Q:

Definition of the Quantum plane and the Yang Baxter Equation

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?

A:

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...

