The Linear Span of Uniform Matrix Product States

The variety of uniform matrix product states arises both in algebraic geometry as a natural generalization of the Veronese variety, and in quantum many-body physics as a model for a translation-invariant system of sites placed on a ring. Using methods from linear algebra, representation theory, and invariant theory of matrices, we study the linear span of this variety.


Introduction
Tensor networks are a powerful tool for the description of tensors living in a high dimensional space.Since their original conception in quantum many-body physics [1,15,24], they have found a wide range of applications in different fields, such as numerical tensor calculus [2,9,17,23], algebraic complexity theory [14], graphical models [28], and machine learning [6,9].In the study of these varieties of tensors, some important general results and developments have been achieved using methods from differential and complex geometry [19], algebraic geometry and representation theory [3,4,5,7,8,18,21].
In this article, we focus on a particular type of tensor networks: uniform matrix product states.From a quantum mechanics perspective, they model translation-invariant physical systems of sites placed on a ring.The geometry of uniform matrix product states has been extensively studied [10,11,19,25], but our understanding of them is still far from complete, and several fundamental mathematical problems remain open.
Our geometric point of view is the following: we fix a tensor space (C n ) ⊗d , and consider the set of all tensors in this space that admit a translation-invariant matrix product state representation, with a given bond dimension m.After taking the closure of this set, we obtain an algebraic variety, which we denote by uMPS(m, n, d).The ultimate goal would be to obtain a complete description of this variety, i.e., to find all defining equations.Phrased in this generality, this question is likely to be intractable.Indeed, even just determining which linear equations, if any, vanish on our variety is poorly understood.This is precisely the goal of this paper: For which parameters m, n, d ∈ N does ⟨uMPS(m, n, d)⟩ fill the ambient space?(Question 2.6) How does ⟨uMPS(m, n, d)⟩ decompose as a GL n -representation?(Question 2.9) Note that since taking the Zariski closure of any set does not change its linear span, the space ⟨uMPS(m, n, d)⟩ is the linear span of all tensors that admit a uMPS representation with bond dimension m.
The variety uMPS(m, n, d) does not only arise from tensor networks, it is also a very natural geometric construction in its own right.Indeed, as we will soon see, uMPS(m, n, d) is the closed image of the polynomial map that takes as input an n-tuple of m × m matrices, and outputs the traces of all d-fold products of the given matrices.In this way, it is a natural generalization of the classical Veronese variety.Our main Problem 1.1 is therefore equivalent to the following: Problem 1.2.Let A 0 , . . ., A n−1 be m×m matrices with generic1 entries.Which linear relations hold among the polynomials where (i 1 , . . ., i d ) ∈ [n] d ?
The ring generated by all polynomials Tr(A i 1 • • • A i d ), where the generic matrices are fixed but we allow d to vary, is known as the trace algebra.In his classical work [27], Procesi described how to obtain all relations between the generators of this ring in principle.Note however, that the question we are asking is slightly different: we are only interested in relations between traces of matrix products of a fixed length d.
This article is divided into three sections.In Section 2, we define the variety of uniform matrix product states, and describe its natural symmetries.In Section 3, we undertake a computational study of the space ⟨uMPS(m, n, d)⟩ in the smallest nontrivial case m = n = 2.We describe an algorithm that can compute this space, viewed as a GL 2 -representation: more precisely, we exploit and compute its decomposition into GL 2 -submodules.Based on this technique, we obtain a conjectured formula for the character and, in particular, the dimension of the weight spaces, and consequently of the whole space ⟨uMPS(2, 2, d)⟩.Section 4 contains our main theoretical results.In Section 4.1, we introduce a powerful method to find linear equations that vanish on ⟨uMPS(m, n, d)⟩, based on the Cayley-Hamilton theorem: Theorem 4.5.Let A 0 , . . ., A m , B be m × m matrices.Then for every ℓ ∈ N it holds that Here S m is the symmetric group acting on {0, 1, . . ., m −1}, and C m+1 is the cyclic group acting on {0, 1, . . ., m}.Theorem 4.5 actually gives nontrivial trace relations, i.e., linear relations that do not follow either from cyclic permutations or reflections of the factors.As a corollary, we show that, already for d at least quadratic in the bond dimension m, the linear span ⟨uMPS(m, n, d)⟩ does not fill its natural ambient space Cyc d (C n ), the space of cyclically symmetric tensors; significantly improving the state of the art.More precisely, the linear span of the space of uniform matrix product states is a proper subspace of the space of cyclically invariant tensors under the following conditions: , then uMPS(m, n, d) is contained in a proper linear subspace of the space of cyclically invariant tensors.
In Section 4.2 we study the special case m = n = 2, based on the computations in Section 4.1, and, using the trace parametrization, we provide an upper bound on the dimension of ⟨uMPS(2, 2, d)⟩ which is close to optimal.Moreover, we take some first steps towards proving our conjectured character formulas: in the first cases, i.e., for the first few weights, we can provide an explicit basis of generators for the weight spaces of the representation, again using our Cayley-Hamilton technique.

Preliminaries
We once and for all fix three parameters m, n, d ∈ N. We will consider tensors in the space (C n ) ⊗d .The standard basis of C n will be written as {e 0 , . . ., e n−1 }.We abbreviate the set {0, . . ., n − 1} to [n].
Definition 2.1.The uniform Matrix Product State parametrization is given by the map (2.1) We denote the image of this map by uMPS • (m, n, d).The closure of uMPS • (m, n, d) in the Euclidean or equivalently Zariski topology over the complex numbers, is the algebraic variety of uniform matrix product states, denoted by uMPS(m, n, d).
Remark 2.2.If we think of (C m×m ) n as the space of m × m × n tensors, then the uMPS parametrization takes a tensor in this space and contracts it d times with itself in a circle; see Figure 1.Remark 2.3.An alternative name for uMPS is translation invariant matrix product states with periodic boundary conditions [10,25].The name "uniform matrix product states" is sometimes reserved for the thermodynamic limit, where the number d of sites approaches infinity.Our terminology is consistent with [11,19].
As mentioned in the introduction, the main question we will try to answer in this article is the following: Question 2.4.Determine the linear span of uMPS(m, n, d); i.e., the smallest vector subspace of (C n ) ⊗d containing uMPS(m, n, d).In particular: what is the dimension of this space?

Cyclic and symmetric invariance
The space (C n ) ⊗d comes equipped with an action of the symmetric group S d : for σ ∈ S d and ω The symmetric group S d naturally contains the cyclic group C d and the dihedral group D 2d as subgroups.To be precise: let r, s ∈ S d be the cyclic permutation and reflection defined respectively by and is the cyclic subgroup generated by r, and D 2d is the subgroup generated by r and s.The cyclically symmetric tensors and dihedrally symmetric tensors are then the elements of (C n ) ⊗d that are invariant under the action of these subgroups: Note that both are linear subspaces of (C n ) ⊗d , and that Dih In other words, we can think of uMPS(m, n, d) as a subvariety of the ambient space Cyc d (C n ).As noted in [11,Corollary 3.18], if we fix n and d and let m grow, the space uMPS(m, n, d) will eventually fill this entire ambient space.It is known that for m = d, equality holds [11,Proposition 3.11].On the other hand, it follows from a dimension count ([11, Theorem 3.14], see also [22]) that if d ≫ m, the inclusion ⟨uMPS(m, n, d)⟩ ⊂ Cyc d (C n ) is strict.In Section 4.1 we will prove that already for d = O m 2 , we have a strict inclusion.
Observation 2.7.In the case m = n = 2, we have the stronger inclusion

This is a consequence of the identity Tr(
, which holds for any pair of 2 × 2 matrices A 0 , A 1 and sequence i 1 , . . ., i d with i j ∈ {0, 1}.See [16,Theorem 1.1] Example 2.8.We consider uMPS (2,2,6).For every A 0 , A 1 ∈ C 2×2 , we have the trace relation Tr , that does not come from invariance of trace under cyclic permutations of the matrices.

GL n -invariance
The general linear group GL n naturally acts on the space (C n ) ⊗d : given g ∈ GL n and ω Clearly, Cyc d (C n ) and Dih d (C n ) are invariant under this action.The following computation shows that uMPS(m, n, d) is invariant under this action as well: This means that the space ⟨uMPS(m, n, d)⟩ we are interested in is naturally a representation of GL n .In the remainder of this subsection, we briefly recall what we will use about representation theory of GL n , and fix notation.
We once and for all fix a torus T ⊂ GL n , consisting of all diagonal matrices; and identify T with (C * ) n .For λ = (λ 0 , . . ., λ n−1 ) ∈ Z n and t = diag(t 0 , . . ., t n−1 ) ∈ T , we write Let V be any representation of GL n .For any λ ∈ Z n , the weight space V λ is defined as It is a well-known fact from representation theory that as vector spaces; and that knowing the dimensions of the weight spaces uniquely determines the representation V up to isomorphism.The polynomial is known as the character of V .If we view our representation as a morphism ρ : GL n → GL(V ), the character χ V is equal to Tr(ρ(diag(t 0 , . . ., t n−1 ))).
So we can refine our main Question 2.4 to: Question 2.9.Let V = ⟨uMPS(m, n, d)⟩, viewed as a GL n -representation.For every weight λ ∈ Z n , determine the dimension of the weight space V λ .

Words, necklaces and bracelets
As a warm-up, let us consider the classical representation , and by expanding we find that dim V λ is equal to the multinomial coefficient d λ 1 ,...,λn if λ i = d, and zero otherwise.We can also see this in terms of coordinates, and this will be useful later: Definition 2.10.A word of length d on the alphabet [n] is just an ordered tuple I = (i 1 , . . ., i d ), with i j ∈ [n].The weight of a word I is a tuple w(I) = (w 0 , . . ., w n−1 ) ∈ Z n , where w i is the number of entries in I that are equal to i.
For every word I = (i 1 , . . ., i d ), we can define a vector e I := e i 1 ⊗ • • • ⊗ e i d .The space (C n ) ⊗d has a basis given by the e I , where I runs over all words of length d on the alphabet [n].In addition, every e I is a weight vector of weight w(I).So the dimension of the weight space V λ is the number of words of weight λ, which is indeed the multinomial coefficient d λ 1 ,...,λn .We move on to the subrepresentations Cyc d (C n ) and Dih d (C n ), the natural ambient spaces for uniform matrix product states.
where φ is Euler's totient function.There is also a formula for for d odd. (2.4) We only state the formula for |B λ (n, d)| in the case of binary bracelets (i.e.n = 2), as that is the only case that is relevant to us: for w even.
(2.5) Equations (2.2)-(2.5)can be derived from the Pólya enumeration theorem [26] applied to the action of the cyclic and the dihedral group, respectively.

Computations
In this section, we describe how to computationally answer Question 2.9 for fixed parameters.We focus on the smallest interesting case m = n = 2.In this case we are dealing with representations of GL 2 , so the weights are in Z 2 .Moreover, the only occurring weights in C a linear equation α i x i vanishes on X if and only if the identity α i p i = 0 holds in the polynomial ring C[y 1 , . . ., y s ].In particular, the dimension of the linear span of X is equal to the dimension of the subspace of C[y 1 , . . ., y s ] spanned by the p i 's.

The trace parametrization
If we directly use Definition 2.1, we see that uMPS(2, 2, d) is the closed image of a polynomial map C 8 → Dih d C 2 .However, in this specific case there is an alternative parametrization by C 5 instead of C 8 : the trace parametrization, which appears to be computationally more efficient in practice.It is based on the connection between uniform matrix product states and invariant theory of matrices.Remark 3.3.The trace algebra is precisely the subring of R consisting of all elements that are invariant under simultaneous conjugation: This is known as the first fundamental theorem in the invariant theory of matrices [27,29].such that for every pair (A 0 , A 1 ) of 2 × 2 matrices, the following equality holds: The above means that uMPS(2, 2, d) is the image of the polynomial map where b runs over all bracelets of length d.This is the trace parametrization.
In order to compute the polynomials P b , for bracelets of length 3, one verifies that For bracelets of length ≥ 4, we can inductively use the following identity [29], which holds for every quadruple (A, B, C, D) of 2 × 2 matrices:

Computing the character
The weight space ⟨uMPS(2, 2, d)⟩ w is the linear span of the image of the map where b ranges over all bracelets of weight w.By Observation 3.1, we need to compute the dimension of the linear subspace of C[T 0 , T 1 , T 00 , T 01 , T 11 ] spanned by the P b 's.This can be computed by putting the coefficients of the P b 's in a matrix and computing its rank.
Putting everything together, we obtain Algorithm 1, which for a given d computes the character (in particular the dimension) of ⟨uMPS(2, 2, d)⟩.We implemented this algorithm in Sage-Math [13].
Based on these computations, we make the following conjecture: This would in particular imply:

Higher degree equations
Equations of tensor varieties such as secant varieties of the Segre variety or Veronese variety are well known to be hard to compute.For uniform matrix product states, Critch and Morton gave a complete description of the ideal of uMPS(2, 2, 4) and uMPS(2, 2, 5) and several linear equation of uMPS(2, 2, d) are given for d until 12, cf.[10].The generators of the ideal of uMPS(2, 2, d) for d = 4, 5, 6 are given in [11].
Our algorithm from the previous section computes the linear span of uMPS(2, 2, d), which is equivalent to computing the degree 1 part of its defining ideal.With some small modifications, one can obtain an algorithm to compute the degree k part of the defining ideal, viewed as a GL 2 -representation.In this paper we will only give a brief sketch of this algorithm.For a more detailed account, including the use of raising operators to speed up the algorithm, we refer the reader to [12].
We first consider a slightly more general setting: let X be any variety that is given as the closed image of a homogeneous polynomial map p : C s → C N as in (3.1), and let I be its (homogeneous) defining ideal.Suppose that p is compatible with a given GL n -action on C s Table 1.Character of ⟨uMPS(2, 2, d)⟩.Since D w = D d−w (more generally: for every GL 2 -representation V we have that V (b0,b1) = V (b1,b0) ), we only list D w for w ≤ d 2 .The dimension of ⟨uMPS(2, 2, d)⟩ equals twice the sum of the D w 's if d is odd, and twice the sum of the D w 's minus the last D w if d is even.
, where ν k is the k'th Veronese embedding and π w is the projection onto the weight space.The latter map commutes with the action of the torus.Then, the weight space I k,w ⊆ S k C N * w is equal to Using this algorithm, we computed I(uMPS(2, 2, d)) 2 for d ≤ 10 and I(uMPS(2, 2, d)) 3 for d ≤ 9. Our code is available at [13], and our results are summarized in Tables 2 and 3.

Linear relations via Cayley-Hamilton
We recall the classical Cayley-Hamilton theorem.We use this result in order to find linear equations for uMPS(m, n, d + k), k ≥ m based on linear equations for uMPS(m, n, N ),   In fact, the only thing we will use is the following statement, which (since deg p A = m) immediately follows from Theorem 4.1.
Corollary 4.2.Let A ∈ C m×m .Then A k , for k ≥ m can be written as a linear combination of its previous powers.Lemma 4.3.Let c = (c 1 , . . ., c s ) ∈ C s be a vector of coefficients and i j ℓ 1≤ℓ≤d,1≤j≤s be indices; with i j ℓ ∈ [n].Assume that for every n-tuple (A 0 , . . ., A n−1 ) of m × m matrices and every k < m the following identity holds: Then the same identity holds for arbitrary k ∈ N.
Proof .We use induction on k ≥ m.By Corollary 4.2, A k 0 can be written as a linear combination of its previous powers A l 0 , for l = 0, . . ., m − 1.Therefore, we have The usefulness of Lemma 4.3 stems from the fact that one can find expressions of the form (4.1) which are trivial for small k, in the sense that they follow from the cyclic invariance of the trace, but nontrivial for large k.We illustrate this in the example below.
Example 4.4.We show that for any 2 × 2 matrices A 0 , A 1 , A 2 , A 3 and any k ≥ 0, the following identity holds By the above argument, it suffices to show the identity for k = 0 and k = 1.But these both follow from cyclic invariance of the trace: This immediately gives us a nontrivial linear relation on uMPS (2, 4, d), for d ≥ 6.But we can also find linear relations on uMPS(2, 2, d).For instance, if we put k = 2, A 2 = A 2 1 and which is the unique linear relation on uMPS(2, 2, 8) that doesn't follow from dihedral symmetry.Theorem 4.5.Let A 0 , . . ., A m , B be m × m matrices.Then for every ℓ ∈ N it holds that Here S m is the symmetric group acting on {0, 1, . . ., m − 1}, and C m+1 is the cyclic group acting on {0, 1, . . ., m}.
Proof .We will first show the statement for ℓ ∈ {0, 1, . . ., m − 1}.So let us fix such an ℓ.We will write Let us write c a for the permutation that cyclically permutes the first a elements.Precisely Step 1.For σ ∈ S m and τ ∈ C m+1 , we define .
From Step 1 we have that T (σ, τ ) = T (σ, τ ).There will be cancellations of terms in (4.2) as sgn(σ) sgn(τ ) = − sgn(σ) sgn(τ ) from Step 3. Finally using Step 2 we ensure that all terms will cancel out therefore establishing the given identity for 0 ≤ l ≤ m − 1.Now using Lemma 4.3, we conclude that the given identity (4.2) holds for all l ∈ N. ■ , then uMPS(m, n, d) is contained in a proper linear subspace of the space of cyclically invariant tensors.
Proof .Let ℓ ≥ m and let S m+1 denote the symmetric group acting on {0, 1, . . ., m − 1, ℓ}.As the trace is invariant under cyclic permutations, for every τ ∈ C m+1 , can be written as fixing all positions of A i 's and permuting B j 's according to some σ ′ ∈ S m+1 .If for a moment we identify {0, 1, . . ., m − 1, ℓ} with [m], we have σ ′ = σ • τ −1 .Therefore, we can rewrite (4.2) as follows: Let X 0 , X 1 , X 2 be m × m matrices, and in (4.3) substitute A 0 = X 0 , B = X 1 , and A i = X 2 for i = 1, . . ., m.Note that even after that substitution, the ternary bracelets corresponding to the (m + 1)! terms in (4.3) are all distinct.Hence no two terms will cancel, and we get a nontrivial linear relation on uMPS(m, 3, d), where With a bit more care, one can also get nontrivial relations on uMPS(m, 2, d) in this way.For instance if we take ℓ = m and in (4. 3) we substitute A 0 = X 0 X m+1 1 X 0 , B = X 1 , and A i = X 0 for i = 1, . . ., m, one verifies that again no terms cancel, and hence we found a nontrivial linear relation on uMPS(m, 2, d), where d = m+3 As in the previous section, we abbreviate "weight λ = (w, d − w) ∈ Z 2 " to "weight w".In the rest of this section, we provide a proof of Conjecture 3.7 in the cases of weight w = 0, 1, 2, 3.
Consider the parametrization of uMPS(2, 2, d) in coordinates It is in particular a polynomial map in the unknown entries of the matrices A 0 , A 1 ∈ C 2×2 , that we denote by We will write The proof of the cases w = 0, 1, 2, 3 consists in determining that a basis of the vector space W w is given by a number of elements that coincides with our conjectured dimension.In order to do this, we guess a set of generators {T b } of the cardinality prescribed by the conjectured dimension.The guess of the T b 's is based on determining a specific subset of the set of binary bracelets of weight w, of the proper cardinality.Then, we prove that the generators are linearly independent for a fixed choice of matrices A 0 and A 1 .
The cases w = 0 and w = 1 are easy: Proposition 4.9.The space W 0 is a 1-dimensional vector space generated by the polynomial T 0,...,0 = Tr A d 0 .The space W 1 is a 1-dimensional vector space generated by the polynomial T Proof .Note that If we make the substitutions Since for the given choice of A 0 , A 1 the polynomials (4.4) are d 2 linearly independent polynomials, the same holds for a generic choice of matrices.■  Proof .We will show that the polynomials are linearly independent even after the following substitution: Then W 3 is spanned by the following polynomials: We now simply have to put the coefficients of these polynomials in a matrix and show it has full rank.For d even the matrix of coefficients is given by

, we get Tr
Reordering the summands, we obtain

Figure 1 .
Figure 1.Graphical representation of map (2.1) defining uMPS(m, n, d).There are in total d tensors of order m × m × n, contracted along the respective edges.The parameters m and n are called bond and physical dimension, respectively.The bond dimensions associated to the edges of the graph represent the dimension of the vector spaces along which the tensor contraction takes place.

5 . 2 . 5 .
where the inclusion is strict unless d ≤ 2 or n = 2 and d ≤ Observation The set uMPS(m, n, d) is a subset of the space of cyclically invariant tensors Cyc d (C n ) ⊂ (C n ) ⊗d because of the trace invariance under cyclic permutations of the matrices: given M 1 , . . ., M d ∈ C m×m then Tr(M 1 • • • M d ) = Tr(M σ(1) • • • M σ(d) ), for σ ∈ C d .

Question 2 . 6 .
For fixed n and d, what is the smallest m such that ⟨uMPS(m, n, d)⟩ = Cyc d (C m )?
Definition 2.11.A necklace (of length d on the alphabet [n]) is an equivalence class of words, where two words are equivalent if they agree up to the action C d .A bracelet (of length d on the alphabet [n]) is an equivalence class of words, where two words are equivalent if they agree up to the action D 2d .For a fixed necklace or bracelet, all words in the equivalence class clearly have the same weight; this is the weight of the necklace or bracelet.We denote by N (n, d), resp.B(n, d), the set of necklaces, resp.bracelets, of length d on [n] and by N λ (n, d) ⊂ N (n, d), resp.B λ (n, d) ⊂ B(n, d), the subset of elements of weight λ ∈ Z n .To every necklace N ∈ N (n, d), we associate a basis vector e N := 1 d σ∈C d σ • e I , where I is any representative of N .Then Cyc d (C n ) has a basis given by {e N : N ∈ N (n, d)}.Moreover, e N is a weight vector of weight w(N ), hence we find that the dimension of the weight space of weight λ is given by the number |N λ (n, d)| of necklaces of weight λ.Remark 2.12.The number |N (n, d)| of necklaces of length d on [n] can be counted by the following formula, see for instance [30, Theorem 7.10]:

. 3 )
To every bracelet b ∈ B(n, d), we associate a basis vector e b := 1 2d σ∈D 2d σ • e I , where I is a representative of b.Then Dih d (C n ) has a basis given by {e b : b ∈ B(n, d)}, and the dimension of the weight space of weight λ is given by the number |B λ (n, d)| of bracelets of weight λ.Remark 2.13.The number of bracelets of length d on [n] is given by

Definition 3 . 2 .
Let R = C a k ij 1≤i,j≤m;1≤k≤n be the polynomial ring in m 2 n variables, and for k = 0, . . ., n − 1, let A k := a k ij 1≤i,j≤m be generic m × m matrices.The trace algebra C m,n is the subalgebra of R generated by the polynomials Tr(A i 1 • • • A is ), where (i 1 , . . ., i s ) runs over all words (or equivalently: necklaces) in [n].

Conjecture 3 . 8 (d 4 − 1 192 d 4 −
Corollary of Conjecture 3.4d 2 + 192d + 192 for d even, 10d 2 + 192d + 201 for d odd. p k w * , see for example [20, Proposition 4.4.1.1].The character of this representation is given by the difference of the characters of ℑ p k w and S k C N .The latter character can be computed from the character of C N , and hence we are left to compute the character of ℑ p k w .In the case X = uMPS(2, 2, d), we can compute the character of ℑ p k w using a minor modification of Algorithm 1, where in the second loop we replace the collection of polynomials {P [b] | weight(b) = w} with their k-fold products k j=1 P [b j ] | j weight(b j ) = w .

Theorem 4 . 1 (
Cayley-Hamilton).Let A ∈ C m×m be a m × m complex matrix and p A (λ) = det(λ Id m − A) its characteristic polynomial.Then p A (A) = 0 where p A (•) is the corresponding matrix-valued polynomial.

0 .Proposition 4 . 10 .
10...0 = Tr A 1 A d−1 Proof .If w = 0 then b = (0 . . .0) ∈ B 0 (2, d) is the only binary bracelet of weight zero, and if w = 1 then b = (10 . . .0) ∈ B 1 (2, d) is the only binary bracelet of weight 1. ■ We now turn to the case w = 2. Then Conjecture 3.7 states that dim W w = d 2 .But d 2 is exactly the number B 2 (2, d) of generators T b of W w ; hence we need to show that they are linearly independent.The polynomials {T b : b ∈ B 2 (2, d)} are linearly independent.

Lemma 4 . 11 . 2 and
The cardinality of B3 equals d − 3.Proof .The cardinality of B3 is the sum of the number of binary bracelets of weight 3 containing 11 and the number of binary bracelets of weight 3 containing 101 but not 11, that are d−2In order to prove the case w = 3 we need to show that T b : b ∈ B3 is a basis of W 3 .We first show linear independence: Lemma 4.12.The polynomials T b : b ∈ B3 are linearly independent.

. 2 × d− 2 2 2 + d−4 2 = 3 : 4 . 13 .
By elementary row operations, we can reduce the left upper part to a diagonal matrix of order d−1 2 .The left lower part is filled with zeros.The (rectangular) right lower block of dimension d−4 can be put in the following upper triangular forms, for d even and odd respectively, for d odd.Both the obtained blocks have rank d−42 .We have that the rank of S is d−1 d − 3, and this concludes the proof.■We finish our proof by showing thatT b : b ∈ B3 spans W Lemma Every polynomial T 10 a 10 b 10 c = Tr A 1 A a 0 A 1 A b 0 A 1 A c 0 , with 1 < a ≤ b ≤ c, a + b + c = d −3 is an element of the linear span T b : b ∈ B3 .Proof .Notice that the elements of B 3 (2, n) \ B3 can be written without loss of generality in the form 10 a 10 b 10 c , with 1 < a ≤ b ≤ c.We use induction on a.If a = 0 and a = 1 then 10 a 10 b 10 c ∈ B3 and we are done.If we substitute A 1 → A 1 A a−1 0 , A 2 → A 1 A b 0 and A 3 → A 1 in the equation given by Theorem 4.5

T 10 a
10 b 10 c = T 10 b 1010 a+c−1 + T 110 a 10 b+c − T 10 b+1 110 a+c−1 − T 1010 a−1 10 b+c + T 10 a−1 10 b+1 10 c .All terms in the parenthesis have as subscript an elements of B3 , and the last term is in T b : b ∈ B3 by the induction hypothesis.This concludes the proof.■