$-designs are identified as maximal systems of equiangular lines,
and then we will provide several examples. We start by reviewing some
preliminaries; the reader is encouraged to see [23] for more information.

Let $q$ be a prime power. Given $a\in\mathbb{F}_{q^{2}}$, we abbreviate
$\overline{a}=a^{q}$ for its image under the Frobenius automorphism fixing
$\mathbb{F}_{q}\leq\mathbb{F}_{q^{2}}$. The conjugate transpose of a matrix
$A$ is denoted by $A^{*}$. We consider $\mathbb{F}_{q^{2}}^{d}$ under the
nondegenerate Hermitian form $\langle x,y\rangle=x^{*}y$, which is notably
conjugate-linear in the first variable. A subspace
$V\leq\mathbb{F}_{q^{2}}^{d}$ is called nondegenerate if $V\cap
V^{\perp}=\\{0\\}$, where

$V^{\perp}:=\\{x\in\mathbb{F}_{q^{2}}^{d}:\langle x,y\rangle=0\text{ for every
}y\in V\\}.$

In that case, every $x\in\mathbb{F}_{q^{2}}^{d}$ can be written uniquely as
$x=Px+Qx$ with $Px\in V$ and $Qx\in V^{\perp}$, where
$P\colon\mathbb{F}_{q^{2}}^{d}\to\mathbb{F}_{q^{2}}^{d}$ is orthogonal
projection onto $V$.

######  Definition 10.

Let $V\leq\mathbb{F}_{q^{2}}^{d}$ be nondegenerate. We say
$\\{x_{k}\\}_{k\in[n]}$ in $V$ is a $c$-tight frame for $V$ with constant
$c\in\mathbb{F}_{q}$ if

  * (i)

$\operatorname{span}\\{x_{k}\\}_{k\in[n]}=V$, and

  * (ii)

$\sum_{k\in[n]}\langle x_{k},y\rangle x_{k}=cy$ for every $y\in V$.

For $a\in\mathbb{F}_{q}$, a $c$-tight frame is an equal-norm tight frame, or
$(a,c)$-NTF, if

  * (iii)

$\langle x_{k},x_{k}\rangle=a$ for every $k\in[n]$.

For $b\in\mathbb{F}_{q}$, an $(a,c)$-NTF is an equiangular tight frame, or
$(a,b,c)$-ETF, if

  * (iv)

$\langle x_{k},x_{\ell}\rangle\langle x_{\ell},x_{k}\rangle=b$ for every
$k,\ell\in[n]$ with $k\neq\ell$.

Meanwhile, an $(a,b)$-equiangular system in $V$ satisfies (iii) and (iv), but
not necessarily (i) or (ii).

Notice that (ii) implies (i) if $c\neq 0$. Furthermore, if $P$ is orthogonal
projection onto $V$, then (ii) is equivalent to

  * (ii′)

$\sum_{k\in[n]}x_{k}x_{k}^{*}=cP$.

We will repeatedly make use of the following basic results from [23]:

######  Proposition 11 (Corollary 3.8 in [23]).

If $V\leq\mathbb{F}_{q^{2}}^{d}$ is nondegenerate and $\\{x_{k}\\}_{k\in[n]}$
is a tight frame for $V$ with constant $c=0$, then $n\geq 2\dim V$.

######  Proposition 12 (Equation (3.2) and Proposition 4.7 in [23]).

  * (a)

If $\\{x_{k}\\}_{k\in[n]}$ is an $(a,c)$-NTF for $V$, then $na=c\dim V$.

  * (b)

If $\\{x_{k}\\}_{k\in[n]}$ is an $(a,b,c)$-ETF for $V$, then $a(c-a)=(n-1)b$.

######  Proposition 13 (Gerzon’s bound, see Theorem 4.1 in [23] and its
proof).

If $\\{x_{k}\\}_{k\in[n]}$ is an $(a,b)$-equiangular system in
$\mathbb{F}_{q^{2}}^{d}$ and $a^{2}\neq b$, then $n\leq d^{2}$. If equality
holds and $a\neq 0$, then $\\{x_{k}x_{k}^{*}\\}_{k\in[n]}$ is a basis for the
$\mathbb{F}_{q}$-linear space

$\\{X\in\mathbb{F}_{q^{2}}^{d\times d}:X=X^{*}\\}.$

If equality holds and $a=0$, then the $\mathbb{F}_{q}$-span of
$\\{x_{k}x_{k}^{*}\\}_{k\in[n]}$ is the subspace

$\\{X\in\mathbb{F}_{q^{2}}^{d\times d}:X=X^{*},~{}\operatorname{tr}X=0\\},$

and $\sum_{k\in[n]}x_{k}x_{k}^{*}=0$ is the unique $\mathbb{F}_{q}$-linear
dependency of $\\{x_{k}x_{k}^{*}\\}_{k\in[n]}$ up to a scalar.

###  3.1 Projective 2-designs

Throughout this subsection, we assume $q$ is odd. Let
$e_{1},\dotsc,e_{d}\in\mathbb{F}_{q^{2}}^{d}$ denote the standard basis. Then
$\\{e_{i}\otimes e_{j}\\}_{i,j\in[d]}$ is a basis for
$(\mathbb{F}_{q^{2}}^{d})^{\otimes 2}$. We write

$(\mathbb{F}_{q^{2}}^{d})^{\otimes
2}_{\operatorname{sym}}:=\bigg{\\{}\sum_{i\in[d]}\sum_{j\in[d]}c_{ij}(e_{i}\otimes
e_{j}):c_{ij}=c_{ji}\text{ for every }i,j\in[d]\bigg{\\}}$

for the subspace of symmetric tensors, and we define

$\Pi_{d}^{(2)}:=\frac{1}{2}\sum_{i\in[d]}\sum_{j\in[d]}\Big{(}e_{i}e_{i}^{*}\otimes
e_{j}e_{j}^{*}+e_{i}e_{j}^{*}\otimes
e_{j}e_{i}^{*}\Big{)}\in(\mathbb{F}_{q^{2}}^{d\times d})^{\otimes 2}.$

######  Lemma 14.

$(\mathbb{F}_{q^{2}}^{d})^{\otimes
2}_{\operatorname{sym}}\leq(\mathbb{F}_{q^{2}}^{d})^{\otimes 2}$ is
nondegenerate, and $\Pi_{d}^{(2)}$ is its orthogonal projection.

###### Proof.

For nondegeneracy, let $y=\sum_{i\in[d]}\sum_{j\in[d]}a_{ij}(e_{i}\otimes
e_{j})\in(\mathbb{F}_{q^{2}}^{d})^{\otimes 2}_{\operatorname{sym}}$ be
nonzero. Then there exist $k,\ell\in[d]$ such that $a_{k\ell}=a_{\ell k}\neq
0$. It follows that

$\displaystyle\langle y,e_{k}\otimes e_{\ell}+e_{\ell}\otimes e_{k}\rangle$
$\displaystyle=\sum_{i\in[d]}\sum_{j\in[d]}\overline{a_{ij}}\langle
e_{i}\otimes e_{j},e_{k}\otimes
e_{\ell}\rangle+\sum_{i\in[d]}\sum_{j\in[d]}\overline{a_{ij}}\langle
e_{i}\otimes e_{j},e_{\ell}\otimes e_{k}\rangle=2\overline{a_{k\ell}},$

which is nonzero by assumption. Thus, $y$ is not orthogonal to
$(\mathbb{F}_{q^{2}}^{d})^{\otimes 2}_{\operatorname{sym}}$.

Next, we show that $\Pi_{d}^{(2)}$ projects orthogonally onto
$(\mathbb{F}_{q^{2}}^{d})^{\otimes 2}_{\operatorname{sym}}$. To this end,
choose any vector $x=\sum_{k\in[d]}\sum_{\ell\in[d]}c_{k\ell}(e_{k}\otimes
e_{\ell})$ and compute

$\displaystyle\Pi_{d}^{(2)}x$
$\displaystyle=\frac{1}{2}\sum_{i\in[d]}\sum_{j\in[d]}\sum_{k\in[d]}\sum_{\ell\in[d]}c_{k\ell}\Big{(}(e_{i}e_{i}^{*}\otimes
e_{j}e_{j}^{*})(e_{k}\otimes e_{\ell})+(e_{i}e_{j}^{*}\otimes
e_{j}e_{i}^{*})(e_{k}\otimes e_{\ell})\Big{)}$
$\displaystyle=\frac{1}{2}\sum_{i\in[d]}\sum_{j\in[d]}\sum_{k\in[d]}\sum_{\ell\in[d]}c_{k\ell}\Big{(}(e_{i}e_{i}^{*})e_{k}\otimes(e_{j}e_{j}^{*})e_{\ell}+(e_{i}e_{j}^{*})e_{k}\otimes(e_{j}e_{i}^{*})e_{\ell}\Big{)}$
$\displaystyle=\frac{1}{2}\sum_{k\in[d]}\sum_{\ell\in[d]}c_{k\ell}\bigg{(}\sum_{i\in[d]}\sum_{j\in[d]}(e_{i}e_{i}^{*})e_{k}\otimes(e_{j}e_{j}^{*})e_{\ell}+\sum_{i\in[d]}\sum_{j\in[d]}(e_{i}e_{j}^{*})e_{k}\otimes(e_{j}e_{i}^{*})e_{\ell}\bigg{)}$
$\displaystyle=\frac{1}{2}\sum_{k\in[d]}\sum_{\ell\in[d]}c_{k\ell}(e_{k}\otimes
e_{\ell}+e_{\ell}\otimes e_{k}),$

which belongs to $(\mathbb{F}_{q^{2}}^{d})^{\otimes 2}_{\operatorname{sym}}$.
Then

$x-\Pi_{d}^{(2)}x=\frac{1}{2}\sum_{k\in[d]}\sum_{\ell\in[d]}c_{k\ell}(e_{k}\otimes
e_{\ell}-e_{\ell}\otimes e_{k}),$

and it is straightforward to check this is orthogonal to every
$y\in(\mathbb{F}_{q^{2}}^{d})^{\otimes 2}_{\operatorname{sym}}$. As such,
$x=(\Pi_{d}^{(2)}x)+(x-\Pi_{d}^{(2)}x)$ gives the desired decomposition of
$x$. ∎

As a consequence of Lemma 14, we may consider tight frames over
$(\mathbb{F}_{q^{2}}^{d})^{\otimes 2}_{\operatorname{sym}}$. This allows us to
define the following analog of Proposition 1(b):

######  Definition 15.

$\\{x_{k}\\}_{k\in[n]}$ in $\mathbb{F}_{q^{2}}^{d}$ is an
$(a,c_{1},c_{2})$-projective $2$-design if $a,c_{1},c_{2}\in\mathbb{F}_{q}$
and

  * (i)

$\langle x_{k},x_{k}\rangle=a$ for every $k\in[n]$,

  * (ii)

$\\{x_{k}\\}_{k\in[n]}$ is a $c_{1}$-tight frame for $\mathbb{F}_{q^{2}}^{d}$,
and

  * (iii)

$\\{x_{k}^{\otimes 2}\\}_{k\in[n]}$ is a $c_{2}$-tight frame for
$(\mathbb{F}_{q^{2}}^{d})^{\otimes 2}_{\operatorname{sym}}$.

With this definition, the finite field setting enjoys an analogy to
Proposition 2:

######  Theorem 16.

If $\\{x_{k}\\}_{k\in[n]}$ in $\mathbb{F}_{q^{2}}^{d}$ is a projective
$2$-design, then $n\geq d^{2}$.

To prove this theorem, we will repeatedly make use of the following:

######  Lemma 17.

If $\\{x_{k}\\}_{k\in[n]}$ in $\mathbb{F}_{q^{2}}^{d}$ is an
$(a,c_{1},c_{2})$-projective $2$-design with $c_{2}\neq 0$, then for every
$A\in\mathbb{F}_{q^{2}}^{d\times d}$, it holds that

$A=\frac{2}{c_{2}}\sum_{k\in[n]}x_{k}x_{k}^{*}Ax_{k}x_{k}^{*}-\operatorname{tr}A\cdot
I_{d}.$

###### Proof.

To prove the result, we multiply both sides of the identity
$\sum_{k\in[n]}(x_{k}^{\otimes 2})(x_{k}^{\otimes
2})^{*}=c_{2}\cdot\Pi_{d}^{(2)}$ by $A\otimes I_{d}$ and then “trace out” the
first subsystem. Explicitly, we define the partial trace
$\operatorname{tr}_{1}\colon\mathbb{F}_{q^{2}}^{d\times
d}\otimes\mathbb{F}_{q^{2}}^{d\times d}\to\mathbb{F}_{q^{2}}^{d\times d}$ by
taking

$\operatorname{tr}_{1}(A\otimes B):=\operatorname{tr}(A)\cdot B$

and extending linearly. Since $\\{x_{k}\\}_{k\in[n]}$ is a projective
$2$-design, we have

$\operatorname{tr}_{1}\bigg{(}\sum_{k\in[n]}(x_{k}^{\otimes 2})(x_{k}^{\otimes
2})^{*}(A\otimes
I_{d})\bigg{)}=\operatorname{tr}_{1}\Big{(}c_{2}\cdot\Pi_{d}^{(2)}\cdot(A\otimes
I_{d})\Big{)}.$ (3)

We cycle the trace to simplify the left-hand side of (3):

$\displaystyle\operatorname{tr}_{1}\bigg{(}\sum_{k\in[n]}(x_{k}^{\otimes
2})(x_{k}^{\otimes 2})^{*}(A\otimes I_{d})\bigg{)}$
$\displaystyle=\operatorname{tr}_{1}\bigg{(}\sum_{k\in[n]}(x_{k}x_{k}^{*})^{\otimes
2}(A\otimes I_{d})\bigg{)}$
$\displaystyle=\operatorname{tr}_{1}\bigg{(}\sum_{k\in[n]}x_{k}x_{k}^{*}A\otimes
x_{k}x_{k}^{*}\bigg{)}$
$\displaystyle=\sum_{k\in[n]}\operatorname{tr}(x_{k}x_{k}^{*}A)\cdot
x_{k}x_{k}^{*}=\sum_{k\in[n]}x_{k}(x_{k}^{*}Ax_{k})x_{k}^{*}.$ (4)

For the right-hand side of (3), we apply the definition of $\Pi_{d}^{(2)}$:

$\displaystyle\operatorname{tr}_{1}\Big{(}c_{2}\cdot\Pi_{d}^{(2)}\cdot(A\otimes
I_{d})\Big{)}$
$\displaystyle=\operatorname{tr}_{1}\bigg{(}\frac{c_{2}}{2}\sum_{i\in[d]}\sum_{j\in[d]}\Big{(}e_{i}e_{i}^{*}\otimes
e_{j}e_{j}^{*}+e_{i}e_{j}^{*}\otimes e_{j}e_{i}^{*}\Big{)}(A\otimes
I_{d})\bigg{)}$
$\displaystyle=\frac{c_{2}}{2}\sum_{i\in[d]}\sum_{j\in[d]}\Big{(}\operatorname{tr}(e_{i}e_{i}^{*}A)\cdot
e_{j}e_{j}^{*}+\operatorname{tr}(e_{i}e_{j}^{*}A)\cdot e_{j}e_{i}^{*}\Big{)}$
$\displaystyle=\frac{c_{2}}{2}\Big{(}\operatorname{tr}A\cdot I_{d}+A\Big{)}.$
(5)

The result follows by equating (4) to (5) and rearranging. ∎

###### Proof of Theorem 16.

Suppose $\\{x_{k}\\}_{k\in[n]}$ in $\mathbb{F}_{q^{2}}^{d}$ is an
$(a,c_{1},c_{2})$-projective $2$-design.

Case I: $c_{2}=0$. Then $\\{x_{k}x_{k}^{*}\\}_{k\in[n]}$ is an $(a^{2},0)$-NTF
for $(\mathbb{F}_{q^{2}}^{d})_{\operatorname{sym}}^{\otimes 2}$. By
Proposition 11,

$n\geq
2\cdot\operatorname{dim}((\mathbb{F}_{q^{2}}^{d})_{\operatorname{sym}}^{\otimes
2})=d^{2}+d>d^{2}.$

Case II: $c_{1}\neq 0$ and $c_{2}\neq 0$. Apply Lemma 17 and the identity
$\sum_{k\in[n]}x_{k}x_{k}^{*}=c_{1}\cdot I_{d}$:

$\displaystyle A$
$\displaystyle=\frac{2}{c_{2}}\sum_{k\in[n]}x_{k}x_{k}^{*}Ax_{k}x_{k}^{*}-\operatorname{tr}A\cdot
I_{d}$
$\displaystyle=\frac{2}{c_{2}}\sum_{k\in[n]}x_{k}x_{k}^{*}Ax_{k}x_{k}^{*}-\operatorname{tr}A\cdot\frac{1}{c_{1}}\sum_{k\in[n]}x_{k}x_{k}^{*}=\sum_{k\in[n]}\bigg{(}\frac{2}{c_{2}}x_{k}^{*}Ax_{k}-\frac{1}{c_{1}}\operatorname{tr}A\bigg{)}x_{k}x_{k}^{*}$

for every $A\in\mathbb{F}_{q^{2}}^{d\times d}$. It follows that the
$\mathbb{F}_{q^{2}}$-span of $\\{x_{k}x_{k}^{*}\\}_{k\in[n]}$ is
$\mathbb{F}_{q^{2}}^{d\times d}$, and so $n\geq d^{2}$.

Case III: $c_{1}=0$ and $c_{2}\neq 0$. Lemma 17 implies that the
$\mathbb{F}_{q^{2}}$-span of $\\{x_{k}x_{k}^{*}\\}_{k\in[n]}\cup\\{I_{d}\\}$
is $\mathbb{F}_{q^{2}}^{d\times d}$, while the identity
$\sum_{k\in[n]}x_{k}x_{k}^{*}=c_{1}\cdot I_{d}=0$ implies that
$\\{x_{k}x_{k}^{*}\\}_{k\in[n]}$ is linearly dependent. Since
$\\{x_{k}x_{k}^{*}\\}_{k\in[n]}\cup\\{I_{d}\\}$ is a linearly dependent
spanning set of size $n+1$, it follows that $n+1\geq d^{2}+1$, i.e., $n\geq
d^{2}$. ∎

######  Theorem 18.

Any two of the following statements together imply the third statement:

  * (a)

$\\{x_{k}\\}_{k\in[n]}$ in $\mathbb{F}_{q^{2}}^{d}$ is a projective
$2$-design.

  * (b)

$n=d^{2}$.

  * (c)

There exist $a,b,c_{1}\in\mathbb{F}_{q}$ such that

    * (i)

$a^{2}\neq b$,

    * (ii)

$a^{2}-b=\frac{bc_{1}}{a}$ if $a\neq 0$,

    * (iii)

$d\equiv-1\bmod p$ if $a=0$, and

    * (iv)

$\\{x_{k}\\}_{k\in[n]}$ in $\mathbb{F}_{q^{2}}^{d}$ is an
$(a,b,c_{1})$-equiangular tight frame.

When (a), (b), and (c) hold, $\\{x_{k}\\}_{k\in[n]}$ is an
$(a,c_{1},c_{2})$-projective $2$-design with $c_{2}=2(a^{2}-b)$.

To prove Theorem 18, we need a method of demonstrating that a collection of
vectors forms a projective $2$-design. For this, we will apply the following:

######  Lemma 19.

Take $\mathbb{F}$ to be $\mathbb{C}$ or $\mathbb{F}_{q^{2}}$. Given
$\\{x_{k}\\}_{k\in[n]}$ in $\mathbb{F}^{d}$, define
$\Psi\colon\mathbb{F}^{d\times d}\to\mathbb{F}^{d\times d}$ by

$\Psi(A):=\sum_{k\in[n]}x_{k}x_{k}^{*}A^{*}x_{k}x_{k}^{*}.$

Then

$\sum_{k\in[n]}(x_{k}^{\otimes 2})(x_{k}^{\otimes
2})^{*}=\sum_{i\in[d]}\sum_{j\in[d]}e_{i}e_{j}^{*}\otimes\Psi(e_{i}e_{j}^{*}).$

To prove Lemma 19, we will use lemmas from the previous section, with the
appropriate interpretation of conjugation in the case
$\mathbb{F}=\mathbb{F}_{q^{2}}$; indeed, the proofs of these results are valid
under this interpretation.

###### Proof of Lemma 19.

Consider the linear map $\overline{\Psi}\colon\mathbb{F}^{d\times
d}\to\mathbb{F}^{d\times d}$ defined by
$\overline{\Psi}(A):=\overline{\Psi(A)}$. Then

$(T\circ\overline{\Psi})(A)=\Psi(A)^{*}=\sum_{k\in[n]}x_{k}x_{k}^{*}Ax_{k}x_{k}^{*}=\sum_{k\in[n]}x_{k}\overline{x}_{k}^{\top}A(x_{k}\overline{x}_{k}^{\top})^{*}.$

Lemma 7 then gives

$\overline{\Psi}(A)=\sum_{k\in[n]}\overline{x}_{k}\overline{x}_{k}^{\top}A(\overline{x}_{k}\overline{x}_{k}^{\top})^{*}.$

Finally, we apply Lemma 8 to get

$\sum_{k\in[n]}(\overline{x}_{k}^{\otimes 2})(\overline{x}_{k}^{\otimes
2})^{*}=\sum_{i\in[d]}\sum_{j\in[d]}e_{i}e_{j}^{*}\otimes\overline{\Psi}(e_{i}e_{j}^{*}),$

and we take conjugates of both sides to obtain the result. ∎

###### Proof of Theorem 18.

First, (a)$\wedge$(c)$\Rightarrow$(b) follows from Theorem 16 and Proposition
13.

Next, we demonstrate (a)$\wedge$(b)$\Rightarrow$(c) by considering each case
in the proof of Theorem 16.

Case I: $c_{2}=0$. This case does not occur since $n\geq d^{2}+d$ implies
$n\neq d^{2}$.

For the remaining cases, we have $c_{2}\neq 0$. For these cases, we will use
the fact that $a=0$ if and only if $c_{1}=0$. To see this, apply Lemma 17 with
$A=I_{d}$ and the identity $\sum_{k\in[n]}x_{k}x_{k}^{*}=c_{1}\cdot I_{d}$ to
get

$I_{d}=\frac{2}{c_{2}}\sum_{k\in[n]}x_{k}x_{k}^{*}I_{d}x_{k}x_{k}^{*}-\operatorname{tr}I_{d}\cdot
I_{d}=\frac{2a}{c_{2}}\sum_{k\in[n]}x_{k}x_{k}^{*}-d\cdot
I_{d}=\bigg{(}\frac{2ac_{1}}{c_{2}}-d\bigg{)}\cdot I_{d}$

If $0\in\\{a,c_{1}\\}$, then the above identity implies $d\equiv-1\bmod p$ and
$n=d^{2}\equiv 1\bmod p$. Furthermore, since $\\{x_{k}\\}_{k\in[n]}$ is an
$(a,c_{1})$-NTF, Proposition 12(a) gives that $na=dc_{1}$. If
$0\in\\{a,c_{1}\\}$, then squaring both sides gives
$a^{2}=n^{2}a^{2}=d^{2}c_{1}^{2}=c_{1}^{2}$. It follows that $a=0$ if and only
if $c_{1}=0$, as claimed. Furthermore, $d\equiv-1\bmod p$ if $a=0$.

Case II: $c_{1}\neq 0$ and $c_{2}\neq 0$. For every
$A\in\mathbb{F}_{q^{2}}^{d\times d}$, Lemma 17 and the identity
$\sum_{k\in[n]}x_{k}x_{k}^{*}=c_{1}\cdot I_{d}$ together imply

$A=\sum_{k\in[n]}\bigg{(}\frac{2}{c_{2}}x_{k}^{*}Ax_{k}-\frac{1}{c_{1}}\operatorname{tr}A\bigg{)}x_{k}x_{k}^{*}.$
(6)

Since $\\{x_{k}x_{k}^{*}\\}_{k\in[n]}$ is a spanning set of
$\mathbb{F}_{q^{2}}^{d\times d}$ of size
$n=d^{2}=\operatorname{dim}(\mathbb{F}_{q^{2}}^{d\times d})$, it is also a
basis. Then the decomposition (6) is unique. For $A:=x_{\ell}x_{\ell}^{*}$,
this implies

$\frac{2}{c_{2}}\langle
x_{k},x_{\ell}\rangle^{q+1}-\frac{a}{c_{1}}=\frac{2}{c_{2}}x_{k}^{*}Ax_{k}-\frac{1}{c_{1}}\operatorname{tr}A=\left\\{\begin{array}[]{cl}1&\text{if
}k=\ell\\\ 0&\text{if }k\neq\ell.\end{array}\right.$ (7)

It follows that

$\langle x_{k},x_{\ell}\rangle^{q+1}=\frac{ac_{2}}{2c_{1}}=:b$

whenever $k\neq\ell$, i.e., $\\{x_{k}\\}_{k\in[n]}$ is an $(a,b,c_{1})$-ETF.
Then (7) gives

$\displaystyle\frac{2}{c_{2}}a^{2}-\frac{a}{c_{1}}$ $\displaystyle=1,$ (8)
$\displaystyle\frac{2}{c_{2}}b-\frac{a}{c_{1}}$ $\displaystyle=0.$ (9)

Subtract (9) from (8) and rearrange to get

$a^{2}-b=\frac{c_{2}}{2}\neq 0.$ (10)

Finally, since $a\neq 0$, we may rearrange (9) to get
$\frac{c_{2}}{2}=\frac{bc_{1}}{a}$, which combined with (10) gives
$a^{2}-b=\frac{bc_{1}}{a}$, as claimed.

Case III: $c_{1}=0$ and $c_{2}\neq 0$. We claim that the
$\mathbb{F}_{q^{2}}$-span of $\\{x_{k}x_{k}^{*}\\}_{k\in[n]}$ equals the
$(d^{2}-1)$-dimensional subspace $\\{X\in\mathbb{F}_{q^{2}}^{d\times
d}:\operatorname{tr}X=0\\}$. Indeed, the inclusion $\subseteq$ follows from
the fact that $\operatorname{tr}(x_{k}x_{k}^{*})=a=0$ for every $k\in[n]$. The
reverse inclusion follows from a dimension count, since Lemma 17 implies that
the $\mathbb{F}_{q^{2}}$-span of
$\\{x_{k}x_{k}^{*}\\}_{k\in[n]}\cup\\{I_{d}\\}$ is all of
$\mathbb{F}_{q^{2}}^{d\times d}$. Next, since $n=(d^{2}-1)+1$, it follows that
the identity $\sum_{k\in[n]}x_{k}x_{k}^{*}=c_{1}\cdot I_{d}$ gives the only
linear dependency of $\\{x_{k}x_{k}^{*}\\}_{k\in[n]}$ up to scalar
multiplication, namely, $\sum_{k\in[n]}x_{k}x_{k}^{*}=0$. Taking
$A:=x_{\ell}x_{\ell}^{*}$ in Lemma 17 gives

$x_{\ell}x_{\ell}^{*}=\frac{2}{c_{2}}\sum_{k\in[n]}x_{k}x_{k}^{*}x_{\ell}x_{\ell}^{*}x_{k}x_{k}^{*}-\operatorname{tr}x_{\ell}x_{\ell}^{*}\cdot
I_{d}=\frac{2}{c_{2}}\sum_{k\in[n]}\langle
x_{k},x_{\ell}\rangle^{q+1}x_{k}x_{k}^{*},$

and rearranging gives

$\sum_{k\in[n]}z_{k\ell}x_{k}x_{k}^{*}=0,$ (11)

where

$z_{k\ell}:=\left\\{\begin{array}[]{ll}\frac{2}{c_{2}}\langle
x_{k},x_{\ell}\rangle^{q+1}&\text{if }k\neq\ell\\\ \frac{2}{c_{2}}\langle
x_{\ell},x_{\ell}\rangle^{q+1}-1&\text{if }k=\ell\\\
\end{array}\right\\}=\left\\{\begin{array}[]{cl}\frac{2}{c_{2}}\langle
x_{k},x_{\ell}\rangle^{q+1}&\text{if }k\neq\ell\\\ -1&\text{if }k=\ell.\\\
\end{array}\right.$

Since $\sum_{k\in[n]}x_{k}x_{k}^{*}=0$ is the unique dependency up to scaling,
the dependency (11) requires $z_{k\ell}=-1$ for every $k\neq\ell$, i.e.,

$\langle x_{k},x_{\ell}\rangle^{q+1}=-\frac{c_{2}}{2}=:b.$

As such, $\\{x_{k}\\}_{k\in[n]}$ is an $(a,b,c_{1})$-ETF. Since $c_{2}\neq 0$,
we have $b\neq 0=a^{2}$, as claimed.

Finally, we demonstrate (b)$\wedge$(c)$\Rightarrow$(a) with the help of Lemma
19. To this end, we consider the linear map $\Psi^{*}\colon
A\mapsto\Psi(A)^{*}$. Since $\\{x_{k}\\}_{k\in[n]}$ is an $(a,b,c_{1})$-ETF,
then

$\displaystyle\Psi^{*}(x_{\ell}x_{\ell}^{*})=\sum_{k\in[n]}x_{k}x_{k}^{*}x_{\ell}x_{\ell}^{*}x_{k}x_{k}^{*}$
$\displaystyle=\sum_{k\in[n]}\langle
x_{k},x_{\ell}\rangle^{q+1}x_{k}x_{k}^{*}$
$\displaystyle=a^{2}x_{\ell}x_{\ell}^{*}+\sum_{\begin{subarray}{c}k\in[n]\\\
k\neq\ell\end{subarray}}bx_{k}x_{k}^{*}=(a^{2}-b)x_{\ell}x_{\ell}^{*}+bc_{1}\cdot
I_{d}.$ (12)

This expression obfuscates the linearity of $\Psi^{*}$, which we elucidate in
two separate cases.

Case I: $a\neq 0$. Since $\operatorname{tr}(x_{\ell}x_{\ell}^{*})=a$, we may
continue (12):

$\Psi^{*}(x_{\ell}x_{\ell}^{*})=(a^{2}-b)x_{\ell}x_{\ell}^{*}+\frac{bc_{1}}{a}\cdot\operatorname{tr}(x_{\ell}x_{\ell}^{*})\cdot
I_{d}.$

Since $0\neq a^{2}\neq b$, it follows from Proposition 13 that the
$\mathbb{F}_{q^{2}}$-span of $\\{x_{k}x_{k}^{*}\\}_{k\in[n]}$ has dimension
$\operatorname{dim}_{\mathbb{F}_{q}}\\{X\in\mathbb{F}_{q^{2}}^{d\times
d}:X=X^{*}\\}=d^{2}$, and so it equals $\mathbb{F}_{q^{2}}^{d\times d}$. Thus,
we may linearly extend the above identity to get

$\Psi^{*}(A)=(a^{2}-b)A+\frac{bc_{1}}{a}\cdot\operatorname{tr}A\cdot I_{d}$

for every $A\in\mathbb{F}_{q^{2}}^{d\times d}$. In particular, we have

$\displaystyle\Psi(e_{i}e_{j}^{*})=\Big{(}\Psi^{*}(e_{i}e_{j}^{*})\Big{)}^{*}$
$\displaystyle=\Big{(}(a^{2}-b)e_{i}e_{j}^{*}+\frac{bc_{1}}{a}\cdot\operatorname{tr}(e_{i}e_{j}^{*})\cdot
I_{d}\Big{)}^{*}$
$\displaystyle=(a^{2}-b)e_{j}e_{i}^{*}+\frac{bc_{1}}{a}\cdot\delta_{ij}\cdot
I_{d}=(a^{2}-b)\cdot\Big{(}e_{j}e_{i}^{*}+\delta_{ij}\cdot I_{d}\Big{)},$

where the last step applies our assumption that $a^{2}-b=\frac{bc_{1}}{a}$.
Then Lemma 19 gives

$\displaystyle\sum_{k\in[n]}(x_{k}^{\otimes 2})(x_{k}^{\otimes 2})^{*}$
$\displaystyle=\sum_{i\in[d]}\sum_{j\in[d]}e_{i}e_{j}^{*}\otimes\Psi(e_{i}e_{j}^{*})$
$\displaystyle=2(a^{2}-b)\sum_{i\in[d]}\sum_{j\in[d]}e_{i}e_{j}^{*}\otimes\frac{1}{2}\Big{(}e_{j}e_{i}^{*}+\delta_{ij}\cdot
I_{d}\Big{)}=2(a^{2}-b)\cdot\Pi_{d}^{(2)}.$

Since $c_{2}:=2(a^{2}-b)\neq 0$ by assumption, it follows that
$\\{x_{k}\\}_{k\in[n]}$ is an $(a,c_{1},c_{2})$-projective $2$-design, as
desired.

Case II: $a=0$. Then $d\equiv-1\bmod p$ by assumption. Furthermore, since
$\\{x_{k}\\}_{k\in[n]}$ is an $(a,c_{1})$-NTF, Proposition 13(a) gives that
$0=na=dc_{1}=-c_{1}$, i.e., $c_{1}=0$. With this, we continue (12):

$\Psi^{*}(x_{\ell}x_{\ell}^{*})=(a^{2}-b)x_{\ell}x_{\ell}^{*}+bc_{1}\cdot
I_{d}=-bx_{\ell}x_{\ell}^{*}.$ (13)

By equality in Gerzon’s bound, Proposition 13 implies that the
$\mathbb{F}_{q^{2}}$-span of $\\{x_{k}x_{k}^{*}\\}_{k\in[n]}$ has dimension
$\operatorname{dim}_{\mathbb{F}_{q}}\\{X\in\mathbb{F}_{q^{2}}^{d\times
d}:X=X^{*},~{}\operatorname{tr}X=0\\}=d^{2}-1$, and therefore equals
$\\{X\in\mathbb{F}_{q^{2}}^{d\times d}:\operatorname{tr}X=0\\}$. By linearity,
(13) implies $\Psi^{*}(A)=-bA$ for every $A\in\mathbb{F}_{q^{2}}^{d\times d}$
with $\operatorname{tr}A=0$. In order to determine $\Psi^{*}(A)$ for all $A$,
we also consider

$\Psi^{*}(I_{d})=\sum_{k\in[n]}x_{k}x_{k}^{*}I_{d}x_{k}x_{k}^{*}=a\sum_{k\in[n]}x_{k}x_{k}^{*}=0.$

Since $\\{x_{k}x_{k}^{*}\\}_{k\in[n]}\cup\\{I_{d}\\}$ spans
$\mathbb{F}_{q^{2}}^{d\times d}$, we may obtain a formula for $\Psi^{*}(A)$ by
extending linearly. To this end, denote

$\hat{A}:=A+\operatorname{tr}A\cdot I_{d}=A-\frac{\operatorname{tr}A}{d}\cdot
I_{d}\in\\{X\in\mathbb{F}_{q^{2}}^{d\times d}:\operatorname{tr}X=0\\}.$

Then

$\Psi^{*}(A)=\Psi^{*}(\hat{A}-\operatorname{tr}A\cdot
I_{d})=\Psi^{*}(\hat{A})-\operatorname{tr}A\cdot\Psi^{*}(I_{d})=-b\hat{A}=-b(A+\operatorname{tr}A\cdot
I_{d}).$

In particular, we have

$\Psi(e_{i}e_{j}^{*})=\Big{(}\Psi^{*}(e_{i}e_{j}^{*})\Big{)}^{*}=\Big{(}-b(e_{i}e_{j}^{*}+\operatorname{tr}(e_{i}e_{j}^{*})\cdot
I_{d})\Big{)}^{*}=-b(e_{j}e_{i}^{*}+\delta_{ij}\cdot I_{d}).$

Then Lemma 19 gives

$\displaystyle\sum_{k\in[n]}(x_{k}^{\otimes 2})(x_{k}^{\otimes 2})^{*}$
$\displaystyle=\sum_{i\in[d]}\sum_{j\in[d]}e_{i}e_{j}^{*}\otimes\Psi(e_{i}e_{j}^{*})$
$\displaystyle=-2b\sum_{i\in[d]}\sum_{j\in[d]}e_{i}e_{j}^{*}\otimes\frac{1}{2}\Big{(}e_{j}e_{i}^{*}+\delta_{ij}\cdot
I_{d}\Big{)}=-2b\cdot\Pi_{d}^{(2)}.$

Since $c_{2}:=-2b\neq-2a^{2}=0$ by assumption, it follows that
$\\{x_{k}\\}_{k\in[n]}$ is an $(a,c_{1},c_{2})$-projective $2$-design, as
desired. ∎

###  3.2 A construction for Gerzon equality

Theorem 18 allows one to easily identify projective $2$-designs over
$\mathbb{F}_{q^{2}}^{d}$. For example, [23] constructs a $(0,1,0)$-ETF of
$d^{2}$ vectors in $\mathbb{F}_{3^{2}}^{d}$ for every $d=2^{2\ell+1}$ with
$\ell\in\mathbb{N}$. Since $2^{2\ell+1}\equiv-1\bmod 3$, Theorem 18 implies
that each of these systems of vectors forms a $(0,0,1)$-projective $2$-design
for $\mathbb{F}_{3^{2}}^{d}$. The following result constructs additional
examples:

######  Theorem 20.

Select any prime $p$, positive integer $k$, and prime power $r$ such that $p$
divides $r-1$ and $r^{2}+r+1$ divides $p^{k}+1$. Put $q:=p^{k}$ and
$d:=r^{2}+r+1$. Let $D\subseteq\mathbb{Z}/d\mathbb{Z}$ denote the Singer
difference set, select a primitive element
$\alpha\in\mathbb{F}_{q^{2}}^{\times}$, put $\omega:=\alpha^{(q^{2}-1)/d}$,
and define translation and modulation operators by

$(Tf)(x):=f(x-1),\qquad(Mf)(x):=\omega^{x}\cdot f(x),\qquad
f\colon\mathbb{Z}/d\mathbb{Z}\to\mathbb{F}_{q^{2}}.$

Then $\\{M^{s}T^{t}\mathbf{1}_{D}\\}_{s,t\in\mathbb{Z}/d\mathbb{Z}}$ is a
$(2,1,2d)$-equiangular tight frame of $d^{2}$ vectors in
$\mathbb{F}_{q^{2}}^{d}$.

The ETF construction in Theorem 20 is a finite field analog of a biangular
Gabor frame that was suggested in [8, 25]. In the finite field setting, one
might view this as a Steiner ETF [18] in which a harmonic ETF [44, 43, 49, 14]
plays the role of a “flat” simplex. Empirically, there are many
$(p,k,r)\in\mathbb{N}^{3}$ that satisfy the constraints that $p$ is prime, $r$
is a prime power, $p$ divides $r-1$, and $r^{2}+r+1$ divides $p^{k}+1$. In
fact, there are infinitely many, conditioned on the Lenstra–Pomerance–Wagstaff
conjecture of the infinitude of Mersenne primes: whenever there exists a
Mersenne prime $r=2^{m}-1$, we may take $p=2$ and $k=3m$. We claim that the
ETF construction in Theorem 20 forms a projective $2$-design for
$\mathbb{F}_{q^{2}}^{d}$ whenever $p>3$. First, we have $a^{2}=4\neq 1=b$ and
$a=2\neq 0$, and so by Theorem 18, it suffices to verify that
$a^{2}-b=\frac{bc_{1}}{a}$. Indeed, $a^{2}-b=4-1=3$ and $\frac{bc_{1}}{a}=d$,
and since $p$ divides $r-1$ by assumption, we have $d=r^{2}+r+1\equiv
1+1+1=3\bmod p$, as desired. The following table lists the smallest dimensions
for which Theorem 20 offers a construction, with gray columns indicating
projective $2$-designs.

$d$ | 13 | 57 | 73 | 307 | 757 | 993 | 1723 | 1723 | 2257 | 2257 | 2451 | 3541 | 3541 | 5113
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---
$p$ | 2 | 2 | 7 | 2 | 2 | 2 | 2 | 5 | 2 | 23 | 2 | 2 | 29 | 2
$k$ | 6 | 9 | 12 | 51 | 378 | 15 | 287 | 287 | 90 | 30 | 63 | 118 | 590 | 213
$r$ | 3 | 7 | 8 | 17 | 27 | 31 | 41 | 41 | 47 | 47 | 49 | 59 | 59 | 71

###### Proof of Theorem 20.

Proposition 6.1 in [23] gives that
$\\{M^{s}T^{t}\mathbf{1}_{D}\\}_{s,t\in\mathbb{Z}/d\mathbb{Z}}$ is an
$(a,da)$-NTF with

$a=\langle\mathbf{1}_{D},\mathbf{1}_{D}\rangle=|D|\equiv 2\bmod p,$

where the last step applies the fact that $|D|=r+1\equiv 2\bmod p$. It remains
to verify equiangularity with parameter $b=1$. To this end, we compute

$\langle M^{s}T^{t}\mathbf{1}_{D},M^{u}T^{v}\mathbf{1}_{D}\rangle^{q+1}$

for every $(s,t)\neq(u,v)$ in two cases.

Case I: $t=v$. Consider the discrete Fourier transform matrix
$\mathcal{F}=[\omega^{ij}]_{i,j\in\mathbb{Z}/d\mathbb{Z}}\in\mathbb{F}_{q^{2}}^{d\times
d}$, and observe that
$\\{M^{s}T^{t}\mathbf{1}_{D}\\}_{s\in\mathbb{Z}/d\mathbb{Z}}$ equals the
columns of $\operatorname{diag}(\mathbf{1}_{D+t})\mathcal{F}$, which in turn
is a zero-padded version of the $|D|\times d$ submatrix $\mathcal{F}_{D+t}$.
Since $D+t$ is a difference set with parameter $\lambda=1$, (the proof of)
Theorem 5.7 in [23] gives that $\mathcal{F}_{D+t}$ is a
$(|D|,|D|-\lambda,d)$-ETF. It follows that

$\langle
M^{s}T^{t}\mathbf{1}_{D},M^{u}T^{t}\mathbf{1}_{D}\rangle^{q+1}=|D|-\lambda\equiv
1\bmod p$

whenever $s\neq u$.

Case II: $t\neq v$. We exploit the well-known fact that the so-called
development

$\\{D+z:z\in G\\}$

of a difference set $D\subseteq G$ with parameter $\lambda$ gives a symmetric
block design in which every pair of blocks intersects in exactly $\lambda$
points (see Theorem 18.6 in [30]). Since the Singer difference set has
parameter $\lambda=1$, we have

$\displaystyle\langle
M^{s}T^{t}\mathbf{1}_{D},M^{u}T^{v}\mathbf{1}_{D}\rangle$
$\displaystyle=\sum_{x\in\mathbb{Z}/d\mathbb{Z}}(M^{s}T^{t}\mathbf{1}_{D})(x)^{q}(M^{u}T^{v}\mathbf{1}_{D})(x)$
$\displaystyle=\sum_{x\in\mathbb{Z}/d\mathbb{Z}}(\omega^{sx}\mathbf{1}_{D}(x-t))^{q}(\omega^{ux}\mathbf{1}_{D}(x-v))$
$\displaystyle=\sum_{x\in\mathbb{Z}/d\mathbb{Z}}\omega^{(qs+u)x}\cdot\mathbf{1}_{D}(x-t)\mathbf{1}_{D}(x-v)=\omega^{(qs+u)x_{0}},$

where $x_{0}$ is the unique member of the block intersection $(D+t)\cap(D+v)$.
Then

$\langle
M^{s}T^{t}\mathbf{1}_{D},M^{u}T^{v}\mathbf{1}_{D}\rangle^{q+1}=(\omega^{(qs+u)x_{0}})^{q+1}=1,$

as desired. ∎

##  4 The quaternionic setting

Consider the following generalization of Proposition 1(c) for
$\mathbb{F}\in\\{\mathbb{R},\mathbb{C},\mathbb{H}\\}$; see [36, 46]. Put
$m:=\frac{1}{2}\cdot[\mathbb{F}:\mathbb{R}]$ and $N=md$, and given
$A,B\in\mathbb{F}^{d\times d}$, define $\langle
A,B\rangle:=\operatorname{Re}\operatorname{tr}(A^{*}B)$. We say unit vectors
$\\{x_{k}\\}_{k\in[n]}$ in $\mathbb{F}^{d}$ form a projective $2$-design if

$\frac{1}{n^{2}}\sum_{k\in[n]}\sum_{\ell\in[n]}\langle
x_{k}x_{k}^{*},x_{\ell}x_{\ell}^{*}\rangle=\frac{m}{N},\qquad\frac{1}{n^{2}}\sum_{k\in[n]}\sum_{\ell\in[n]}\langle
x_{k}x_{k}^{*},x_{\ell}x_{\ell}^{*}\rangle^{2}=\frac{m(m+1)}{N(N+1)}.$

We say a projective $2$-design $\\{x_{k}\\}_{k\in[n]}$ for $\mathbb{F}^{d}$ is
tight if it is also equiangular, which occurs precisely when equality is
achieved in the lower bound $n\geq d+m(d^{2}-d)$ [3]. In this section, we use
tight projective $2$-designs in quaternionic spaces to form nice arrangements
of real subspaces. Given $r$-dimensional subspaces $\\{S_{k}\\}_{k\in[n]}$ of
a real vector space $V$, select an orthonormal basis $\\{x_{ki}\\}_{i\in[r]}$
for each $S_{k}$ and compute the cross-Gramians $G_{k\ell}:=[\langle
x_{ki},x_{\ell j}\rangle]_{i,j\in[r]}$. We say $\\{S_{k}\\}_{k\in[n]}$ is
equi-isoclinic if there exists $\alpha\geq 0$ such that
$G_{k\ell}^{*}G_{k\ell}=\alpha I_{r}$ for every $k,\ell\in[n]$ with
$k\neq\ell$. We say $\\{S_{k}\\}_{k\in[n]}$ forms a tight fusion frame if

$\frac{1}{n^{2}}\sum_{k\in[n]}\sum_{\ell\in[n]}\|G_{k\ell}\|_{F}^{2}=\frac{r^{2}}{\operatorname{dim}V}.$

An equi-isoclinic tight fusion frame is an optimal code in the Grassmannian
$\operatorname{Gr}(r,V)$ under the chordal distance, as it achieves equality
in the simplex bound [12]. It is also an optimal packing in terms of the
spectral distance, where we view each $S_{k}$ as a subset of projective space
and seek to maximize the minimum distance between these subsets of this metric
space [13]. One may rightly view such objects as analogs of equiangular tight
frames.

Since quaternion multiplication is noncommutative (e.g.,
$\mathrm{i}\mathrm{j}=-\mathrm{j}\mathrm{i}$), we start by carefully verifying
a few simple facts that may be unfamiliar to the reader:

######  Lemma 21.

$\mathbb{H}^{d\times d}$ is a real Hilbert space with inner product $\langle
A,B\rangle=\operatorname{Re}\operatorname{tr}(A^{*}B)$.

###### Proof.

First, given $u,v\in\mathbb{H}$, we observe that
$\operatorname{Re}(\overline{u}v)$ equals the dot product between the
corresponding real coordinate vectors:

$\displaystyle\operatorname{Re}((a-b\mathrm{i}-c\mathrm{j}-d\mathrm{k})(e+f\mathrm{i}+g\mathrm{j}+h\mathrm{k}))=ae+bf+cg+dh=(a,b,c,d)\cdot(e,f,g,h).$

Next, given $A\in\mathbb{H}^{d\times d}$, let
$\operatorname{vec}(A)\in\mathbb{R}^{4d^{2}}$ denote the vector of real
coordinates of the entries of $A$. Then the above observation gives

$\operatorname{Re}\operatorname{tr}(A^{*}B)=\operatorname{Re}\sum_{j\in[d]}(A^{*}B)_{jj}=\operatorname{Re}\sum_{j\in[d]}\sum_{i\in[d]}(A^{*})_{ji}B_{ij}=\sum_{i\in[d]}\sum_{j\in[d]}\operatorname{Re}\overline{A_{ij}}B_{ij}=\operatorname{vec}(A)\cdot\operatorname{vec}(B).$

The result follows. ∎

######  Lemma 22.

Suppose $A\in\mathbb{H}^{m\times n}$ and $B\in\mathbb{H}^{n\times m}$. Then
$\operatorname{Re}\operatorname{tr}(AB)=\operatorname{Re}\operatorname{tr}(BA)$.

###### Proof.

Consider the algebra homomorphism $f\colon\mathbb{H}\to\mathbb{C}^{2\times 2}$
defined by

$f(a+b\mathrm{i}+c\mathrm{j}+d\mathrm{k})=\left[\begin{array}[]{rr}a+b\mathrm{i}&c+d\mathrm{i}\\\
-c+d\mathrm{i}&a-b\mathrm{i}\end{array}\right].$

Observe that $\operatorname{tr}f(z)=2\operatorname{Re}z$. Given
$M\in\mathbb{H}^{n\times n}$, one may apply $f$ entrywise to obtain
$f(M)\in\mathbb{C}^{2n\times 2n}$, and then
$\operatorname{Re}\operatorname{tr}M=\frac{1}{2}\operatorname{tr}f(M)$.
Applying this to $AB$ and $BA$ gives

$\operatorname{Re}\operatorname{tr}(AB)=\tfrac{1}{2}\operatorname{tr}(f(AB))=\tfrac{1}{2}\operatorname{tr}(f(A)f(B))=\tfrac{1}{2}\operatorname{tr}(f(B)f(A))=\tfrac{1}{2}\operatorname{tr}(f(BA))=\operatorname{Re}\operatorname{tr}(BA),$

as claimed. ∎

######  Lemma 23.

Suppse $A,B\in\mathbb{H}^{d\times d}$ satisfy $A^{*}=A$ and $B^{*}=-B$. Then
$\langle A,B\rangle=0$.

###### Proof.

Symmetry of the real inner product gives

$\langle A,B\rangle=\langle
B,A\rangle=\operatorname{Re}\operatorname{tr}(B^{*}A)=\operatorname{Re}\operatorname{tr}(-BA^{*})=-\operatorname{Re}\operatorname{tr}(A^{*}B)=-\langle
A,B\rangle.$

Rearrange to get the result. ∎

In fact, the real subspace of anti-Hermitian matrices is the orthogonal
complement of the real subspace of Hermitian matrices. This can be seen by
dimension counting: Hermitian matrices have dimension $d+4\binom{d}{2}$, while
anti-Hermitian matrices have dimension $3d+4\binom{d}{2}$, and the sum of
these is $4d^{2}$, i.e., the dimension of $\mathbb{H}^{d\times d}$.

Given $x\in\mathbb{H}^{d}\setminus\\{0\\}$, let $S(x)$ denote the
$3$-dimensional real subspace of $\mathbb{H}^{d\times d}$ defined by

$S(x):=\\{xzx^{*}:z\in\mathbb{H},~{}\operatorname{Re}z=0\\}.$

For each $A\in S(x)$, it holds that

$A^{*}=(xzx^{*})^{*}=x\overline{z}x^{*}=-xzx^{*}=-A.$

As such, $S(x)$ is actually a subspace of the anti-Hermitian matrices. Observe
that for each $u,v\in\\{\mathrm{i},\mathrm{j},\mathrm{k}\\}$, it holds that

$\displaystyle\langle xux^{*},xvx^{*}\rangle$
$\displaystyle=\operatorname{Re}\operatorname{tr}((xux^{*})^{*}xvx^{*})$
$\displaystyle=\operatorname{Re}\operatorname{tr}(x\overline{u}x^{*}xvx^{*})$
$\displaystyle=\operatorname{Re}\operatorname{tr}(\overline{u}x^{*}xvx^{*}x)=\|x\|^{4}\cdot\operatorname{Re}(\overline{u}v)=\left\\{\begin{array}[]{cl}\|x\|^{4}&\text{if
}u=v\\\ 0&\text{otherwise.}\end{array}\right.$

Thus, $\\{x\mathrm{i}x^{*},x\mathrm{j}x^{*},x\mathrm{k}x^{*}\\}$ is an
orthogonal basis for $S(x)$. With this, we prove the following:

######  Theorem 24.

Consider unit vectors $\\{x_{k}\\}_{k\in[n]}$ in $\mathbb{H}^{d}$.

  * (a)

If $\\{x_{k}\\}_{k\in[n]}$ is equiangular, then $\\{S(x_{k})\\}_{k\in[n]}$ is
equi-isoclinic.

  * (b)

If $\\{x_{k}\\}_{k\in[n]}$ is a projective $2$-design, then
$\\{S(x_{k})\\}_{k\in[n]}$ is tight in anti-Hermitian space.

###### Proof.

Given unit vectors $x,y\in\mathbb{H}^{d}$, we are interested in the cross-
Gramian $G_{xy}$ between
$\\{x\mathrm{i}x^{*},x\mathrm{j}x^{*},x\mathrm{k}x^{*}\\}$ and
$\\{y\mathrm{i}y^{*},y\mathrm{j}y^{*},y\mathrm{k}y^{*}\\}$. The entry indexed
by $(u,v)\in\\{\mathrm{i},\mathrm{j},\mathrm{k}\\}^{2}$ is given by

$\displaystyle\langle
xux^{*},yvy^{*}\rangle=\operatorname{Re}\operatorname{tr}((xux^{*})^{*}yvy^{*})$
$\displaystyle=\operatorname{Re}\operatorname{tr}(x\overline{u}x^{*}yvy^{*})$
$\displaystyle=\operatorname{Re}\operatorname{tr}(\overline{u}x^{*}yvy^{*}x)=\operatorname{Re}(\overline{u}x^{*}yvy^{*}x).$
(14)

If $x^{*}y=0$, then $G_{xy}=0$. Otherwise, define $z:=\frac{x^{*}y}{|x^{*}y|}$
and continue (14):

$\langle
xux^{*},yvy^{*}\rangle=\operatorname{Re}(\overline{u}x^{*}yvy^{*}x)=|x^{*}y|^{2}\cdot\operatorname{Re}(\overline{u}zvz^{-1}).$

It follows that $G_{xy}$ equals $|x^{*}y|^{2}$ times a matrix representation
of the special orthogonal map $q\mapsto zqz^{-1}$ over imaginary
$q\in\mathbb{H}$. This implies (a).

For (b), we apply the facts that $G_{xy}=|x^{*}y|^{2}\cdot Q$ for some
$Q\in\operatorname{SO}(3)$ and

$\displaystyle|x^{*}y|^{2}=\overline{x^{*}y}x^{*}y=y^{*}xx^{*}y=\operatorname{Re}\operatorname{tr}(y^{*}xx^{*}y)=\operatorname{Re}\operatorname{tr}(xx^{*}yy^{*})=\langle
xx^{*},yy^{*}\rangle$

in order to compute the frame potential of $\\{S(x_{k})\\}_{k\in[n]}$:

$\displaystyle\frac{1}{n^{2}}\sum_{k\in[n]}\sum_{\ell\in[n]}\|G_{x_{k}x_{\ell}}\|_{F}^{2}$
$\displaystyle=\frac{1}{n^{2}}\sum_{k\in[n]}\sum_{\ell\in[n]}3|x_{k}^{*}x_{\ell}|^{4}$
$\displaystyle=\frac{3}{n^{2}}\sum_{k\in[n]}\sum_{\ell\in[n]}\langle
x_{k}x_{k}^{*},x_{\ell}x_{\ell}^{*}\rangle^{2}=3\cdot\frac{m(m+1)}{N(N+1)}=\frac{9}{d(2d+1)}.$

It follows that $\\{S(x_{k})\\}_{k\in[n]}$ forms a tight fusion frame in the
$d(2d+1)$-dimensional real vector space of $d\times d$ quaternionic anti-
Hermitian matrices. ∎

Theorem 24 implies that every tight projective $2$-design for $\mathbb{H}^{d}$
corresponds to an equi-isoclinic tight fusion frame of $d(2d-1)$ subspaces of
$\mathbb{R}^{d(2d+1)}$ of dimension $3$. To date, such designs are only known
to exist for $d\in\\{2,3\\}$. First, $\mathbb{HP}^{1}$ is isometric to
$S^{4}$, and so the six vertices of a $5$-dimensional regular simplex easily
deliver a tight projective $2$-design for $\mathbb{H}^{2}$; this in turn
determines $6$ subspaces of $\mathbb{R}^{10}$ of dimension $3$. The $d=3$ case
is resolved by Theorem 4.12 in [11], which uses a variant of the
Newton–Kantorovich theorem to obtain a computer-assisted proof of the
existence of a $15$-point simplex in $\mathbb{HP}^{2}$; this determines $15$
subspaces of $\mathbb{R}^{21}$ of dimension $3$. The authors are not aware of
any other constructions of equi-isoclinic tight fusion frames with these
parameters. Also, it is open whether tight projective $2$-designs exist for
$\mathbb{H}^{d}$ with $d>3$.

## Acknowledgments

The first part of this paper was inspired by a beautiful talk [39] given by
Vern Paulsen at the Codes and Expansions online seminar. Much of this work was
conducted during the SOFT 2020: Summer of Frame Theory virtual workshop. DGM
was partially supported by AFOSR FA9550-18-1-0107 and NSF DMS 1829955.

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