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I was reading this on the quantum plane and the Yang Baxter equation. John Baez says that imposing $$ R(X\otimes X)= X\otimes X $$ $$ R(Y\otimes Y)= Y\otimes Y $$ $$ R(X\otimes Y)=q Y\otimes X $$ $$ R(Y\otimes X)=q X\otimes Y + (1-q^2) Y\otimes X $$ the resulting R-matrix satisfy the YBE, i.e. $$ \left(R\otimes id\right)\left(id\otimes R\right)\left(R\otimes id\right)=\left(id\otimes R\right)\left(R\otimes id\right)\left(id\otimes R\right) $$ I then wrote the matrix in the following base {$X\otimes X,Y\otimes Y,X\otimes Y,Y\otimes X$} obtaining $$ R=\left(\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & q\\ 0 & 0 & q & 1-q^{2} \end{array}\right) $$ But it doesn't seem to satisfy the YBE. What did I do wrong or what did I miss? And what is the relation between the definition of the quantum plane and the Yang Baxter Equation?

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  • $\begingroup$ The equations look correct to me. How did you make your computation of YBE? $\endgroup$ – David Hill Mar 9 '18 at 18:24
  • $\begingroup$ In Mathematica I defined the R matrix as the one before and then r1 = KroneckerProduct[id, R]; r2 = KroneckerProduct[R, id]; r1.r2.r1 - r2.r1.r2 // MatrixForm $\endgroup$ – Dac0 Mar 9 '18 at 22:20
  • $\begingroup$ I take it the operation for $\otimes$ here is matrix multiplication? $\endgroup$ – Kevin Apr 23 '18 at 13:04
  • $\begingroup$ Not really, you can find more information on this equation here en.wikipedia.org/wiki/Yang%E2%80%93Baxter_equation $\endgroup$ – Dac0 Apr 23 '18 at 13:26
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Let us consider $X_0=X$ and $X_1=Y$, then introduce $$X_{abc}=X_a\otimes X_b\otimes X_c\ .$$ The Yang-Baxter relation deals with $R\otimes 1$ and $1\otimes R$, (the one being the identity here,) so we need a choice of the order of the basis, that is best compatible with these associations.

If we write $R$ with respect to the basis with indices $00$, $11$, $01$, $10$, then "adding an index" in front, respectively behind, leads to the triple indices: $$ 000\ ,\ 011\ ,\ 001\ ,\ 010\ ;\ 100\ ,\ 111\ ,\ 101\ ,\ 110\ ; $$ and respectively $$ 000\ ,\ 110\ ,\ 010\ ,\ 100\ ;\ 001\ ,\ 111\ ,\ 011\ ,\ 101\ ; $$ and we have to "rematch" the indices by using a corresponding permutation matrix to recover the order.

To have a better control, i prefer to write the matrix $R$ w.r.t. $00$, $01$, $10$, $11$, the binary order. So appending a digit in front of the doubles leads to two blocks, and appending a digit behind of them gives an interlacing. The related matrices are then: \begin{align} r &= 1-q^2\ ,\\\\ R &= \begin{bmatrix} 1 & & &\\ & 0 & q & \\ & q & r & \\ & & & 1 \end{bmatrix} \\\\ 1\otimes R &= \begin{bmatrix} 1 & & & & & & & \\ & 0 & q & & & & & \\ & q & r & & & & & \\ & & & 1 & & & & \\\hline & & & & 1 & & &\\ & & & & & 0 & q & \\ & & & & & q & r & \\ & & & & & & & 1 \end{bmatrix} \\\\ R \otimes 1 & = \begin{bmatrix} 1 & & & & & & & \\ & 1 & & & & & & \\\hline & & 0 & & q & & & \\ & & & 0 & & q & & & \\\hline & & q & & r & & \\ & & & q & & r & \\\hline & & & & & & & 1 & \\ & & & & & & & & 1 \end{bmatrix} \end{align} The corresponding computer check using sage is as follows:

S.<q> = PolynomialRing( QQ )
r   = 1-q^2
R   = matrix( S, 4, 4, [ 1,0,0,0, 0,0,q,0, 0,q,r,0, 0,0,0,1 ] )#
R23 = block_diagonal_matrix( [R,R] )
S8  = SymmetricGroup(8)
s   = S8( [1,3,5,7,2,4,6,8] )
P   = s.matrix() 
R12 = P.inverse() * R23 * P
print "R12 * R23 * R12 is the matrix\n%s\n" % (R12 * R23 * R12)
print "R23 * R12 * R23 is the matrix\n%s\n" % (R23 * R12 * R23)
print "Are the two matrices equal? %s" % bool( R12 * R23 * R12 == R23 * R12 * R23 )

Results:

R12 * R23 * R12 is the matrix
[       1        0        0        0        0        0        0        0]
[       0        0        0        0      q^2        0        0        0]
[       0        0      q^2        0 -q^3 + q        0        0        0]
[       0        0        0        0        0        0      q^2        0]
[       0      q^2 -q^3 + q        0 -q^2 + 1        0        0        0]
[       0        0        0        0        0      q^2 -q^3 + q        0]
[       0        0        0      q^2        0 -q^3 + q -q^2 + 1        0]
[       0        0        0        0        0        0        0        1]

R23 * R12 * R23 is the matrix
[       1        0        0        0        0        0        0        0]
[       0        0        0        0      q^2        0        0        0]
[       0        0      q^2        0 -q^3 + q        0        0        0]
[       0        0        0        0        0        0      q^2        0]
[       0      q^2 -q^3 + q        0 -q^2 + 1        0        0        0]
[       0        0        0        0        0      q^2 -q^3 + q        0]
[       0        0        0      q^2        0 -q^3 + q -q^2 + 1        0]
[       0        0        0        0        0        0        0        1]

Are the two matrices equal? True

So it may be that a proof for the Yang-Baxter relation by writing matrices and multiplying them is not the "simplest" one (to write and check). Personally, i would do the computations as follows, and observe how the braid relations are propagated through the calculus -- where all tensor product signs are omitted: \begin{align} R\ XX &= XX\\\\ R\ YY &= YY\\\\ R\ XY &= q\; YX\\\\ R\ YX &= q\; XY + (1-q^2)\; YX\\\\ &\text{so we have:}\\\\ % ========== (R1)(1R)(R1)\ XXX &= (R1)(1R)\ XXX = (R1)\ XXX = XXX\ ,\\\\ (1R)(R1)(1R)\ XXX &= (1R)(R1)\ XXX = (1R)\ XXX = XXX\ ,\\\\ &\\\\ (R1)(1R)(R1)\ XXY &= (R1)(1R)\ XXY = q\;(R1)\ XYX = q^2\; YXX\ ,\\\\ (1R)(R1)(1R)\ XXY &= q\; (1R)(R1)\ XYX = q^2\; (1R)\ YXX = q^2\; YXX\ ,\\\\ &\\\\ (R1)(1R)(R1)\ XYX &= q\; (R1)(1R)\ YXX = q\; (R1)\ YXX = q\;( q\;XYX + (1-q^2)\;YXX )\ ,\\\\ (1R)(R1)(1R)\ XYX &= (1R)(R1)\ (q\;XXY + (1-q^2)\;XYX) = (1R)\ (q\;XXY + q(1-q^2)\;YXX) \\\\ &= (q^2\;XYX + q(1-q^2)\;YXX)\ ,\\\\ &\\\\ (R1)(1R)(R1)\ XYY &= q\; (R1)(1R)\ YXY = q^2\; (R1)\ YYX = q^2\; YYX\ ,\\\\ (1R)(R1)(1R)\ XYY &= (1R)(R1)\ XYY = q\; (1R)\ YXY = q^2\; YYX\ ,\\\\ &\\\\ % ==================================================================================================== (R1)(1R)(R1)\ YXX &= (R1)(1R)\ (q\;XYX + (1-q^2)\;YXX) \\\\ &= (R1)\ (q^2\;XXY + q(1-q^2)\;XYX + (1-q^2)\;YXX) \\\\ &= q^2\;XXY + q^2(1-q^2)\;YXX + (1-q^2)\;( q\;XYX + (1-q^2)\;YXX)\ ,\\\\ (1R)(R1)(1R)\ YXX &= (1R)(R1)\ YXX \\\\ &= (1R)\ (q\; XYX + (1-q^2)\; YXX) \\\\ &= q\;(q\; XXY + (1-q^2)\; XYX)+ (1-q^2)\; YXX\ ,\\\\ &\\\\ (R1)(1R)(R1)\ YXY &= (R1)(1R)\ (q\; XYY + (1-q^2)\; YXY) = (R1)\ (q\; XYY + q(1-q^2)\; YYX) \\\\ &= q^2\; YXY + q(1-q^2)\; YYX\ ,\\\\ (1R)(R1)(1R)\ YXY &= q\; (1R)(R1)\ YYX = q\; (1R)\ YYX = q\; (q\; YXY + (1-q^2)\; YYX)\ ,\\\\ &\\\\ (R1)(1R)(R1)\ YYX &= (R1)(1R)\ YYX = (R1)\ (q\; YXY + (1-q^2)\; YYX) \\\\ &= q\;(q\; YXY + (1-q^2)\; YYX) + (1-q^2)\; YYX\ ,\\\\ (1R)(R1)(1R)\ YYX &= (1R)(R1)\ (q\;YXY + (1-q^2)\;YYX) \\\\ &= (1R)\ (q\;(q\; XYY + (1-q^2)\; YXY) + (1-q^2)\;YYX) \\\\ &= q\;(q\; XYY + q\;(1-q^2)\; YYX) + (1-q^2)\;(q\; YXY + (1-q^2)\; YYX)\ ,\\\\ &\\\\ (R1)(1R)(R1)\ YYY &= (R1)(1R)\ YYY = (R1)\ YYY = YYY\ ,\\\\ (1R)(R1)(1R)\ YYY &= (1R)(R1)\ YYY = (1R)\ YYY = YYY\ . \end{align} Maybe i should have used $r$ instead of $(1-q^2)$ to keep it simple and digestible.

Yes, the relation to the quantum plane. I think, the next node

The Quantum Plane

in the provided link does more than i can do here...

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  • $\begingroup$ Ok I think the misteke I made was that I used the 4 dimensional identity instead of the 2 dimensional identity :D :D :D $\endgroup$ – Dac0 Apr 26 '18 at 6:38

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