Identities in an algebra

From: Kevin Buzzard (buzzard_at_imperREMial.aOVEc.uk)
Date: 02/11/05


Date: Fri, 11 Feb 2005 21:06:51 +0000 (UTC)

A student of mine is finishing up writing his thesis; if he can prove
that a certain diagram commutes then he will be done!

We spent some time yesterday trying to work out what needs to be done.
The question below isn't equivalent (the original question is about
p-adic Siegel modular forms), but it suffices to finish the argument.
The question below is basically equivalent to a messy combinatorial
identity that nowadays could be checked by computer, but there is
probably a neater way of finishing the job. Can anyone help?

Let L be the 4-dimensional Lie algebra with basis D,E,F,G, and
satisfying

[D,E]=F, [D,F]=G, [D,G]=0,
[E,F]=[E,G]=[F,G]=0

(this is a Lie algebra; for example D could be d/dX and E could
be X^2d/dY acting on R[X,Y])

Is it true that for all integers a,b>=0 the following identity
is true in the universal enveloping algebra of L:

D^a E^(a+b) D^(a+2b) E^b = E^b D^(a+2*b) E^(a+b) D^a

?

It's true if one of a,b is zero and it's true for a,b<=3 (computer
calculation). I tried proving it by hand for a=b=1 and got very bogged
down; I still don't "know why" it's true. On the other hand, for
experts in this sort of associative algebra it might all be easy.
The question is about identites in the free algebra generated
by D,E,F,G, modulo the bi-ideal generated by the 6 relations above.
A computer algebra package like magma can check instances of the
equality e.g.

> P<D,E,F,G>:=FreeAlgebra(Rationals(),4);
> I:=ideal<P|D*E-E*D-F,D*F-F*D-G,D*G-G*D,E*F-F*E,E*G-G*E,F*G-G*F>;
> D*E^2*D^3*E-E*D^3*E^2*D in I;
true

but this isn't giving me any feeling as to what's going on.

Cheers,

Kevin Buzzard



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