# Unanswered Questions in Charmed Baryon Physics

John M. Yelton  University of Florida, Gainesville, FL, USA, 32611

###### Abstract

This is an experimentalist’s list of questions concerning the physics of the
charmed baryon sector which have no satisfactory answer.

##  I Introduction

In many conferences experimentalists talk about carefully justified results
rather than analyses they are actively pursuing, and theorists often present
work on models which pertain to measurements that cannot be performed in the
near future. Sometimes this leads to a disappointing lack of real
communication between the two groups. I give this talk as an experimentalist,
pointing out features of the data that I do not believe have had satisfactory
explanations, but which might in fact be easily comprehensible. All the data
presented is freely available on the web and I am not representing any
experiment.

##  II The $\Lambda_{c}(2880)^{+}$

In $\Lambda_{c}^{+}$ spectroscopy many states have been found, and can all be
explained from a model where the $\Lambda_{c}^{+}$ comprises a heavy (charm)
quark and a light di-quark. This configuration predicts a plethora of states.
For instance, the addition of one unit of orbital angular monentum can be
placed between the heavy quark and light di-quark (a “$\lambda$” excitation)
or between the light quarks in the di-quark (a “$\rho$” excitation). Once two
units of orbital angular momentum are allowed, there are 3 different
possiblities of their placement. Each of these combinations, with its
particular “light-quark degrees of freedoms”, can then can combine with the
heavy-quark spin to make doublets.

Figure 1 shows the M($\Lambda_{c}^{+}\pi^{+}\pi^{-}$) mass in $e^{+}e^{-}$
annihilation data from Belle Thesis . The $\Lambda_{c}(2880)$ sticks out from
the background even though there must be many states around that mass - this
feature caused a surprise when first found CLEO2880 . The state is generally
considered to have spin-parity of $J^{P}=\frac{5}{2}^{+}$ based on
measurements by Belle B2880  and LHCb LHC2880 . The natural explanation is
that this is a $\lambda=2$ excitation. My question is, why is this one state
so narrow? Data from LHCb LHC2880  implies that its $J^{P}\ =\
\frac{3}{2}^{+}$ partner is 20 ${\rm MeV}/c^{2}$ lower and is too wide to be
seen clearly in a plot such as this. In addition, the $\Lambda_{c}(2880)^{+}$
state has a higher branching fraction into $\Sigma_{c}(2455)\pi$ than into
$\Sigma_{c}(2520)\pi$, whereas the rules of decay for particles of particular
spin-parities, the latter should be preferred as it can occur as a “P-wave”,
whereas the former is restricted to “F-wave”. It is as if the matrix element
for the former tends towards zero and suppresses the decay to
$\Sigma_{c}(2520)\pi$ so that the state is narrow. Why is this?

Figure 1: The mass difference,
$M(\Lambda_{c}^{+}\pi^{+}\pi^{-})-M(\Lambda_{c}^{+})$, Belle data. There are
four clear signals, from right-to-left, the $\Lambda_{c}(2880)^{+}$,
$\Lambda_{c}(2765)^{+}$, $\Lambda_{c}(2625)^{+}$, and $\Lambda_{c}(2593)^{+}$.

##  III The $\Lambda_{c}(2593)$

As we go down to lower masses in the same spectrum (Fig. 1), there is a very
large $\Lambda_{c}(2765)$ resonance LC2765  which is only given one “star” in
the Particle Data Book PDG  but clearly exists, and is generally considered to
be a “Roper”-like radial excitation. Next is the prominent $\Lambda_{c}(2625)$
which extends beyond the top of the plot, and then the $\Lambda_{c}(2593)$
peak which is distorted by its proximity to the $\Sigma_{c}(2455)\pi$
threshold.

Figure 2: The excitation energy (i.e. mass above ground state) of the eight
heavy-quark baryons identified as $\lambda=1$ excitations. The red lines show
the rough threshold for preferred decays of the states. Data from the PDG PDG
.

The identification of the quark structure of the $\Lambda_{c}(2593)$ was made
on discovery C2595  as being the $J^{P}=\frac{1}{2}^{-}$ lowest-mass
(“$\lambda$”) orbital excitation. Based on this identification, together with
that of the $\Lambda_{c}(2625)$ as its $J^{P}=\frac{3}{2}^{-}$ partner, the
scale was set for the excitation energy ($\approx 300\ {\rm MeV}/c^{2}$) for a
$\lambda\ =\ 1$ orbital excitation and the splitting between the states of
$\approx 30\ {\rm MeV}/c^{2}$. According to the standard potential model, the
first should be roughly independent of heavy-quark mass, and the latter
approximately inversely proportional to the heavy-quark mass. Thus, this model
thus predicted the masses (as well as the properties) of more than six other
states; the $\Xi_{c}(2790)$ and $\Xi_{c}(2815)$ iso-doublets, and the
$\Lambda_{b}(5912)$ and $\Lambda_{b}(5920)$ have all been subsequently
observed. Figure 2 shows how the expected pattern of masses has been
confirmed.

Despite the success of these predictions, recent papers have conjectured that
the $\Lambda_{c}(2593)$ can be explained as a molecular state MS  or a
dynamically generated meson-baryon state DGS . The former is motivated by the
fact that its mass is very close to the sum of the $\Sigma_{c}(2455)$ and
$\pi$ masses. However, it can be seen in Fig. 2 that of the eight states, four
have masses above the kinematic threshold of their preferred decays, three
have masses below (which keeps the states very narrow) - it is hardly
surprising that one has a mass that coincides, within the fuzziness caused by
the isospin mass-splitting of the transition pions, with its particular
threshold. My questions are, if the $\Lambda_{c}(2593)$ is not the $\lambda=1$
orbital excitation, how do you explain that the model correctly predicted the
other six states? Secondly, what parameter of the $\Lambda_{c}(2593)$ should
we be measuring in order to discriminate between the models? The resonance
shape is clearly complicated by the threshold, but as yet there is nothing
observed inconsistent with the quark content being dominated by the simplest
model. We rate scientific theories by the validity of their predictions, and
the (heavy quark)/(light di-quark) model for heavy baryons has served us very
well in this respect. Of course, in any conversation on the inner structure of
baryons, we must always be aware that no one simple model will give the
complete story.

##  IV $\Xi_{c}$ Isospin splitting

There has been considerable progress in recent years on the $\Xi_{c}$
spectrum. There are six iso-doublets with well-measured masses, plus several
other states not as well defined. When tabulating the masses of the six (Tab.
1), it appears that the isospin mass splitting divides into two groups, one
with around 3.5 ${\rm MeV}/c^{2}$, and one with less than 1 ${\rm MeV}/c^{2}$.
The first group all have a spin-0 di-quark, and the latter group have a spin-1
di-quark. My question, is this a rule that can then be applied to help
identify higher mass states that are being discovered?

Table 1: Isospin mass splitting in the $\Xi_{c}$ iso-doublets. Data taken from PDG  and Yelton . Particle |  $M(\Xi_{c}^{+})-M(\Xi_{c}^{0})$ ${\rm MeV}/c^{2}$
---|---
$\Xi_{c}$ | $-3.3\pm 0.4$
$\Xi_{c}(2645)$ | $-0.9\pm 0.5$
$\Xi_{c}(2815)$ | $-3.5\pm 0.5$
$\Xi_{c}(2980)$ | $-4.8\pm 0.5$
$\Xi_{c}^{\prime}$ | $-0.8\pm 0.5$
$\Xi_{c}(2790)$ | $-3.3\pm 0.6$

##  V $\Omega_{c}^{0}$ Decays

Of the four weakly decaying charmed baryons, the least is known about the
$\Omega_{c}$ (css). The general knowledge of its decays seemed uncontroversial
until recently. The lifetime of the $\Omega_{c}^{0}$ had been measured by
three experiments to be very short, and this agreed with early ideas on the
expected lifetime hierarchy of the four lifetimes OMT . However, of the three
experiments measuring the lifetimes WA ; 687 ; FOCUS , one had signals in
various modes but never managed to publish an $\Omega_{c}$ mass WA , and
another used a decay mode that has never been seen by other experiments 687 ,
despite a search OC . I note that if you over-estimate your signal yields you
tend to produce a short lifetime measurement. Recently LHCb found a much
longer lifetime LHCOl , showing that the knowledge of the $\Omega_{c}^{0}$
weak decays was less definitive than had been thought. (I note that one paper
CHENG  brought attention to this problem before the experimental measurement).
Information on the $\Omega_{c}^{0}$ branching fractions is comparatively
scarce, but one aspect that stands out is that the branching ratio
$B(\Omega_{c}^{0}\to\Omega^{-}\pi^{+}\pi^{-}\pi^{+})/B(\Omega_{c}^{0}\to\Omega^{-}\pi^{+})$
is much less than the analagous ratios for the other charmed baryons (Tab. 2).
In the most simple decay models, these should all be similar - it easy to see
how factors of two or three can arise from differences in phase-space etc. but
the difference of a factor of ten seems very large. There is a similar trend
(though not as marked) with the decay into $\Omega^{-}\pi^{+}\pi^{0}$. My
question is whether these branching fractions tell us anything basic about the
differences in the decays of the weakly decaying charmed baryons.

Table 2: Branching ratios comparison for charmed baryons. Data taken from PDG . $\Omega_{c}^{0}$ | $\Lambda_{c}^{+}$ | $\Xi_{c}^{0}$ | $\Xi_{c}^{+}$
---|---|---|---
$\Omega^{-}\pi^{+}\pi^{0}/\Omega^{-}\pi^{+}$ | $\Lambda\pi^{+}\pi^{0}/\Lambda\pi^{+}$ | $\Xi^{-}\pi^{+}\pi^{0}/\Xi^{-}\pi^{+}$ | $\Xi^{0}\pi^{+}\pi^{0}/\Xi^{0}\pi^{+}$
$1.97\pm 0.17$ | $5.46\pm 0.42$ | Not measured | $4.2\pm 1.5$
$\Omega^{-}\pi^{+}\pi^{-}\pi^{+}/\Omega^{-}\pi^{+}$ | $\Lambda\pi^{+}\pi^{-}\pi^{+}/\Lambda\pi^{+}$ | $\Xi^{-}\pi^{+}\pi^{-}\pi^{+}/\Xi^{-}\pi^{+}$ | $\Xi^{0}\pi^{+}\pi^{-}\pi^{+}/\Xi^{0}\pi^{+}$
$0.29\pm 0.04$ | $2.84\pm 0.34$ | $3.3\pm 0.5$ | $3.1\pm 1.0$

##  VI Excited $\Omega_{c}^{0}$ Spectroscopy

This is one subject where, rather than too few, there are too many answers.
The LHCb OMCS  discovery of five narrow states brought an immediate flood of
explanations. The most obvious (though naive) explanation is that the first
$\lambda=1$ excitation states for the $\Omega_{c}^{0}$ will in fact be a
quintuplet of states, and given that higher spin usually indicates higher
mass, the states can be labeled as
$J^{P}=(\frac{1}{2}^{-},\frac{1}{2}^{-},\frac{3}{2}^{-},\frac{3}{2}^{-},\frac{5}{2}^{-})$
The Belle analysis BOMCS  showed four of the states, but the last one is
missing. This did not strike people as surprising given that the fifth (i.e.
highest mass) of the LHCb quintuplet was their smallest peak. However, in
$e^{+}e^{-}$ continuum production, the production rates of states within a
family occurs via the “$2J+1$” rule. This has been verified in a variety of
systems NII . Even considering mass suppression, and the possibility of decays
to $\Xi_{c}^{\prime}$, it would seem surprising that the highest spin state
would not be seen by Belle. The plot thickens with the (low statistics) search
for these states by LHCb LBD  in $\Lambda_{b}$ decays, which again shows the
fifth state missing. My question is this, when considering the possible
assignment of states, why do people not take into account the production
ratios? For $e^{+}e^{-}$ the ratio between the different members of a group
(for instance, the $\lambda=1$ quintuplet) is clear, but no theorist has
mentioned them in their attempts to explain the data. In LHCb’s first paper,
the situation is more confusing as different production mechanisms are mixed
together, and in their second they had no theoretical prediction of the
production of the various states in $\Lambda_{b}$ decays. Surely these ratios
could have been predicted in advance of the experiment.

##  VII Excited $\Sigma_{c}$ Spectroscopy

It was way back in 2004 that Belle reported the discovery SCSBe  of a new
$\Sigma_{c}$ resonance decaying into $\Lambda_{c}^{+}\pi$. The natural
explanation is that it is an orbital excitation of the $\Sigma_{c}$. However,
like the $\Omega_{c}^{0}$, the $\Sigma_{c}$ should have a family of a
quintuplet of $\lambda=1$ low-mass states, so why is there only one peak? It
is true that some of the five might be too wide to be seen, but I do not
believe that any model predicts one to be narrow and four to be wide. This
certainly indicates that the peak seen may be due to overlapping states. Very
similar results are shown in an (unpublished) analysis using BaBar data Ahmed
. I note that BaBar SCSB  also see a peak in the same mode, but at a different
mass, in $B$ decays. To disentangle these observations what we need is not
just predictions of the masses and widths of states in this range, but also
estimates of the relative production rates, particularly in $B$ decays. I also
cannot help asking the question of why LHCb have not shown information on this
mass spectrum.

##  VIII Summary

I have highlighted several questions which I believe could be fruitful lines
of research for theorists. They are just a few of many I have! The charmed
baryon sector is already very rich in information, and the next few years we
will see more results, so there are many opportunities to make predictions
that can be directly tested.

###### Acknowledgements.

This work is supported in part be DOE contract DE-SC0009824.

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[by]Jianxin Chen, Joel Klassen, and Bei ZengSimone Severini (chair) & Fernando
Brandao (co-chair)2The 8th Conference on the Theory of Quantum Computation,
Communication and Cryptography111 10.4230/LIPIcs.xxx.yyy.p

# Universal Entanglers for Bosonic and Fermionic Systems111This work was
partially supported by NSERC and CIFAR

Joel Klassen  Department of Physics, University of Guelph
50 Stone Road East, Guelph, Ontario, Canada
<EMAIL_ADDRESS>Institute for Quantum Computing
200 University Avenue West, Waterloo, Ontario, Canada  Jianxin Chen
Department of Mathematics & Statistics, University of Guelph
50 Stone Road East, Guelph, Ontario, Canada
<EMAIL_ADDRESS>Institute for Quantum Computing
200 University Avenue West, Waterloo, Ontario, Canada  UTS-AMSS Joint Research
Laboratory for Quantum Computation and Quantum Information Processing
Academy of Mathematics and Systems Science, Chinese Academy of Sciences,
Beijing, China  Bei Zeng  Department of Mathematics & Statistics, University
of Guelph
50 Stone Road East, Guelph, Ontario, Canada
<EMAIL_ADDRESS>Institute for Quantum Computing
200 University Avenue West, Waterloo, Ontario, Canada

###### Abstract.

A universal entangler (UE) is a unitary operation which maps all pure product
states to entangled states. It is known that for a bipartite system of
particles $1,2$ with a Hilbert space
$\mathbb{C}^{d_{1}}\otimes\mathbb{C}^{d_{2}}$, a UE exists when
$\min{(d_{1},d_{2})}\geq 3$ and $(d_{1},d_{2})\neq(3,3)$. It is also known
that whenever a UE exists, almost all unitaries are UEs; however to verify
whether a given unitary is a UE is very difficult since solving a quadratic
system of equations is NP-hard in general. This work examines the existence
and construction of UEs of bipartite bosonic/fermionic systems whose wave
functions sit in the symmetric/antisymmetric subspace of
$\mathbb{C}^{d}\otimes\mathbb{C}^{d}$. The development of a theory of UEs for
these types of systems needs considerably different approaches from that used
for UEs of distinguishable systems. This is because the general entanglement
of identical particle systems cannot be discussed in the usual way due to the
effect of (anti)-symmetrization which introduces “pseudo entanglement” that is
inaccessible in practice. We show that, unlike the distinguishable particle
case, UEs exist for bosonic/fermionic systems with Hilbert spaces which are
symmetric (resp. antisymmetric) subspaces of
$\mathbb{C}^{d}\otimes\mathbb{C}^{d}$ if and only if $d\geq 3$ (resp. $d\geq
8$). To prove this we employ algebraic geometry to reason about the different
algebraic structures of the bosonic/fermionic systems. Additionally, due to
the relatively simple coherent state form of unentangled bosonic states, we
are able to give the explicit constructions of two bosonic UEs. Our
investigation provides insight into the entanglement properties of systems of
indisitinguishable particles, and in particular underscores the difference
between the entanglement structures of bosonic, fermionic and distinguishable
particle systems.

###### Key words and phrases:

Universal Entangler, Bosonic States, Fermionic States

###### 1991 Mathematics Subject Classification:

J.2 Physics

##  1\. Introduction

Entanglement sits at the core of the counterintuitive and useful properties of
quantum mechanics. At its inception Schrödinger labeled entanglement “the
characteristic trait of quantum mechanics, the one that enforces its entire
departure from classical lines of thought.” [19] This observation remains true
today, and with the advent of quantum computing, its practical consequences
have never before been more real. However after decades of effort,
entanglement remains poorly understood [13, 1, 11]. A promising avenue for
furthering our understanding of entanglement is cataloguing and analyzing the
various means of generating it. There is a sense that those mechanisms which
generate maximal amounts of entanglement, or most consistently generate
entanglement, are especially enlightening because they serve as bounds on what
can and can not be done, thus restricting our domain of inquiry.

One outcome of this line of thought is the concept of a universal entangler
(UE). A UE is a unitary operator which maps any non-entangled state to an
entangled state [3]. A UE can act as a useful tool, both theoretically and
experimentally, due to its generality. This generality is derived from the
fact that a UE admits any non-entangled quantum states. However this
generality also makes demonstrating the properties of UEs very difficult. For
instance, while it has been shown that UEs do exist for a system with Hilbert
space $\mathbb{C}^{d_{1}}\otimes\mathbb{C}^{d_{2}}$ when
$\min{(d_{1},d_{2})}\geq 3$ and $(d_{1},d_{2})\neq(3,3)$, proving this fact
has been nontrivial, requiring techniques from algebraic geometry [3]. To date
no elementary method is known which can achieve the same results.
Additionally, although it has been shown that whenever UEs exist almost all
unitaries are UEs [4], explicit constructions of UEs remain ellusive. This is
due to the fact that the problem of verifying whether a given unitary is a UE
is in general intractable since the verification is equivalent to solving a
quadratic system of equations which is hard in general [6]. So far the only
explicitly known UE is an example for the $(d_{1},d_{2})=(3,4)$, from an order
$12$ Hadamard matrix [4]. In general more advanced methods may be needed in
order to construct UEs, as well as to verify their universality.

The theory of entanglement of systems of indistinguishable particles has
garnered much attention during the past decade [17, 16, 12, 14, 5, 1]. The
entanglement of systems of indistinguishable particles cannot necessarily be
approached in the same way as the distinguishable particle case because the
symmetry requirement of the wave functions (i.e. symmetrization for bosonic
system and antisymmetrization for fermionic system) may introduce ‘pseudo
entanglement’ which is not accessible in practice [5, 16, 17, 12, 14]. It is
now widely agreed that non-entangled states correspond to the coherent states
$|v\rangle^{\otimes N}$ [15] for indistinguishable bosonic systems and to
Slater determinants for indistinguishable fermionic systems [5, 1]. A natural
line of inquiry is to identify the existence and construction of UEs for
systems of indistinguishable particles. Indistinguishable bipartite
bosonic/fermionic states are symmetric/antisymmetric states of the Hilbert
space $\mathbb{C}^{d}\otimes\mathbb{C}^{d}$. This does not necessarily mean
that the theory of UEs of distinguishable particles is readily generalizable
to UEs for indistinguishable particles. Some obvious reasons for this are: 1.
although almost all unitaries are UEs when $d>3$, the lack of understanding of
explicit constructions prevents us from directly verifying whether there exist
any UEs which are symmetric under particle permutation; 2. the definition of a
non-entangled state for systems of indistinguishable fermions is dramatically
different from that of systems of distinguishable particles (in fact, a single
Slater determinant, when viewed as an antisymmetric distinguishable particle
state, is indeed entangled).

This paper discusses the existence and construction of UEs for both
indistinguishable bipartite bosonic (BUE) and fermionic (FUE) systems.
Employing techniques in algebraic geometry, considering the different
algebraic structures of the bosonic and fermionic systems, we show that, in
contrast to the distinguishable particle case, BUEs exist for bosonic systems
if and only if the single particle Hilbert space has dimension $d\geq 3$, and
FUEs exist for fermionic systems if and only if the single particle Hilbert
space has dimension $d\geq 8$. We also show, similarly to the distinguishable
particle case, that for dimensions where BUEs/FUEs exist, almost all unitaries
are BUEs/FUEs. Finally, because the unentangled states of indistinguishable
bosonic systems are of a relatively simple coherent state form
$|v\rangle\otimes|v\rangle$, which implies a hidden linear structure for the
product states (i.e. the set of all single particle states $|v\rangle$ form a
vector space), the construction of BUEs becomes significantly simpler. We have
found a simple explicit construction of a BUE based on permutation matrices
which holds for all $d\geq 3$, and another one based on Householder-type gates
[10] which holds for all $d\geq 5$. Unfortunately the explicit construction
and verification of FUEs, like distinguishable particle UEs, remains a
significantly more intractable problem.

We believe that our investigation provides insight into the entanglement
properties of identical particle systems, and in particular the different
entanglement structures between bosonic, fermionic and distinguishable
particle systems.

We organize our paper as follows. In section 2 we review some previously
established results about UEs and provide some preliminaries about bosonic and
fermionic systems to help establish our main results. In section 3 we give a
proof for the existence and prevalence of BUEs, and give two explicit examples
of their construction. In Section 4 we give a proof for the existence and
prevalence of FUEs. Finally, in section 5, we provide a brief summary of our
results and a discussion of future directions.

##  2\. Preliminaries

This section provides preliminaries to help establish our main results for
BUEs and FUEs. We first briefly review UEs for distinguishable particle
systems established in [3]. We then further briefly review basic entanglement
theory for bosonic and fermionic systems.

###  2.1. Universal entanglers

For the case of distinguishable particles, it is known that any given quantum
system is identified with some finite (or infinite) Hilbert space
$\mathcal{H}$. Moreover, two unit vectors are indistinguishable if they differ
only by a global phase factor. Hence, distinct pure states can be put in
correspondence with “rays” in $\mathcal{H}$, or equivalently, points in the
projective Hilbert space $\mathbb{P}(\mathcal{H})$.

We consider pure states for bipartite systems, whose Hilbert space is
$\mathbb{C}^{d_{1}}\otimes\mathbb{C}^{d_{2}}$. A bipartite quantum state is a
product state if $|\psi\rangle=|\psi_{1}\rangle\otimes|\psi_{2}\rangle$ for
some $|\psi_{1}\rangle\in\mathbb{C}^{d_{1}}$ and
$|\psi_{2}\rangle\in\mathbb{C}^{d_{2}}$. Otherwise, it is an entangled state.
It is straightforward to see that the set of all the product states do not
form a linear vector space, so one does not expect that the UE problem can be
examined using basic tools from linear algebra.

Instead, it is observed that the set of normalized product states in a
composite system associated with $\mathbb{C}^{d_{1}}\otimes\mathbb{C}^{d_{2}}$
is isomorphic to a projective variety in $\mathbb{P}^{d_{1}d_{2}-1}$, a well
studied object in algebraic geometry. Before continuing, we need some basic
notations and necessary background materials from algebraic geometry [8].

For any positive integer $n$, the set of all $n$-tuples from $\mathbb{C}$ is
called an $n$-dimensional affine space over $\mathbb{C}$. An element of
$\mathbb{C}^{n}$ is called a point, and if point
$P=(a_{1},a_{2},\cdots,a_{n})$ with $a_{i}\in\mathbb{C}$, then the $a_{i}$’s
are called the coordinates of $P$. Informally, an affine space is what is left
of a vector space after forgetting its origin.

We define projective $n$-space, denoted by $\mathbb{P}^{n}$, to be the set of
equivalence classes of $(n+1)-$tuples $(a_{0},\cdots,a_{n})$ from
$\mathbb{C}$, not all zero, under the equivalence relation given by
$(a_{0},\cdots,a_{n})\sim(\lambda a_{0},\cdots,\lambda a_{n})$ for all
$\lambda\in\mathbb{C}$, $\lambda\neq 0$. We use $[a_{0}:\cdots:a_{n}]$ to
denote the projective coordinates of this point.

The polynomial ring in $n$ variables, denoted by
$\mathbb{C}[x_{1},x_{2},\cdots,x_{n}]$, is the set of polynomials in $n$
variables with coefficients in field $\mathbb{C}$.

A subset $Y$ of $\mathbb{C}^{n}$ is an algebraic set if it is the common zeros
of a finite set of polynomials $f_{1},f_{2},\cdots,f_{r}$ with
$f_{i}\in\mathbb{C}[x_{1},x_{2},\cdots,x_{n}]$ for $1\leq i\leq r$, which is
also denoted by $Z(f_{1},f_{2},\cdots,f_{r})$.

One may observe that the union of a finite number of algebraic sets is an
algebraic set, and the intersection of any family of algebraic sets is again
an algebraic set. Therefore, by taking the open subsets to be the complements
of algebraic sets, we can define a topology, called the Zariski topology on
$\mathbb{C}^{n}$.

A nonempty subset $Y$ of a topological space $X$ is called irreducible if it
cannot be expressed as the union of two proper closed subsets. The empty set
is not considered to be irreducible.

An affine algebraic variety is an irreducible closed subset of
$\mathbb{C}^{n}$, with respect to the induced topology.

A notion of algebraic variety may also be introduced in projective spaces,
called projective algebraic variety: a subset $Y$ of $\mathbb{P}^{n}$ is an
algebraic set if it is the common zeros of a finite set of homogeneous
polynomials $f_{1},f_{2},\cdots,f_{r}$ with
$f_{i}\in\mathbb{C}[x_{0},x_{1},\cdots,x_{n}]$ for $1\leq i\leq r$. We call
open subsets of irreducible projective varieties quasi-projective varieties.

Observe that a product state in $\mathbb{C}^{d_{1}}\otimes\mathbb{C}^{d_{2}}$
can be written as the Kronecker product of a vector
$v_{1}\in\mathbb{C}^{d_{1}}$ and another vector $v_{2}\in\mathbb{C}^{d_{2}}$.
Let’s further write these vectors in the computational basis, say
$v_{1}=\left(\begin{array}[]{cccc}x_{1},&x_{2},&\cdots&,x_{d_{1}}\end{array}\right)$
and
$v_{2}=\left(\begin{array}[]{cccc}y_{1},&y_{2},&\cdots&,y_{d_{2}}\end{array}\right)$.
Their product state is a $d_{1}d_{2}$-dimensional vector

$\displaystyle\left(\begin{array}[]{ccccccc}z_{1},&z_{2},&\cdots&z_{d_{2}},&z_{d_{2}+1},&\cdots&,z_{d_{1}d_{2}}\end{array}\right)$
$\displaystyle=$
$\displaystyle\left(\begin{array}[]{ccccccc}x_{1}y_{1},&x_{1}y_{2},&\cdots&x_{1}y_{d_{2}},&x_{2}y_{1},&\cdots&,x_{d_{1}}y_{d_{2}}\end{array}\right)$

Hence $z_{(i-1)d_{2}+j}=x_{i}y_{j}$ for any $1\leq i\leq d_{1},1\leq j\leq
d_{2}$. It follows that

$\displaystyle z_{(i_{1}-1)d_{2}+j_{1}}z_{(i_{2}-1)d_{2}+j_{2}}$
$\displaystyle=$ $\displaystyle
z_{(i_{1}-1)d_{2}+j_{2}}z_{(i_{2}-1)d_{2}+j_{1}}$

for any $1\leq i_{1},i_{2}\leq d_{1},1\leq j_{1},j_{2}\leq d_{2}$. On the
other hand, any $d_{1}d_{2}$-dimensional vector $(z_{k})_{k=1}^{d_{1}d_{2}}$
satisfying the above polynomials can be written as the tensor product of
$v_{1}\in\mathbb{C}^{d_{1}}$ and $v_{2}\in\mathbb{C}^{d_{2}}$ [8]. This
implies that the set of normalized product states in
$\mathbb{C}^{d_{1}}\otimes\mathbb{C}^{d_{2}}$ is isomorphic to a projective
variety in $\mathbb{P}^{d_{1}d_{2}-1}$ which is called a “Segre variety” and
denoted as $\Sigma_{d_{1},d_{2}}$. This simple observation provides an
algebraic geometric description of product states and entangled states.

Therefore, a unitary operator $U$ acting on
$\mathbb{C}^{d_{1}}\otimes\mathbb{C}^{d_{2}}$ is a UE if and only

$U(\Sigma_{d_{1},d_{2}})\bigcap\Sigma_{d_{1},d_{2}}=\emptyset.$

From the geometric point of view, a UE will rotate the set of product states
to another set which is completely void of product states.

In [3], it is proved that UEs exist if and only if $\min\\{d_{1},d_{2}\\}\geq
3$ and $(d_{1},d_{2})\neq(3,3)$. Surprisingly, it is further illustrated that
a random unitary operator acting on such a bipartite system will even rotate
the set of product states to another set which contains nothing but nearly
maximally entangled states [4].

Although it has been shown that a random unitary gate will almost surely be a
UE of a bipartite quantum system $\mathbb{C}^{d_{1}}\otimes\mathbb{C}^{d_{2}}$
if $\min\\{d_{1},d_{2}\\}\geq 3$ and $(d_{1},d_{2})\neq(3,3)$, constructing an
explicit UE for any bipartite quantum system is not that easy. One simple
strategy is to randomly pick a unitary gate acting on
$\mathbb{C}^{d_{1}}\otimes\mathbb{C}^{d_{2}}$, and then verify whether it is a
UE by solving a family of polynomial equations. Unfortunately, there is no
known efficient way to solve quadratic polynomial systems [6]. So far,
explicit UEs are only known for $(d_{1},d_{2})=(3,4)$ [4].

###  2.2. Bosonic systems

It is known that bosonic states lie in the $2$nd symmetric tensor power of
$\mathbb{C}^{d}$, denoted by $\vee^{2}\mathbb{C}^{d}$. A state in
$\vee^{2}\mathbb{C}^{d}$ is a product state if it can be written as some
$|\alpha\rangle\otimes|\alpha\rangle$, i.e. it is a coherent state [14, 5].
Any state which cannot be written as such a symmetric product form does
demonstrate correlation which can be potentially used in quantum information
processing [14], and hence is considered entangled.

Any bipartite bosonic pure state is local unitarily equivalent to
$\sum_{\alpha}\lambda_{\alpha}|\alpha\rangle\otimes|\alpha\rangle$ [12, 14].
This then indicates a hidden linear structure for bipartite bosonic pure
states because the single particles states $|\alpha\rangle$ form a vector
space.

From the algebraic geometric point of view, any bosonic product state
$|\alpha\rangle\otimes|\alpha\rangle$ can be written as a vector with
projective coordinates

$[a_{1}a_{1}:a_{1}a_{2}:\cdots:a_{1}a_{d}:a_{2}a_{1}:a_{2}a_{2}:\cdots:a_{2}a_{d}:a_{3}a_{1}:\cdots:a_{d}a_{d}]$

where $[a_{1}:\cdots:a_{d}]$ are the projective coordinates of
$|\alpha\rangle$.

Such points can be characterized by a family of polynomials again. In fact,
the set of projective points with coordinates

$[a_{1}a_{1}:a_{1}a_{2}:\cdots:a_{1}a_{d}:a_{2}a_{1}:a_{2}a_{2}:\cdots:a_{2}a_{d}:a_{3}a_{1}:\cdots:a_{d}a_{d}]$

is obviously isomorphic to the set of the following points

$[a_{1}^{2}:a_{2}^{2}:\cdots:a_{d}^{2}:a_{1}a_{2}:a_{1}a_{3}:\cdots:a_{1}a_{d}:a_{2}a_{3}:\cdots:a_{d-1}a_{d}]$

which is known as the Veronese variety in algebraic geometry [7].

Hence the set of bosonic product states corresponds to a special case of
Veronese variety whose dimension is $d-1$. This fact will be used in our
further investigation.

###  2.3. Fermionic systems

Consider the pure states of a bipartite fermionic system whose Hilbert space
is the antisymmetric subspace of $\mathbb{C}^{d}\otimes\mathbb{C}^{d}$. The
Pauli exclusion principle requires that $d\geq 2$. We denote the $2$nd
exterior power of $\mathbb{C}^{d}$, i.e. the antisymmetric subspace of
$\mathbb{C}^{d}\otimes\mathbb{C}^{d}$ by $\wedge^{2}\mathbb{C}^{d}$. For any
$|\alpha\rangle,|\beta\rangle\in\mathbb{C}^{d}$, we use the notation

$|\alpha\rangle\wedge|\beta\rangle=\frac{1}{\sqrt{2}}(|\alpha\rangle\otimes|\beta\rangle-|\beta\rangle\otimes|\alpha\rangle),$
(3)

to denote a single Slater determinant.

A quantum state $|\psi\rangle$ in $\wedge^{2}\mathbb{C}^{d}$ is said to be
decomposable if it can be written as an exterior product of individual vectors
from $\mathbb{C}^{d}$, i.e. there exists
$|\alpha\rangle,|\beta\rangle\in\mathbb{C}^{d}$ such that
$|\psi\rangle=|\alpha\rangle\wedge|\beta\rangle$. Decomposable states are
considered unentangled, as any correlation results purely from the fermionic
statistics, and so is not useful for quantum information processing [17, 16].
Any state which cannot be written in such a decomposable form does demonstrate
correlation which can be potentially used in quantum information processing
[17, 16] , and hence is considered to be entangled.

Any bipartite fermionic pure state is local unitarily equivalent to
$\sum_{\alpha}\lambda_{i}|\alpha_{i}\rangle\wedge|\beta_{i}\rangle$ [17, 16],
where $|\alpha_{i}\rangle,|\beta_{i}\rangle\in\mathbb{C}^{d}$,
$\langle{\alpha_{i}}|\beta_{j}\rangle=0$,
$\langle\alpha_{i}|\alpha_{j}\rangle=\delta_{ij}$, and
$\langle\beta_{i}|\beta_{j}\rangle=\delta_{ij}$. This is an analogue of the
Schmidt decomposition of a distinguishable particle system and hence is called
the Slater decomposition. Similarly to the distinguishable particle case, the
set of all decomposable states do not form a linear vector space, so one does
not expect that the FUE problem can be examined using basic tools from linear
algebra.

Again, let’s look over the decomposable (or fermionic product) states from the
algebraic geometric point of view. As we showed before, a decomposable state
can be written as $|\psi\rangle=|\alpha\rangle\wedge|\beta\rangle$ where
$|\alpha\rangle$ and $|\beta\rangle$ are two vectors in $\mathbb{C}^{d}$. Let
$S_{\psi}$ be the $2$-dimensional subspace spanned by $|\alpha\rangle$ and
$|\beta\rangle$. A different basis for $S_{\psi}$ will give a different
exterior product, but the two exterior products will differ only by a nonzero
scale. Ignoring the nonzero scale, any decomposable state corresponds to a
$2$-dimensional subspace in $\mathbb{C}^{d}$ and vice versa. Hence the set of
decomposable states is isomorphic to the set of $2$-dimensional subspaces
which is known as a Grassmannian $G(2,d)$ [7]. It is not that obvious that
$G(2,d)$ can be characterized by a set of polynomials, but it can be. The
correspondence we have just shown is known as the Plücker embedding of a
Grassmannian into a projective space:

$\tau:G(2,d)\rightarrow\mathbb{P}(\wedge^{2}\mathbb{C}^{d}).$

This embedding satisfies certain simple quadratic polynomials and is called
the Grassmann-Plücker relations (see e.g. p. A III.172 Eq. (84-(J,H)) in [2],
Prop 11-32 in [9], and [5]). This implies the Grassmannian embeds as an
algebraic variety of $\mathbb{P}(\wedge^{2}\mathbb{C}^{d})$.

##  3\. Bosonic Universal Entanglers

###  3.1. Existence and Prevalence

Recall that a bosonic state in $\vee^{2}\mathbb{C}^{d}$ is a product state if
it can be written as $|\alpha\rangle\otimes|\alpha\rangle$ for some
$|\alpha\rangle$. A quantum gate acting on $\vee^{2}\mathbb{C}^{d}$ is said to
be a bosonic universal entangler (BUE) if it will map every product state to
some entangled state.

Note that the set of product states of a bosonic system can also be
characterized by a set of polynomials. Indeed, let
$\Lambda=\\{|\alpha\rangle\otimes|\alpha\rangle:|\alpha\rangle\in\mathbb{C}^{d}\\}$,
this is a precisely the Veronese variety [7]. Furthermore, $\Lambda$ is
isomorphic to $\mathbb{C}^{d}$. For any
$|\psi\rangle\in\vee^{2}\mathbb{C}^{2}$, let us denote
$\text{rank}|\psi\rangle\equiv\min\\{r:|\psi\rangle=\sum\limits_{i=1}^{r}|a_{i}\rangle|a_{i}\rangle\\}$.

######  Theorem 3.1.

There is a BUE acting on $\vee^{2}\mathbb{C}^{d}$ if and only if $d\geq 3$.
Furthermore, when $d\geq 3$, almost every quantum gate acting on
$\vee^{2}\mathbb{C}^{d}$ is a BUE.

######  Proof 3.2.

For $d\leq 2$, we have

$\displaystyle\dim U(\Lambda)+\dim\Lambda=2\dim\Lambda=2(d-1)\geq{d+1\choose
2}-1=\dim\mathbb{P}(\vee^{2}\mathbb{C}^{d}).$ (4)

This implies there is no BUE for $\vee^{2}\mathcal{C}^{d}$. This assertion
follows from the dimension counting theorem which states that the intersection
of any two projective varieties $\mathcal{A}$ and
$\mathcal{B}\subseteq\mathbb{P}^{m}$ is nonempty if
$\dim\mathcal{A}+\dim\mathcal{B}\geq m$. More specifically, we have
$U(\Lambda)\bigcap\Lambda\neq\emptyset$.

On the other hand, consider the set of quantum gates acting on a system of two
indistinguishable bosons. Any quantum gate acting on this system should be a
symmetric gate, i.e., $SUS=U$, where $S$ is the swap operator. Equivalently,
$U$ is a quantum gate acting on $\vee^{2}\mathbb{C}^{d}$.

Let
$\mathcal{X}=\\{\Phi|\Phi\in\mathcal{U}(\vee^{2}\mathbb{C}^{d}),\Phi(\Lambda)\cap\Lambda\neq\emptyset\\}$.
Our aim is to show that $\mathcal{X}$ is a proper subset of
$\mathcal{U}(\vee^{2}\mathbb{C}^{d})$. If this is so, then the existence of
BUEs will be automatically guaranteed.

Let’s consider the Zariski topology on the projective space. In this setting,
the unitary group $\mathcal{U}(\vee^{2}\mathbb{C}^{d})$ is Zariski dense in
the general linear group $GL(\vee^{2}\mathbb{C}^{d})$ [18]. We further define
$\mathcal{X}^{\prime}=\\{\Phi|\Phi\in
GL(\vee^{2}\mathbb{C}^{d}),\Phi(\Lambda)\cap\Lambda\neq\emptyset\\}$. It is
easy to see that $\mathcal{X}\subseteq\mathcal{X}^{\prime}$.

The dimension of its Zariski closure $\dim\overline{\mathcal{X}^{\prime}}$ is
bounded by ${d+1\choose 2}^{2}-({d+1\choose 2}-1)+2(d-1)$. See Lemma C.1 in
Appendix C for details.

Now we prove the existence of a BUE as follows. If $U$ is not a BUE,
$\mathcal{U}(\vee^{2}\mathbb{C}^{d})\subset\mathcal{X}^{\prime}$, then
$GL(\vee^{2}\mathbb{C}^{d})=\overline{\mathcal{U}(\vee^{2}\mathbb{C}^{d})}\subset{\overline{\mathcal{X}^{\prime}}}$.
However, $\dim(\overline{\mathcal{X}^{\prime}})\leq{d+1\choose
2}^{2}-({d+1\choose 2}-1)+2(d-1)<{d+1\choose 2}^{2}=\dim
GL(\vee^{2}\mathbb{C}^{d})$. This is a contradiction. So
$\mathcal{U}(\vee^{2}\mathbb{C}^{d})\not\subset\mathcal{X}^{\prime}$, i.e. a
unitary operator $\Phi\in\mathcal{U}(\vee^{2}\mathbb{C}^{d})$ with universal
entangling power exists.

We will now show that $\mathcal{X}$ is not only a proper subset, but also a
negligible subset of $\mathcal{U}(\vee^{2}\mathbb{C}^{d})$.

$\mathcal{U}(\vee^{2}\mathbb{C}^{d})$ is a locally compact Lie group of
dimension ${d+1\choose 2}^{2}$. Recall that
$\dim(\overline{\mathcal{X}^{\prime}})$ is at most ${d+1\choose
2}^{2}-({d+1\choose 2}-1)+2(d-1)<{d+1\choose
2}^{2}=\dim(\mathcal{U}(\vee^{2}\mathbb{C}^{d}))$.

We have shown $\dim(\overline{\mathcal{X}^{\prime}})<{d+1\choose
2}^{2}=\dim(\mathcal{U}(\vee^{2}\mathbb{C}^{d}))$.
$\overline{\mathcal{X}^{\prime}}$ is Noetherian (i.e. any descending sequence
of its closed subvarieties is stationary), then
$\overline{\mathcal{X}^{\prime}}$ is a union of finitely many smooth
subvarieties of $GL(\vee^{2}\mathbb{C}^{d})$ with lower dimensions. Hence
$\overline{\mathcal{X}^{\prime}}\cap\mathcal{U}(\vee^{2}\mathbb{C}^{d})$
(which contains $\mathcal{X}^{\prime}\cap\mathcal{U}(\vee^{2}\mathbb{C}^{d})$,
the set of our main interest) is a union of finite many submanifolds of
$\mathcal{U}(\vee^{2}\mathbb{C}^{d})$ with lower dimensions. Therefore,
$\mathcal{X}^{\prime}\cap\mathcal{U}(\vee^{2}\mathbb{C}^{d})$ is measure zero
in $\mathcal{U}(\vee^{2}\mathbb{C}^{d})$ which implies that a random unitary
operator $U$ is almost surely a BUE.

###  3.2. Explicit Construction

As we have shown in Theorem 3.1, a random unitary acting on
$\vee^{2}\mathbb{C}^{d}$ will almost surely be a BUE. Hence we can pick an
arbitrary unitary acting on $\vee^{2}\mathbb{C}^{d}$ and verify whether it
will map some product state to another product state. Recall that the set of
product states in a bosonic system is isomorphic to $\mathbb{C}^{d}$. This
will make it easier to verify whether a unitary is a BUE. Here we provide
verifications of two different classes of BUEs.

####  3.2.1. Householder-type Bosonic Universal Entanglers

For $d\geq 5$ and any subspace $S\subset\vee^{2}\mathbb{C}^{d}$, let’s
consider the following gate $U=\mathbb{I}_{\vee^{2}\mathbb{C}^{d}}-2P_{S}$
where $P_{S}$ is a projection to some subspace $S$. These gates are known as
Householder matrices in linear algebra [10] and they are widely used to
perform QR decomposition.

A gate $U$ constructed in this way will be a BUE if the subspace $S$ is chosen
properly to satisfy the following two constraints:

  1. 1.

There is no product state in $S^{\perp}$.

  2. 2.

$\text{rank}|\psi\rangle\geq 3$ for any $|\psi\rangle\in S$.

This claim can be proved by contradiction. Assume there are two product states
$|\psi\rangle|\psi\rangle$ and $|\phi\rangle|\phi\rangle$ such that
$(\mathbb{I}_{\vee^{2}\mathbb{C}^{d}}-2P_{S})|\psi\rangle|\psi\rangle=|\phi\rangle|\phi\rangle$,
we have
$2P_{S}|\psi\rangle|\psi\rangle=|\psi\rangle|\psi\rangle-|\phi\rangle|\phi\rangle$.
$P_{S}|\psi\rangle|\psi\rangle\neq 0$ since there is no product state in
$S^{\perp}$. On the other hand, $P_{S}|\psi\rangle|\psi\rangle$ is a vector in
$S$ which is a subspace completely void of states with rank no more than $2$.
This contradicts our assumption.

In this subsection, we will construct a subspace $S$ to satisfy the above two
constraints for any $d\geq 5$. A family of BUEs will follow immediately.

Let $S$ be the span of the following vectors.

$\displaystyle|11\rangle+|23\rangle+|32\rangle,$
$\displaystyle|22\rangle+|34\rangle+|43\rangle,$ $\displaystyle\textrm{\ \ \ \
\ \ \ \ \ }\cdots,$
$\displaystyle|d-2,d-2\rangle+|d-1,d\rangle+|d,d-1\rangle,$
$\displaystyle|d-1,d-1\rangle+|d,1\rangle+|1,d\rangle,$
$\displaystyle|d,d\rangle+|12\rangle+|21\rangle.$

We first show there is no product state in $S^{\perp}$. Assume
$|\psi\rangle|\psi\rangle\perp S$ where
$|\psi\rangle=\sum\limits_{i=1}^{d}a_{i}|i\rangle$. The orthogonality implies
the following equations.

$\displaystyle(E1)\left\\{\begin{array}[]{lll}a_{1}^{2}+2a_{2}a_{3}&=&0,\\\
a_{2}^{2}+2a_{3}a_{4}&=&0,\\\ &\vdots&\\\
a_{d}^{2}+2a_{1}a_{2}&=&0.\end{array}\right.$

The only common solution to the above equations is
$(a_{1},a_{2},\cdots,a_{d})=(0,0,\cdots,0)$ when $d\geq 3$. See Appendix A for
details.

Hence, there is no product state in $S^{\perp}$.

Next, we will verify that $\text{rank}|\psi\rangle\geq 3$ for any
$|\psi\rangle\in S$.

Assume there is some state $|\psi\rangle\in S$ with rank no more than $2$.
Let’s say

$\displaystyle c_{1}(|11\rangle+|23\rangle+|32\rangle),$ (6) $\displaystyle+$
$\displaystyle c_{2}(|22\rangle+|34\rangle+|43\rangle),$ (7) $\displaystyle+$
$\displaystyle\textrm{\ \ \ \ \ \ \ \ \ }\cdots,$ (8) $\displaystyle+$
$\displaystyle c_{d}(|d,d\rangle+|12\rangle+|21\rangle),$ (9) $\displaystyle=$
$\displaystyle(x_{1}|1\rangle+x_{2}|2\rangle+\cdots+x_{d}|d\rangle)(x_{1}|1\rangle+x_{2}|2\rangle+\cdots+x_{d}|d\rangle),$
(10) $\displaystyle+$
$\displaystyle(y_{1}|1\rangle+y_{2}|2\rangle+\cdots+y_{d}|d\rangle)(y_{1}|1\rangle+y_{2}|2\rangle+\cdots+y_{d}|d\rangle).$
(11)

Then we have the following equations.

$\displaystyle(E2)\left\\{\begin{array}[]{lll}x_{1}^{2}+y_{1}^{2}&=&c_{1},\\\
x_{2}^{2}+y_{2}^{2}&=&c_{2},\\\ &\vdots&\\\ x_{d}^{2}+y_{d}^{2}&=&c_{d},\\\
x_{1}x_{2}+y_{1}y_{2}&=&c_{d},\\\ x_{2}x_{3}+y_{2}y_{3}&=&c_{1},\\\
&\vdots&\\\ x_{d}x_{1}+y_{d}y_{1}&=&c_{d-1},\\\
x_{i}x_{j}+y_{i}y_{j}&=&0\forall|i-j|\geq 2.\end{array}\right.$ (21)

There is no nonzero $(c_{0},c_{2},\cdots,c_{d})$ satisfying the above
equations when $d\geq 5$. See Appendix B for details. Hence
$\text{rank}|\psi\rangle\geq 3$ for any $|\psi\rangle\in S$.

This implies that $U=I-2P_{S}$ is a bosonic universal entangler for any $d\geq
5$.

####  3.2.2. Permutation Universal Entanglers

Any product state can be written as the following.

$\displaystyle|\phi\rangle|\phi\rangle$ $\displaystyle=$
$\displaystyle(\sum\limits_{i=1}^{d}a_{i}|i\rangle)(\sum\limits_{j=1}^{d}a_{j}|j\rangle)$
(22) $\displaystyle=$
$\displaystyle\sum\limits_{i,j=1}^{d}a_{i}a_{j}|ij\rangle$ (23)
$\displaystyle=$
$\displaystyle\sum\limits_{i=1}^{d}a_{i}^{2}|ii\rangle+\sum\limits_{1\leq
i<j\leq d}\sqrt{2}a_{i}a_{j}(\frac{|ij\rangle+|ji\rangle}{\sqrt{2}}).$ (24)

Any bosonic state $|\psi\rangle\in\vee^{2}\mathbb{C}^{d}$ can be denoted as a
${d+1\choose 2}$-dimensional vector

$(x_{11},x_{22},\cdots,x_{dd},x_{12},\cdots,x_{1d},x_{21},\cdots,x_{d-1d})$

since we can always write $|\psi\rangle$ as a linear combination of bosonic
basis states

$x_{11}|11\rangle+x_{22}|22\rangle+\cdots+x_{dd}|dd\rangle+x_{12}\frac{|12\rangle+|21\rangle}{\sqrt{2}}+x_{13}\frac{|13\rangle+|31\rangle}{\sqrt{2}}+\cdots+x_{d-1,d}\frac{|d-1,d\rangle+|d,d-1\rangle}{\sqrt{2}}.$

$|\psi\rangle$ is a product state if and only if there exists some nonzero
vector $(a_{1},a_{2},\cdots,a_{d})$ such that

$\displaystyle(x_{11},\cdots,x_{dd},x_{12},x_{13},\cdots,x_{1d},x_{23},\cdots,x_{d-1d})$
(25) $\displaystyle=$
$\displaystyle(a_{1}^{2},\cdots,a_{d}^{2},\sqrt{2}a_{1}a_{2},\sqrt{2}a_{1}a_{3},\cdots,\sqrt{2}a_{1}a_{d},\sqrt{2}a_{2}a_{3},\cdots,\sqrt{2}a_{d-1}a_{d}).$
(26)

A permutation matrix $U$ acting on the ${d+1\choose 2}$-dimensional vector
space is certainly a bosonic quantum gate.

For any $d\geq 3$, let’s define a permutation matrix $U$ as the following:

$\displaystyle U$ $\displaystyle=$
$\displaystyle\sum\limits_{i=1}^{d}(\frac{|i,i+1\rangle+|i+1,i\rangle}{\sqrt{2}})\langle
ii|+\sum\limits_{i=1}^{d}|ii\rangle(\frac{\langle i,i+1|+\langle
i+1,i|}{\sqrt{2}})$ $\displaystyle+\sum\limits_{1\leq i<i+1<j\leq
d}\frac{(|ij\rangle+|ji\rangle)(\langle ij|+\langle ji|)}{2}.$

Here the addition and subtraction are all modulo $d$, but the results range
from $1$ to $d$.

$U$ is a unitary matrix since it is simply a rotation of the ${d+1\choose
2}$-dimensional vector space. Let’s assume $U$ will map some (bosonic) product
state to another (bosonic) product state. Without loss of generality, let’s
assume

$\displaystyle U(\sum\limits_{i=1}^{d}a_{i}^{2}|ii\rangle+\sum\limits_{1\leq
i<j\leq
d}\sqrt{2}a_{i}a_{j}(\frac{|ij\rangle+|ji\rangle}{\sqrt{2}}))=\sum\limits_{i=1}^{d}b_{i}^{2}|ii\rangle+\sum\limits_{1\leq
i<j\leq d}\sqrt{2}b_{i}b_{j}(\frac{|ij\rangle+|ji\rangle}{\sqrt{2}}).$

It follows that

$\displaystyle a_{1}^{2}$ $\displaystyle=$ $\displaystyle\sqrt{2}b_{1}b_{2},$
$\displaystyle a_{2}^{2}$ $\displaystyle=$ $\displaystyle\sqrt{2}b_{2}b_{3},$
$\displaystyle\vdots$ , $\displaystyle a_{d}^{2}$ $\displaystyle=$
$\displaystyle\sqrt{2}b_{d}b_{1},$ $\displaystyle\sqrt{2}a_{1}a_{2}$
$\displaystyle=$ $\displaystyle b_{1}^{2},$ $\displaystyle\sqrt{2}a_{2}a_{3}$
$\displaystyle=$ $\displaystyle b_{2}^{2},$ $\displaystyle\vdots$ ,
$\displaystyle\sqrt{2}a_{d}a_{1}$ $\displaystyle=$ $\displaystyle b_{d}^{2}.$

Hence we have
$\prod\limits_{i=1}^{d}a_{i}^{2}=(\sqrt{2})^{d}\prod\limits_{i=1}^{d}b_{i}b_{i+1}=\sqrt{2}^{d}\prod\limits_{i=1}^{d}b_{i}^{2}$.
Similarly,
$\prod\limits_{i=1}^{d}b_{i}^{2}=\sqrt{2}^{d}\prod\limits_{i=1}^{d}a_{i}^{2}$.
The above two equations imply that there exists some $1\leq t\leq d$ such that
$a_{t}=0$.

The equation $b_{t}^{2}=\sqrt{2}a_{t}a_{t+1}$ implies $b_{t}=0$. Then
$a_{t-1}^{2}=\sqrt{2}b_{t-1}b_{t}=0$ will implies $a_{t-1}=0$. By repeating
the above procedure, we will eventually have $a_{i}=0$ for any $1\leq i\leq
d$. This contradicts our assumption that $U$ will map some (bosonic) product
state to another (bosonic) product state. Hence $U$ is a bosonic universal
entangler.

##  4\. Fermionic Universal Entanglers

Given a bipartite system of indisitinguishable fermions
$\wedge^{2}\mathbb{C}^{d}$, a $2$-vector in $\wedge^{2}\mathbb{C}^{d}$ is said
to be decomposable if it can be written as an exterior product of individual
vectors from $\mathbb{C}^{d}$. Decomposable $2$-vectors are also considered to
be unentangled states in this fermionic system.

We say a quantum gate $U$ is a fermionic universal entangler (FUE) if $U$ will
transform every product state to some entangled state.

######  Theorem 4.1.

There is some FUE acting on a bipartite system of indisitinguishable fermions
$\wedge^{2}\mathbb{C}^{d}$ if and only if $d\geq 8$. Furthermore, almost every
quantum gate acting on $\wedge^{2}\mathbb{C}^{d}$is an FUE when $d\geq 8$.

######  Proof 4.2.

Let
$\Gamma_{d}=\\{|\phi\rangle\in\wedge^{2}\mathbb{C}^{d}:|\phi\rangle=|\psi_{1}\rangle\wedge|\psi_{2}\rangle\textrm{\
\ for some\ }|\psi_{1}\rangle,|\psi_{2}\rangle\in\mathbb{C}^{d}\\}$. A quantum
gate $U$ is an FUE if and only if

$\displaystyle U(\Gamma_{d})\bigcap\Gamma_{d}=\emptyset.$ (27)

Observe that decomposable $2$-vectors in $\wedge^{2}\mathbb{C}^{d}$ correspond
to weighted $2$-dimensional linear subspaces of $\mathbb{C}^{d}$. If we ignore
the phase factor, decomposable $2$-vectors can be characterized by the
Grassmannian of $2$-dimensional subspaces of $\mathbb{C}^{d}$, an algebraic
subvariety of the projective space $\mathbb{P}(\wedge^{2}\mathbb{C}^{d})$[8].
We will denote the Grassmannian of $r$-dimensional subspaces of
$\mathbb{C}^{d}$ as $G(r,d)$.

First, we examine the necessary condition.

According to the intersection theorem, if $\dim
U(\Gamma_{d})+\dim\Gamma_{d}\geq\dim\mathbb{P}(\wedge^{2}\mathbb{C}^{d})$, or
equivalently, $2\times 2(d-2)=2\dim G(2,d)\geq{d\choose 2}-1$, then for any
$U$, $U(\Gamma_{d})\bigcap\Gamma_{d}\neq\emptyset$. This inequality holds only
for $2\leq d\leq 7$ which implies the fermionic universal entangling device
does not exist for $d\leq 7$.

Now, let’s look into the sufficient condition.

The set of quantum gates acting on a bipartite system of indisitinguishable
fermions $\wedge^{2}\mathbb{C}^{d}$ is the unitary group acting on
$\wedge^{2}\mathbb{C}^{d}$, denoted as
$\mathcal{U}(\wedge^{2}\mathbb{C}^{d})$.

Similarly, let
$\mathcal{Y}=\\{\Phi|\Phi\in\mathcal{U}(\wedge^{2}\mathbb{C}^{d}),\Phi(\Gamma_{d})\cap\Gamma_{d}\neq\emptyset\\}$.
We will show that $\mathcal{Y}$ is a proper subset in
$\mathcal{U}(\wedge^{2}\mathbb{C}^{d})$.

Again, let’s consider the Zariski topology on the projective space. In this
setting, the unitary group $\mathcal{U}(\wedge^{2}\mathbb{C}^{d})$ is Zariski
dense in the general linear group $GL(\wedge^{2}\mathbb{C}^{d})$[18]. We
further define $\mathcal{Y}^{\prime}=\\{\Phi|\Phi\in
GL(\wedge^{2}\mathbb{C}^{d}),\Phi(\Gamma_{d})\cap\Gamma_{d}\neq\emptyset\\}$.
It is easy to see $\mathcal{X}\subseteq\mathcal{Y}^{\prime}$.

Similar to the proof of Lemma C.1 in Appendix C, the dimension of
$\mathcal{Y}^{\prime}$’s Zariski closure $\dim\overline{\mathcal{Y}^{\prime}}$
is bounded by ${d\choose 2}^{2}-({d\choose 2}-1)+2\times 2(d-2)$.

Now we prove the existence of an FUE $U$ as follows. If it does not exist,
$\mathcal{U}(\wedge^{2}\mathbb{C}^{d})\subset\mathcal{Y}^{\prime}$, then
$GL(\wedge^{2}\mathbb{C}^{d})=\overline{\mathcal{U}(\wedge^{2}\mathbb{C}^{d})}\subset{\overline{\mathcal{Y}^{\prime}}}$.
However, $\dim(\overline{\mathcal{Y}^{\prime}})\leq{d\choose 2}^{2}-({d\choose
2}-1)+4(d-2)<{d\choose 2}^{2}=\dim GL(\wedge^{2}\mathbb{C}^{d})$. This is a
contradiction. So
$\mathcal{U}(\wedge^{2}\mathbb{C}^{d})\not\subset\mathcal{Y}^{\prime}$, i.e.
an FUE $\Phi\in\mathcal{U}(\wedge^{2}\mathbb{C}^{d})$ exists.

Following the lines of the proof of Theorem 3.1, we can prove that
$\mathcal{Y}$ is not only a proper subset, but also a neglectable subset in
$\mathcal{U}(\wedge^{2}\mathbb{C}^{d})$.

##  5\. Summary and Discussion

Employing properties of algebraic geometry, we have shown that for bipartite
systems of indistinguishable bosons with a Hilbert space that is the symmetric
subspace of $\mathbb{C}^{d}\otimes\mathbb{C}^{d}$, bosonic universal
entanglers (BUEs) exist if and only if $d\geq 3$. Similarly, we have shown
that for bipartite systems of indistinguishable fermions with a Hilbert space
that is the antisymmetric subspace of $\mathbb{C}^{d}\otimes\mathbb{C}^{d}$,
fermionic universal entanglers (FUEs) exist if and only if $d\geq 8$. These
two results are in contrast to previous results regarding bipartite systems of
distinguishable particles with a Hilbert space
$\mathbb{C}^{d_{1}}\otimes\mathbb{C}^{d_{2}}$, for which universal entanglers
exist if and only if min $(d_{1},d_{2})\geq 3$ and $(d_{1},d_{2})\neq(3,3)$.
This illustrates some of the important differences between the entanglement of
systems of indistinguishable particles to the entanglement of systems of
distinguishable particles.

In contrast, we have illustrated one feature which holds for both systems of
distinguishable and indistinguishable particles. Previous work has shown that,
for systems of distinguishable particles, if a universal entangler exists for
some Hilbert space, then almost all unitaries operating on that space are
universal entanglers. We have shown that this result also holds for systems of
indistinguishable bosons and fermions. However to verify whether or not a
given bipartite unitary is a universal entangler is in general an intractable
problem for both distinguishable particle systems and fermionic systems. This
intractability arises from the fact that solving a system of quadratic
equations is, in general, NP-hard.

Bosonic systems turn out to be special though. Because the set of all product
states is isomorphic to a linear vector space, it is possible to use
elementary methods to verify bosonic universal entanglers. We have given
explicit constructions of two types of BUE, one is of the Householder type
which is valid for $d\geq 5$ and the other is of a permutation type which is
valid for $d\geq 3$. Both are very simple constructions.

It is our hope that our success in finding explicit constructions of BUEs will
help inform the search for explicit constructions of both FUEs and UEs,
problems which remain intractable in general. We can not rule out the
possibiliy that there might be some other structure, beyond just the
corresponding general algebraic varieties, which would provide some special
family of explicitly verifiable UEs or FUEs. In fact, the explicit
construction for the $(3,4)$ system from an order $12$ Hadamard matrix
demonstrated in [4] provides a hint of the possibility of such families.

Another natural direction of inquiry is to explore the entangling power of
these BUEs and FUEs. As demonstrated in [4], a random unitary is not only
almost surely a UE, but it also almost surely maps the set of product states
to another set which contains nothing but nearly maximally entangled states,
with respect to almost any kind of entanglement measure. One would expect
similar properties for BUEs and FUEs. However to go further in that direction
one would need to first establish reasonable entanglement measures for bosonic
and fermionic systems (see, e.g. entanglement measures discussed in [5]).

Finally, it would be useful to generalize these results to multipartite
bosonic and fermionc systems. Our guess is that the bosonic systems might
remain easy to solve since they retain the nice property that the set of all
product states is isomorphic to a linear vector space. The fermionic case is
expected to be much more complicated given that in the multipartite case even
the Grassmann-Plücker relations themselves are harder to describe [2, 9, 5].
We would leave these cases for future investigation.

##### Acknowledgements

JK is supported by NSERC. JC is supported by NSERC, UTS-AMSS Joint Research
Laboratory for Quantum Computation and Quantum Information Processing and NSF
of China (Grant No. 61179030). BZ is supported by NSERC and CIFAR.

##  Appendix A There Is No Nonzero Solution For Polynomial System (E1)

Here we will show there is no nonzero solution $(a_{1},\cdots,a_{d})$
satisfying the following equations.

$\displaystyle(E1)\left\\{\begin{array}[]{lll}a_{1}^{2}+2a_{2}a_{3}&=&0,\\\
a_{2}^{2}+2a_{3}a_{4}&=&0,\\\ &\vdots&\\\
a_{d}^{2}+2a_{1}a_{2}&=&0.\end{array}\right.$

Assume $a_{i}\neq 0$, then $a_{i+1},a_{i+2}$ are nonzero. This follows all
$a_{i}$’s are nonzero.

$\displaystyle\mathop{\Pi}\limits_{i=1}^{d}a_{i}^{2}=\mathop{\Pi}\limits_{i=1}^{d}(-2a_{i+1}a_{i+2})=(-2)^{d}\mathop{\Pi}\limits_{i=1}^{d}a_{i}^{2}.$
(29)

This implies $\mathop{\Pi}\limits_{i=1}^{d}a_{i}^{2}=0$. Hence it is a
contradiction.

Therefore, the only solution to this polynomial system is
$(a_{1},\cdots,a_{d})=(0,\cdots,0)$.

##  Appendix B There Is No Nonzero Solution For Polynomial System (E2)

Here we will show there is no nonzero solution $(c_{0},c_{1},\cdots,c_{d})$
satisfying the following equations.

$\displaystyle(E2)\left\\{\begin{array}[]{lll}x_{1}^{2}+y_{1}^{2}&=&c_{1},\\\
x_{2}^{2}+y_{2}^{2}&=&c_{2},\\\ &\vdots&\\\ x_{d}^{2}+y_{d}^{2}&=&c_{d},\\\
x_{1}x_{2}+y_{1}y_{2}&=&c_{d},\\\ x_{2}x_{3}+y_{2}y_{3}&=&c_{1},\\\
&\vdots&\\\ x_{d}x_{1}+y_{d}y_{1}&=&c_{d-1},\\\
x_{i}x_{j}+y_{i}y_{j}&=&0\forall|i-j|\geq 2.\end{array}\right.$ (39)

For $d\geq 5$, let’s assume there is some $1\leq i\leq d$ such that $x_{i}=0$
and $y_{i}\neq 0$.

It follows from $x_{i}x_{j}+y_{i}y_{j}=0$ for any $|j-i|\geq 2$ that $y_{j}=0$
for any $|j-i|\geq 2$. So, $y_{i+2}=y_{i+3}=0$.

Then, $0\neq
x_{i}^{2}+y_{i}^{2}=x_{i+1}x_{i+2}+y_{i+1}y_{i+2}=x_{i+1}x_{i+2}$. This
implies $x_{i+1},x_{i+2}\neq 0$.

From $x_{i+1}x_{i+3}+y_{i+1}y_{i+3}=0$, we have $x_{i+3}=0$. So, $0\neq
x_{i+2}^{2}+y_{i+2}^{2}=x_{i+3}x_{i+4}+y_{i+3}y_{i+4}=0$. This is a
contradiction.

So, for any $1\leq i\leq d$, we have $x_{i}=y_{i}=0$ or $x_{i}y_{i}\neq 0$.

Let’s look into the various situations.

  1. 1.

There is some $i$ such that $x_{i}=y_{i}=0$. Then we have
$x_{i-1}^{2}+y_{i-1}^{2}=0$. If $x_{i-1}=y_{i-1}=0$, we consider
$x_{i-2}^{2}+y_{i-2}^{2}=0$. By repeating this procedure, if all
$x_{j}$’s,$y_{j}$’s are not all zero, we will find some $i^{\prime}$ such that
$y_{i^{\prime}}=ix_{i^{\prime}}$ or $y_{i^{\prime}}=-ix_{i^{\prime}}$ and
$x_{i^{\prime}+1}=y_{i^{\prime}+1}=0$. We will further have
$y_{i^{\prime}+k}=\pm x_{i^{\prime}+k}$ for any $k=2,\cdots,d-1$. This implies
that $(c_{1},\cdots,c_{d})=0$.

  2. 2.

All $x_{i}$’s, $y_{i}$’s are nonzero. For any fixed $i$,
$\frac{y_{j}}{x_{j}}=-\frac{x_{i}}{y_{i}}$ for any $j=i+2,\cdots,i+d-2$. This
implies $\frac{y_{k}}{x_{k}}$ is a constant $i$ or $-i$. This also implies
$(c_{1},\cdots,c_{d})=0$.

##  Appendix C Proof of Lemma C.1

######  Lemma C.1.

$\dim(\overline{\mathcal{X}^{\prime}})\leq{d+1\choose 2}^{2}-({d+1\choose
2}-1)+2(d-1)$, where $\overline{\mathcal{X}^{\prime}}$ is the Zariski closure
of $\mathcal{X}^{\prime}$.

The following technical lemmas will be needed.

######  Lemma C.2 ([20]).

If $Z_{1}$ and $Z_{2}$ are both irreducible varieties over $\mathbb{C}$, and
$\phi:Z_{1}\rightarrow Z_{2}$ is a dominant morphism, then
$\dim(Z_{2})\leq\dim(Z_{1})$. Here, dominant means $\Phi(Z_{1})$ is dense in
$Z_{2}$.

######  Lemma C.3 ([20]).

If $Z_{1}$ and $Z_{2}$ are both varieties over $\mathbb{C}$, and
$\phi:Z_{1}\rightarrow Z_{2}$ is a morphism, then
$\dim(Z_{1})\leq\dim(Z_{2})+\max\limits_{z\in Z_{2}}{\dim(\phi^{-1}(z))}$.

Lemma C.2 and Lemma C.3 establish a connection between the dimensions of
domain and codomain of a variety morphism.

######  Proof C.4.

We have a morphism $F:GL(\vee^{2}\mathbb{C}^{d})\times\mathbb{P}^{{d+1\choose
2}-1}\rightarrow\mathbb{P}^{{d+1\choose 2}-1}$ which is just the left action
of $GL(\vee^{2}\mathbb{C}^{d})$ on $\mathbb{P}^{{d+1\choose 2}-1}$, defined by
$F(g,[w])=[g\cdot w]$.

We let $y_{0}=(1,0,\cdots,0)$ be a row vector with $d+1\choose 2$ entries, and
for any given $y_{1}$, $y_{2}\in\mathbb{P}^{{d+1\choose 2}-1}$, we choose
proper $g_{1}$ and $g_{2}\in GL(\vee^{2}\mathbb{C}^{d})$, such that
$[g_{1}\cdot y_{0}]=[y_{1}]$ and $[g_{2}\cdot y_{0}]=[y_{2}]$. Then we have

$\displaystyle[g\cdot y_{2}]=[y_{1}]\iff[gg_{2}\cdot y_{0}]=[gg_{1}\cdot
y_{0}]\iff$ $\displaystyle[g_{1}^{-1}gg_{2}\cdot y_{0}]=[y_{0}].$ (40)

From the above observations, F has the following property: for any $y_{1}$,
$y_{2}\in\mathbb{P}^{{d+1\choose 2}-1}$,
$F^{-1}(y_{2})\cap\\{GL(\vee^{2}\mathbb{C}^{d})\times\\{y_{1}\\}\\}\cong\\{\left(\begin{array}[]{cc}z_{1}&\alpha\\\
0&g^{\prime}\end{array}\right):z_{1}\in\mathbb{C}\backslash\\{0\\},g^{\prime}\in
GL({d+1\choose 2}-1),\alpha\in\mathbb{C}^{{d+1\choose 2}-1}\ is\ a\ row\
vector.\\}$. Hence
$\dim(F^{-1}(y_{2})\cap{GL(\vee^{2}\mathbb{C}^{d})\times\\{y_{1}\\}})={d+1\choose
2}^{2}-({d+1\choose 2}-1)$.

Let $P_{1}$, $P_{2}$ be projections of
$GL(\vee^{2}\mathbb{C}^{d})\times\mathbb{P}^{{d+1\choose 2}-1}$ to
$GL(\vee^{2}\mathbb{C}^{d})$, $\mathbb{P}^{{d+1\choose 2}-1}$ respectively.
Now we only look at $GL(\vee^{2}\mathbb{C}^{d})\times\Lambda\subseteq
GL(\vee^{2}\mathbb{C}^{d})\times\mathbb{P}^{{d+1\choose 2}-1}$, to get
$F:GL(\vee^{2}\mathbb{C}^{d})\times\Lambda\rightarrow\mathbb{P}^{{d+1\choose
2}-1}$. Then we have a characterization of $\mathcal{X}^{\prime}$:
$\mathcal{X}^{\prime}=P_{1}F^{-1}(\Lambda)$. In fact

$\displaystyle g\in\mathcal{X}^{\prime}$ $\displaystyle\iff$ $\displaystyle
g(\Lambda)\cap\Lambda\neq\emptyset$ $\displaystyle\iff$ $\displaystyle\exists
z_{1},z_{2}\in\Lambda,s.t.g(z_{1})=z_{2}$ $\displaystyle\iff$
$\displaystyle\exists z_{1},z_{2}\in\Lambda,s.t.(g,z_{1})\in F^{-1}(z_{2})$
$\displaystyle\iff$ $\displaystyle\exists z_{2}\in\Lambda,s.t.g\in
P_{1}F^{-1}(z_{2})$ $\displaystyle\iff$ $\displaystyle g\in
P_{1}F^{-1}(\Lambda).$

So $\overline{\mathcal{X}^{\prime}}\subseteq GL(\vee^{2}\mathbb{C}^{d})$ is
the Zariski closure of $\mathcal{X}^{\prime}$, which is also an algebraic
variety.

Next, we assert that
$P_{1}:F^{-1}(\Lambda)\rightarrow\overline{\mathcal{X}^{\prime}}$ is a
dominant morphism.

Furthermore, consider $\Psi:F^{-1}(\Lambda)\rightarrow\Lambda\times\Lambda$
given by $\Psi(g,[z])=([z],[g\cdot z])$.

For $\forall z_{1}\in\Lambda$, $z_{2}\in\Lambda$, we have
$\Psi^{-1}(z_{1},z_{2})=(g_{2}Tg_{1}^{-1},z_{1})$, where
$T=\\{\left(\begin{array}[]{cc}z_{0}&\alpha\\\
0&g^{\prime}\end{array}\right):z_{0}\in\mathbb{C}\backslash\\{0\\},g^{\prime}\in
GL({d+1\choose 2}-1),\alpha\in\mathbb{C}^{{d+1\choose 2}-1}\ is\ a\ row\
vector\\}$, and $g_{1},g_{2}\in GL(\vee^{2}\mathbb{C}^{d})$, s.t.
$g_{1}(y_{0})=z_{1}$, $g_{2}(y_{0})=z_{2}$. So this is a dominant morphism.
Then we obtain

$\displaystyle\dim(F^{-1}(\Lambda))\leq$
$\displaystyle\dim(T)+\dim(\Lambda\times\Lambda)$ $\displaystyle=$
$\displaystyle{d+1\choose 2}^{2}-({d+1\choose
2}-1)+\dim(\Lambda)+\dim(\Lambda).$

It is required in Lemma C.2 that varieties $Z_{1}$ and $Z_{2}$ be irreducible.
Actually, this condition can be weakened. Lemma C.2 is still true for the more
general case that $Z_{1}$ and $Z_{2}$ are closed subsets of irreducible
varieties[8]. Through this approach, we can fill out the gap and apply this
lemma without danger of confusion. Indeed, the irreduciblity of $Z_{1}$ and
$Z_{2}$ really holds, but verification of this is not easy.

Then from Lemma C.2 and Lemma C.3, we will have

$\displaystyle\dim(\overline{\mathcal{X}^{\prime}})\leq$
$\displaystyle\dim(F^{-1}(\Lambda))$ $\displaystyle\leq$
$\displaystyle{d+1\choose 2}^{2}-({d+1\choose
2}-1)+\dim(\Lambda)+\dim(\Lambda)$ $\displaystyle=$ $\displaystyle{d+1\choose
2}^{2}-({d+1\choose 2}-1)+2(d-1).$

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