Re: Identities in an algebra
From: Kevin Buzzard (buzzard_at_imperREMial.aOVEc.uk)
Date: 02/13/05
- Previous message: Bill Dubuque: "Re: Identities in an algebra"
- In reply to: Bill Dubuque: "Re: Identities in an algebra"
- Next in thread: Jyrki Lahtonen: "Re: Identities in an algebra"
- Reply: Jyrki Lahtonen: "Re: Identities in an algebra"
- Messages sorted by: [ date ] [ thread ]
Date: Sun, 13 Feb 2005 21:51:02 +0000 (UTC)
Bill Dubuque <wgd@nestle.csail.mit.edu> wrote:
> With D = d/dX it suffices to verify the following
>
> a 2a+2b a+2b 2b 2b a+2b 2a+2b a
> D X D X = X D X D
Setting D=d/dX and E=X^2 gives a representation of the universal
enveloping algebra, and you're checking that the identity I want
is true in the image. But I don't think that the representation is
injective; it sends F to 2X and G to 2, so it sends 2EG-F^2 to zero.
On the other hand setting D=d/dX and E=X gives another representation
of the algebra in which 2EG-F^2 is not sent to zero, so 2EG-F^2
is not zero in the algebra itself. Hence I don't see why it
suffices to do this, unfortunately.
I can quite believe that an argument of this nature might work though.
Proving that the image of the identity in a representation
is zero seems to frequently amount to verifying combinatorial
identities; the example above results in a rather simple
identity but I am not so sure that it's strong enough to prove
what I need. I know of more complicated representations which I can
prove are injective; however the resulting combinatorial identities
are also quite complicated (double sums of products of 6 binomial
coefficients!). I am sure that one could prove such identities
nowadays following Zeilberger et al, but I didn't really want to
go down that path, I am still optimistic that there is a simpler
solution [At the end of the day the real reason I don't want
to push this approach through is that the identities above
are in some sense a theorem (hopefully) about Sp_4, and I would one day
like to prove analogous ones for an arbitrary reductive group, when
brute force calculations would be doomed.]
Kevin
- Previous message: Bill Dubuque: "Re: Identities in an algebra"
- In reply to: Bill Dubuque: "Re: Identities in an algebra"
- Next in thread: Jyrki Lahtonen: "Re: Identities in an algebra"
- Reply: Jyrki Lahtonen: "Re: Identities in an algebra"
- Messages sorted by: [ date ] [ thread ]
Relevant Pages
|