Ask Your Question
0

How can I specify a (possibly non-associative) algebra over a finite (small) set of generators by means of structure constants?

asked 2017-05-09 14:09:47 -0500

I am looking for something similar to the Axiom/FriCAS domain AlgebraGivenByStructuralConstants which implements finite rank algebras over a commutative ring, given by the structural constants with respect to a fixed basis [a1,..,an] or equivalently in terms of generating equations of the form

a_i * a_j = gamma_ij1 * a_1 + ... + gamma_ijn * a_n

where gamma is a vector/list of length n of n by n matrices.

In particular I would like to be able to easily compute various properties of such algebras such as the conditions for idempotents etc. For example:

edit retag flag offensive close merge delete

1 answer

Sort by ยป oldest newest most voted
1

answered 2017-05-09 20:10:14 -0500

dan_fulea gravatar image

One possibility is to use FiniteDimensionalAlgebra . (Or similar "magmatic" constructors.)

For example:

F = GF(7)
A = [ Matrix( GL(3,F).random_element().list() )
      for _ in range(3) ]
M = FiniteDimensionalAlgebra( F, A, names='s' )
M

s0, s1, s2 = M.gens()
s = s0, s1, s2

for k in (0,1,2):
    for n in (0,1,2):
        print "%s * %s = %s" % ( s[k], s[n], s[k]*s[n] )
    print

for k in (0,1,2):
    print "A[%s] is\n%s\n" % ( k, A[k] )

Here $F$ is the base field, taken with $7$ elements to make the multiplication constraints simple to follow. The data $A$ is the data of three matrices. (It collects the $\gamma$-constraints from the post.) It is randomly constructed here. (The GL has no structural meaning, it came first into the fingers.)

The above evaluates to:

Finite-dimensional algebra of degree 3 over Finite Field of size 7
s0 * s0 = 2*s0 + 5*s1 + 3*s2
s0 * s1 = s0 + 3*s1 + 4*s2
s0 * s2 = 5*s0 + s1 + s2

s1 * s0 = 4*s1 + s2
s1 * s1 = 5*s0 + 6*s1 + 3*s2
s1 * s2 = 4*s0 + 2*s1 + 5*s2

s2 * s0 = 3*s0 + 3*s1 + 2*s2
s2 * s1 = 3*s0 + 5*s2
s2 * s2 = 5*s0 + 3*s1

A[0] is
[2 5 3]
[0 4 1]
[3 3 2]

A[1] is
[1 3 4]
[5 6 3]
[3 0 5]

A[2] is
[5 1 1]
[4 2 5]
[5 3 0]

I think it is clear how the entries in the three matrices contribute to the multiplication constraints. I tried also...

sage: M.is_associative()
False
sage: M.is_commutative()
False
edit flag offensive delete link more

Comments

This is nice. There is a lot of useful documentation: here but so far I could not easily find how compute the center of the algebra (elements that commute with all other elements) but perhaps this is defined in some other way?

Bill Page _ again gravatar imageBill Page _ again ( 2017-05-10 08:41:48 -0500 )edit

Your Answer

Please start posting anonymously - your entry will be published after you log in or create a new account.

Add Answer

Question Tools

1 follower

Stats

Asked: 2017-05-09 14:09:47 -0500

Seen: 120 times

Last updated: May 09