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Defect of a Kronecker product of unitary matrices

The generalized defect D(U) of a unitary NxN matrix U with no zero entries is the dimension of the real space of directions, moving into which from U we do not disturb the moduli |U_ij| as well as the Gram matrix U'*U in the first order. Then the defect d(U) is equal to D(U) - (2N-1), that is the generalized defect diminished by the dimension of the manifold {Dr*U*Dc : Dr,Dc unitary diagonal}. Calculation of d(U) involves calculating the dimension of the space in R^(N^2) spanned by a certain set of vectors associated with U. We split this space into a direct sum, assuming that U is a Kronecker product of unitary matrices, thus making it easier to perform calculations numerically. Basing on this, we give a lower bound on D(U) (equivalently d(U)), supposing it is achieved for most unitaries with a fixed Kronecker product structure. Also supermultiplicativity of D(U) with respect to Kronecker subproducts of U is shown.


Introduction
The paper is a sequel to [1] where the defect of a unitary matrix was introduced and investigated in a general case.
Let U ∈ U, where by U we shall denote the set of all unitaries N × N. Consider the problem: find all directions (complex matrices) guaranteeing no first order change of |U i,j | for all i, j when moving from U in such a direction, and at the same time no first order disturbance of unitarity of U. The first requirement implies that these directions are of the form iR • U where • is the Hadamard product and R is a real matrix. The second means that the directions in question must belong to the space tangent, at U, to the N 2 dimensional manifold U. This tangent space is the set {EU : E antihermitian} or, if one likes, {UF : F antihermitian}.
Note that the set of directions in question forms a real linear space and it is overparameterized by R in the case of U having zeros among its entries. If U i,j = 0 then the equality i(R + αe i (e j ) T ) • U = iR • U, with e k being the standard basis column vectors, exposes the fact that two different R's can be used to parameterize any single direction. Having this in mind we thus concentrate on the space of R ′ s parameterizing the space of directions in question: {R : iR • U = EU for some antihermitian E} (1) and ask about its dimension. We define the defect of U as this dimension diminished by 2N − 1.
If U has no zero entries, the defect plus 2N − 1 is precisely the dimension of the space of directions in question. The reduction by 2N −1 is motivated by applications of the notion of defect. For most unitaries, especially those with no zero entries, the 2N directions: Other directions in question may lead to unitaries inequivalent to U, but with the same pattern of moduli, provided they exist in a neighbourhood of U. A special case of this property is the existence of a smooth family of inequivalent complex Hadamard matrices stemming from a given complex Hadamard matrix H. Here by a complex Hadamard matrix we understand, unlike in combinatorics community, a unitary with all moduli equal to 1/ √ N , although matrices rescaled to have unimodular entries are more often considered in this context. As far as such complex Hadamard matrices are concerned, they have many applications especially in quantum information theory. We recommend the reader to consult the online catalogue of complex Hadamard matrices available at [3].
The smallest matrix size for which the known collection of inequivalent complex Hadamards is presumably incomplete is 6. Current investigations are concentrated on the search for a 4-dimensional smooth family of inequivalent complex Hadamards. This search is partly motivated by the fact that the value of the defect calculated for most known 6×6 complex Hadamards, including the unitary Fourier matrix F 6 , is equal to 4. This is because the defect of a complex Hadamard H provides an upper bound for the dimension of a smooth family of inequivalent complex Hadamards stemming from H. We suggest the reader to throw a look into our previous work [1] to find more details about this application of the defect. There he can also find that the defect of unitary U (complex Hadamard H) being equal to zero implies that there are no unitaries (complex Hadamards) inequivalent to U (H) in its neighbourhood. At this point it is worth to recall that analogous (see [1]) result was obtained by Remus Nicoara in his work [4], under the name of 'span condition', but in a different context of commuting squares of algebras, in which complex Hadamard matrices are used to define commuting squares of matrix algebras.
This above mentioned value 4, as we now understand, causes one to suppose that all these 6 ×6 complex Hadamards (or complex Hadamards equivalent to these) with such a defect are connected by a single 4-dimensional family. The old 2-dimensional family stemming from the Fourier F 6 was reported, (among other even earlier sources [5,6]) already in our catalogue [7,3]. Other 1-dimensional families were constructed in [8,9], 2-dimensional in [10,11], 3 dimensional in [12], and the first attempt to build a 4-dimensional family of 6 × 6 inequivalent Hadamards has to be attributed to Ferenc Szöllősi due to his succesful work [13], inspired also by some numerical evidence of the existence of such a family contained in [14].
From the above collection of results we conclude that it is of value to study the notion of defect, to search for effective ways to calculate it or estimate it for unitary matrices of a large size.
In the following we employ another, more expedient characterization of the defect. To this end let us start with rewriting the definition (1), where * denotes the Hermitian conjugate.
This further leads to has only its i-th and j-th rows potentially filled with nonzeros. Here A U . Let us convert (5) into purely real conditions: The form trA(B) T is a Hilbert-Schmidt type inner product on the space of real N × N matrices, hence (7) amounts to orthogonality conditions in this space. Consequently, and in accordance with our initial definition the defect of U can be calculated as where () ⊥ is used to indicate the orthogonal complement of the space Å U spanned by where, The properties of orthogonal complements allow now for the following alternative definition of the defect: It was shown by Karabegov [2] that the set of those U for which the defect is greater than 0 is of the Haar measure zero within the set of all unitaries N × N. In this article we give a lower bound on the defect of a Kronecker product of unitary matrices: where the unitaries are of size n 1 × n 1 , n 2 × n 2 , ..., n r × n r , respectively. The bound computed according to our formulas agrees with the defect obtained numerically for Kronecker products of random unitary matrices drawn according to the Haar measure on the unitary group.
However, we do not have here a result concerning measures, analogous to that of Karabegov. Namely, that for unitary matrices U (k) lying beyond a certain set of the Haar measure zero, within the set of all unitary n k × n k matrices, the defect of the resulting Kronecker product (12) is equal to our bound.
Note that the above Kronecker product structure forms a key tool in studying quantum composite systems. For instance, any local unitary operation acting on a system composed of r particles has the form (12). Thus our results on the defect describe algebraic properties of generic local unitary operations, which are also called quantum gates.

Preparation
Let us introduce vector indices into Kronecker product (12), where and In the following, if not stated otherwise, i, j will correspond to the vector indices Let us try for a while to understand vector indices better. The relation between ordinary and vector indices could be defined in a different, but completely equivalent way. For two distinct elements in {1..n 1 } × . . . × {1..n r } let us introduce a relation: Then the relation between ordinary indices and vector indices we choose to be the only bijection φ from {1, . . . , r l=1 n l } into {1..n 1 } × . . . × {1..n r } (for which we write φ(i) = [i 1 ..i r ]), which satisfies: i < j =⇒ φ(i) < φ(j).
Later in this article we talk about subrows and submatrices, built from elementary objects (entries of a row for a subrow, rows or columns of a matrix for a submatrix) indexed by those i, whose k-th subindex i k is fixed at some value: corresponding to i. We can introduce the relation < for the reduced vector indices, and then we note that: (18) Ordering the reduced vector indices induces, again, their relation with reduced ordinary indices i ′ , j ′ ∈ {1, . . . , r l=1,l =k n l } uniquely, that is it is the same as if we used appropriate versions of formulas (14), (15).
Let us analyze, as an example, the case of a submatrix A ′ of matrix A of size N × N, N = r l=1 n l . Let A ′ be built from rows of A indexed by i ∈ A, with the order of rows preserved as is the habit when taking submatrices. Because for i, j indexing rows of A used to build A ′ and for i ′ , j ′ corresponding to reduced vector indices for i and j there holds:  (12 ) are of size at least 2. This we do for convenience of thinking, though we do not exclude the possibility that some or even all of the below arguments remain correct without this assumption. We will address this issue later after announcing the main results at the end of this and at the beginning of the next section.
The corresponding subspaces will be denoted according to the following Note that in both of the above definitions the notation refers to the assumed Kronecker product structure of U. Later we will meet U deprived of one or more of its Kronecker factors. Accordingly, the resulting U ′ will be assumed to be built from a fewer number of factors, and this will be reflected in the notation used for subsets of M U ′ and subspaces of Å U ′ . Any vector in Å U (see (9)) can be written as a linear combination of components each of which belongs to one of the above defined subspaces, hence Å U is the algebraic sum of all of them, Our aim is to show that (24) is in fact a direct sum, which will allow to estimate its dimension. In the meantime a little preparation is needed.
Using notation similar to that of MATLAB, let, with y at the l-th position, be a subrow of the [i 1 , .., i r ]-th row of matrix X of the size identical with the size of the Kronecker product (12). The subrow is composed of entries of [j 1 , .., j r ]-th columns for which j l = y. The horizontal order of entries is preserved.
., j r ] be vector indices of the Kronecker product (12) corresponding to ordinary indices i,j. where c k is at the k-th position.
for any [b 1 , .., b r ]-th row such that b k = i k = j k , c k as above. On the right hand sides of (28) and (29) stand [b 1 , .., b k−1 , b k+1 , .., b r ]-th rows of matrices A constructed from the Kronecker product (12) deprived of its k-th factor: Ordinary indices i ′ , j ′ correspond to the accordingly reduced vector indices Proof a) is of course true for any row of A indexed by a number different from i, j, which is in fact a zero row. As the i-th and j-th rows of A so its real and imaginary parts satisfy (26) and (27) respectively.
In this situation both sides of (28), (29) are zero rows.
The properties stated in the above lemma extend to linear combinations of matrices A

Proof
The part a) of the lemma is of course true for all rows indexed by .,br],: .,br],: , and analogously for the c ks -th subrow designated by the value of the k s -th index, where we ommit the ranges for α (i,j) and σ (i,j) :  .
To prove the part b) we use again (35) and rewrite it for the subrows corresponding to k ′ / ∈ {k 1 ..k p }: Here i, j are such that i k ′ = j k ′ . We may thus use Lemma 2.3 a) to find that summation of subrows (39) produces a zero vector: Related is the following result which we will also need to prove the theorem on a direct sum that comes next.
. It can be chosen in such a way that ., j r ] of the reduced Kronecker product U ′ (30), i.e., are given by the appropriate versions of (14) and (15).
We will be more precise about the above mentioned relation later, but at first let us state it without too many details. Any A contains, as its only nonzero and disjoint submatrices, n k copies of the corresponding A where each copy is multiplied by one of n k multipliers. Column positions of the copies do not depend on i, j. The multipliers, of which at least one is nonzero, depend only on the column positions of the copies they act on. The same applies to S , with the pattern of submatrices and values of the multipliers identical with that for A For a more detailed explanation let us consider the i-th row of A which is equal to where U (k) can also be written as . As a result we get .
The equality of the far left hand side of (41) and the far right hand side of (43) is an entry by entry equality between two rows. Now, if we change the sign of Re in the expressions in (41), (42) and (43) we will obtain a relation between the d k -th subrow of the j-th row in A If we used Im instead of Re, we would get a relation between the d k -th subrow of the i-th or j-th row of S (i,j) U and, correspondingly, the i ′ -th . This is summarized below: One of the n k numbers µ k is nonzero since U (k) is unitary.
where Since all µ k add up to 1 , i.e. the norm of the v-th row of U (k) , we can choose an isomorphism Ψ between M U k,v and Å U ′ to be defined as (again assuming that v stands at k-th position): .., i kp = j kp = v p , and finally i l = j l for l / ∈ {k 1 , .., k p }. Note that according to (48) the isomorphism Ψ maps these matrices into all matrices spanning The only other pair of equal subindices, namely i kq = j kq = v, is absent from the reduced vector indices corresponding to i ′ , j ′ . At other positions we get any possible pair of different values as we take any pair i, j allowed here, that is indexing A Our most important result in this section is the following The algebraic sum of subspaces in (24) is a direct sum.

Proof
We will prove the following statement equivalent to the above theorem.
.,vp) for all possible choices of (k 1 , .., k p ) and (v 1 , .., v p ) be matrices whose sum is equal to a zero matrix. Then all these matrices (we will call them X-matrices) are zero matrices.
The proof will be by induction with respect to r, the number of factors in Kronecker product (12).
If we have only one factor, Assume now that the theorem is true for any (r − 1) factor Kronecker product (12). In particular that what we stated above as eguivalent to our theorem is true for such a product.
Let our X-matrices add up to a zero matrix and let one of the summands, X (k 1 ,..,kp),(v 1 ,..,vp) be nonzero. Let −X (k 1 ,..,kp),(v 1 ,..,vp) be simply denoted by X. Thus X is the sum of the remaining X-matrices. We also need to introduce k ′ ., r}. The remaining X-matrices, forming X in a sum, can be split into two groups: • Into one group we put these belonging to subspaces (Å U ) (l 1 ,..,lq),(w 1 ,..,wq) such that ., k p } is nonempty. The other situation we consider at the end of the proof.) Thanks to that choice we can apply Lemma 2.4 b) to all of them, as well as to their sum, which will be denoted by Y : • Into the second group we put all the X-matrices which belong to (Å U ) (l 1 ,..,lq),(w 1 ,..,wq) such that k ′ 1 ∈ {l 1 , .., l q }, that is l s = k ′ 1 for some s ∈ {1, .., q}, and then w s = v ∈ {1, .., n k ′ 1 }. The group can further be split into subgroups corresponding to different values of v. When we add all the matrices in the whole group to form their sum Since, as we have just said, matrices forming Z v in a sum are responsible for formation of only those b-th rows for which b k ′ 1 = v, we can write for the sum Z of all Z v :  At this moment recall that, using our symbols, X = Y + Z. Also note that if one had to, one would classify X as belonging to the first group since X ∈ (Å U ) (k 1 ,..,kp),(v 1 ,..,vp) and k ′ 1 / ∈ {k 1 , .., k p }. So X satisfies (49) with Y replaced by X and analogous property holds for Z through the linear combination Z = X − Y . The last statement implies that the sums on the left hand side of (51) are zero rows, which in effect gives Z = 0. Further, let us return to component Z v of Z. As Z is a zero matrix and Z v have nonzero rows in disjoint regions, all Z v must be zero matrices.
If one of these , call it X ′ , is nonzero and belongs to (Å U ) ( kg 1 ,..,kg t ) ,(w 1 ,..,wt) with t < p, that is {k g 1 , .., k gt } = {k 1 , .., k p }, it can be expressed with the use of the remaining ones and X. We can repeat the whole reasoning with X ′ playing the role of X and k ′ 1 being replaced by some k ′ ∈ {1, .., r} \ {k g 1 , .., k gt } and simultaneously k ′ ∈ {k 1 , .., k p }. We then classify summand X as belonging to the second group of matrices, since X ∈ (Å U ) (k 1 ,..,kp),(v 1 ,..,vp) and k ′ ∈ {k 1 , .., k p }. We show as above that this second group of matrices consists of zero matrices which contradicts our assumption that X = 0.
Note that at this stage (Å U ) (),() could appear as (Å U ) (kg 1 ,..,kg t ),(w1,..,wt) . The short version of our reasoning for X (the role of X ′ at this stage) belonging to (Å U ) (),() is provided at the end of the proof. Also, precise repetition of our reasoning is always possible because no components in matrix sums are missing -even if we write that a matrix is formed only by some other matrices this means that the remaining ones are there as zero matrices.
Having X ∈ (Å U ) (k 1 ,..,kp),(v 1 ,..,vp) on the left, on the right we can only have nonzero components belonging to (Å U ) (k 1 ,..,kp),(w 1 ,..,wp) with (w 1 , .., w p ) = (v 1 , .., v p ). Obviously (w 1 , .., w p ) are also distinct, since any component (Å U ) (k 1 ,..,kp),(w 1 ,..,wp) in what we aim to prove to be a direct sum, occurs only once. Collecting both sides together we get that the sum of X-matrices (−X is one of them) belonging to (Å U ) (k 1 ,..,kp),(w 1 ,..,wp) , for sequence (k 1 , .., k p ) fixed by the choice of X and for all possible sequences of values (w 1 , .., w p ), is a zero matrix. Since regions where these last candidates for nonzero X-matrices have nonzero rows do not overlap, also they must be zero matrices. This is exactly what we wanted to prove, hence indeed (24) is indeed the direct sum. Now, as promised twice in this proof, we consider the case of X ∈ (Å U ) (),() , that is the case of {k 1 , ..k p } being empty. Splitting the remaining summands into two groups, as previously, for any k ′ ∈ {k ′ 1 , .., k ′ r } = {1, .., r}, we conclude that any X-matrix belonging to (Å U ) (l 1 ,..,lq),(w 1 ,..,wq) such that k ′ ∈ {l 1 , .., l q } is a zero matrix. Taking all k ′ under consideration we find that all X-matrices forming X must be zero matrices. So X must be a zero matrix too.
Let us rewrite (24) in a direct sum fashion: Calculation of the defect of U requires calculation of the dimension of the Å U , which luckily can be decomposed into the dimensions of the direct sum components. We will make this decomposition more informative in a moment.

Estimating the defect
Calculation of all the dimensions of the direct sum components in (52) can be reduced to calculation of dimensions of subspaces like (Å U ) (),() considered for subproducts of the Kronecker product U in (12).
Lemma 3.1 There holds: where is the original Kronecker product (12) deprived of its k 1 -th, ..., k p -th factors.

Proof
In the proof we use Lemma 2.5 repeatedly, assigning appropriate values to U, k and v.
Thus both subspaces mentioned in the lemma are isomorphic through the composition of the above listed isomorphisms, so they are of equal dimension.
Let us simplify our notation by introducing, for U being the product (12): Using the fact that Å U is the direct sum (52)  At this point note also, that for a (r − 1)-element set {k 1 , ..., k r−1 } = {1..r} \ {s} there holds d U {k 1 ,..,k r−1 } = dim (Å U (s) ) (),() = dim (Å U (s) ). We do not assume any Kronecker product structure for U (s) , that is vector indices are identical with ordinary indices here. Using such vector indices we cannot fix their only component, (M U (s) ) (1),(v) would be empty.
Theorem 3.2 Let U be a Kronecker product of unitary matrices (12). Then the dimension of the space Å U constructed for U according to (9) reads: Maybe it is a good moment to wonder about applicability of our formulas when there are 1 × 1 Kronecker factors in product (12) forming U. This has been promised in the paragraph preceding Definition 2.1.
First, recalling this definition, consider subset (M U ) (k 1 ,k 2 ,..,kp),(v 1 ,v 2 ,..,vp) of set M U . If k ∈ {1, .., r} \ {k 1 , .., k p } and n k = 1, this subset is empty. The reason is that you cannot find any A belonging to it, that is indexed by i, j satisfying i k = j k , because i k , j k ∈ {1, .., n k } = {1} . In this case let the corresponding subspace (Å U ) (k 1 ,k 2 ,..,kp),(v 1 ,v 2 ,..,vp) of space Å U be defined as the zero space {0}, which is a space of dimension 0. In particular, when there are 1 × 1 factors in product (12) do not change with multiplication of U by a unimodular number. The answer to our question is thus: yes, we can use (57) in presence of 1 × 1 factors, provided that appropriate d U {k 1 ,..,kp} are assigned 0's. This has some practical importance. For example, if we wanted to calculate defects for a large number of Kronecker products of at most r factors, we could store the sequences of sizes as rows of a matrix with r columns, filling rows with a certain number of 1's where the number of factors would be less than r. These rows would then be the input data for a procedure calculating the defect of an r factor Kronecker product of unitaries.
We can go even furher. Let us also, in this context, return to the direct sum formula (52). We have adopted above a convention, according to which (Å U ) (k 1 ,k 2 ,..,kp),(v 1 ,v 2 ,..,vp) is the zero space if there are 1's in sequence (n 1 , .., n r ) after removing from it the k 1th, ..., k p -th entries. Let us throw out these zero spaces from (52). We are left with (Å U ) (k 1 ,k 2 ,..,kp),(v 1 ,v 2 ,..,vp) where all k associated with n k = 1 sit in {k 1 , .., k p }. But every such (Å U ) (k 1 ,k 2 ,..,kp),(v 1 ,v 2 ,..,vp) can be replaced by an equal space where: • U ′ is U deprived of its 1 × 1 Kronecker factors, with the order of the remaining factors preserved, • (k ′ 1 , k ′ 2 , .., k ′ p ′ ) contains the positions, of subsequent entries of the subsequence obtained from (k 1 , .., k p ) by throwing out all k's such that n k = 1, in the subsequence obtained from (1, 2, .., r) in the same way (This is cumbersome, being a consequence of our notation, and we met such formulation in Lemma 2.5.), is the result of throwing out the entries (of the only possible value 1) out of (v 1 , v 2 , .., v p ) corresponding to entries k of (k 1 , .., k p ) for which n k = 1.
(Both spaces are equal because they are spanned by the same matrices. In general, A , where i ′ , j ′ correspond to the reduced vector indices obtained from the vector indices corresponding to i, j, respectively, by removing those k'th entries of these vector indices (entries necessarily equal to 1), for which k satisfies n k = 1.) After this truncation and replacement (52) looks as if it were written for U ′ , in which case we accept the validity of the direct sum formula because U ′ does not contain 1 × 1 Kronecker factors. Since Å U = Å U ′ , the only difference between U and U ′ being a unimodular factor, the starting point direct sum with all those zero spaces appears to lead to the correct direct sum we would write if we first removed the 1 × 1 Kronecker factors. So, we can go either way, removing or leaving these factors.
It is interesting to give the lower bound for the defect of U being a Kronecker product, because, as it was pointed in the Introduction, for most unitaries the defect is zero. The lower bound we give here is associated with upper bounds on the dimensions d U {k 1 ,..,kp} of (Å U ′ ) (),() for the Kronecker subproducts U ′ as defined in (55), all of which are in fact bounds on d U {} with U replaced by an appropriate U ′ . All we need to know about these bounds is contained in the next lemma. Note that although we assume that the sizes of Kronecker factors are greater than 1, any factor of size 1 -a unimodular number -can be absorbed into one of the factors of size greater than 1. Then the dimension of the space (Å U ) (),() constructed for U (Definition 2.2) is bounded in one of the following ways: • If n k > 2 for any k ∈ {1..r}, then • If n k 1 = . . . = n kp = 2 for some distinct values k s ∈ {1..r}, and for k / ∈ {k 1 ..k p } there holds n k > 2, then Next we show that the b-th row of B filled using the above algorithm, starting from those initially known (n 1 −1)·. . .·(n r −1) entries, satisfies (60) for any k ∈ {1..r}. That is, that the following equality holds for any k ∈ {1..r} and c s ∈ {1..n s } for any s ∈ {1..r} \ {k}: Let in the above expression c k 1 = n k 1 , ..., c kp = n kp and c s = n s for s ∈ {1..r} \ {k 1 , .., k p } with k 1 < k 2 < . . . < k p . By this we mean that n k = c k = c kq = n kq for some q ∈ {1..p}.
(Do not mistake k 1 , ..., k p for those from the second item of the lemma.) Then the left hand side of (61), where some of the entries are obtained using the algorithm from the dotted list above, reads: and equals thus to the right hand side of (61) In this way each row of B ∈ (Å U ) (),() could be potentially parametrized by no more than (n 1 − 1) · . . . · (n r − 1) parameters -if there were no other restrictions caused by the structure of U (s) -but we need to parameterize only N −1 rows in general. This is caused by the property that all column sums in B are zeros, just as is the case with the spanning matrices A (i,j) U . Hence, in general, we have no more than (N −1)·(n 1 −1)·. . .·(n r −1) free parameters to determine B. In other words: all B ∈ (Å U ) (),() belong to a single space of dimension (N − 1)(n 1 − 1) · . . . · (n r − 1).
The case when some n k are equal to 2 is a separate one, but what really counts is when there are more than one n k equal to 2.
Observe now that the rows of B from this group are linear combinations of respective nonzero rows of only those A (i,j) (from the collection of those spanning (Å U ) (),() ) which are indexed by i, j satisfying: Since subsets of (i, j)'s (of the set of all (i, j)'s indexing A Thus, through a linear combination, in the considered group of rows in B one of the rows depends on others. We have 2 p /2 = 2 p−1 possible choices of [s 1 , .., s p ], hence 2 p−1 groups (of the specified type) of rows in B with 2 p−1 dependent rows altogether. Using the previous method of parameterizing rows, we can parameterize no more than (N − 2 p−1 ) rows, introducing no more than (n 1 − 1) · . . . · (n r − 1) parameters into each row. In other words: B being contained in the described above (N − 2 p−1 )(n 1 − 1) · . . . · (n r − 1) dimensional space is a necessary condition for B being a member of (Å U ) (),() .
• Å U is contained in the space tangent to the set of all N × N doubly stochastic matrices, that is the space of all real N × N matrices in which entries in any row or column add up to 1. This space is of dimension (N − 1) 2 .
In what follows Theorem 3.2 there are arguments behind defining (Å U ) (),() as the zero space if there are 1 × 1 Kronecker factors in (12) not absorbed into larger factors, and that d U {} has to be assigned 0 then. In this case we can also use the bound (58) or (59) which produces the correct dimension 0.
Having the above result of Lemma 3.3 let us replace the values d U {k 1 ,..,kp} in Theorem 3.2 by their upper bounds being the appropriate right hand sides of (58) and (59). We will get an upper bound on dim (Å U ), equivalently a lower bound on the defect d( What is more, this can be done successfully also in the presence of unabsorbed 1 × 1 Kronecker factors. Since the bounds (58), (59) yield 0 for d U {k 1 ,..,kp} corresponding to zero spaces in the direct sum (52), whose presence is associated with the 1×1 factors, bounding the dimensions of these spaces does not affect the total upper bound on dim (Å U ) at all, that is we obtain the value we would get if the 1 ×1 factors were first absorbed into larger factors. This can be better understood if one carefully follows the reasoning following Theorem 3.2. The practical consequence is this: if one had some other bounds on d U {k 1 ,..,kp} , yielding zeros for spaces that are necessarily zero spaces because of 1 × 1 factors, one could leave the 1 × 1 factors unabsorbed.
To better expose properties of our bound, it is convenient to introduce another notion, the generalized defect D(U) of a unitary U, defined as When we recall expressions in (8) and the preceding formulas it will be clear that D(U) is the dimension of the space (1), that is the space: The corresponding lower bound on D(U) will be, thanks to Theorem 3.2 and Lemma 3.3 and definitions (55), (56). : where for a nonempty subset A = {l 1 , ..., l s } of {1, ..., r} expression ♯ 2 A denotes • the number of 2's in sequence (n l 1 , ..., n ls ), if there are any, • 1 otherwise.
Although the above expression looks very complicated, below we show that it is, almost, a mere product of trivial functions of the sizes n 1 , ..., n r of factors of the considered Kronecker product.
Because in Lemma 3.3 it was assumed that all Kronecker factors are of size greater than 1, formally (67) is a valid lower bound only in this situation, therefore we confine ourselves in the formulation of the theorem below. But, in the paragraph preceding the one containing the definition (65) of the generalized defect, we argued that even in the presence of 1 × 1 Kronecker factors the total bound (67), obtained by substituting (58, 59) into (57), will be equal to the value (67) obtained for the Kronecker product with absorbed 1 × 1 factors (this can be also guessed purely by analyzing (67) as a function of sequence of sizes (n 1 , .., n r )). Let us leave it as a comment that (67) must, in the presence of 1 × 1 factors, lead to (68) or (69 ) deprived of its factors (2n l − 1) corresponding to n l = 1, but since then 2n l − 1 = 1, in Theorem 3.4 one does not need to assume that all n k > 1.
Theorem 3.4 The lower bound (67) on the generalized defect (definition (65)) of U being the Kronecker product (12) with factors U (1) , ..., U (r) of size n 1 × n 1 , ..., n r × n r respectively, where all n k > 1, is equal to if there is at least one 2 in the sequence (n 1 , ..., n r ), where ♯ 2 {1..r} is the number of 2's in this sequence.

Proof
We need to show that there holds an equality between two values, one being a relatively simple polynomial expression: if n k = 2 for some k ∈ {1, ..., r}. and the other being the long expression in the most outer bracket in (67). This last expression will be simply referred to as (67)'. We will compare the above mentioned two quantities component by component.
We start from the p-th component of (67)', by which we mean: (72) Step 1 The, call it, left part being the result of taking only the left product from the inner bracket in (72) is equal to: The above sum can also be found in the expansion of the left product of (70) or (71). Any component of the above sum is the result of choosing, in the right subproduct of the left product of (70, 71), p brackets at positions k 1 , ..., k p from which 1's will be taken to be multiplied with (n l − 1)'s from the remaining r − p brackets. Thus every left part (73) for p = 1..(r − 1) has its counterpart in the left product of (70, 71).
On the other hand, in the expansion of this product there are two components, namely (n 1 · . . . · n r )(n 1 − 1) · . . . · (n r − 1) and (n 1 · . . . · n r ) so far unexplained. We easily see that the first one is the left part of the 0-th component of (67)': The second one still have to be found.
if there are no 2's in (n l 1 , ..., n l r−p ) and there are some in (n k 1 , .., n kp ), where {l 1 , .., l r−p } is as above. Note that in this situation the additional 2 at the end is caused by the fact that ♯ 2 {l 1 , .., l r−p } = 1 due to the definition of ♯ 2 .
So, in the sum (75) we add expressions like (76) or (77) depending on how 2's are scattered among the considered subsequences of (n 1 , ..., n r ). And we do this over all p = 1...(r − 1) in the total sum in (67)'.
Next we will add only those summands in the sum of all right parts (75) over p = 1...(r − 1), which are associated with sequences (k 1 , ..., k p ) containing a given fixed subsequence (k q 1 , ..., k qt ) = (s 1 , ..., s t ) such that n s 1 , ..., n st are all greater than 2, and which (these (k 1 , .., k p )'s) are such that for any k j not belonging to this subsequence n k j = 2. The sizes n s 1 , ..., n st play the role n kq of (76) and (77). All sequences are assumed to be increasing ones.

1
, that is the number of ways in which we can extend (k q 1 , ..., k qt ) to (k 1 , ..., k p ) by such k that n k = 2. For a summand the formula (76) is used if there are at least two 2's in (n 1 , .., n r ), that is if there is at least one 2 outside (n k 1 , ..., n kp ). Otherwise formula (77) is used for a summand. • If there are no 2's in the sequence (n 1 , .., n r ) In other words, outside the subsequence (n s 1 , ..., n st ) there are only 2's in the sequence (n 1 , ..., n r ). Every summand associated with (k 1 , ..., k p ) produced out of such (s 1 , ..., s t ) is expressed with the use of (76). Note that in the case of only one 2 in (n 1 , ..., n r ) the sum of binomials containes only 1 0 . The sum (83) takes the final form: The reader should realize, that the Case 2 occures only once, when (s 1 , ..., s t ) is of maximal length. Then we get an additional component, the second part of (84). In general we add expressions like (82), over all possible nonempty sequences (s 1 , ..., s t ), that is over all t ranging from 1 to r − (♯ 2 {1..r}). So let us first concentrate on them. As in the case without 2's in (n 1 , ..., n r ) we note that any expression like (82) has its counterpart in the right product of (71). Again it is a matter of choosing brackets from which we take n l 's, at the positions corresponding to the positions of s 1 , ..., s t in the increasing subsequence (k : n k > 2) of (1, ..., r). Note that in this process we also deal with the situation when all n k > 2 are used in (82), in fact in the 'standard' part of (84) ((s 1 , ..., s t ) of maximum length, Case 2 above), which corresponds to taking only n l 's from brackets in (71). What remains in an expansion of (71) are the values: where the last product was mentioned as unexplained at the end of Step 1.
On the other side, in (67)', we have still left the right part of the 0-th component (74) of (67)', which can be written as: the additional component in (84) occuring only for (s 1 , ..., s t ) of maximum length, therefore written as and finally all summands in the left parts (75) for p = 1, ..., r − 1 which are associated with sequences (k 1 , ..., k p ) such that there are only 2's in (n k 1 , ..., n kp ). We add these summands in Step 3 which completes the proof.
Now we calculate how many summands of this type can be found in the sum of all the right parts (75). It is assumed that at least one 2 can be found in (n 1 , ..., n r ). The other case has been completely resolved in the paragraph following formula (80).
• For p = 1 the number of summands is equal to ♯ 2 {1..r} 1 , the number of ways we choose a one element subsequence, containing only 2's, from (n 1 , ..., n r ). We use (90) for a summand if there is only one 2 in (n 1 , ..., n r ), otherwise we use (89). We add all the considered summands below, where the negative sign is inherited from (75). In the first sum we still use the form of a summand originating from (75): This is the end. What is left for us to say is that: • (92) together with (87), both within (67)', compensate for (85) being part of (71).
Our calculation of the bound (68, 69) on the generalized defect D(U) of U given by (12) is based on the formula for the dimension of Å U given by Theorem 3.2, which in turn is based on the decomposition of Å U into direct sum components provided by Theorem 2.6. As far as the case with no 2's in the sequence of sizes (n 1 , ..., n r ) is concerned, we can use a simpler reasoning leading to the corresponding bound (68). To this end let us consider the following lemma, in which we use spaces defined in (66).
for some antihermitian matrices E, F . Then from which we get where (−iE ⊗ F ) is antihermitian because E ⊗ F is hermitian. So R ⊗ S ∈ Î U ⊗V .
The space Î U (k) constructed for a unitary U (k) of size n k × n k contains at least 2n k − 1 independent real matrices, denoted further by R and consequently, by Corollary 3.6, This is equal to (68), which is also valid if there is only one 2 in the sequence of sizes (n 1 , ..., n r ).
If there is more than one matrix of size 2 × 2 among U (k) , in (99) we get a lower bound which is smaller than the lower bound (69). This is caused by the fact that, for x denoting the number of 2's in (n 1 , ..., n r ), there holds an equality for natural x > 1: which is equivalent to The second thing we have achieved in this work is the calculation, based on the above result, of the lower bound on the generalized defect D(U) (defined in (65), being the defect plus 2N −1) when U is a Kronecker product. The generalized defect is supermultiplicative with respect to Kronecker subproducts of U (Corollary 3.6) and this allows us to trivially retrieve the lower bound of Theorem 3.4 when the number of 2 ×2 Kronecker factors does not exceed 1. In the other case Theorem 3.4 gives a better bound than that obtained using supermultiplicativity of D(U). We conjecture that, in either case, the lower bound is attained by most matrices with a fixed Kronecker product structure (fixed, up to an order, sequence of sizes of Kronecker factors).
All the formulas, the one expressing the direct sum forming Å U (see (52)), the resulting one expressing the dimension of Å U (see (57) The author believes that the first of the above mentioned achievements will open the way to easier calculation of the defect for large matrices U with a Kronecker product structure. Such calculations may be performed in order to asess whether a given U gives rise to a smooth family of inequivalent unitaries (i.e. not obtained one from another by multiplying rows and columns by unimodular numbers) with the moduli of entries fixed at the values sitting in [|U i,j |] i,j=1..N , and how large its dimension can be. This is a part of a question about unitary preimagies V to a doubly stochastic matrix |U i,j | 2 i,j=1..N , that is about V such that |V i,j | = |U i,j | for all i, j. This motivation is wider described in a prequel [1] to this paper. A special case, |U i,j | 2 i,j=1..N = [1/N] i,j=1..N , the search for complex Hadamard matrices, attracts even more interest, and it was dicussed in the Introduction.

Acknowledgements
The author would like to thank KarolŻyczkowski and Marek Kuś from Center for Theoretical Physics (Polish Academy of Sciences) in Warsaw, for numerous discussions on the subject and reading this article.
Financial support by the grant number N N202 090239 of Polish Ministry of Science and Higher Education is gratefully acknowledged.