[GAP Forum] Repsn matrices appear to be not homomorphic to group (for A5)

Joris Vergeest jorisvergeest at hotmail.com
Fri Mar 8 21:28:10 GMT 2019


Dear Forum,

I tried to verify, for the A5 group, whether the group of 60 matrices produced by Repsn is homomorphic to A5.
They appear to be not

One example:

The 60 group elements g1, g2, ..., g60 are sort-listed using List(G).

3D representation matrices Mi are obtained using Repsn: Mi = gi^IrreducibleAffordingRepresentation(selChar), for some fixed character selChar.

It is expected that Mi * Mj = Mk, where k is chosen such that gi * gj = gk; then we are dealing with a homomorphism.

For group A5 take i = 2, j = 3. Then:

g2 = (1,5,4), 
g3 = (1,4,5),
g2 * g3 = g1 = () , the identity element. That is g2 and g3 are inverses of each other.

Now from Repsn we obtain:

M2 = 
[ [ -3/5*E(5)-2/5*E(5)^2-2/5*E(5)^3-3/5*E(5)^4, 2/5*E(5)-2/5*E(5)^2-2/5*E(5)^3+2/5*E(5)^4,
        2/5*E(5)+3/5*E(5)^2+3/5*E(5)^3+2/5*E(5)^4 ], 
   [ 0, 0, -1 ],
   [ -2/5*E(5)+2/5*E(5)^2+2/5*E(5)^3-2/5*E(5)^4, -2/5*E(5)-3/5*E(5)^2-3/5*E(5)^3-2/5*E(5)^4, 
      3/5*E(5)+2/5*E(5)^2+2/5*E(5)^3+3/5*E(5)^4 ] ]

M3 = [ [ -E(5)-E(5)^4, E(5)^2+E(5)^3, -2 ], [ E(5)^2+E(5)^3, -1, E(5)^2+E(5)^3 ], 
  [ 1, -E(5)^2-E(5)^3, -E(5)^2-E(5)^3 ] ]

M1 = [ [ 1, 0, 0 ], [ 0, 1, 0 ], [ 0, 0, 1 ] ]

So M2 * M3 should be equal to M1.

In Gap we find:

M2 * M3 = 
[ [ 2/5*E(5)+8/5*E(5)^2+8/5*E(5)^3+2/5*E(5)^4, -2/5*E(5)+7/5*E(5)^2+7/5*E(5)^3-2/5*E(5)^4, 
      11/5*E(5)+14/5*E(5)^2+14/5*E(5)^3+11/5*E(5)^4 ],
            [ -1, E(5)^2+E(5)^3, E(5)^2+E(5)^3 ], 
             [ -2/5*E(5)+2/5*E(5)^2+2/5*E(5)^3-2/5*E(5)^4, -3/5*E(5)-2/5*E(5)^2-2/5*E(5)^3-3/5*E(5)^4, 
      4/5*E(5)+1/5*E(5)^2+1/5*E(5)^3+4/5*E(5)^4 ] ]

which is not equal to M1.

All products Mi * Mj for which neither Mi nor Mj are the identity appear inconsistent with a homomorphism.

BTW: I know that for A5 "correct" representations have been found.  However, I need a reliable method to generate representations for automatic processing of many groups.

Any advise is welcome,

Joris




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