Brown-Klosin_calculations

3152 days ago by jimlb

This is an easy way to do the calculations to make sure they are correct as well as to save typing into latex.  Make sure to click to typeset above before evaluating if you want to see the results nicely.  A couple tips in case you don't use SAGE much:  it uses python, so things are numbered starting at 0, if you are unsure of a command you can type command? and it will bring up usage.

var('a11,a12,a21,a22,c11,c12,c21,c22,b11,b12,b21,b22,d11,d12,d21,d22,e11,e12,e21,e22,delta1,delta2,delta3,delta4,x,z') 
       

Here I do the calculation for the $\mathcal{B}_3$ case.  The augmented matrix $[\Upsilon:\mathcal{D}]$ has the following form.

Gamma = Matrix(SR,[[a11-x*c11,a12-x*c12,0,x*d11-b11],[0,a11-x*c11,a12-x*c12,x*d12-b12],[c21,c22,0,-d21],[0,c21,c22,-d22]]);Gamma 
       

We start with the assumption that $c_{21} \neq 0$.

Gamma.rescale_row(2,1/c21);Gamma 
       
Gamma.swap_rows(2,0);Gamma 
       
Gamma.add_multiple_of_row(2,0,x*c11-a11);Gamma 
       
Gamma.swap_rows(3,1);Gamma 
       
Gamma.rescale_row(1,1/c21);Gamma 
       
Gamma.add_multiple_of_row(3,1,x*c11-a11);Gamma 
       
Gamma.add_multiple_of_row(0,1,-c22/c21);Gamma 
       
Gamma.add_multiple_of_row(2,1,x*c12 - (x*c11-a11)*c22/c21 - a12);Gamma 
       
Gamma.add_multiple_of_row(2,3,c22/c21);Gamma 
       
Gamma.swap_rows(3,2);Gamma 
       

Looking at the bottom right entry we multiply by $c_{21}$ and simplify to get:

$X = x(c_{21}d_{11} - c_{11}d_{21} + c_{22} d_{12} - c_{12}d_{22}) + (a_{12}d_{22} + a_{11} d_{21} - b_{12}c_{22} - b_{11}c_{21})$.

g = Matrix(SR,[[a11,a12,b11,b12],[a21,a22,b21,b22],[c11,c12,d11,d12],[c21,c22,d21,d22]]);g; 
       

Defining Taylor's version of $J$.

J=Matrix(SR,[[0,0,1,0],[0,0,0,1],[-1,0,0,0],[0,-1,0,0]]);J 
       
g*J*g.transpose(); 
       

For $g$ to be symplectic, we must have the $(4,3)$ entry is 0, which says the coefficient of $x$ in $X$ must vanish. Similarly, the $(1,4)$ entry must vanish, which gives the non-$x$ term of $X$ must vanish.  Thus, $X =0$.  It still works! whew..

Now suppose that $c_{21} = 0$.

Gamma = Matrix(SR,[[a11-x*c11,a12-x*c12,0,x*d11-b11],[0,a11-x*c11,a12-x*c12,x*d12-b12],[0,c22,0,-d21],[0,0,c22,-d22]]);Gamma 
       

Assume $c_{22} \neq 0$.

Gamma.rescale_row(2,1/c22);Gamma 
       
Gamma.add_multiple_of_row(1,2,x*c11- a11);Gamma 
       
Gamma.add_multiple_of_row(0,2,x*c12- a12);Gamma 
       
Gamma.add_multiple_of_column(3,1,d21/c22);Gamma 
       
Gamma.swap_columns(0,1);Gamma 
       
Gamma.swap_rows(0,2);Gamma 
       
Gamma.swap_columns(1,2);Gamma 
       
Gamma.swap_rows(1,3);Gamma 
       
Gamma.rescale_row(1,1/c22);Gamma 
       
Gamma.add_multiple_of_column(3,1,d22/c22);Gamma 
       
Gamma.add_multiple_of_row(3,1,c12*x-a12);Gamma 
       
Gamma = Matrix(SR,[[a11-x*c11,a12-x*c12,0,x*d11-b11],[0,a11-x*c11,a12-x*c12,x*d12-b12],[0,0,0,-d21],[0,0,0,-d22]]);Gamma 
       
Gamma= Matrix(SR,[[a11-x*c11,a12-x*c12,0,x*d11-b11],[0,a11-x*c11,a12-x*c12,x*d12-b12]]);Gamma 
       

We now consider the $\mathcal{B}_2$ calculation.

var('a11,a12,a21,a22,c11,c12,c21,c22,b11,b12,b21,b22,d11,d12,d21,d22,e11,e12,e21,e22,delta1,delta2,delta3,delta4,x,z') 
       
Gamma = Matrix(SR,[[c11+x*c21, c12+x*c22 , 0 ,-d11-x*d21],[0 , c11+x*c21 , c12 + x*c22 , -d12 - x*d22],[a21-x*a11+z*c21 , a22-x*a12+z*c22, 0 , -b21+x*b11-z*d21],[0 , a21 - x*a11+z*c12 , a22 - x*a12 + z*c22 , -b22+x*b12 - z*d22]]);Gamma 
       

Assume that $c_{21}x + c_{11} \neq 0$.

Gamma.rescale_row(0,1/(c21*x+c11));Gamma 
       
Gamma.add_multiple_of_row(2,0,a11*x-c21*z-a21);Gamma 
       
Gamma.add_multiple_of_column(1,0,-(c22*x+c12)/(c21*x+c11));Gamma 
       
Gamma.add_multiple_of_column(3,0,(d21*x+d11)/(c21*x+c11));Gamma 
       
Gamma.rescale_row(1,1/(c21*x+c11));Gamma 
       
Gamma.add_multiple_of_column(2,1,-(c22*x+c12)/(c21*x+c11));Gamma 
       
Gamma.add_multiple_of_column(3,1,(d22*x+d12)/(c21*x+c11));Gamma 
       
Gamma.add_multiple_of_row(2,1,a12*x -c22*z -a22 - (c22*x+c12)*(a11*x-c21*z-a21)/(c21*x+c11));Gamma 
       
Gamma.add_multiple_of_row(3,1,a11*x-c12*z-a21);Gamma 
       

If $c_{22}x + c_{12} = 0$, we should get a contradiction as in the write-up.  (It will have rank 2, which gives too many $E$'s that map to $x,z$.)  If we assume $c_{22}x+c_{12} \neq 0$, then we have:

Gamma.add_multiple_of_row(2,3,(c22*x+c12)/(c21*x+c11));Gamma 
       
Gamma.swap_rows(2,3);Gamma 
       

Note the entry in the (4,3) spot is 0; I am not sure why it is not simplifying it to be that.