Matrix multiplication is defined by the multiplication of rows of a left-hand matrix by (successive) columns of a right-hand matrix, where the left-hand rows are transposed (successively). So that there are already defined partial operations: \( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x & y \\ z & v \end{pmatrix} = \begin{pmatrix} ax+bz & ay+bv \\ cx+dz & cy +dv \end{pmatrix} = \begin{pmatrix} ax & ay \\ cx & cy \end{pmatrix}+ \begin{pmatrix} bz & bv \\ dz & dv \end{pmatrix} = \begin{pmatrix} ax & ay \\ 0 & 0 \end{pmatrix}+ \begin{pmatrix} 0 & 0 \\ cx & cy \end{pmatrix}+ \begin{pmatrix} bz & bv \\ 0 & 0 \end{pmatrix}+ \begin{pmatrix} 0 & 0 \\ dz & dv \end{pmatrix}\) \( = \begin{pmatrix} a & 0 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} x & y \\ z & v \end{pmatrix} + \begin{pmatrix} 0 & b \\ 0 & 0 \end{pmatrix} \begin{pmatrix} x & y \\ z & v \end{pmatrix} + \begin{pmatrix} 0 & 0 \\ c & 0 \end{pmatrix} \begin{pmatrix} x & y \\ z & v \end{pmatrix} + \begin{pmatrix} 0 & 0 \\ 0 & d \end{pmatrix} \begin{pmatrix} x & y \\ z & v \end{pmatrix} \) Let \( \rho_1 = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \rho_2 = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}, \overline {\rho_1} = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}, \overline {\rho_2} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \) Then \( \rho_1 + \rho_2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \overline {\rho_1} + \overline {\rho_2} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \) the last pair of matrices is isomorphic to \( S_2 \), the symmetric group on two letters. (Apologies, life intervenes, back later)
On the notation I'm using. \( \rho_i \) indicates a matrix with a 1 in the ith column; \( \overline {\rho_i} \) indicates a matrix with a 1 in the ith column, row-transposed relatively to its unbarred partner. The matrices are all 2x2, so \( i \in \{1,2\} \). Now I can do this: \( P = \{ \rho_1, \rho_2, \overline {\rho_1}, \overline {\rho_2} \} \) Then if \( i,j \in \{1,2\}; i \ne j \) \( \{ \rho_i, \overline {\rho_i}\}\{\rho_j, \overline {\rho_j} \} = P \times P \) And we find that \( \rho_i\rho_j = 0,\; \rho_i\overline {\rho_j} = \overline {\rho_j},\; \overline {\rho_i}\rho_j = \overline {\rho_i},\; \overline {\rho_i}\overline {\rho_j} = \rho_j \) But when \( i = j \), we get: \( \{ \rho_i, \overline {\rho_i}\}\{ \rho_i, \overline {\rho_i}\} = \{ \rho_i\rho_i, \; \rho_i \overline {\rho_i},\; \overline {\rho_i}\rho_i, \; \overline {\rho_i}\overline {\rho_i}\} = \{ \rho_i, \; 0. \; \overline {\rho_i}, \; 0 \} \) More succinctly, we have \( \{ \rho_i, \overline {\rho_i} \}\rho_i = \{ \rho_i, \overline {\rho_i} \} \), and \( \{ \rho_i, \overline {\rho_i} \}\overline {\rho_j} = \{ \overline {\rho_j}, \rho_j \} \) iff \( i \ne j \); otherwise multiplication gives the 0-matrix. So \( \rho_i \) acts like the identity on (elements of) P with the same index, when multiplying on the right. However \( \overline {\rho_j} \) permutes the indices, and changes the "sign" of elements when acting on the right.
Ok, now define the identity matrix recursively: \( I_1 = \begin{pmatrix} 1 \end{pmatrix},\; I_2 = \begin{pmatrix} I_1 & 0 \\ 0 & I_1 \end{pmatrix},\; I_4 = \begin{pmatrix} I_2 & 0 \\ 0 & I_2 \end{pmatrix},\; . . . \) And \( I_3 = \begin{pmatrix} I_2 & 0 \\ 0 & I_1 \end{pmatrix},\; I_5 = \begin{pmatrix} I_4 & 0 \\ 0 & I_1 \end{pmatrix},\; . . . \). The reason for separating odd from even (except for I_1) is the pattern you can see. Also define another matrix operation (on P): \( \rho_1 \otimes \rho_1 = \begin{pmatrix} \rho_1 & 0 \\ 0 & 0 \end{pmatrix}; \rho_1 \otimes \{\rho_j, \overline {\rho_j} \} =\biggl\{ \begin{pmatrix} \{\rho_j, \overline {\rho_j} \} & 0 \\ 0 & 0 \end{pmatrix}\biggr\}; i,j \in \{1,2\} \), and similarly for the rest of P. Then you also have that: \( I_4 = \begin{pmatrix} I_2 & 0 \\ 0 & I_2 \end{pmatrix} = \begin{pmatrix} \rho_1 & 0 \\ 0 & \rho_1 \end{pmatrix} + \begin{pmatrix} \rho_2 & 0 \\ 0 & \rho_2 \end{pmatrix} = \rho_1 \otimes (\rho_1 + \rho_2) + \rho_2 \otimes (\rho_1 + \rho_2) \). So you can have: \( \overline{\rho_1} \otimes (\rho_1 + \rho_2) + \overline{\rho_2} \otimes (\rho_1 + \rho_2) = \begin{pmatrix} 0 & I_2 \\ I_2 & 0 \end{pmatrix} \) and so on. In fact P with the "\( \otimes \)" operation defines all of \( S_4 \), the symmetric group with twice as many letters as \( S_2 \). The space has "expanded". Actually, I should really state that as a proposition: the whole of \( S_4 \), in its 4 x 4 matrix representation, can be decomposed, each element is a sum of two non-zero matrices, one with "diagonal" elements, one with "off diagonal" elements, as a 2 x 2 block matrix of elements from P. So the operation (matrix product or Kronecker product) also defines (induces) a blocking operation which takes 4 x 4 permutation matrices to 2 x 2 matrices composed of elements of P. All that is done in reverse, above.
I think where I'm going here is, if matrix multiplication corresponds to a sum of partial matrix operations, then if a matrix is a permutation matrix, there are sums of partial permutations. That doesn't sound like such a big deal, I guess.
More about the notation: I hope it's clear that \( \biggl\{ \begin{pmatrix} \{\rho_i, \overline {\rho_j} \} & 0 \\ 0 & 0 \end{pmatrix}\biggr\}; i,j \in \{1,2\} \) means the set of 4 x 4 matrices with an element of P in the \( \rho_1 \) "position". I guess I should really then have this as \( \biggl\{ \begin{pmatrix} \{\rho_i, \overline {\rho_j} \} & 0 \\ 0 & 0 \end{pmatrix}\Bigg | i,j \in \{1,2\} \biggr\} \), so I can write \( \biggl\{ \begin{pmatrix} P & 0 \\ 0 & 0 \end{pmatrix} \biggr\} \), to mean the same thing since it isn't confusing given a context. Then I can construct a set of partial permutation matrices with elements of P "on the diagonal": \( \rho_1 \otimes P \bigg( =\biggl\{ \begin{pmatrix} P & 0 \\ 0 & 0 \end{pmatrix}\biggr\}\bigg ) + \rho_2 \otimes P \bigg ( = \biggl\{ \begin{pmatrix} 0 & 0 \\ 0 & P \end{pmatrix}\biggr\} \bigg ) = \biggl\{ \begin{pmatrix} P & 0 \\ 0 & P \end{pmatrix}\biggr\}\). There are 16 of these, and so there are 16 more with "off-diagonal" elements of P. If you add the two sets together only a subset will be elements of \( S_4 \).
We also have that \( \rho_1^{ \intercal} = \rho_1;\;\rho_2^{ \intercal} = \rho_2;\; \overline {\rho_1}^{\intercal} = \overline {\rho_2};\; \overline {\rho_2}^{\intercal} = \overline {\rho_1} \). So for example: \( \begin{pmatrix} \rho_1 & \overline {\rho_1} \\ \overline {\rho_2} & \rho_2 \end{pmatrix}^{\intercal} = \begin{pmatrix} \rho_1^{\intercal} & \overline {\rho_1}^{\intercal} \\ \overline {\rho_2}^{\intercal} & \rho_2^{\intercal} \end{pmatrix} = \begin{pmatrix} \rho_1 & \overline {\rho_2} \\ \overline {\rho_1} & \rho_2 \end{pmatrix} \). But \( \begin{pmatrix} \overline{\rho_1} & \rho_1 \\ \rho_2 & \overline{\rho_2} \end{pmatrix}^{\intercal} \ne \begin{pmatrix} \overline{\rho_1}^{\intercal} & \rho_1^{\intercal} \\ \rho_2^{\intercal} & \overline{\rho_2}^{\intercal} \end{pmatrix} \).
And, finally \( ( \rho_1 + \rho_2) _{ij} = \delta_{ij}\; | i,j \in \{1,2\} \). Hence \( ( \rho_1 + \rho_2) \otimes ( \rho_1 + \rho_2) = (\rho_1\otimes \rho_1 + \rho_1\otimes \rho_2 + \rho_2 \otimes \rho_1 + \rho_2 \otimes \rho_2)_{ijkl} = \delta_{ij}\delta_{kl}| i,j,k,l \in \{1,2\} \) Edit. Notice I forgot to put indices on the product on the left of the = in the last eqn. Also notice these aren't "really" needed, all the nonzero products have indices already. Now a question: If I distinguish strands in a braid group from strands in a permutation group, which mathematically is the quotient of a braid group (so think of the strands in a permutation group as also being quotiented), is there a difference also between the smallest matrix representations for a crossing in either group? I think so.
Sorry, this is a mistake: \( \begin{pmatrix} \rho_1 & \overline {\rho_1} \\ \overline {\rho_2} & \rho_2 \end{pmatrix}^{\intercal} = \begin{pmatrix} \rho_1^{\intercal} & \overline {\rho_1}^{\intercal} \\ \overline {\rho_2}^{\intercal} & \rho_2^{\intercal} \end{pmatrix} = \begin{pmatrix} \rho_1 & \overline {\rho_2} \\ \overline {\rho_1} & \rho_2 \end{pmatrix} \). The number of matrices in \( S_4 \) which do transpose this way appear to be those composed only from the subset of P \( = \{ \overline {\rho_i}\} \). The matrix in the incorrect example is the same as Kauffman's R-matrix, sorry again.
The reason I'm doing this is to see if it flies (after posting it and staring at it for a bit). I'm not necessarily trying to prove anything except maybe to myself. Anyway, this theme of partial operations on matrices goes to quantum computing and the operations of partial trace and partial transpose on matrices (tensors) over Hilbert spaces (a qubit, recall, is mathematically a Riemann sphere). Also, the partial transpose (as a function) partitions permutation spaces like \( S_4 \) into at least two subsets, those elements with a partial transpose of 'block' sub-matrices, and those without one under closure (whether the result is an element of \( S_n \), i.e. a permutation matrix). Further, those elements with a partial transpose equal to the matrix transpose are another partition, I wasn't really considering that when I admitted my error in the last post. The error is with the first equality, corrected, it's like this: \( \begin{pmatrix} \rho_1 & \overline {\rho_1} \\ \overline {\rho_2} & \rho_2 \end{pmatrix}^{\intercal} \ne \begin{pmatrix} \rho_1^{\intercal} & \overline {\rho_1}^{\intercal} \\ \overline {\rho_2}^{\intercal} & \rho_2^{\intercal} \end{pmatrix} = \begin{pmatrix} \rho_1 & \overline {\rho_2} \\ \overline {\rho_1} & \rho_2 \end{pmatrix} \).
The problem is that \( \begin{pmatrix} \rho_1 & \overline {\rho_2} \\ \overline {\rho_1} & \rho_2 \end{pmatrix} \) isn't a permutation matrix; the indices of the P-elements are the same in each column. In fact given a partial matrix like \( \begin{pmatrix} \rho_1 & 0 \\ 0 & \rho_2 \end{pmatrix} \), the off diagonal elements are already determined (by closure), so there is only one choice for the second partial matrix in the sum, which is: \( \begin{pmatrix} 0 & \overline {\rho_1} \\ \overline {\rho_2} & 0 \end{pmatrix} \). The indices also determine this. There must be 16 permutation matrices with this form, the remaining 8 are sums of two partial matrices which have the same form (are both diagonal or both off diagonal).
Probably my notation and somewhat loose mathematics has a few people cringing, so I'll try to tidy up a bit. P is a set, but it isn't closed under matrix multiplication because some products are the 0-matrix. So along with the "rules" \( \{\rho_i, \overline{\rho_i}\}\rho_i = \{\rho_i, \overline{\rho_i}\} \), \(\{\rho_i, \overline{\rho_i}\}\overline{\rho_j} = \{\overline{\rho_j}, \rho_j\}\), there are: \( \{\rho_i, \overline{\rho_i}\} \overline{\rho_i} = \{\rho_i, \overline{\rho_i}\}\rho_j = \{0,0\} \). These are all the right cosets of P under composition (multiplication); P has no identity element so is a semigroup. (Actually that should be P ∪ {0}, a semigroup with five elements) There is a 1-1 map between partial matrices of the form: \( \begin{pmatrix} P & 0 \\ 0 & P \end{pmatrix} \) and \( \begin{pmatrix} 0 & P \\ P & 0 \end{pmatrix} \).Which means if you choose one of either form, the second choice is determined by the first if you want their sum to be a permutation matrix. A bit of investigation shows that the (partial) matrices: \( \begin{pmatrix} 0 & \rho_i \\ \rho_i & 0 \end{pmatrix}, \; \begin{pmatrix} 0 & \overline{\rho_i} \\ \overline{\rho_i} & 0 \end{pmatrix} \) are the only forms with a transpose equal to their partial transpose (I've left the braces out, but these are sets of matrices). They have off diagonal P-elements with the same sign and the same index. Hence there are four elements of \( S_4 \) with this symmetry. Everything seems to be partial here: I've just defined the partial transpose of a partial permutation matrix. Oops, you also get four more permutations with partial transpose symmetry, these are: \( \begin{pmatrix} 0 & \rho_i + \rho_j \\ \rho_i + \rho_j & 0 \end{pmatrix}, \; \begin{pmatrix} 0 & \overline{\rho_i} + \overline{\rho_j} \\ \overline{\rho_i} + \overline{\rho_j} & 0 \end{pmatrix}, \; \begin{pmatrix} \rho_i + \rho_j & 0 \\ 0 & \rho_i + \rho_j \end{pmatrix}, \; \begin{pmatrix} \overline{\rho_i} + \overline{\rho_j} & 0 \\ 0 & \overline{\rho_i} + \overline{\rho_j} \end{pmatrix}\; \bigg| i \ne j \), one of which is, of course, I.
Dammit, I must be too distracted lately, this is something obvious that I knew about yonks ago, which is if you choose or select a partial 4 x 4 matrix with either a diagonal or off diagonal form there are two other choices because there are two "unfilled" rows or columns in the first choice, so the map isn't 1-1. But there are 1-1 maps between subsets, depending on whether you want the final matrix to have entries with the same sign or different signs. And I left out that the rules listed apply only when \( i \ne j; i,j \in \{1,2\} \).
Heh, so what I said about the subsets \( \begin{pmatrix} P & 0 \\ 0 & P \end{pmatrix} \) and \( \begin{pmatrix} 0 & P \\ P & 0 \end{pmatrix} \) is correct,there is a 1-1 map between these subsets and you get a 16 element subset of \( S_4 \) when you add the corresponding partial matrices together. The rest of the group is indeed defined by injective maps from each subset to itself: \( \begin{pmatrix} 0 & \rho_i + \rho_j \\ \rho_i + \rho_j & 0 \end{pmatrix}, \; \begin{pmatrix} 0 & \overline{\rho_i} + \overline{\rho_j} \\ \overline{\rho_i} + \overline{\rho_j} & 0 \end{pmatrix}, \; \begin{pmatrix} \rho_i + \rho_j & 0 \\ 0 & \rho_i + \rho_j \end{pmatrix}, \; \begin{pmatrix} \overline{\rho_i} + \overline{\rho_j} & 0 \\ 0 & \overline{\rho_i} + \overline{\rho_j} \end{pmatrix}\; \bigg| i \ne j \). And \( \begin{pmatrix} 0 & \rho_i + \rho_j \\ \overline{\rho_i} + \overline{\rho_j} & 0 \end{pmatrix}, \; \begin{pmatrix} 0 & \overline{\rho_i} + \overline{\rho_j} \\ \rho_i + \rho_j & 0 \end{pmatrix}, \; \begin{pmatrix} \rho_i + \rho_j & 0 \\ 0 & \overline{\rho_i} + \overline{\rho_j} \end{pmatrix}, \; \begin{pmatrix} \overline{\rho_i} + \overline{\rho_j} & 0 \\ 0 & \rho_i + \rho_j \end{pmatrix}\; \bigg| i \ne j \).
The matrices \( \rho_1 + \rho_2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \overline {\rho_1} + \overline {\rho_2} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \) as mentioned, are the matrix representation of (the elements of) \( S_2 \). If we consider a two letter string like ab, we need to write this as a 2 x 1 matrix, a vector, so the elements of \( S_2 \) can act on it. Instead, let \( I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, X = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\), and define operations on the string ab like \( I(ab) = ab,\; X(ab) = ba \). Now suppose the string is really like: aa'bb'. If ab is like a pair of points with a descending strand from each point, aa'bb' is like a pair of bands, a and a' lie on the edges of the first band, and b and b' on the edges of the second band. Now we have: \( \Bigg\{ \begin{pmatrix} 0 & \rho_i + \rho_j \\ \rho_i + \rho_j & 0 \end{pmatrix}, \; \begin{pmatrix} \rho_i + \rho_j & 0 \\ 0 & \rho_i + \rho_j \end{pmatrix}, \; \bigg| i \ne j \Bigg\}\) acting on aa'bb'. If we ignore where the "sign" is in each of aa', bb', that is we make a'a = aa', and b'b = bb', we get the rest of the eight element subgroup ( the last eight matrices in post #13) isomorphic to elements in \( S_2 \). If we don't ignore the sign's position in the string, then one or both bands will be twisted by some elements from this subset. Oh yeah. If you chose \( I(aa'bb') = a'a b'b\), and \(X(aa'bb') = b'ba'a\), for instance, both operations twist each band by 180°, but this twist is "in" a third dimension. If you quotient the subgroup (represented by the eight matrices in post #13) by the equivalence, which effectively ignores "the twist", everything "collapses" to a single diagonal and a single off diagonal pair of matrices equivalent to \( S_2 \). Nice. Instead of two one-dimensional strands (a mathematical abstraction), we cross two-dimensional bands (an abstraction of the first abstraction) over each other, with or without twisting them (depending on which of the eight matrices we choose, but one has diagonal, the other off diagonal, P-elements). So, can we count the number of twists in some permutation this way? You'd think so. But why would you? Well, why wouldn't you if you had some idea it was connected to particle interactions (or rather, an abstraction of same), in quantum mechanics? Huh?
However, \(S_4\) has subgroups of 12 (Alternating group, \(A_4\)) and 8 elements (three overlapping representations of \(Dih_4\)), but none of size 16. Moreover, I find only 8 elements of \(S_4\) in this pattern: \( M_0 = { \tiny \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} }, M_1 = { \tiny \begin{pmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} }, M_6 = { \tiny \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{pmatrix} }, M_7 = { \tiny \begin{pmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{pmatrix} }, \\ M_{16} = { \tiny \begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix} }, M_{17} = { \tiny \begin{pmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix} }, M_{22} = { \tiny \begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{pmatrix} }, M_{23} = { \tiny \begin{pmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{pmatrix} } \) Which is indeed a representation of \(Dih_4\) as the multiplication table shows: \(\begin{array}{c|cc} \times & M_0 & M_1 & M_6 & M_7 & M_{16} & M_{17} & M_{22} & M_{23} \\ \hline \\ M_0 & M_0 & M_1 & M_6 & M_7 & M_{16} & M_{17} & M_{22} & M_{23} \\ M_1 & M_1 & M_0 & M_7 & M_6 & M_{17} & M_{16} & M_{23} & M_{22} \\ M_6 & M_6 & M_7 & M_0 & M_1 & M_{22} & M_{23} & M_{16} & M_{17} \\ M_7 & M_7 & M_6 & M_1 & M_0 & M_{23} & M_{22} & M_{17} & M_{16} \\ M_{16} & M_{16} & M_{22} & M_{17} & M_{23} & M_0 & M_6 & M_1 & M_7 \\ M_{17} & M_{17} & M_{23} & M_{16} & M_{22} & M_1 & M_7 & M_0 & M_6 \\ M_{22} & M_{22} & M_{16} & M_{23} & M_{17} & M_6 & M_0 & M_7 & M_1 \\ M_{23} & M_{23} & M_{17} & M_{22} & M_{16} & M_7 & M_1 & M_6 & M_0 \end{array}\) This subgroup of 8 is generated by \(M_{17}\) and \(M_{23}\) with the remaining elements given by: \( M_0 = M_{17}^4 = M_{23}^2, M_1 = M_{17}^3 M_{23}, M_6 = M_{17} M_{23}, M_7 = M_{17}^2, M_{16} = M_{17}^2 M_{23}, M_{22} = M_{17}^3 = M_{17}^{-1} \) https://en.wikiversity.org/wiki/Symmetric_group_S4#Order_8 https://en.wikipedia.org/wiki/Examp...group_of_a_square_-_dihedral_group_of_order_8
Quite, the set of 16 elements isn't closed, so can't be a group. However each of the 16 elements in that subset of \( S_4 \) can be decomposed into two partial matrices in a way that the 8 element set (which is a subgroup as you point out) cannot be decomposed. I'm just playing around with permutation groups, but I want to be able to visualise certain things in my own, perhaps inimitable way. For instance, using bands and twisting them is just something I thought I could do, or that I'm "allowed" to take pairs instead of single letters and use only the 8 element subset. I realise that so far, there is no way to tell which way a band gets twisted, I need to represent that somehow, that somehow seems to be complex numbers.
Isn't that just: \( \begin{array}{cc} M_{16} & \quad & M_{23} & \quad & M_0 & \quad & M_7 \\ M_{22} & & M_{17} & & M_6 & & M_1 \end{array}\) in the notation I introduced for \(S_4\). What are the other 8 elements? Have you double-counted? Because addition of matrices is commutative, \(\rho_1 + \rho_2 = \rho_2 + \rho_1\) and because of that fact: \(\left\{ P | \exists i, j \in \{ 1, 2 \} \; \wedge \; i \lt j \; \wedge P = \rho_i + \rho_j \right\} = \left\{ P | \exists i, j \in \{ 1, 2 \} \; \wedge \; i \gt j \; \wedge P = \rho_i + \rho_j \right\} \) and \( \left\{ P | \exists i, j \in \{ 1, 2 \} \; \wedge \; i \neq j \; \wedge P = \rho_i + \rho_j \right\} \\ = \left\{ P | \exists i, j \in \{ 1, 2 \} \; \wedge \; i \lt j \; \wedge P = \rho_i + \rho_j \right\} \cap \left\{ P | \exists i, j \in \{ 1, 2 \} \; \wedge \; i \gt j \; \wedge P = \rho_i + \rho_j \right\} \\ = \left\{ P | \exists i, j \in \{ 1, 2 \} \; \wedge \; i \lt j \; \wedge P = \rho_i + \rho_j \right\} \) so it looks like you are double-counting.
I see what you mean, each matrix written the way I did it can be composed two ways thanks to commutativity of addition, but there are only 8 matrices in the set you quoted. And I've been careless with "set" and "group", so forget about the group operation for now. Working backwards from the set of 4 x 4 permutation matrices, without any operation defined or any order, each matrix is decomposed into two partials as outlined. Then we see a pattern: 8 of these decompose into two partials with nonzero 2 x 2 blocks (which blocks are sums of elements from P), each lying on the "block diagonal", or the "block off diagonal". These are (using your notation) in the set \( \Bigg\{ M_0,\;M_1\;,M_6,\;M_7,\;M_{16},\;M_{17},\;M_{22},\;M_{23} \Bigg\} \), and I can see that the subscripts you're using correspond to the numbers I've seen in a Cayley table. The remainder: \( \Bigg\{ M_2,\;M_3\;,M_4,\;M_5,\;M_8,\;M_9,\;M_{10}\;M_{11},\;M_{12},\;M_{13},\;M_{14},\;M_{15},\;M_{18},\;M_{19},\;M_{20},\;M_{21} \Bigg\} \), decompose into one of each type (of partial matrix) with no sums, and so there must be a 1-1 map (pretty obvious) from one type to the other under summation. That's all really, but then, what does this pattern mean? Maybe not a whole lot, but is it, for instance a general thing over all the symmetric groups?
Ok, I've decided that some of the notation I've been using would probably get red-inked (redinked?) if I used it in an assignment. So those sets of partial matrices should be notated like: \( \Bigg\{ \begin{pmatrix} a & 0 \\ 0 & b \end {pmatrix} \bigg| a,b \in P \Bigg\},\;\Bigg\{ \begin{pmatrix} 0 & c \\ d & 0 \end {pmatrix} \bigg| c,d \in P \Bigg\} \). Both sets contain 16 elements. There is a 1-1 map from one set to the other which takes pairs of partial matrices to 16 elements of \( S_4 \). There are also 1-1 maps from each set to itself where each map takes the 16 elements in each set to 4 elements of \( S_4 \) (so there are four distinct pairs in each set mapped to the same permutation matrix), giving 16 + 4 + 4 = 24 distinct matrices.
Correct. However, all 32 elements are only of rank 2. They all have determinant of 0. Since there are 16!, almost 21 trillion one-to-one mappings between two sets of size 16 and \(\frac{16! \; \times \; 24! }{8!} \approx 3.2 × 10^{32}\) ways to associate those pairs with a 16-element subset of \(S_4\), that seems more like a question than a claim. Since the two sets of 16 matrices are all of rank 2, you can't mean to multiply them together to get an element of \(S_4\) because all of those have rank 4. So addition looks like a logical choice for how to associated pairs of matrices with elements of \(S_4\). With that hint, it's like 2x2 Sudoku. a and d must be subscripted with different numbers. b and c must be subscripted with different numbers. a and c must differ in exactly one of subscript and overbar status. b and d must differ in exactly one of subscript and overbar status. Then \(M_X = \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix} + \begin{pmatrix} 0 & c \\ d & 0 \end{pmatrix} \) has 16 unique solutions, all indexed by the 16 possible choices of a and b from P. The 4 cases where \(a = \rho_1 \) : \(M_{14} = { \tiny \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix} } = \begin{pmatrix} \rho_1 & 0 \\ 0 & \rho_1 \end{pmatrix} + \begin{pmatrix} 0 & \rho_2 \\ \rho_2 & 0 \end{pmatrix} \) \(M_2 = { \tiny \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} } = \begin{pmatrix} \rho_1 & 0 \\ 0 & \rho_2 \end{pmatrix} + \begin{pmatrix} 0 & \overline{\rho_1} \\ \overline{\rho_2} & 0 \end{pmatrix} \) \(M_8 = { \tiny \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{pmatrix} } = \begin{pmatrix} \rho_1 & 0 \\ 0 & \overline{\rho_1} \end{pmatrix} + \begin{pmatrix} 0 & \rho_2 \\ \overline{\rho_2} & 0 \end{pmatrix} \) \(M_{12} = { \tiny \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \end{pmatrix} } = \begin{pmatrix} \rho_1 & 0 \\ 0 & \overline{\rho_2} \end{pmatrix} + \begin{pmatrix} 0 & \overline{\rho_1} \\ \rho_2 & 0 \end{pmatrix} \)
