Entangled subspaces and generic local state discrimination with pre-shared entanglement

*]Benjamin Lovitz
$\dag$$\ddagger$]Nathaniel Johnston

[*]Institute for Quantum Computing and Department of Applied Mathematics, University of Waterloo, Canada
[$\dag$]Department of Mathematics & Computer Science, Mount Allison University, Canada
[$\ddagger$]Department of Mathematics & Statistics, University of Guelph, Canada

Walgate and Scott have determined the maximum number of generic pure quantum states in multipartite space that can be unambiguously discriminated by an LOCC measurement [41]. In this work, we determine this number in a more general setting in which the local parties have access to pre-shared entanglement in the form of a resource state. We find that, for an arbitrary pure resource state, this number is equal to the Krull dimension of (the closure of) the set of pure states obtainable from the resource state by SLOCC. This dimension is known for several resource states, for example the GHZ state.

Local state discrimination is closely related to the topic of entangled subspaces, which we study in its own right. We introduce $r$-entangled subspaces, which naturally generalize previously studied spaces to higher multipartite entanglement. We use algebraic geometric methods to determine the maximum dimension of an $r$-entangled subspace, and present novel explicit constructions of such spaces. We obtain similar results for symmetric and antisymmetric $r$-entangled subspaces, which correspond to entangled subspaces of bosonic and fermionic systems, respectively.

§ INTRODUCTION

An LOCC measurement is a quantum measurement on multipartite space that can be implemented by local operations and classical communication (LOCC). A set of $n$ pure states $\{[v_1],\dots, [v_n]\}$ in multipartite space is locally (unambiguously) discriminable if there exists an LOCC measurement with $n+1$ outcomes $\{1,\dots, n, ?\}$ that, when performed on any $[v_a]$, outputs either $a$ or $?$, with non-zero probability to output $a$.

Walgate and Scott determined that, for $m$ local spaces of (affine) dimensions $d_1,\dots, d_m$, almost all sets of ${\sum_{j=1}^m (d_j-1)+1}$ pure states are locally discriminable, and that this is the largest number for which this holds [41]. We extend Walgate and Scott's result to determine this number in a more general setting in which the parties are given access to pre-shared entanglement, in the form of a resource state, that they can use to implement their LOCC measurement. For example, we prove that almost all sets of ${r\sum_{j=1}^m (d_j-1)+r}$ pure states are locally discriminable with the tensor rank $r$ GHZ state ${[\tau_{r,m}]=\left[\frac{1}{\sqrt{r}}\sum_{a=1}^r e_a^{\otimes m}\right]}$, where $e_1,\dots, e_r$ are standard basis vectors. Furthermore, this is often the largest number for which this holds. For $m=2$, this number can be increased: almost all sets of ${d_1d_2- (d_1-\min\{d_1,r\})(d_2-\min\{d_2,r\})}$ pure states are locally discriminable with $[\tau_{r,2}]$ (or any other Schmidt rank $r$ state), and this is the largest number for which this holds.

More generally, we characterize this number for an arbitrary pure resource state $[w]$. The SLOCC image of $[w]$, denoted by $\Ima([w])$, is the set of pure states obtainable from $[w]$ by stochastic local operations and classical communication (SLOCC). This set is also known as the downward closure of $[w]$ with respect to SLOCC, and can alternatively be characterized as the set of pure states that have a Tucker decomposition with core state $[w]$ [39, 34]. Letting $d=\dim(\overline{\Ima([w])})$ be the Krull dimension of the closure of $\Ima([w])$, we prove that almost all sets of $d+1$ pure states are locally discriminable with $[w]$, and that this is the largest number for which this holds. To recover Walgate and Scott's result, observe that the SLOCC image of a trivial resource state is the set of unentangled (or, product) pure states. This set is already closed, and has dimension $d=\sum_{j=1}^m (d_j-1)$.

There are two different notions of a property holding for “almost all” sets of pure states (and also, subspaces) used for different results in [41]: One, that the property holds for a full measure subset with respect to the unitarily invariant probability measure (in which case we will say it holds for a generic such object); and two, that the property holds for a Zariski open dense subset (in which case we will say it holds for a general such object). Walgate and Scott characterize local unambiguous state discrimination (or, LUSD) under the generic notion, but state a relevant algebraic geometry result (reviewed in the next paragraph) under the general notion. We prove that both of these results hold under either notion.

To prove our characterization of general LUSD with a resource state, we invoke a recent result of Bandyopadhyay et al., which states that a set of pure states $\{[v_1],\dots, [v_n]\}$ is locally discriminable with $[w]$ if and only if there exist pure states $\{[u_1],\dots, [u_n]\}\subseteq \Ima([w])$ for which $({u_a^\t v_b \neq 0 \iff a=b})$ [8]. In particular, ${\Ima([w]) \cap \spn\{[v_1],\dots, [v_{n-1}]\}^\perp\neq \{\}}$, because the intersection contains $[u_n]$. Thus, if a general set of $n$ pure states is locally discriminable with $[w]$, then the orthogonal complement of the span of a general set of $n-1$ pure states has non-trivial intersection with $\Ima([w])$. We prove this implies that a general (projective) linear subspace of codimension $n$ intersects $\Ima([w])$ non-trivially. A theorem in algebraic geometry states that, for a quasiprojective variety $X$, the smallest codimension of a general (projective) linear subspace disjoint from $X$ is $\dim(X)+1$. Although $\Ima([w])$ is not always quasiprojective, it is constructible, from which we can conclude $n \leq d+1$. The converse, that a general set of $d+1$ pure states is locally discriminable with $[w]$, is obtained by similar reasoning. Generic LUSD with a resource state is handled similarly, with the crucial observation that the algebraic geometry result also holds with “general” replaced by “generic." Walgate and Scott characterized generic LUSD by proving a special case of this observation, and invoking Bandyopadhyay et al.'s result in the case of a trivial resource state (this case was first proven by Chefles [12]). Our methods are similar those of Walgate and Scott, with some added technicalities to handle both the general and generic cases, and to deal with the fact that $\Ima([w])$ is not always a variety.

The tensor rank of a pure state is the minimum number $r$ for which that state can be written as a superposition of $r$ product states. It is known that the set of pure states of tensor rank at most $r$ is precisely $\Ima([\tau_{r,m}])$. We define an $r$-entangled subspace to be a projective linear subspace that avoids $\overline{\Ima([\tau_{r,m}])}$ (this closure is known as the set of pure states of border rank at most $r$). In bipartite space, the tensor rank is equal to the Schmidt rank, and $\Ima([\tau_{r,2}])$ is already closed. Parthasarathy determined the maximum dimension of a 1-entangled subspace [33], and Bhat explicitly constructed a 1-entangled subspace of maximum dimension [7]. Cubitt et al. proved analogous results for $r$-entangled subspaces of bipartite space [13]. A theorem in algebraic geometry states that if $X$ is a projective variety, then the minimum codimension of a projective linear subspace disjoint from $X$ is $\dim(X)+1$ (and, by the algebraic geometry result of the previous paragraph, almost all subspaces of this codimension avoid $X$). It follows, from a known bound on $\dim(\overline{\Ima([\tau_{r,m}])}),$ that there always exists an $r$-entangled subspace of dimension

\begin{align}
d_1 \cdots d_m - r\sum_{j=1}^m (d_j-1)-r-1,
\end{align}

whenever this quantity is non-negative. Furthermore, this is often the maximum dimension of such a subspace.

Using this bound, we explicitly construct $2$-entangled subspaces of maximum dimension in tripartite space with local (affine) dimensions $d_1, d_2 \in \{2,3\}$ and $d_3=2$ (i.e. qubits and qutrits); and in quadripartite space with local (affine) dimensions $d_1=\dots=d_4=2$ (i.e. all qubits). To prove that these subspaces are indeed 2-entangled, we solve an equivalent ideal membership problem using the Macaulay2 software package [19, 31]. While ideal membership problems are notoriously intractable in general, our positive results reveal that this may not be the case for verifying $r$-entangled subspaces (at least for small $r$).

We also define $r$-entangled subspaces of the symmetric and antisymmetric spaces, which correspond to bosonic and fermionic entangled subspaces, respectively [18]. We determine the maximum dimension of such subspaces. We also explicitly construct maximal symmetric and antisymmetric $r$-entangled subspaces of bipartite space for arbitrary $r$, and of multipartite space for $r=1$, which matches the cases of standard $r$-entangled subspace constructions presented in [7, 13].

It is known that, under various notions of entanglement, the maximum dimension of an entangled subspace is precisely the maximum number of negative eigenvalues of an entanglement witness [4, 28, 27]. The number of negative eigenvalues quantifies "how good" the witness is at detecting entanglement. We prove that this connection between subspaces and negative eigenvalues holds under a much more general notion of witness, including multipartite $r$-entanglement witnesses.

There are other types of entangled subspaces that have been studied in previous works: non-positive partial transpose subspaces, for which every mixed state supported on that subspace has non-positive partial transpose [28, 27]; genuinely entangled subspaces, for which every element is non-product with respect to every bipartition [13, 15]; and subspaces of bipartite space with high entropy of entanglement [23]. Entangled subspaces are connected to unextendible product bases, and have found applications, for example, in quantum error correction [20, 36] and quantum tomography [24].

This work is organized as follows. In Section <ref> we review some mathematical preliminaries for this work, including projective notation; the Segre, Veronese, and Grassmannian varieties; relevant results in algebraic geometry; the SLOCC image; and LUSD. In Section <ref> we state and prove our characterization of generic/general LUSD with a resource state, and in Section <ref> we present our results on $r$-entangled subspaces and entanglement witnesses.

§ PRELIMINARIES

In this section, we review several necessary preliminaries for this work. In this preamble, we review some basic objects in affine and projective space, including the crucial observation that pure states can be identified with elements of projective space. In Section <ref> we review the Segre, Veronese, and Grassmannian varieties, which correspond to the sets of unentangled pure states in standard, symmetric, and antisymmetric space, respectively. We also define the $r$-th secants to these varieties, which we will use to define $r$-entangled subspaces in these three settings. In Section <ref> we describe a useful alternative interpretation of the Grassmannian as the set of projective planes of a fixed dimension, and prove some necessary preliminaries to our characterization of generic/general LUSD with a resource state. In Section <ref> we review a known result in algebraic geometry on the dimension of a linear subspace disjoint from a variety, and prove some extensions that we will require. In Section <ref> we review the SLOCC image. In Section <ref> we review LUSD.

Let $[m]=\{1,\dots, m\}$ when $m$ is a positive integer. For a non-zero vector space $\V$ (which we always take to be over $\complex$), let $\proj (\V)$ (or more briefly, $\proj \V$) be the set of $1$-dimensional linear subspaces of $\V$, and let $\proj^{D}=\proj(\complex^{D+1})$. We abuse the $[\cdot]$ notation and also write ${[v] \in \proj \V}$ for the span of a non-zero vector $v \in \V$. For a subset $X \subseteq \proj \V$, let $\hat{X} \subseteq \V$ be the affine cone over $X$. Explicitly, $\hat{X}=\{x \in \V : [x] \in X\}\cup \{0\}$. In the other direction, for a subset $Z \subseteq \V$ that forms a cone (i.e. for all $z \in Z$ and $\alpha \in \complex$ it holds that $\alpha z \in Z$), we let $\check{Z}= \{ [z] \in \proj \V : z \in Z \setminus \{0\}\}$ be the projectivization of $Z$.

We use projective notation throughout most of this work. In particular, we define a projective $n-1$ plane (or alternatively, a projective linear subspace of dimension $n-1$) in $\proj \V$ to be the projectivization of an $n$-dimensional linear subspace of $\V$. By convention, a projective linear subspace has dimension one less than its affine cone. For example, if $\{v_1,\dots, v_n\} \subseteq \V$ is linearly independent, then $\dim\spn \{v_1,\dots, v_n\} =n$, and $\dim\spn\{[v_1],\dots, [v_n]\}=n-1$. As a result, the numbers we present as the  maximum dimension of a (projective) entangled subspace will be one less than the numbers presented in previous works (e.g. [33, 13]), which used affine notation.

Let $\braket{\cdot}{\cdot}: \V \times \V \rightarrow \complex$ be the standard Hermitian inner product given by ${\braket{u}{v}=u^* v}$, where $(\cdot)^*$ denotes the conjugate transpose. Let $\Sp(\V) \subseteq \V$ denote the unit sphere of $\V$ with respect to the inner product $\braket{\cdot}{\cdot}$. In quantum physics, the unit sphere modulo complex phase represents the pure quantum states of $\V$. Crucially, we identify $\proj \V$ with the set of pure quantum states in $\V$ under the canonical bijection which identifies a 1-dimensional linear subspace with the equivalence class of unit vectors in that subspace. This is a useful identification, as it allows us to regard certain sets of pure quantum states as algebraic varieties, and study them using the machinery of algebraic geometry. For brevity, we will refer to pure quantum states simply as “states,” and we will refer to mixed quantum states (defined and treated only in Section <ref>) as “mixed states.”

For vector spaces $\W$ and $\V$, let $\Lin(\W,\V)$ denote the space of linear maps from $\W$ to $\V$, and let $\Lin(\V)=\Lin(\V,\V)$. Note that any multipartite space $\V=\bigotimes_{j=1}^m \V_j$ is canonically isomorphic to $\Lin(\bigotimes_{j \in S} \V_j^*, \bigotimes_{j \in [m]\setminus S} \V_j)$ for any subset $S \subseteq [m]$. The flattening rank of a state $[v] \in \proj\V$ is the maximum, taken over all subsets $S \subseteq [m]$ of size $1 \leq \abs{S} \leq m-1$, of the matrix rank of $[v]$ as an element of (the projectivization of) $\Lin(\bigotimes_{j \in S} \V_j^*, \bigotimes_{j \in [m]\setminus S} \V_j)$. Let $\Pos(\V)\subseteq \Lin(\V)$ be the set of positive semi-definite operators on $\V$, let $\Un(\V)\subseteq \Lin(\V)$ be the set of unitary operators on $\V$, and let $\GL(\V)\subseteq \Lin(\V)$ be the set of invertible operators on $\V$. Let $\{e_1,\dots, e_d\}$ be the standard basis of $\complex^d$.

§.§ The Segre, Veronese, and Grassmannian varieties (and their secants)

We will be particularly interested in the algebraic varieties known as the Segre, Veronese, and Grassmannian varieties, as these correspond to the sets of unentangled states in a space of distinguishable, bosonic, and fermionic particles, respectively [18]. In this section, we briefly describe these varieties, as well as their $r$-th secants, which we will use to study $r$-entangled subspaces. We refer the reader to [10, 30, 21, 5] for more in-depth treatments of these objects.

The Segre variety

\begin{align}\label{seg}
Y=\setft{Seg}(\proj^{d_1-1}\times \dots\times \proj^{d_m-1})\subseteq \textstyle{\proj (\bigotimes_{j=1}^m \complex^{d_j})}
\end{align}

is the image of the Segre embedding, and is equal to the set of states of the form ${[x_1 \otimes \dots \otimes x_m]}$, where $x_j \in \complex^{d_j}$ for each $j \in [m]$. The Segre variety corresponds to the unentangled (or, product) states in the space $\proj(\bigotimes_{j=1}^m \complex^{d_j})$ of distinguishable particles.

For each permutation $\sigma \in S_m$, let $P_{\sigma} \in \Lin(\bigotimes^m \complex^d)$ be the linear map defined on the standard product basis as

\begin{align}
P_{\sigma}(e_{a_1}\otimes\dots\otimes e_{a_m})=e_{a_{\sigma^{-1}(1)}}\otimes \dots \otimes e_{a_{\sigma^{-1}(m)}}.
\end{align}

Let $\proj(\bigvee^m \complex^d)\subseteq \proj(\bigotimes^m \complex^d)$ denote the symmetric subspace, i.e, the set of states ${[x]\in \proj (\bigotimes^m \complex^d)}$ such that $P_{\sigma} x=x$ for all $\sigma \in S_m$. This space is spanned by states of the form $[x_1 \vee \dots \vee x_m]$, where

\begin{align}
[x_1 \vee \dots \vee x_m]=\left[ \sum_{\sigma \in S_m} x_{\sigma(1)}\otimes \dots \otimes x_{\sigma(m)}\right].
\end{align}

In quantum physics, $\proj(\bigvee^m \complex^d)$ represents a bosonic space of indistinguishable particles. (This space can also be thought of as the set of homogeneous polynomials of degree $m$ in $d$ variables.) The Veronese variety

\begin{align}
\nu_m(\proj^{d-1})\subseteq \textstyle{\proj({\bigvee}^m \complex^d)}
\end{align}

is the image of the $m$-th Veronese embedding, and is equal to the set of states in $\proj(\bigvee^m \complex^d)$ of the form $[x^{\vee m}]=[x^{\otimes m}]$. The Veronese variety corresponds to the set of unentangled states in the bosonic space $\proj({\bigvee}^m \complex^d)$.

Let $\proj(\bigwedge^m \complex^d)\subseteq \proj(\bigotimes^m \complex^d)$ denote the antisymmetric subspace, i.e. the set of states $[x]\in \proj (\bigotimes^m \complex^d)$ such that $P_{\sigma} x= (-1)^{\setft{sgn} (\sigma)} x$ for all $\sigma \in S_m$. This space is spanned by the set of states of the form $[x_1 \wedge \dots \wedge x_m]$, where

\begin{align}
[x_1 \wedge \dots \wedge x_m]=\left[ \sum_{\sigma \in S_m} (-1)^{\sgn(\sigma)} x_{\sigma(1)}\otimes \dots \otimes x_{\sigma(m)}\right].
\end{align}

In quantum physics, the antisymmetric subspace represents a fermionic space of indistinguishable particles. The Grassmannian variety

\begin{align}
\Gr(m-1,\proj^{d-1}) \subseteq \textstyle{\proj({\bigwedge}^m \complex^d)}
\end{align}

is the set of states in $\proj(\bigwedge^m \complex^d)$ of the form $[x_1 \wedge \dots \wedge x_m]$. The Grassmannian variety corresponds to the set of unentangled states in the fermionic space $\proj({\bigwedge}^m \complex^d)$. The Grassmannian can also be viewed as the variety of projective $m-1$ planes in $\proj^{d-1}$. We develop this perspective further in Section <ref>, as we will make frequent use of it.

Now we introduce secant varieties. For a projective variety $X \subseteq \proj^D$, let

\begin{align}\label{sigma_r}
\sigma_r(X)=\overline{\bigcup_{[x_1],\dots, [x_r] \in X} \spn\{[x_1],\dots, [x_r]\}} \subseteq \proj^D
\end{align}

be the r-th secant variety to $X$, where the closure can equivalently be taken with respect to either the Zariski or Euclidean topology.

It is a standard result that

\begin{align}
\dim(\sigma_r(X)) \leq \min\{D, r\dim(X)+r-1\}.
\end{align}

If equality holds in this expression, then $\sigma_r(X)$ is said to have the expected dimension, and otherwise it is said to be defective.

For the Segre variety $Y\subseteq \proj(\bigotimes_{j=1}^m \complex^{d_j})$, we have $\dim(Y)=\sum_{j=1}^m (d_j-1)$ and $\dim(\proj(\bigotimes_{j=1}^m \complex^{d_j}))=d_1\cdots d_m-1$, so

\begin{align}
\dim(\sigma_r(Y))\leq \min\{d_1\cdots d_m-1, r \sum_{j=1}^m (d_j-1)+r-1\}.
\end{align}

The border rank of a state $[v] \in \proj(\bigotimes_{j=1}^m \complex^{d_j})$ is the smallest positive integer $r$ for which $[v] \in \sigma_r(Y)$. The tensor rank of $[v]$ is the smallest $r$ for which $[v]$ is in the span of $r$ elements of $Y$ (i.e. $[v]$ is contained in the pre-closed set of  (<ref>) with $X=Y$).

A conjecturally complete set of defective $\sigma_r(Y)$ have been proposed in [3], which are nicely summarized in <cit.>. For example,

\begin{align}\label{eq:bipartite_sr_dim}
\dim(\sigma_r( \seg(\proj^{d_1-1}\times \proj^{d_2-1})))=d_1 d_2-(d_1-\min\{d_1,r\})(d_2-\min\{d_2,r\})-1,
\end{align}

so $\sigma_r( \seg(\proj^{d_1-1}\times \proj^{d_2-1}))$ is defective in many cases. Under the identification ${\complex^{d_1}\otimes \complex^{d_2} \cong \Lin((\complex^{d_1})^*,\complex^{d_2})}$, the variety $\sigma_r( \seg(\proj^{d_1-1}\times \proj^{d_2-1}))$ corresponds to the set of (projective) $d_2 \times d_1$ matrices of rank at most $r$.

For the Veronese variety $\nu_m(\proj^{d-1})\subseteq \proj({\bigvee}^m(\complex^d))$, we have $\dim(\nu_m(\proj^{d-1}))=d-1$ and $\dim(\proj({\bigvee}^m(\complex^d)))=\binom{d+m-1}{m-1}-1$, so

\begin{align}
\dim(\sigma_r(\nu_m(\proj^{d-1})))\leq \min\left\{\binom{d-1+m}{m}-1,rd-1\right\}.
\end{align}

A complete set of defective Veronese secants are known [1]; see also Theorem 2 in [5]. In particular, $\sigma_r(\nu_2(\proj^{d-1}))$ is defective whenever $2 \leq r \leq d-1$, in which case

\begin{align}
\dim(\sigma_r(\nu_2(\proj^{d-1})))=\min\left\{\binom{d+1}{2}-1,rd-\binom{r}{2}-1\right\}.
\end{align}

Note that there is a typo in the expression of this dimension in Theorem 2 of [5]. Under the identification $\complex^{d}\otimes \complex^{d} \cong \Lin((\complex^{d})^*,\complex^{d})$, the variety $\sigma_r(\nu_2(\proj^{d-1}))$ corresponds to the set of (projective) symmetric $d \times d$ matrices of rank at most $r$.

For the Grassmannian variety $\Gr(m-1,\proj^{d-1}) \subseteq \proj({\bigwedge}^m(\complex^d))$, we have

\begin{align}
\end{align}

and $\dim(\proj({\bigwedge}^m(\complex^d)))=\binom{d}{m}-1$, so

\begin{align}
\dim(\sigma_r(\Gr(m-1,\proj^{d-1})))\leq \min\left\{\binom{d}{m}-1, r m(d-m)+r-1\right\}.
\end{align}

As with the Segre variety, there are a conjecturally complete set of defective Grassmannian secants; see [6] and also Conjecture 7 in [5]. Similarly to the Segre and Veronese varieties, $\sigma_r(\Gr(1,\proj^{d-1}))$ is defective whenever $2 \leq r < \floor{\frac{d}{2}}$, in which case

\begin{align}
\dim(\sigma_r(\Gr(1,\proj^{d-1})))=\binom{d}{2}-\binom{d-2r}{2}-1=2r(d-r)-r-1.
\end{align}

Under the identification $\complex^{d}\otimes \complex^{d} \cong \Lin((\complex^{d})^*,\complex^{d})$, the variety $\sigma_r(\Gr(1,\proj^{d-1}))$ corresponds to the set of projective antisymmetric $d\times d$ matrices of rank at most $2r$.

§.§ The set of projective $n-1$ planes as a projective variety and a probability space

In Section <ref> we defined, for a non-zero vector space $\V$, the Grassmannian variety ${\Gr(n-1,\proj \V)}$ to be the set of decomposable elements of $\proj({\bigwedge}^n \V)$. In this section, we observe that ${\Gr(n-1,\proj \V)}$ is canonically bijective to the set of projective $n-1$ planes in $\proj \V$, and thus endows this set with the structure of a projective variety. We also describe this set as a probability space via its unitarily invariant probability measure. In an abuse of notation, we will use ${\Gr(n-1,\proj \V)}$ to refer to any one of the following: the projective variety of decomposable elements of $\proj({\bigwedge}^n \V)$, the projective variety of projective $n-1$ planes in $\proj \V$, and the probability space of projective $n-1$ planes in $\proj \V$. It is important to keep in mind the canonical bijection between the set of projective $n-1$ planes in $\proj \V$ (namely, $\Gr(n-1, \proj \V)$), and the set of $n$-dimensional linear subspaces of $\V$ (typically denoted $\Gr(n, \V)$). While $\Gr(n, \V)$ is perhaps more standard, we prefer $\Gr(n-1, \proj \V)$, as we would like the elements of each subspace to be states. Note that we have replaced the symbol $m$ with the symbol $n$ in this section, to match later notation in which the Grassmannian is viewed as a space of projective $n-1$ planes.

The two different views of $\Gr(n-1, \proj \V)$, as a projective variety and a probability space, afford two different notions of “almost all” projective $n-1$ planes, which are used for different results in [41]. We say that a property holds for a general projective $n-1$ plane if there exists a Zariski open dense subset of $\setft{Gr}(n-1,\proj \V)$ on which the property holds. We say that a property holds for a generic projective $n-1$ plane if there exists a full measure subset of $\setft{Gr}(n-1,\proj \V)$ on which the property holds. Sets of $n$ pure states can similarly be viewed in two different ways. We say a property holds for a general set of $n$ pure states in $\proj \V$ if there exists a Zariski open dense subset $U \subseteq \proj(V)^{\times n}$ such that for every $([v_1], \dots, [v_n]) \in U$, the property holds for $\{[v_1],\dots, [v_n]\}$; and and we say that a property holds for a “generic” set of $n$ pure states in $\proj \V$ if the property holds on a full measure subset of $\proj \V^{\times n}$. As Zariski open dense subsets are of full measure, every general property is generic.

In this section, we describe in more detail the Grassmannian as a projective variety and probability space, and review several results that we will use to prove our characterization of generic/general LUSD with a resource state. Namely, a general (generic) set of $n$ states spans a general (generic) projective $n-1$ plane, and a general (generic) projective $n-1$ plane is spanned by a general (generic) set of $n$ states. In addition, we observe that the bijection ${\Gr(n-1,\proj^d)\cong \Gr(d-n,\proj^d)}$, which sends a subspace to its orthogonal complement, defines an isomorphism of projective varieties and probability spaces.

We first describe a canonical bijection between the Grassmannian $\setft{Gr}(n-1,\proj \V)$ of decomposable elements of $\proj({\bigwedge}^n \V)$ and the set of projective $n-1$ planes in $\proj \V$, hence endowing this set with the structure of a projective variety. Consider the map

\begin{align}
\pi : \V^{\times n} &\dashrightarrow \setft{Gr}(n-1,\proj \V),
\end{align}

defined on the open subset $U \subseteq \V^{\times n}$ of linearly independent $n$-tuples, and given by $\pi(v_1,\dots , v_n)=[v_1 \wedge \dots \wedge v_n]$. Observe that $\pi$ is constant on tuples of vectors that form a basis for the same subspace, and hence is a well-defined map from the set of projective $n-1$ planes to $\setft{Gr}(n-1,\proj \V).$ This induced map (known as the Plücker embedding) is easily seen to be bijective, and hence endows the set of projective $n-1$ planes with the structure of the projective variety $\setft{Gr}(n-1,\proj \V).$

Observe that the map $\pi$ defines a quotient (in the sense of Section II.6.3 in [9]) of $U$ by the algebraic group $\GL(\complex^n)$ under the action $A \cdot (v_1, \dots , v_n)=(v_1,\dots,v_n) A^{-1}$ for all $A \in \GL(\complex^n)$ and $(v_1,\dots, v_n)\in U$, where the product on the right is matrix multiplication. This fact is certainly known, but we prove it for completeness. To prove that $\pi$ is a quotient map, note that both $U$ and $\setft{Gr}(n-1,\proj \V)$ are irreducible and smooth, so it suffices to verify that $\pi$ is a surjective, open map by <cit.>. Surjectivity is obvious. To prove that $\pi$ is open, it suffices to prove that it is flat by <cit.>, which in turn follows from the fact that $U$ and ${\setft{Gr}(n-1,\proj \V)}$ are smooth, and the fibers of $\pi$ are equidimensional <cit.>. It follows that the Zariski topology on $\setft{Gr}(n-1,\proj \V)$ is precisely the quotient topology of $U$ by the equivalence relation induced by the group action. In other words, a subset $V\subseteq \setft{Gr}(n-1,\proj \V)$ is open if and only if $\pi^{-1}(V)$ is open.

Passing to $\proj(\V)^{\times n}$, it follows that a subset $V\subseteq \setft{Gr}(n-1,\proj \V)$ is open if and only if $\tilde{\pi}^{-1}(V)$ is open, where

\begin{align}\label{pi_tilde}
\tilde{\pi}: \proj(\V)^{\times n} \dashrightarrow \setft{Gr}(n-1,\proj \V)
\end{align}

is defined by $\tilde{\pi}([v_1],\dots,[v_n])=\pi(v_1,\dots,v_n)$, and $\proj(\V)^{\times n}$ is viewed as a projective variety via the Segre embedding, as usual. This proves that a general set of $n$ states spans a general projective $n-1$ plane, and vice versa.

The bijection $\Gr(n-1,\proj^d)\cong \Gr(d-n,\proj^d)$, which sends a subspace to its orthogonal complement with respect to some non-degenerate bilinear form $\ip{\cdot}{\cdot}$, defines an isomorphism of projective varieties (see e.g. Section 7 in [25], or Lemma 1.11 (ii) in [29] and the discussion thereafter). For completeness, we briefly describe how to verify this when $\ip{u}{v}=u^\t v$, which is easily extended to the general case. For any element $[v]=[v_1 \wedge \dots \wedge v_n]\in \Gr(n-1,\proj^d)$, there exists an element of $\pi^{-1}([v])$ in reduced column echelon form $\left(\frac{\I_{n}}{A}\right)$ for some $A \in \Lin(\complex^{n},\complex^{d+1-n})$. It is straightforward to verify that $\left(\frac{A^\t}{-\I_{d+1-n}}\right)\in \pi^{-1}([v]^\perp)$, where $[v]^\perp$ denotes the orthogonal complement of the projective $n-1$ plane $[v]$. One can also verify that, up to sign, the $n\times n$ minor of $\left(\frac{\I_{n}}{A}\right)$ corresponding to a column index set $S \subseteq [d+1]$ of size $\abs{S}=n$ is precisely the $(d+1-n) \times (d+1-n)$ minor of the matrix $\left(\frac{A^\t}{-\I_{d+1-n}}\right)$ corresponding to the column index set $[d+1] \setminus S$. The result follows from the fact that these minors are exactly the coordinates of $[v] \in \proj(\bigwedge^n \complex^d)$ and $[v]^\perp \in \proj(\bigwedge^{d+1-n} \complex^d)$ in the Plücker embedding, respectively [21].

Now we describe $\setft{Gr}(n-1, \proj \V)$ as a probability space, following [41], and review the desired statements in this setting. Let $\nu$ denote the unit Haar measure on $\Un(\V)$. Note that $\Un(\V)$ acts on $\setft{Gr}(n-1, \proj \V)$ by

\begin{align}
{A \cdot [v_1 \wedge \cdots \wedge v_n]= [A v_1 \wedge \cdots \wedge A v_n]}
\end{align}

for all $A \in \Un(\V)$ and ${[v_1 \wedge \cdots \wedge v_n] \in \setft{Gr}(n-1, \proj \V)}$. Fix an element $[v] \in \setft{Gr}(n-1,\proj \V)$, and define $\mu_n: \setft{Gr}(n-1,\proj \V) \rightarrow [0,1]$ as

\begin{align}\label{measure}
\mu_n(E) = \nu(\{A \in \Un(\V) : A \cdot [v] \in E\})
\end{align}

for any Borel subset $E \subseteq \setft{Gr}(n-1, \proj \V)$. The probability measure induced on ${\proj \V=\Gr(0,\proj \V)}$ in this way will be denoted by $\mu$. Note that for any Borel subset ${E \subseteq \setft{Gr}(n-1,\proj \V)}$, it holds that

\begin{align}
\mu_n(E)=(\mu \times \dots \times \mu)(\tilde{\pi}^{-1}(E)),
\end{align}

as the righthand side also defines a unitarily invariant probability measure on ${\setft{Gr}(n-1,\proj \V)}$, and such a measure is unique. As a result, a subset $S \subseteq \setft{Gr}(n-1,\proj \V)$ is full measure with respect to $\mu_n$ if and only if $\tilde{\pi}^{-1}(S)$ is full measure with respect to $\mu \times \dots \times \mu$. This proves that a generic set of $n$ states spans a generic projective $n-1$ plane, and vice versa. It is straightforward to verify that $\mu_n(E) = \mu_{d-n+1}(E^\perp)$, and hence $\Gr(n-1,\proj^d)\cong \Gr(d-n,\proj^d)$ as probability spaces.

§.§ The maximum dimension of a projective linear subspace disjoint from a variety

The following classical result in algebraic geometry, along with the subsequent extensions, will be instrumental in proving our characterization of generic/general LUSD with a resource state. We will also use it to
determine the maximum dimension of an entangled subspace.

Let $X \subseteq \proj^D$ be a quasi-projective variety. Then $X$ has dimension $d$ if and only if $d$ is the smallest non-negative integer such that a general projective $D-d-1$ plane is disjoint from $X$.

If $X$ is projective, then $X$ has dimension $d$ if and only if $d$ is the smallest non-negative integer such that there exists a projective $D-d-1$ plane disjoint from $X$.

This characterization is taken to be the definition of the dimension of an irreducible variety in <cit.>, and is shown to be equivalent to other standard notions of dimension, e.g. the Krull dimension. The above extension to the reducible case is straightforward: let ${X=X_1\cup\dots\cup X_t}$ be the irreducible decomposition of $X$, let $d=\dim(X)$, and let $d_i=\dim(X_i)$ for each $i \in [t]$. Then $d=\max_{i\in[t]} d_i$ by definition. Since a general $D-d-1$ plane is disjoint from each $X_i$, it follows that a general $D-d-1$ plane is disjoint from $X$. Furthermore, $d$ is the smallest non-negative integer that satisfies this property, because for any $\tilde{d}<d$ it holds that $\tilde{d}<d_i$ for some $i\in [t]$, and hence a general $D-\tilde{d}-1$ plane intersects $X_i$ (and thus, $X$). The refined statement when $X$ is projective is handled similarly.

We now observe that, in Theorem <ref>, the term “general” can equivalently be replaced by “generic.” This fact, observed in <cit.> in the special case that $X$ is the Segre variety, will be instrumental in proving that our results hold under either notion of “almost all.”

Let $X \subseteq \proj^D$ be a quasiprojective variety. A general projective $n-1$ plane is disjoint from $X$ if and only if a generic projective $n-1$ plane is disjoint from $X$.

The statement asserts that there exists an open dense subset $U\subseteq \setft{Gr}(n-1,\proj^D)$ such that every element of $U$ is disjoint from $X$ if and only if there exists a full measure subset $S \subseteq \setft{Gr}(n-1,\proj^D)$ satisfying the same condition. The first statement trivially implies the second, as every open dense set is of full measure. For the reverse implication, suppose that a full measure subset $S \subseteq \setft{Gr}(n-1,\proj^D)$ is disjoint from $X$. Let $\mathcal{C}_{n-1}(X)\subseteq \setft{Gr}(n-1,\proj^D)$ denote the set of projective $n-1$ planes intersecting $X$. It can be shown that this set is constructible <cit.>, and hence contains an open dense subset $V$ of its closure. If $\overline{\mathcal{C}_{n-1}(X)}\subsetneq \setft{Gr}(n-1,\proj^D)$, then the open dense subset $U=\setft{Gr}(n-1,\proj^D) \setminus \overline{\mathcal{C}_{n-1}(X)}$ satisfies the desired conditions. If $\overline{\mathcal{C}_{n-1}(X)}=\setft{Gr}(n-1,\proj^D)$, then $V$ and $S$ are both of full measure in $\setft{Gr}(n-1,\proj^D)$, hence they intersect non-trivially, a contradiction.

§.§ The SLOCC image

LOCC channels are quantum channels (completely positive, trace preserving maps) on multipartite space that can be implemented by local operations and classical communication (LOCC). SLOCC maps are completely positive, trace non-increasing maps that can be implemented with non-zero probability by an LOCC channel. In other words, SLOCC maps represent LOCC channels with postselection [40].

Let $\V=\bigotimes_{j=1}^m \complex^{d_j}$ and $\W=\bigotimes_{j=1}^{m} \complex^{c_j}$ be multipartite spaces. For a state $[w] \in \proj{\W}$, we define the SLOCC image of $[w]$ in $\proj{\V}$, $\Ima_{\proj{\V}}([w]) \subseteq \proj{\V}$, to be the set of states in $\proj{\V}$ obtainable from $[w]$ by SLOCC. Equivalently, this is the set of states in $\proj{\V}$ that can be obtained from $[w]$ by LOCC with non-zero probability. A related notion is the SLOCC orbit of a state $[v]\in \proj \V$, denoted $\mathcal{O}_{[v]}\subseteq \proj{\V}$, which is the set of states in $\Ima_{\proj{\V}}([v])$ that can be converted back to $[v]$ by SLOCC, i.e.

\begin{align}
\mathcal{O}_{[v]} = \{[u] \in \Ima_{\proj{\V}}([v]) : [v] \in \Ima_{\proj{\V}}([u])\}.
\end{align}

In this section, we observe several properties of $\Ima_{\proj{\V}}([w])$ that will be useful later on, and also mention some useful properties of $\mathcal{O}_{[v]}$ for good measure.

We can describe these sets mathematically as

\begin{align}
\Ima_{\proj{\V}}([w])=\{[(A_1 \otimes \dots \otimes A_m)w] : A_i \in \Lin(\complex^{c_i},\complex^{d_i}) \quad \!\text{for all} &\quad \! i\in[m]\\
&\text{and}\quad \! (A_1 \otimes\dots \otimes A_m) w \neq 0\},
\end{align}

\begin{align}
\mathcal{O}_{[v]}=\{[(A_1 \otimes\dots \otimes A_m) v] : A_i \in \setft{GL}(\complex^{d_i})\quad\text{for all}\quad  i \in [m] \};
\end{align}

see [16]. True to its name, $\mathcal{O}_{[v]}$ is the orbit of $[v]$ by the product (projective) general linear group.

If $\V=\complex^{d_1}\otimes\complex^{d_2}$ and $\W=\complex^{c_1}\otimes \complex^{c_2}$ are bipartite spaces, then $\Ima_{\proj \V}([w])$ is the set of states in $\proj \V$ of Schmidt rank less than or equal to the Schmidt rank of $[w]$, and $\mathcal{O}_{[v]}$ is the set of states of Schmidt rank equal to the Schmidt rank of $[v]$. In multipartite space, consider the tensor rank $r$ GHZ state

\begin{align}\label{GHZ}
[\tau_{r,m}]=\left[\sum_{a=1}^r e_a^{\otimes m} \right] \subseteq \proj((\complex^r)^{\otimes m}).
\end{align}

It is straightforward to verify that $\Ima_{\proj{\V}}([\tau_{r,m}])$ is the set of states in $\proj{\V}$ of tensor rank at most $r$, and hence ${\overline{\Ima_{\proj{\V}}([\tau_{r,m}])}=\sigma_r}(Y)$, where $Y$ is the Segre variety of $\proj \V$ defined in (<ref>). It is clear that $\mathcal{O}_{[\tau_{r,m}]}$ is the set of states of the form $[\sum_{a=1}^r x_{a,1} \otimes \dots \otimes x_{a,m}]$, where $\{x_{1,j},\dots, x_{r,j}\}\subseteq \complex^{d_j}$ is linearly independent for all $j \in [m]$.

We conclude this section by observing that $\Ima_{\proj{\V}}([w])$ and $\mathcal{O}_{[v]}$ are both irreducible and constructible subsets in the Zariski topology. Note that $\Ima_{\proj{\V}}([w])$ is the image of the irreducible quasiprojective variety

\begin{align*}
Z=\{[A_1 \otimes \dots \otimes A_m] : A_i \in \Lin(\complex^{c_i},\complex^{d_i}) \quad \!\text{for all} \quad \! i\in[m]\quad\text{and}\quad \! (A_1 \otimes\dots \otimes A_m) w \neq 0\}
\end{align*}

under the morphism $w: Z \rightarrow \proj{\V}$ that sends $(A_1 \otimes\dots \otimes A_m)$ to $[(A_1 \otimes\dots \otimes A_m) w]$. Since $Z$ is constructible, $\Ima_{\proj{\V}}([w])$ is constructible by Chevalley's theorem <cit.>.

Since $Z$ is irreducible, $\Ima_{\proj{\V}}([w])$ is irreducible in the subspace topology. Similar arguments show that $\mathcal{O}_{[v]}$ is also constructible and irreducible. Since both  $\Ima_{\proj{\V}}([w])$ and $\mathcal{O}_{[v]}$ are constructible, they each contain an open dense subset of their closure <cit.>. In fact, $\mathcal{O}_{[v]}$ is itself an open dense subset of its closure (i.e. $\mathcal{O}_{[v]}$ is locally closed) <cit.>. It can be shown that $\overline{\mathcal{O}_{[v]}}=\overline{\Ima_{\proj{\V}}([v])}$.

§.§ Local unambiguous state discrimination (LUSD)

We conclude this section by reviewing unambiguous state discrimination (USD), and its local counterpart, local unambiguous state discrimination (LUSD). We use these notions in Section <ref> to characterize generic/general LUSD with a resource state.

A set of states $\{[v_1],\dots,[v_n]\} \subseteq \proj{\V}$ is (unambiguously) discriminable if there exists a quantum measurement with $n+1$ outcomes $\{1,\dots,n,?\}$ that, when performed on any $[v_a]$, outputs either $a$ or $?$, with nonzero probability to output $a$. Mathematically, this is equivalent to the existence positive semi-definite operators $M_1,\dots,M_n, M_? \in \Pos(\V)$ such that $M_1+\dots+M_n+M_?=\I$, and $(\braket{v_b}{M_a v_b} \neq 0 \iff a = b)$.  Note that $\{[v_1],\dots,[v_n]\}$ is discriminable if and only if it is linearly independent.

Let $\V = \bigotimes_{j=1}^m \complex^{d_j}$ be a multipartite space. We say that a set of states ${\{[v_1],\dots,[v_n]\} \subseteq \proj{\V}}$ is locally (unambiguously) discriminable if it is discriminable by a measurement implementable by an LOCC channel, with local subsystems $\proj \complex^{d_j}$. We say that a set of states $\{[v_1],\dots,[v_n]\}$ is locally (unambiguously) discriminable with resource state $[w] \in \proj{\W}$ if $\{[v_1 \otimes w], \dots, [v_n \otimes w]\}$ is locally discriminable, where the local subsystem $j\in [m]$ is now the composite system $\proj (\complex^{d_j} \otimes \complex^{c_j})$.

§ GENERIC LOCAL STATE DISCRIMINATION WITH PRE-SHARED ENTANGLEMENT

In this section, we characterize the maximum number of generic pure states that can be locally discriminated with a fixed resource state $[w]$. We furthermore prove that this is equal to the maximum number of general pure states that can be locally discriminated with $[w]$.

To prove this characterization, we require the following mathematical description of LUSD with a resource state, which was proven by Bandyopadhyay et al.

Let $\V=\bigotimes_{j=1}^m \complex^{d_j}$ and $\W=\bigotimes_{j=1}^m \complex^{c_j}$. A set of states $\{[v_1],\dots, [v_n]\} \subseteq \proj{\V}$ is locally discriminable with resource state $[w]\in \proj{\W}$ if and only if there exist states

\begin{align}
[u_1],\dots, [u_n] \in \Ima_{\proj \V} ([w])
\end{align}

such that $(u_a^\t {v_b} \neq 0 \iff a= b)$, where the transpose can equivalently be taken with respect to any product basis of $\V$.

Note that this statement indeed does not depend on the choice of product basis over which the transpose is taken, since any two product bases are related by a product change of basis $A=A_1 \otimes \dots \otimes A_m \in \GL(\V)$, and

\begin{align}
[u_a]\in \Ima_{\proj \V} ([w]) \iff [A u_a] \in \Ima_{\proj \V} ([w]).
\end{align}

Also note that, if the Hermitian inner product is preferred, an alternative (equivalent) statement is that there exist $[u_1],\dots,[u_n] \in \Ima_{\proj \V} ([\overline{w}])$ such that ${({\braket{u_a}{v_b}}\neq 0 \iff a= b)}$, where $\overline{w}$ denotes the complex conjugate of $w$ with respect to any product basis of $\W$.

Note that the SLOCC image of a trivial (i.e. non-existent) resource state is simply the Segre variety $Y$ of product states, defined in (<ref>). This case of Theorem <ref>, first proven by Chefles, was used to characterize generic LUSD in [41], and we use the above generalization to characterize generic/general LUSD with a resource state.

Let $\V=\bigotimes_{j=1}^m \complex^{d_j}$ and $\W=\bigotimes_{j=1}^m \complex^{c_j}$. A generic set of $n$ states in $\proj \V$ is locally discriminable with resource state $[w] \in \proj \W$ if and only if ${n \leq \dim (\overline{\Ima_{\proj{\V}} ([w])})+1}$. The same statement holds if the term “generic” is replaced with “general.”

Let $X= \overline{\Ima_{\proj \V}(w)}$ and $d=\dim(X)$. We will make use of the map $\tilde{\pi}$ defined in (<ref>), as well as several observations made in Section <ref>. Suppose that a full measure subset ${T \subseteq \proj(\V)^{\times n}}$ is locally discriminable with $[w] \in \proj \W$. Let $S=\omega_{\hat{n}}(T)$, where $\omega_{\hat{n}}: \proj(\V)^{\times n} \rightarrow \proj(\V)^{\times n-1}$ is the projection onto the first $n-1$ copies of $\proj(\V)$. Then $S$ is of full measure in $\proj(\V)^{\times n-1}$, so $\tilde{\pi}(S)^\perp$ is of full measure in $\setft{Gr}(d_1\dots d_m-n, \proj \V)$, and every element of $\tilde{\pi}(S)^\perp$ intersects $\Ima_{\proj \V}([w])$ non-trivially by Theorem <ref>. By Theorem <ref> and Proposition <ref>, $n \leq d+1$. As every open dense subset is full measure, this proves the “only if” direction for both the general and generic versions of the statement.

Conversely, suppose $n \leq d+1$. It suffices to prove the general version for this direction. By Theorem <ref>, the desired result is equivalent to the existence of an open dense subset of $\proj(\V)^{\times n}$ contained in $\bigcap_{a=1}^n S_a$, where

\begin{align*}
S_a= \{([v_1],\dots,[v_n]): \text{there exists} \quad [u] \in \Ima_{\proj \V}([w]) \quad\text{such that} \quad (u^\t v_b \neq 0 \iff a= b) \}
\end{align*}

for each $a \in [n]$. Since $S_a$ is constructible, it contains an open dense subset of its closure. To complete the proof, it suffices to show that $\overline{S_a}=\proj(\V)^{\times n}$ for each $a \in [n]$. We do so by constructing a subset of $S_a$ that is dense in $\proj(\V)^{\times n}$.

We take $a=n$ to ease the notation (the other $a\in [n]$ follow by symmetry). Let ${U \subseteq {\Ima_{\proj \V}([w])}}$ be an open dense subset of $X$. Then $U$ is an (irreducible) quasiprojective variety of dimension $d$. By Theorem <ref> and the inequality $n \leq d+1$, there exists an open dense subset $W \subseteq \Gr(d_1\cdots d_m-n, \proj \V)$ for which every subspace contained in $W$ intersects $U$ non-trivially. Therefore, the set $Z=\tilde{\pi}^{-1}(W^\perp)\subseteq \proj(\V)^{\times n-1}$ is open dense, and is such that for each $v=(v_1,\dots,v_{n-1}) \in Z$ there exists $u_v \in U$ with $u_v^\t{v_b}=0$ for all $b\in [n-1]$. Let

\begin{align}
T_v = \{ [v_n] \in \proj \V : u_v^\t v_n\neq 0\}.
\end{align}

The set

\begin{align}
V_n=\bigcup_{v \in Z} \{v\} \times T_v
\end{align}

is clearly contained in $S_n$. To complete the proof, we show that $\overline{V_n}=\proj(\V)^{\times n}$. For any nonempty open subset $S \subseteq \proj(\V)^{\times n}$, there exists $v \in Z$ such that the set $(\{v\} \times \proj(\V)) \cap S$ is open dense inside $\{v\} \times \proj(\V)$. Since $T_v \subseteq \proj \V$ is open dense, it follows that ${\{v\} \times T_v \subseteq \{v\} \times \proj \V}$ is open dense, so

\begin{align}
(\{v\} \times T_v) \cap S\subseteq \{v\} \times \proj(\V)
%=(\{v\} \times T_v)\cap\left[(\{v\} \times \proj(V)) \cap S\right]
\end{align}

is open dense. Thus, $V_n \cap S \neq \{\}$. Since $V_n$ has non-trivial intersection with every non-empty open subset $S \subseteq (\proj \V)^{\times n}$, it follows that $\overline{V_n}=(\proj\V)^{\times n}$. This completes the proof.

Let $\V=\bigotimes_{j=1}^m \complex^{d_j}$. A generic (or equivalently, general) set of $n$ states in $\proj \V$ is locally discriminable with the tensor rank $r$ GHZ state $[\tau_{r,m}]$ defined in (<ref>) if and only if $n \leq \dim(\sigma_r(Y))+1$, where $Y$ is the Segre variety of product states defined in (<ref>).

Let $\V=\bigotimes_{j=1}^m \complex^{d_j}$. A generic (or equivalently, general) set of $n$ states in $\proj \V$ is locally discriminable if and only if $n \leq \sum_{j=1}^m (d_j-1)+1$.

Note that the generic version of Corollary <ref> is Theorem 4.3 in [41].

Let $\V=\complex^{d_1}\otimes\complex^{d_2}$, $\W=\complex^{c_1}\otimes\complex^{c_2}$. A generic (or equivalently, general) set of $n$ states in $\proj \V$ is locally discriminable with a Schmidt rank $r$ resource state $[w]\in \proj \W$ if and only if $n \leq d_1d_2-(d_1-\min\{d_1,r\})(d_2-\min\{d_2,r\})$.

§ ENTANGLED SUBSPACES

For a multipartite space $\V=\bigotimes_{j=1}^m \complex^{d_j}$, we define a (standard) $r$-entangled (projective linear) subspace of $\proj \V$ to be a projective linear subspace disjoint from $\sigma_r(Y)$, where $Y$ is the Segre variety of product states (<ref>). In other words, an $r$-entangled subspace is one that does not contain any states of border rank at most $r$. Similarly, we define a symmetric $r$-entangled (projective linear) subspace of $\proj({\bigvee}^m \complex^d)$ to be a projective linear subspace disjoint from $\sigma_r(\nu_m(\proj^{d-1}))$, and an antisymmetric $r$-entangled (projective linear) subspace of $\proj({\bigwedge}^m \complex^d)$ to be a projective linear subspace disjoint from $\sigma_r(\Gr(m-1,\proj^{d-1}))$. These three types of $r$-entangled subspaces correspond to entangled subspaces in systems of distinguishable, bosonic, and fermionic particles, respectively [18].

In this section, we use Theorem <ref> to characterize the maximum dimension of these entangled subspaces. We then explicitly construct maximal symmetric and antisymmetric entangled subspaces in the cases $r=1$ and $m=2$, which matches the cases for which constructions of (standard) $r$-entangled subspaces are known [7, 13]. We also construct maximal $2$-entangled subspaces of $\proj(\complex^{d_1}\otimes \complex^{d_2} \otimes \complex^2)$ for $d_1,d_2 \in \{2,3\}$, and also of $\proj(\complex^2 \otimes \complex^2 \otimes \complex^2 \otimes \complex^2)$.

§.§ The maximum dimensions of entangled subspaces

In this section, we use Theorem <ref> to determine the maximum dimensions of entangled subspaces, invoking the known dimensions of secant varieties reviewed in Section <ref>. Corollary <ref> generalizes results of [33] and [13].

Let $\V=\bigotimes_{j=1}^m \complex^{d_j}$, and let

\begin{align}
Y=\setft{Seg}(\proj^{d_1-1}\times \dots\times \proj^{d_m-1})\subseteq \proj(\V)
\end{align}

denote the Segre variety of product states. The maximum dimension of an $r$-entangled subspace of $\proj(\V)$ is

\begin{align}
d_1\cdots d_m-\dim(\sigma_r(Y))-2,
\end{align}

and a general (or equivalently by Proposition <ref>, generic) projective linear subspace of this dimension is $r$-entangled.

As a result, there always exists an $r$-entangled subspace of dimension

\begin{align}
d_1\cdots d_m- r\sum_{j=1}^m (d_j-1)-r-1,
\end{align}

whenever this quantity is non-negative. Furthermore, this is often the maximum dimension of an $r$-entangled subspace, with a conjecturally complete set of exceptions [3, 5]. If $r=1$ then this is the maximum dimension, which gives Parthasarathy's result [33].

In the bipartite case $\V=\complex^{d_1} \otimes \complex^{d_2}$, combining Corollary <ref> with Equation (<ref>) gives that the maximum dimension of an $r$-entangled subspace is

\begin{align}
\end{align}

whenever $r\leq \min\{d_1,d_2\}$ (this is <cit.>). Under the isomorphism ${\complex^{d_1} \otimes \complex^{d_2} \cong \Lin((\complex^{d_1})^*,\complex^{d_2})}$, this is the maximum dimension of a projective linear subspace of $d_2 \times d_1$ matrices of rank greater than $r$.

If $r=2$ and $m\geq 3$, then $\sigma_r(Y)$ has the expected dimension, so the largest dimension of a $2$-entangled subspace in this case is

\begin{align}\label{eq:2entangleddim}
d_1 \cdots d_m-2\sum_{j=1}^m (d_j-1)-3.
\end{align}

We explicitly construct maximal $2$-entangled subspaces in Section <ref>.

Let $m$ and $d$ be positive integers, and let

\begin{align}
\nu_m(\proj^{d-1})\subseteq \textstyle{\proj({\bigvee}^m \complex^d)}
\end{align}

denote the Veronese variety of unentangled states in the symmetric space. The maximum dimension of a symmetric $r$-entangled subspace of $\proj({\bigvee}^m \complex^d)$ is

\begin{align}
\binom{d-1+m}{m}-\dim(\sigma_r(\nu_m(\proj^{d-1})))-2
\end{align}

and a general (or equivalently by Proposition <ref>, generic) projective linear subspace of this dimension is symmetric $r$-entangled.

As a result, there always exists a symmetric $r$-entangled subspace of dimension

\begin{align}\label{symmetric_dim}
\binom{d-1+m}{m}-rd-1,
\end{align}

whenever this quantity is non-negative. This is the maximum dimension of a symmetric $r$-entangled subspace in many cases, with a known set of exceptions [1, 5]. If $r=1$, then (<ref>) is the maximum dimension. If $m=2$, then the maximum dimension is given by

\begin{align}\label{symmetric_bipartite_dim}
\binom{d-r+1}{2}-1.
\end{align}

We explicitly construct maximal symmetric $r$-entangled subspaces in these two cases in Section <ref>.

Let $m$ and $d$ be positive integers with $m \geq d$, and let

\begin{align}
\Gr(m-1,\proj^{d-1}) \subseteq \textstyle{\proj({\bigwedge}^m \complex^d)}
\end{align}

denote the Grassmannian variety of unentangled states in the antisymmetric space. The maximum dimension of an antisymmetric $r$-entangled subspace of  $\proj({\bigwedge}^m(\complex^d))$ is

\begin{align}
\binom{d-1+m}{m}-\dim(\sigma_r(\Gr(m-1,\proj^{d-1})))-2,
\end{align}

and a general (or equivalently by Proposition <ref>, generic) projective linear subspace of this dimension is antisymmetric $r$-entangled.

As a result, there always exists an antisymmetric $r$-entangled subspace of dimension

\begin{align}\label{antisymmetric_dim}
\binom{d-1+m}{m}-1-rm(d-m)-r,
\end{align}

whenever this quantity is non-negative. This is often the maximum dimension of an antisymmetric $r$-entangled subspace, with a conjecturally complete set of exceptions [6, 5]. If $r=1$, then (<ref>) is the maximum dimension. If $m=2$, then the maximum dimension is given by

\begin{align}\label{eq:antisy_bipartite}
\binom{d-2r}{2}-1.
\end{align}

We explicitly construct maximal antisymmetric $r$-entangled subspaces in these two cases in Section <ref>.

§.§ Explicit constructions of entangled subspaces of maximum dimension

Corollaries <ref>, <ref>, and <ref> establish the maximum possible dimension of standard $r$-entangled, symmetric $r$-entangled, and antisymmetric $r$-entangled subspaces, respectively, and show that almost all subspaces of this dimension are $r$-entangled. Despite the abundance of these subspaces, explicit constructions of them are only known for standard $r$-entangled subspaces in the $r = 1$ case [7] and the $m = 2$ case [13]. In this section, we match these results for symmetric and antisymmetric $r$-entangled subspaces by providing explicit constructions in these two cases. We also construct maximal $2$-entangled subspaces of $\proj(\complex^{d_1}\otimes \complex^{d_2}\otimes \complex^2)$ for $d_1,d_2 \in \{2,3\}$, and of $\proj(\complex^2 \otimes \complex^2 \otimes \complex^2 \otimes \complex^2)$.

Many of these constructions (and indeed, many of the already known constructions of similar subspaces from [13, 24, 11]) are based on totally non-singular matrices, which are matrices with the property that all of their minors (i.e., determinants of square submatrices) are non-zero. For example, every Vandermonde matrix

\begin{align}\label{eq:vandermonde}
\begin{bmatrix}
1 & \alpha_1 & \alpha_1^2 & \cdots & \alpha_1^{d-1} \\
1 & \alpha_2 & \alpha_2^2 & \cdots & \alpha_2^{d-1} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & \alpha_d & \alpha_d^2 & \cdots & \alpha_d^{d-1}
\end{bmatrix}
\end{align}

with $\alpha_i \neq \alpha_j \neq 0$ for all $i \neq j$ is totally non-singular (a slightly stronger property of these matrices, called total positivity, was proved in the case when $0 < \alpha_1 < \cdots < \alpha_d$ in [17], but the same proof works for total non-singularity in general). In fact, total non-singularity is a generic phenomenon: the set of matrices that are not totally non-singular has measure zero.

The following result, which was proved in <cit.>, provides the reason that totally non-singular matrices are of use to us.

Let $M$ be an $n \times n$ totally non-singular matrix with $n \geq k$, and let $v \in \complex^n$ be a linear combination of $k$ of the columns of $M$. Then $v$ contains at least $n-k+1$ non-zero entries.

§.§.§ Maximal symmetric $r$-entangled subspaces of bipartite space

To construct an $r$-entangled subspace of $\proj(\complex^d \vee \complex^d)$ attaining the bound (<ref>), we first note that the isomorphism $\complex^d \otimes \complex^d \cong \Lin((\complex^d)^*,\complex^d)$ shows that it is equivalent to construct a projective linear subspace of symmetric $d\times d$ matrices of rank greater than $r$ of dimension

\[
\binom{d-r+1}{2}-1.
\]

We construct such a subspace by placing columns of totally non-singular matrices along the super- and sub-diagonals of those symmetric matrices. More specifically, for each ${0 \leq i \leq d-r-1}$ and $1 \leq j \leq d-r-i$, let $M_j^i$ be the matrix that has the $j$-th column of some $(d-i) \times (d-i)$ totally non-singular matrix along its $i$-th super-diagonal (where for each fixed $i$, the same totally non-singular matrix is used for all $j$). For example, if $d = 6$, $r = 2$, $i = 1$, and we choose the totally non-singular matrix to be the Vandermonde matrix (<ref>) with $\alpha_j = j$ for $1 \leq j \leq d-i = 5$, then

\[
M_1^1 = \begin{bmatrix}
0 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 & 0 & 0 \\
\end{bmatrix}, \ M_2^1 = \begin{bmatrix}
0 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 2 & 0 & 0 & 0 \\
0 & 0 & 0 & 3 & 0 & 0 \\
0 & 0 & 0 & 0 & 4 & 0 \\
0 & 0 & 0 & 0 & 0 & 5 \\
0 & 0 & 0 & 0 & 0 & 0 \\
\end{bmatrix}, \ M_3^1 = \begin{bmatrix}
0 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 4 & 0 & 0 & 0 \\
0 & 0 & 0 & 9 & 0 & 0 \\
0 & 0 & 0 & 0 & 16 & 0 \\
0 & 0 & 0 & 0 & 0 & 25 \\
0 & 0 & 0 & 0 & 0 & 0 \\
\end{bmatrix}.
\]

We claim that the following set of symmetric matrices is a basis of an $r$-entangled subspace of $\proj(\complex^d \vee \complex^d)$ (once we convert the matrices back into states in the canonical way):

\begin{align*}
B = \Big\{ \left[M_j^i + \big(M_j^i\big)^\t \right]: 0 \leq i \leq d-r-1, 1 \leq j \leq d-r-i \Big\}.
\end{align*}

The fact that the set $B$ is linearly independent (and thus a basis of its span) follows immediately from Lemma <ref>: every non-zero linear combination of those basis matrices has, for each $0 \leq i \leq d-r-1$, at least $(d-i) - (d-r-i) + 1 = r+1 \geq 1$ non-zero entries along its $i$-th super-diagonal, and thus does not equal the zero matrix.

In fact, this argument also shows why this subspace is $r$-entangled: every non-zero diagonal of a matrix $M \in \mathrm{span}(B)$ contains at least $r+1$ non-zero entries, so there is an $(r+1) \times (r+1)$ submatrix of $M$ that is upper triangular with non-zero diagonal entries (and is thus invertible), so $\mathrm{rank}(M) \geq r+1$. Since the rank of a matrix corresponds to the symmetric tensor rank in $\proj(\complex^d \vee \complex^d)$, the result follows.

All that remains is to count the number of vectors in $B$:

\begin{align*}
|B| = \sum_{i=0}^{d-r-1}\big(d-r-i\big) & = (d-r)^2 - \sum_{i=0}^{d-r-1}i \\
& = (d-r)^2 - \frac{1}{2}(d-r)(d-r-1) = \binom{d-r+1}{2}.
\end{align*}

Since the projective dimension of the subspace is $|B|-1$, this completes the proof.

§.§.§ Maximal antisymmetric $r$-entangled subspaces of bipartite space

To construct an $r$-entangled subspace of $\proj(\complex^d \wedge \complex^d)$ attaining the bound (<ref>), we note that the isomorphism $\complex^d \otimes \complex^d \cong \Lin((\complex^{d})^*,\complex^{d})$ shows that it is equivalent to construct a projective linear subspace of antisymmetric $d \times d$ matrices with rank greater than $2r$ (not $r$) of dimension

\[
\binom{d-2r}{2}-1.
\]

The construction of this subspace is identical to the symmetric construction from Section <ref>, except we omit the $M_j^i$ matrices with non-zero entries on the main diagonal (i.e., the ones with $i = 0$), and we subtract in the lower-triangular portion of each matrix instead of adding. That is, a basis of this subspace is

\begin{align*}
B = \Big\{ \left[M_j^i - \big(M_j^i\big)^\t\right] : 1 \leq i \leq d-2r-1, 1 \leq j \leq d-2r-i \Big\},
\end{align*}

so the dimension of this subspace is

\begin{align*}
|B|-1 = \binom{d-2r+1}{2} - (d-2r) -1= \binom{d-2r}{2}-1,
\end{align*}

as desired.

§.§.§ Maximal symmetric $1$-entangled subspaces of multipartite space

By Equation (<ref>) and the subsequent discussion, the maximum dimension of a symmetric $1$-entangled subspace of $\proj({\bigvee}^m \complex^d)$ is

\begin{align}
\binom{m+d-1}{m}-d-1.
\end{align}

In this section, we construct such a subspace.

Consider the subspace spanned by the linearly independent set

\begin{align}
\{[e_{a_1} \vee \cdots \vee e_{a_m}] : (a_1,\dots,a_m)\in [d]^{\times n} \setminus \Delta_d^n\},
\end{align}

where $\Delta_d^n=\{(a,\dots, a) : a  \in [d]\}$.
This subspace clearly has the correct dimension, and if

\begin{align}
\sum_{a \in [d]^{\times n} \setminus \Delta_d^n} \alpha_a (e_{a_1} \vee \cdots \vee e_{a_m})= x^{\otimes m}
\end{align}

for some $x =\sum_{b=1}^d \beta_b e_b$, then for each $b \in [d]$ the coefficient of $e_b^{\otimes m}$ in the expansion of $x^{\otimes m}$ is zero, and hence $\beta_b=0$. It follows that $x=0$, a contradiction.

§.§.§ Maximal antisymmetric $1$-entangled subspaces of multipartite space

By Equation (<ref>) and the subsequent discussion, the maximum dimension of an antisymmetric 1-entangled subspace of $\proj (\bigwedge^m \complex^d)$ is

\begin{align}
\binom{d}{m}-m(d-m)-2,
\end{align}

whenever $d \geq m$ (otherwise, $\bigwedge^m \complex^d=0$). We construct a subspace that attains this bound in a somewhat similar manner to the non-positive partial transpose subspaces constructed in [28, 27]. Let

\begin{align}
J=\left\{\binom{m}{2}+m-1,\binom{m}{2}+m,\dots, dm-\binom{m}{2}-1, dm-\binom{m}{2}\right\}.
\end{align}

For each $s \in J$, let

\begin{align}
I_s=\left\{(a_1,\dots, a_m) \in [d]^m : 1\leq a_1<\dots< a_m\leq d \quad \text{and} \quad \sum_{j=1}^m a_j =s\right\}.
\end{align}

\begin{align}
\proj \W=\{ [v] \in \textstyle{\proj ( {\bigwedge}^m \complex^d)} : \sum_{a\in I_s} v_a =0\},
\end{align}

where $v_a$ is the coefficient of $e_{a_1} \wedge \dots \wedge e_{a_m}$ in the expansion of $v$ with respect to the standard basis

\begin{align}
\{[e_{a_1}\wedge \dots \wedge e_{a_m}] : 1\leq a_1<\dots< a_m\leq d\}
\end{align}

of $\proj ( {\bigwedge}^m \complex^d)$. We first observe that $\proj \W$ has the correct dimension. Note that ${\abs{J}=m(d-m)+1}$, and one linear constraint is placed on $\W$ for each $s \in J$, so

\begin{align}
\dim(\proj \W)=\dim(\textstyle{\proj ( {\bigwedge}^m \complex^d)})-\abs{J}=\binom{d}{m}-m(d-m)-2,
\end{align}

as desired.

To complete the proof, we need only show that $\proj \W$ does not contain any state of the form $[x_1 \wedge \dots \wedge x_m]$. Proving this property is quite technical, so we begin by proving it in the special case $m=2$ as a warm-up.

Let $[v] \in \proj \W$ be arbitrary, and let

\begin{align}
t= \min \{s \in J : v_a \neq 0 \quad \text{for some} \quad a \in I_s\}.
\end{align}

Let $a,b \in I_t$ be any two multi-indices for which $a \neq b$ and $v_a, v_b \neq 0$. Since ${a_1+a_2=b_1+b_2=t}$, there exists a permutation $\sigma \in S_2$ for which

\begin{align}
b_{\sigma(1)} < a_{\sigma(1)}<a_{\sigma(2)}<b_{\sigma(2)}.
\end{align}

Under the inclusion $\wedge^2 (\complex^d)\subseteq \V_1 \otimes \V_2$, where $\V_1, \V_2 \cong \complex^d$, we can regard $v$ as an element of $\Lin(\V_2^*,\V_1)$.

Under this identification, consider the $4 \times 4$ submatrix of $v$ corresponding to the column index $\{{b_{\sigma(1)}}, {a_{\sigma(1)}},{a_{\sigma(2)}},{b_{\sigma(2)}}\}$ and row index $\{{b_{\sigma(2)}},{a_{\sigma(2)}},{a_{\sigma(1)}},{b_{\sigma(1)}}\}$. It is straightforward to verify that this matrix takes the form

\begin{align}
\begin{blockarray}{*{4}{c} l}
\begin{block}{*{4}{>{$\footnotesize}c<{$}} l}
$b_{\sigma(2)}$ & $a_{\sigma(2)}$ & $a_{\sigma(1)}$ & $b_{\sigma(1)}$ & \\
\end{block}
\begin{block}{[*{4}{c}]>{$\footnotesize}l<{$}}
\pm v_b&0&0& 0& $b_{\sigma(1)}$ \\
*&\pm v_a& 0&0 & $a_{\sigma(1)}$ \\
*& * & \pm v_a& 0 & $a_{\sigma(2)}$ \\
* & * & * & \pm v_b& $b_{\sigma(2)}$\\
\end{block}
\end{blockarray},
\end{align}
where an asterisk $(*)$ denotes an entry we don't care about, and a $\pm$ denotes a sign we don't care about. It follows that the rank of $v$ under this identification is at least four. This proves that $[v] \notin \Gr(1, \proj^{d-1})$, since any such state has rank two under this identification. This completes the proof that $\proj \W$ is antisymmetric 1-entangled in the case $m=2$.

We now proceed to prove that $\proj \W$ is antisymmetric 1-entangled for arbitrary $m$. As before, let $[v] \in \proj \W$ be arbitrary, let
\begin{align}
t= \min \{s \in J : v_a \neq 0 \quad \text{for some} \quad a \in I_s\},
\end{align}
and let $a,b \in I_t$ be any two multi-indices for which $a \neq b$ and $v_a, v_b \neq 0$. Let $A=\{a_1,\dots, a_m\}$, $B=\{b_1,\dots, b_m\}$, and let $S=A \cap B$. Let $q=m-\abs{S}$, and let ${s_{q+1}<\dots<s_m \in [d]}$ be such that $\{s_{q+1},\dots, s_{m}\}=S$.

Suppose toward contradiction that $[v]=[x_1 \wedge \dots \wedge x_m]$ for some $x_1,\dots, x_m \in \complex^d$. Writing $x_{j,a} \in \complex$ for the $a$'th coordinate of $x_j$ in the standard basis, we may assume that for any $j\in [m]$ and $i \in \{q+1,\dots,m\}$, it holds that $x_{j,s_i}\neq 0 \iff i=j$. This can be observed by considering the element of $\pi^{-1}[v]$ that is in reduced row echelon form with respect to the re-ordered standard basis in which the elements $e_{s_{q+1}},\dots, e_{s_{q+m}}$ come first. In this nice form, $v_a \neq 0$ implies
\begin{align}
w_{A\setminus S}\neq 0,
\end{align}
where $w=x_1 \wedge \dots \wedge x_q$, so $[w] \in \Gr(q-1,\proj^{d-1})$. The notation $w_{A\setminus S}$ denotes the coefficient of $w$ with respect to the standard basis element $e_{c_1} \wedge \dots \wedge e_{c_q}$, where ${c_1<\dots<c_q \in [d]}$ are such that $\{c_1,\dots, c_q\}=A\setminus S$. Similarly, $v_b \neq 0$ implies $w_{B \setminus S} \neq 0$. Let $Q=(A \cup B)\setminus S$, and let $\tilde{c}_1<\dots<\tilde{c}_{2q} \in [d]$ be such that $\{\tilde{c}_1,\dots, \tilde{c}_{2q}\}=Q$. Regarding $w$ as an element of $\Lin(\bigotimes_{k=2}^q \V_k^*,\V_1)$, consider the submatrix of $w$ with column index $Q$ and row index $\{\tilde{c}_{\hat{1}},\dots,\tilde{c}_{\hat{2q}}\}\subseteq [d]^{\times q-1}$, where

\begin{align}
\tilde{c}_{\hat{j}}=
\begin{cases}
(A \setminus S)\setminus \{\tilde{c}_{j}\} ,& \tilde{c}_j \in A \setminus S\\
(B \setminus S)\setminus \{\tilde{c}_{j}\} ,& \tilde{c}_j \in B \setminus S
\end{cases}\quad \text{for each} \quad j \in [2q].
\end{align}
Similarly to before, we have implicitly identified the set $(A \setminus S)\setminus \{\tilde{c}_{j}\}$ with a strictly increasing $q-1$-tuple, and likewise for $B$. By the minimality of $t$, it follows that this submatrix is lower-triangular with each diagonal entry equal to $\pm w_{A \setminus S}$ or $\pm w_{B \setminus S}$, and hence has rank $2q$. This is a contradiction, as any element of $\Gr(q-1,\proj^{d-1})$ has rank $q$ when viewed in this way. This completes the proof that $\proj \W$ is antisymmetric 1-entangled.

\subsubsection{Some maximal $2$-entangled subspaces of multipartite space}\label{sec:multipartite_higher_rank}

There are a handful of higher-rank multipartite cases where it is straightforward to construct a $2$-entangled subspace of $\proj(\bigotimes_{j=1}^m\mathbb{C}^{d_j})$ of maximum dimension, and we consider these well-known.

For example, if $m = 3$ and $d_1 = d_2 = d_3 = 2$ then the maximum dimension of a $2$-entangled subspace is zero, and an explicit example is simply given by
\begin{align}
[e_2 \otimes e_1 \otimes e_1+e_1\otimes e_2 \otimes e_1+e_1 \otimes e_1 \otimes e_2].
\end{align}
Similarly, if $m = 3$ and $d_1 = 3$, $d_2 = d_3 = 2$ then we see that $2$-entangled subspaces cannot be any larger than $1$-dimensional, since we can identify $\complex^3 \otimes \complex^2 \otimes \complex^2 \cong \complex^3 \otimes \complex^4$ in the natural way and then use the bipartite bound from [13]. Furthermore, we can construct a $2$-entangled subspace of this dimension just by using the fact that a state has border rank $\geq 3$ if and only if it has flattening rank $\geq 3$~\cite[Theorem 5.1]{landsberg2004ideals}, so the explicit construction of a $2$-entangled subspace of $\proj(\complex^3 \otimes \complex^4)$ of dimension $1$ from [13] also works in $\proj(\complex^3 \otimes \complex^2 \otimes \complex^2)$.

In general, however, explicit constructions of $r$-entangled subspaces in this multipartite higher-rank setting are rather ad-hoc. Our contribution here is to present explicit examples of $2$-entangled subspaces of $\proj(\complex^3 \otimes \complex^3 \otimes \complex^2)$ and of $\proj(\complex^2 \otimes \complex^2 \otimes \complex^2\otimes \complex^2)$ (i.e., the smallest non-trivial and previously unknown cases) that are $5$-dimensional, which is maximal by Equation~\eqref{eq:2entangleddim}.

In $\proj(\complex^3 \otimes \complex^3 \otimes \complex^2)$, we first let $\delta,\epsilon,\theta,\kappa \in \complex^4$ and then consider the following set of $6$ states:
\begin{align}% first tensor factor is row, second is column, third is block column
B=\big\{ & \left[ \big(e_1 \otimes e_1+e_2 \otimes e_2 + e_3 \otimes e_3\big)\otimes e_1 \right],\\
& \left[\big(e_1 \otimes e_1+e_2 \otimes e_2+e_3 \otimes e_3 \big)\otimes e_2 \right],\\
& \left[\delta_1 (e_1 \otimes e_2 \otimes e_1)+\delta_2 (e_2 \otimes e_1 \otimes e_1)+\delta_3 (e_3 \otimes e_1 \otimes e_2)+\delta_4 (e_1 \otimes e_3 \otimes e_2)\right],\\
&\left[\epsilon_1 (e_1 \otimes e_2 \otimes e_1)+\epsilon_2 (e_2 \otimes e_1 \otimes e_1)+\epsilon_3 (e_3 \otimes e_1\otimes e_2)+\epsilon_4 (e_1 \otimes e_3 \otimes e_2)\right],\\
&\left[\theta_1 (e_1 \otimes e_2 \otimes e_1)+\theta_2 (e_3 \otimes e_1 \otimes e_1)+\theta_3 (e_3\otimes e_2 \otimes e_2)+\theta_4 (e_2 \otimes e_3 \otimes e_2)\right],\\
&\left[\kappa_1 (e_1 \otimes e_2 \otimes e_1)+\kappa_2 (e_3 \otimes e_1 \otimes e_1)+\kappa_3 (e_3\otimes e_2 \otimes e_2)+\kappa_4 (e_2 \otimes e_3 \otimes e_2)\right]
\big\}\label{eq:qutrittritbit_basis}
\end{align}
%	B = \big\{ & \big(\ket{00} + \ket{11} + \ket{22}\big)\ket{0}, \big(\ket{00} + \ket{11} + \ket{22}\big)\ket{1}, \\ % z, then w
%	& \delta_1\ket{010} + \delta_2\ket{100} + \delta_3\ket{201} + \delta_4\ket{021}, \ \theta_1\ket{020} + \theta_2\ket{200} + \theta_3\ket{211} + \theta_4\ket{121}, \\% x and then y
%	& \epsilon_1\ket{010} + \epsilon_2\ket{100} + \epsilon_3\ket{201} + \epsilon_4\ket{021}, \kappa_1\ket{020} + \kappa_2\ket{200} + \kappa_3\ket{211} + \kappa_4\ket{121}\big\}.% x and then y

It is straightforward to show that if $\{\delta, \epsilon,\theta,\kappa\}$ is linearly independent then so is the set~\eqref{eq:qutrittritbit_basis}, so it is a basis of its span. Under the choice $\delta = (0,1,1,1)$, $\epsilon = (1,1,2,0)$, $\theta = (1,1,1,0)$, and $\kappa = (0,2,1,1)$; we have verified using the Macaulay2 software package [19] that $\spn(B)$ is indeed 2-entangled (i.e. all of its members have border rank $\geq 3$). In Appendix~A we provide an alternate, non-computer proof that $\spn(B)$ is 2-entangled. The details of this proof are somewhat more complicated than in the other constructions that we have considered (we elaborate on this point at the end of the appendix). We have developed a heuristic algorithm in MATLAB which suggests that most choices of $\{\delta, \epsilon,\theta,\kappa\}$ produce a 2-entangled subspace. Our Macaulay2 and MATLAB codes are available on github.com [31].

In $\proj(\complex^2 \otimes \complex^2 \otimes \complex^2\otimes \complex^2)$, we similarly let $\phi,\psi \in \complex^4$ and then consider the following set of $6$ states:
\begin{align}
\Big\{ & \left[e_1 \otimes e_1 \otimes e_1 \otimes e_2+e_1 \otimes e_2\otimes e_2 \otimes e_1+e_2 \otimes e_2 \otimes e_1 \otimes e_2\right],\\
&\left[e_1 \otimes e_1 \otimes e_2 \otimes e_1+e_2 \otimes e_1 \otimes e_1 \otimes e_2+e_2 \otimes e_2 \otimes e_2 \otimes e_1\right],\\
&\left[e_1 \otimes e_2 \otimes e_1 \otimes e_1+e_2 \otimes e_1  \otimes e_1 \otimes e_1+e_2\otimes e_1 \otimes e_2 \otimes e_2+e_2 \otimes e_2 \otimes e_1 \otimes e_2\right],\\
&\left[e_1\otimes e_1 \otimes e_2 \otimes e_1+e_1 \otimes e_2\otimes e_1 \otimes e_1+e_1 \otimes e_2 \otimes e_2\otimes e_2+e_2 \otimes e_1 \otimes e_2 \otimes e_2\right],\\
&\left[\phi_1 (e_1^{\otimes 4})+\phi_2(e_1\otimes e_1 \otimes e_2\otimes e_2)+\phi_3(e_2\otimes e_2 \otimes e_1\otimes e_1)+\phi_4(e_2^{\otimes 4})\right],\\
&\left[\psi_1 (e_1^{\otimes 4})+\psi_2(e_1\otimes e_1 \otimes e_2\otimes e_2)+\psi_3(e_2\otimes e_2 \otimes e_1\otimes e_1)+\psi_4(e_2^{\otimes 4})\right]
\Big\}\label{eq:fourqubit_basis}
\end{align}

%\begin{align*}\begin{split}\label{eq:fourqubi\phi_basis}% block row, block col, row, col
%	B = \big\{ & \ket{0001} + \ket{0110} + \ket{1101}, \ \ket{0010} + \ket{1001} + \ket{1110}, \\
%	& \ket{0100} + \ket{1000} + \ket{1011} + \ket{1101}, \ \ket{0010} + \ket{0100} + \ket{0111} + \ket{1011}, \\
%	& \phi_1\ket{0000} + \phi_2\ket{0011} + \phi_3\ket{1100} + \phi_4\ket{1111}, \\
%	& \psi_1\ket{0000} + \psi_2\ket{0011} + \psi_3\ket{1100} + \psi_4\ket{1111}\big\}.
% MATLAB says maximum overlap of a tensor rank-2 state with this subspace is approximately 0.9996201

It is straightforward to show that if $\{\phi,\psi\}$ is linearly independent then so is the set~\eqref{eq:fourqubit_basis}, so it is a basis of its span. Furthermore, if $\phi = (0,1,1,1)$ and $\psi = (1,2,1,0)$ then we have shown via Macaulay2 that its span is $2$-entangled. As before, our numerics suggest that most choices of $\{\phi, \psi\}$ produce a 2-entangled subspace [31].

\subsection{Witnesses and multipartite Schmidt number}\label{sec:witnesses}

This is the only section in which we consider mixed states, so we introduce notation and related definitions for them here. For a vector space $\V$ (which we always take to be over $\complex$), let $\density{\V}\subseteq \Lin(\V)$ be the set of positive semi-definite operators on $\V$ with trace one, which we refer to as the set of \textit{mixed quantum states}, or \textit{density operators}. Let $\herm{\V}\subseteq \Lin(\V)$ be the set of Hermitian operators on $\V$. For a positive semi-definite operator ${P \in \pos{\V}}$, we define the \textit{support} of $P$ to be $\im(P)$. For a subset $Z \subseteq \V$ we say that $\rho$ is \textit{supported on} $Z$ if $\im(P) \subseteq {Z}$. We say that $P,Q \in \pos{\V}$ have \textit{orthogonal support} if $\braket{v}{u}=0$ for all $v \in \im(P)$ and $u \in \im(Q)$.

The notion of an operator that ``witnesses'' a particular property of a (mixed) quantum state is an important one in quantum information theory---it provides a way of demonstrating that property via a single measurement, without the need to have complete information about the state.

%The \textit{support} of a mixed state $\rho \subseteq \density{\V}$ is its the image $\im(\rho) \subseteq \V$, and equivalently is the set of vectors that appear in a decomposition of $P$ as a positive combination of rank-one operators.

\begin{definition}\label{defn:schmidt_number_multi}
Given a subset $Z \subseteq \V$, a \emph{not-${Z}$ witness} is a Hermitian matrix $W \in \herm{\V}$ for which
\begin{enumerate}
\item $\tr(W\rho) \geq 0$ for all (mixed) states $\rho \in \density{\V}$ supported on $Z$, and

\item there exists some (mixed) state $\sigma \in \density{\V}$ such that $\tr(W\sigma) < 0$.
\end{enumerate}
\end{definition}

%We have used $\hat{X}$, instead of just $X$, to emphasize the fact that $X$ is a subset of the affine space $\V$, and not of $\proj \V$. As mentioned in the beginning of Section~\ref{mp}, for a projective subset $X \subseteq \proj \V$, $\hat{X}\subseteq \V$ denotes the affine cone over $X$.

For example, if $\hat{Y} \subseteq \complex^{d_1} \otimes \complex^{d_2}$ is (the affine cone over) the set of product states defined in~\eqref{seg}, then not-$\hat{Y}$ witnesses are called \emph{entanglement witnesses} (since any mixed state $\sigma$ for which $\tr(W\sigma) < 0$ is then guaranteed to be entangled). Slightly more generally, if ${\hat{Z} \subseteq \complex^{d_1} \otimes \complex^{d_2}}$ is (the affine cone over) the set of states of tensor rank (i.e., Schmidt rank) at most $r$, then the not-$\hat{Z}$ witnesses are called $r$-entanglement witnesses [35] (or \emph{$r$-block positive} [37]), and any mixed state $\sigma$ for which $\tr(W\sigma) < 0$ is said to have \emph{Schmidt number} [38] greater than $r$.

Even more generally, if $\hat{X} \subseteq \bigotimes_{j=1}^m \complex^{d_j}$ is (the affine cone over) the set of states with tensor rank at most $r$, then we still refer to not-$\hat{X}$ witnesses as $r$-entanglement witnesses, and the witnessed (mixed) states with support not in $\hat{X}$ are said to have (multipartite) Schmidt number larger than $r$ [14]. However, a hiccup that occurs in this multipartite case that does not in the bipartite case is that, since the set of states with (multipartite) Schmidt number $\leq r$ is not closed, it is not true that every mixed state with Schmidt number $> r$ can be detected by some $r$-entanglement witness. Indeed, only the mixed states that are outside of the \emph{closure} of that set of states can be detected, so these $r$-entanglement witnesses are better thought of as witnesses for border rank, not tensor rank.

It is well-known, at least in the bipartite case, that entangled subspaces can be used to construct entanglement witnesses with the maximum number of negative eigenvalues (see [28, 27], for example). We now show that the same is true in the multipartite case, and even for not-$Z$ witnesses in general, as long as $Z$ is a Euclidean closed cone.

\begin{theorem}\label{thm:neg_evals_subspace_corr}
Suppose $Z \subseteq \V$ is a Euclidean closed cone. The maximum number of negative eigenvalues that a not-$Z$ witness can have is equal to the maximum dimension of a linear subspace $\W \subseteq \V$ for which $\W \cap Z = \{0\}$.
\end{theorem}

\begin{proof}
Throughout this proof, we let $n$ be the maximum dimension of a linear subspace of $\V$ that trivially intersects $Z$.

Suppose $W\in \herm{\V}$ is a not-$Z$ witness with $s$ negative eigenvalues. To see that $s \leq n$, write $W = W_{+} - W_{-}$ where $W_{+},W_{-}$ are positive semidefinite with orthogonal support (i.e., they come from the spectral decomposition of $W$). Letting $\mathcal{W}=\im(W_{-})$, we have $\dim(\W) = s$. Furthermore, for all $v \in \mathcal{W}\setminus \{0\}$, it holds that
\begin{align*}
\tr\big(W v v^* \big) = \tr\big(W_{+}v v^*- W_{-}v v^*\big) = -\tr(W_{-} v v^*) < 0,
\end{align*}
from which it follows that $\mathcal{W}$ has trivial intersection with $Z$, so $s \leq n$.

Conversely, to see that there is a not-$Z$ witness with $n$ negative eigenvalues, let $P$ be the orthogonal projection onto some linear subspace $\mathcal{W} \subseteq \V$ of dimension $n$ that trivially intersects $Z$. Then $\mathrm{Tr}(Pvv^*) < \braket{v}{v}$ for all $v \in Z$, and the fact that $Z$ is a closed cone implies that there exists a real constant $0 < \epsilon < 1$ such that $\mathrm{Tr}(Pvv^*) \leq \epsilon \braket{v}{v}$ for all $v \in Z$. If we define the matrix $W = \I - \tfrac{1}{\epsilon}P$, then $W$ has exactly $\dim(\W) = n$ negative eigenvalues. Furthermore, $\mathrm{Tr}(Wvv^*) \geq 0$ for all $v \in Z$, so $W$ is a not-$Z$ witness.
\end{proof}

Combining Theorem~\ref{thm:neg_evals_subspace_corr} with Theorem~\ref{dimension_equivalence}, we get that for any projective variety ${X \subseteq \proj^{D}}$, the maximum number of negative eigenvalues of a not-$\hat{X}$ witness is $D-\dim(X)$. We close this section by applying this result to entanglement witnesses.

\begin{cor}\label{cor:neg_evals_subspace_rent}
The maximum number of negative eigenvalues that an $r$-entanglement witness can have is exactly
\[
d_1\cdots d_m - \dim(\sigma_r(Y)) - 1,
\]
where $Y = \mathrm{Seg}(\mathbb{P}^{d_1-1} \times \cdots \times \mathbb{P}^{d_m-1})$, as usual.
\end{cor}

For example, if $P \in \Pos(\complex^3 \otimes \complex^3 \otimes \complex^2)$ is the orthogonal projection onto the $6$-dimensional subspace spanned by (the affine cone over) the set $B$ in~\eqref{eq:qutrittritbit_basis}, then there exists some scalar $\alpha > 1$ such that $W = I - \alpha P$ is a $2$-entanglement witness (here we have set $\alpha = 1/\epsilon$, where $\epsilon < 1$ is as in the proof of Theorem~\ref{thm:neg_evals_subspace_corr}), and furthermore there is no $2$-entanglement witness with more negative eigenvalues than this one (which has $6$). Importantly, this gives (at least in principle) a \emph{measurable} way of showing that a multipartite state has large Schmidt number: if measuring a state $\rho \in \Pos(\complex^3 \otimes \complex^3 \otimes \complex^2)$ produces an expectation value $\tr(W\rho)$ that is strictly negative, then $\rho$ must have Schmidt number at least $3$.

Unfortunately, finding an explicit value of $\alpha > 1$ that actually works to make $W$ a $2$-entanglement witness is a non-trivial task, which we have not been able to solve analytically. Numerics performed in MATLAB, however, strongly suggest that the optimal choice of $\alpha$ is approximately $1.0113$.

A similar construction with $P \in \Pos(\complex^2 \otimes \complex^2 \otimes \complex^2 \otimes \complex^2)$ being the orthogonal projection onto the $6$-dimensional subspace spanned by (the affine cone over) the set $B$ in~\eqref{eq:fourqubit_basis} gives a $2$-entanglement witness $W = I - \alpha P \in \Lin(\complex^2 \otimes \complex^2 \otimes \complex^2 \otimes \complex^2)$ with $6$ negative eigenvalues (for some value of $\alpha > 1$).

We can analogously define a symmetric (or antisymmetric) $r$-entanglement witness to be a Hermitian operator $W \in \herm{\bigvee^m\mathbb{C}^{d}}$ (or $W \in \herm{\bigwedge^m\mathbb{C}^{d}}$) such that ${\tr(W\rho) \geq 0}$ for all $\rho \in\density{\bigvee^m\mathbb{C}^{d}}$ (or $\rho \in \density{\bigwedge^m\mathbb{C}^{d}}$) with rank $\leq r$, and for which there exists some $\sigma \in\density{\bigvee^m\mathbb{C}^{d}}$ (or $\sigma \in \density{\bigwedge^m\mathbb{C}^{d}}$) such that $\tr(W\sigma) < 0$.
\begin{cor}\label{cor:neg_evals_subspace_sym}
The maximum number of negative eigenvalues that a symmetric $r$-entanglement witness can have is exactly
\[
\binom{d-1+m}{m}-\dim(\sigma_r(\nu_m(\proj^{d-1})))-1.
\]
\end{cor}

\begin{cor}\label{cor:neg_evals_subspace_asym}
The maximum number of negative eigenvalues that an antisymmetric $r$-entanglement witness can have is exactly
\[
\binom{d-1+m}{m}-\dim(\sigma_r(\Gr(m-1,\proj^{d-1})))-1.
\]
\end{cor}

For example, in the bipartite $m = 2$ case, we see that the maximum number of negative eigenvalues that symmetric and antisymmetric $r$-entanglement witnesses can have is
\[
\binom{d-r+1}{2} \quad \text{and} \quad \binom{d-2r}{2},
\]

\section*{Acknowledgments}
We thank William Slofstra for his help on numerous occasions with the geometric arguments in this work. We thank Debbie Leung, Micha\l{} Oszmaniec and John Watrous for helpful discussions. B.L. was supported by NSERC Discovery Grant number RGPIN-2019-03949. N.J. was supported by NSERC Discovery Grant number RGPIN-2016-04003.

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\section*{Appendix: A 2-Entangled Qutrit-Qutrit-Qubit Subspace}

In this appendix, we provide a non-computer proof that the span of the set $B$ from Equation~\eqref{eq:qutrittritbit_basis} is $2$-entangled. In order to show this, we prove that each member of that span has a flattening with rank~$3$. Indeed, since $\mathbb{C}^3 \otimes \mathbb{C}^3 \otimes \mathbb{C}^2$ is naturally isomorphic to the space of $3 \times 6$ matrices, we can think of this subspace as consisting of block matrices of the form
\[
M = \left[\begin{array}{ccc|ccc}
\lambda & \alpha_1 & \beta_1 & \gamma & 0 & \alpha_4 \\
\alpha_2 & \lambda & 0 & 0 & \gamma & \beta_4 \\
\beta_2 & 0 & \lambda & \alpha_3 & \beta_3 & \gamma
\end{array}\right],
\]
where $\alpha_j = \delta_j + \epsilon_j$ and $\beta_j = \theta_j + \kappa_j$ for all $1 \leq j \leq 4$ (and $\{\delta, \epsilon,\theta,\kappa\}$ is as in Section~\ref{sec:multipartite_higher_rank}). The partial transpose of $M$ is
\[
M^\Gamma = \left[\begin{array}{ccc|ccc}
\lambda & \alpha_2 & \beta_2 & \gamma & 0 & \alpha_3 \\
\alpha_1 & \lambda & 0 & 0 & \gamma & \beta_3 \\
\beta_1 & 0 & \lambda & \alpha_4 & \beta_4 & \gamma
\end{array}\right],
\]
which is another flattening of this same state, so our goal is to show that $\mathrm{rank}(M) \geq 3$ or $\mathrm{rank}(M^\Gamma) \geq 3$.

Importantly, because of how we chose $\{\delta, \epsilon,\theta,\kappa\}$, we know if that if $\alpha_i\neq 0$ for any $i \in [4]$, then $\alpha_i \neq 0$ for at least three $i \in [4]$, and similarly for the $\beta_j$'s. Indeed, we saw the desirability of this property in Section~\ref{sec:construct}, where we repeatedly used Lemma~\ref{lem:tot_nonsing_cols}.

To show that $\mathrm{rank}(M) \geq 3$ or $\mathrm{rank}(M^\Gamma) \geq 3$, we now split into several cases depending on which of $\lambda$, $\gamma$, $\alpha_i$ and $\beta_j$ equal $0$.

\begin{itemize}
\item[Case 1(a):] $\lambda = 0$, $\alpha_1,\alpha_2,\alpha_3 \neq 0$.

The submatrix of $M$ corresponding to its $1$st, $2$nd, and $4$th columns, up to permutation similarity, has the form
\begin{align*}
\begin{bmatrix}
\alpha_2 & 0 & 0\\
\beta_2 & \alpha_3 & 0\\
0 & \gamma & \alpha_1\\
\end{bmatrix}
\end{align*}
which clearly has rank $3$ since it is triangular with non-zero diagonal entries.
\end{itemize}

The above case contains the flavor of most of the cases that we will consider, so from now on we just list which columns of $M$ or $M^\Gamma$ give rise to a submatrix that is (up to permutation similarity) triangular with non-zero diagonal entries, and thus has rank~$3$. For example, for Case~1(a) we would just now just say ``$M(1,2,4)$''.

\begin{itemize}
\item[Case 1:] $\lambda = 0$.
\begin{itemize}
\item[(a):] $\alpha_1,\alpha_2,\alpha_3 \neq 0$. $M(1,2,4)$.

\item[(b):] $\alpha_1,\alpha_2,\alpha_4 \neq 0$. $M^\Gamma(1,2,4)$.

\item[(c):] $\alpha_1,\alpha_3,\alpha_4 \neq 0$.
\begin{itemize}
\item[(i):] $\gamma = 0$. $M^\Gamma(1,4,6)$.

\item[(ii):] $\gamma \neq 0$. $M(2,4,5)$.
% x1  w  0
%  0 x3 y3
%  0  0  w
\end{itemize}

\item[(d):] $\alpha_2,\alpha_3,\alpha_4 \neq 0$.
\begin{itemize}
\item[(i):] $\gamma = 0$. $M(1,4,6)$.
%  0  0 x4
% x2  0 y4
% y2 x3  0
\item[(ii):] $\gamma \neq 0$. $M^\Gamma(2,4,5)$.
\end{itemize}

\item[(e):] $\alpha_1 = \alpha_2 = \alpha_3 = \alpha_4 = 0$.
\begin{itemize}
\item[(i):] $\gamma = 0$.
This case is identical to Case~1(a--d) via symmetry (just rotate $M$ by $180$ degrees, which does not change its rank).

\item[(ii):] $\gamma \neq 0$, $\beta_1 \neq 0$. $M^\Gamma(1,4,5)$.

\item[(iii):] $\gamma \neq 0$, $\beta_2,\beta_3,\beta_4 \neq 0$. $M(1,4,6)$.
\end{itemize}
\end{itemize}

\item[Case 2:] $\lambda \neq 0$, $\alpha_1 = \alpha_2 = \alpha_3 = \alpha_4 = 0$.
\begin{itemize}
\item[(a):] $\beta_3 = \beta_4 = 0$ (and thus $\beta_1 = \beta_2 = 0$ too). $M(1,2,3)$.

\item[(b):] $\beta_3 \neq 0$.
\begin{itemize}
\item[(i):] $\gamma = 0$. $M(1,2,5)$.

\item[(ii):] $\gamma \neq 0$. $M(2,4,6)$.
\end{itemize}

\item[(c):] $\beta_4 \neq 0$.
\begin{itemize}
\item[(i):] $\gamma = 0$. $M^\Gamma(1,2,5)$.

\item[(ii):] $\gamma \neq 0$. $M(2,4,6)$.
\end{itemize}
\end{itemize}

\item[Case 3:] $\lambda \neq 0$ and $\alpha_i \neq 0$ for at least three $i\in [4]$.
\begin{itemize}
\item[(a):] $\beta_3 = \beta_4 = 0$.
\begin{itemize}
\item[(i):] $\alpha_3 \neq 0$. $M^\Gamma(2,3,6)$.
%  z  0  0
% x2  x3 0
%  0  w  z

\item[(ii):] $\alpha_4 \neq 0$. $M(2,3,6)$.
%  z  0  0
% x1  x4 0
%  0  w  z
\end{itemize}

\item[(b):] $\beta_3 \neq 0$, $\beta_4 = 0$.
\begin{itemize}
\item[(i):] $\alpha_4 = 0$, $\gamma = 0$. $M(1,3,4)$.

\item[(ii):] $\alpha_4 = 0$, $\gamma \neq 0$. $M^\Gamma(4,5,6)$.

\item[(iii):] $\alpha_4 \neq 0$, $\gamma = 0$. $M(2,3,6)$.

\item[(iv):] $\alpha_4 \neq 0$, $\gamma \neq 0$, $\alpha_2 \neq 0$. $M^\Gamma(2,3,5)$.
% w  z  0
% 0 x2 y2
% 0  0  z

\item[(v):] $\alpha_4 \neq 0$, $\gamma \neq 0$, $\alpha_2 = 0$. This case is much more difficult, so we leave it until after the remaining cases are dealt with.
\end{itemize}

\item[(c):] $\beta_3 = 0$, $\beta_4 \neq 0$. This case is identical to Case~3(b) by taking the partial transpose (i.e., replace $M$ with $M^\Gamma$ and vice-versa).

\item[(d):] $\beta_3, \beta_4 \neq 0$.
\begin{itemize}
\item[(i):] $\alpha_4 = 0$, $\gamma = 0$. $M(1,4,6)$.

\item[(ii):] $\alpha_4 = 0$, $\gamma \neq 0$. $M^\Gamma(3,4,5)$.
% w  0  0
% y4 z  0
% 0  y2 w

\item[(iii):] $\alpha_4 \neq 0$, $\beta_2 = 0$, $\gamma \neq 0$. $M^\Gamma(3,4,5)$.

\item[(iv):] $\alpha_4 \neq 0$, $\beta_2 = 0$, $\gamma = 0$, $\alpha_3 = 0$. $M^\Gamma(1,3,6)$.

\item[(v):] $\alpha_4 \neq 0$, $\beta_2 = 0$, $\gamma = 0$, $\alpha_3 \neq 0$. $M(2,3,4)$.

\item[(vi):] $\alpha_4 \neq 0$, $\beta_2 \neq 0$, $\gamma = 0$. $M^\Gamma(2,3,4)$.

\item[(vii):] $\alpha_4 \neq 0$, $\beta_2 \neq 0$, $\gamma \neq 0$. This is another difficult case that we deal with separately.
\end{itemize}
\end{itemize}
\end{itemize}

The only two remaining cases from above are 3(b)(v) and 3(d)(vii). These cases require a more intricate argument to demonstrate that at least one of $M$ or $M^\Gamma$ has rank~$3$, which we now provide.

\subsection*{Case 3(b)(v)}\label{sec:case3bv}

For this case, $\gamma,\lambda \neq 0$, $\alpha_2 = \beta_4 = 0$, $\alpha_1,\alpha_3,\alpha_4 \neq 0$, and $\beta_1,\beta_2,\beta_3 \neq 0$. Let ${\alpha=(\alpha_1,\alpha_2,\alpha_3, \alpha_4)}$ and $\beta=(\beta_1,\beta_2,\beta_3,\beta_4)$. Because of how we chose $\{\delta, \epsilon,\theta,\kappa\}$, we have
$\alpha = (\alpha_1,\alpha_1+\alpha_4,2\alpha_1+\alpha_4,\alpha_4)$ and $\beta =  (\beta_1,\beta_1+2\beta_4,\beta_1+\beta_4,\beta_4)$. Furthermore, using the facts that $0 = \alpha_2 = \alpha_1+\alpha_4$ and $0 = \beta_4$ shows that $\alpha = (\alpha_1,0,\alpha_1,-\alpha_1)$ and $\beta = (\beta_1,\beta_1,\beta_1,0)$. The matrix $M^\Gamma$ thus has the form
\begin{align*}
M^\Gamma & = \left[\begin{array}{ccc|ccc}
\lambda & 0 & \beta_1 & \gamma & 0 & \alpha_1 \\
\alpha_1 & \lambda & 0 & 0 & \gamma & \beta_1 \\
\beta_1 & 0 & \lambda & -\alpha_1 & 0 & \gamma
\end{array}\right].
\end{align*}
We can then see that $\mathrm{rank}(M^\Gamma) = 3$ as follows. If it had rank $\leq 2$ then it would be the case that $\det(M^\Gamma(1,2,3)) = \lambda(\lambda^2 - \beta_1^2) = 0$, which implies $\beta_1 = \pm \lambda$. We would similarly have $\det(M^\Gamma(4,5,6)) = \gamma(\gamma^2 + \alpha_1^2) = 0$, which implies $\alpha_1 = \pm i \gamma$. Finally, we would also have $\det(M^\Gamma(3,4,6)) = -\beta_1(\gamma\lambda + \alpha_1\beta_1) = 0$, which (since $\alpha_1 = \pm i \gamma$ and $\beta_1 = \pm \lambda$) implies $\gamma\lambda \pm i \gamma\lambda = 0$, which is impossible (recall that all of these variables are non-zero). It follows that at least one of these determinants is non-zero, so $\mathrm{rank}(M^\Gamma) = 3$, which completes this case.

\subsection*{Case 3(d)(vii)}

For this case, $\gamma,\lambda \neq 0$, $\alpha_4 \neq 0$, and $\beta_2,\beta_3,\beta_4 \neq 0$. We now split into four subcases:

\begin{itemize}
\item $\alpha_1 = \beta_3 = 0$. This case is identical to Case~3(b)(v) from Section~\ref{sec:case3bv} by taking the partial transpose, swapping the roles of $\gamma$ and $\lambda$, and swapping the roles of $\alpha$ and $\beta$.

\item $\alpha_1 = 0$, $\beta_3 \neq 0$. $M(2,4,5)$.

\item $\alpha_1 \neq 0$, $\beta_3 = 0$. $M(2,3,5)$.

% This difficult subcase does not depend on the particular choice of Vandermonde columns! Rewrite a bit without simplifying x2, x3, y2, or y3.
\item $\alpha_1,\beta_3 \neq 0$. This is the difficult subcase. Similarly to Case 3(b)(v),
we have ${\alpha= (\alpha_1,\alpha_1+\alpha_4,2\alpha_1+\alpha_4,\alpha_4)}$ and $\beta= (\beta_1,\beta_1+2\beta_4,\beta_1+\beta_4,\beta_4)$, so $M$ and $M^\Gamma$ have the form
\begin{align*}
M & = \left[\begin{array}{ccc|ccc}
\lambda & \alpha_1 & \beta_1 & \gamma & 0 & \alpha_4 \\
\alpha_1+\alpha_4 & \lambda & 0 & 0 & \gamma & \beta_4 \\
\beta_1+2\beta_4 & 0 & \lambda & 2\alpha_1+\alpha_4 & \beta_1+\beta_4 & \gamma
\end{array}\right] \quad \text{and} \\
M^\Gamma & = \left[\begin{array}{ccc|ccc}
\lambda & \alpha_1+\alpha_4 & \beta_1+2\beta_4 & \gamma & 0 & 2\alpha_1+\alpha_4 \\
\alpha_1 & \lambda & 0 & 0 & \gamma & \beta_1+\beta_4 \\
\beta_1 & 0 & \lambda & \alpha_4 & \beta_4 & \gamma
\end{array}\right].
\end{align*}
Now suppose that $\rank(M) = \rank(M^\Gamma) = 0$. Then we would have
\begin{align}
{\det(M(2,3,6)) = \lambda(\lambda\alpha_4 - \alpha_1\beta_4 - \gamma\beta_1) = 0},
\end{align}
which implies $\lambda\alpha_4 - \gamma\beta_1 = \alpha_1\beta_4$. We would also have
\begin{align}
\det(M^\Gamma(1,4,5)) = \gamma(\gamma\beta_1 - \lambda\alpha_4 - \alpha_1\beta_4) = 0,
\end{align}
which implies $\lambda\alpha_4 - \gamma\beta_1 = - \alpha_1\beta_4$. Combining these two expressions for $\lambda\alpha_4 - \gamma\beta_1$ shows that $\alpha_1\beta_4 = 0$, which contradicts the fact that we are assuming that $\alpha_1,\beta_4 \neq 0$ in this case. It follows that at least one of these determinants is non-zero, which completes this final subcase and the proof.
\end{itemize}

Numerics suggest that almost all choices of $\{\delta, \epsilon,\theta,\kappa\}$ lead to this subspace being $2$-entangled. Indeed, the primary property of $\{\delta, \epsilon,\theta,\kappa\}$ that we made use of was that a non-zero linear combination of $\delta$ and $\epsilon$ must never have more than one $0$ entry, and similarly for $\theta$ and $\kappa$ (this property is generic). The only other place where the particular entries of these vectors was used was in Case~3(b)(v) (and the symmetric first subcase of Case~3(d)(vii)), where non-invertibility of a $3 \times 3$ submatrix of $M$ or $M^\Gamma$ was more delicate.

To illustrate why Case~3(b)(v) is more delicate, notice that if we had instead chosen $\delta = (1,1,1,2)$, $\epsilon = (0,-1,1,1)$, $\theta = (2,1,1,0)$, and $\kappa = (1,1,0,1)$, then it would still be the case that any non-zero linear combination of $\delta$ and $\epsilon$ would never have more than one $0$ entry, and similarly for $\theta$ and $\kappa$ (so all of the other cases still work fine). However, the proof would fall apart in Case~3(b)(v), since we would get the pair of matrices
\begin{align*}
M = \left[\begin{array}{ccc|ccc}
1 & -1 & 1 & 1 & 0 & 1 \\
0 & 1 & 0 & 0 & 1 & 0 \\
1 & 0 & 1 & 1 & 1 & 1
\end{array}\right] \quad \text{and} \quad M^\Gamma = \left[\begin{array}{ccc|ccc}
1 & 0 & 1 & 1 & 0 & 1 \\
-1 & 1 & 0 & 0 & 1 & 1 \\
1 & 0 & 1 & 1 & 0 & 1
\end{array}\right],
\end{align*}
both of which have rank $2$.
\end{document}