Message Boards Message Boards

0
|
2425 Views
|
1 Reply
|
0 Total Likes
View groups...
Share
Share this post:

Calculate the unitary irreducible representation of a matrix group.

Posted 1 year ago

I created a matrix group as follows:

In[33]:= gensSG141ITA1={
 {{1, 0, 0, 0}, {0, 1, 0, 0}, {0, 0, 1, 0}, {0,0,0,1}},
 {{-1, 0, 0, 1/2}, {0, -1, 0, 1/2}, {0, 0, 1, 1/2}, {0,0,0, 1}},
 {{0, -1, 0, 0}, {1, 0, 0, 1/2}, {0, 0, 1, 1/4}, {0,0,0, 1}},
 {{-1, 0, 0, 1/2}, {0,  1, 0,  0}, {0,  0,  -1, 3/4}, {0,0,0, 1}},
 {{-1, 0, 0, 0}, {0, -1, 0, 1/2}, {0, 0, -1, 1/4}, {0,0,0, 1}},
 {{1, 0, 0, 1/2}, {0, 1, 0, 1/2}, {0, 0, 1, 1/2}, {0,0,0, 1}}
}; 

newmat = DeleteDuplicates[#[[1 ;; 3, 1 ;; 3]] & /@ gensSG141ITA1];

(*
Apply the following procedure to construct a group from generating elements (matrices)
*)
gengroupn[ge1_] := Module[{},
  ge = ge1;
  ne = Length[ge];
  l1 = 1;
  ne = Length[ge];
  While[l1 <= ne,
   l2 = 1;
   While[l2 <= ne,
    res = FullSimplify[Together[ge[[l1]] . ge[[l2]]]];
    If[! MemberQ[ge, res],
     ne++; AppendTo[ge, res]];
    l2++];
   l1++];
  ge
  ]

gg = gengroupn[newmat];
ne = Length[gg]
gg

Out[37]= 16

Out[38]= {{{1, 0, 0}, {0, 1, 0}, {0, 0, 1}}, {{-1, 0, 0}, {0, -1, 
   0}, {0, 0, 1}}, {{0, -1, 0}, {1, 0, 0}, {0, 0, 1}}, {{-1, 0, 
   0}, {0, 1, 0}, {0, 0, -1}}, {{-1, 0, 0}, {0, -1, 0}, {0, 
   0, -1}}, {{0, 1, 0}, {-1, 0, 0}, {0, 0, 1}}, {{1, 0, 0}, {0, -1, 
   0}, {0, 0, -1}}, {{1, 0, 0}, {0, 1, 0}, {0, 0, -1}}, {{0, -1, 
   0}, {-1, 0, 0}, {0, 0, -1}}, {{0, 1, 0}, {-1, 0, 0}, {0, 
   0, -1}}, {{0, 1, 0}, {1, 0, 0}, {0, 0, -1}}, {{0, -1, 0}, {1, 0, 
   0}, {0, 0, -1}}, {{1, 0, 0}, {0, -1, 0}, {0, 0, 1}}, {{-1, 0, 
   0}, {0, 1, 0}, {0, 0, 1}}, {{0, -1, 0}, {-1, 0, 0}, {0, 0, 
   1}}, {{0, 1, 0}, {1, 0, 0}, {0, 0, 1}}}

Now, I want to calculate the irreducible unitary representation of it. Any hints for this question?

Regards, Zhao

POSTED BY: Hongyi Zhao

Hello Zhao,

gg is already unitary, means as there are no complex numbers it is hermitian.

Map[MatrixForm, Transpose[#].# & /@ gg, 1]

You can create another subgroup of gg from the matrices with gg[[3,3]] == 1 (I did not (yet) test whether the other matrices with IdentiyMatrix[3] form a group as well)

g31 = Select[gg, #[[3, 3]] == 1 &]
Map[MatrixForm, g31, 1]
Map[MatrixForm, g31, 1]
gg31 = gengroupn[g31]
gg31 - g31

This again can be decomposed in a one- and a two-dimensional representation of gg

g312dim = Flatten[{#[[1 ;; 2, 1 ;; 2]]}, 1] & /@ g31;
Map[MatrixForm, g312dim, 1]

I think this cannot be further "reduced", but I am by no means sure.

POSTED BY: Hans Dolhaine
Reply to this discussion
Community posts can be styled and formatted using the Markdown syntax.
Reply Preview
Attachments
Remove
or Discard

Group Abstract Group Abstract