Date: Thu, 11 Aug 2011 20:32:14 -0700 (PDT)
From: John H Palmieri <jhpalmier...@gmail.com>
Reply-To: sage-algebra@googlegroups.com
To: sage-algebra@googlegroups.com
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In-Reply-To: <8bb9c5b3-bee2-4d1f-be1c-242eb7b7f200@d8g2000prf.googlegroups.com>
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Subject: Re: Infinite dimensional Lie algebras
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On Thursday, August 11, 2011 8:05:39 PM UTC-7, Simon Wood wrote:
>
> Hello everyone 
>
> I'm new to Sage and unfortunately still completely clueless. I'm 
> hoping to be able to use Sage for some computations involving infinite 
> dimensional Lie algebras and their representations. Does Sage have 
> useful tools for defining algebras with an infinite number of 
> generators? I see there is an "AlgebrasWithBasis" function, but as far 
> as I can tell it requires the number of generators to be finite. 
>
> What would be the best way to say for example construct the Virasoro 
> algebra?


My first inclination would be to base it on CombinatorialFreeModule: if you 
know a basis, this is a good way to go.  It will take care of addition and 
scalar multiplication, so you'll have to write a bracket method.  If you're 
feeling ambitious, you could write something for the Sage categories 
framework, implementing "Lie algebras with basis", by imitating "algebras 
with basis" (sage/categories/algebras_with_basis.py).  Then for any 
particular example, you should just have to define a method 
"bracket_on_basis" which takes a pair of basis elements and computes their 
bracket; the category framework (which you will set up separately) will then 
define a "bracket" method for dealing with arbitrary elements of your Lie 
algebra.

I used the existing framework for Hopf algebras for the current 
implementation (as of prerelease versions of Sage 4.7.1) of the Steenrod 
algebra -- see <http://trac.sagemath.org/sage_trac/ticket/10052>.  I 
basically only had to define products, antipodes, and coproducts on basis 
elements, and the framework took care of everything else.  (Note, by the 
way, that the Steenrod algebra is infinite-dimensional, and this is not a 
problem for working with CombinatorialFreeModules.)

-- 
John


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<br><br>On Thursday, August 11, 2011 8:05:39 PM UTC-7, Simon Wood wrote:<bl=
ockquote class=3D"gmail_quote" style=3D"margin: 0;margin-left: 0.8ex;border=
-left: 1px #ccc solid;padding-left: 1ex;">Hello everyone
<br>
<br>I'm new to Sage and unfortunately still completely clueless. I'm
<br>hoping to be able to use Sage for some computations involving infinite
<br>dimensional Lie algebras and their representations. Does Sage have
<br>useful tools for defining algebras with an infinite number of
<br>generators? I see there is an "AlgebrasWithBasis" function, but as far
<br>as I can tell it requires the number of generators to be finite.
<br>
<br>What would be the best way to say for example construct the Virasoro
<br>algebra?</blockquote><div><br>My first inclination would be to base it =
on CombinatorialFreeModule: if you know a basis, this is a good way to go.&=
nbsp; It will take care of addition and scalar multiplication, so you'll ha=
ve to write a bracket method.&nbsp; If you're feeling ambitious, you could =
write something for the Sage categories framework, implementing "Lie algebr=
as with basis", by imitating "algebras with basis" (sage/categories/algebra=
s_with_basis.py).&nbsp; Then for any particular example, you should just ha=
ve to define a method "bracket_on_basis" which takes a pair of basis elemen=
ts and computes their bracket; the category framework (which you will set u=
p separately) will then define a "bracket" method for dealing with arbitrar=
y elements of your Lie algebra.<br><br>I used the existing framework for Ho=
pf algebras for the current implementation (as of prerelease versions of Sa=
ge 4.7.1) of the Steenrod algebra -- see &lt;http://trac.sagemath.org/sage_=
trac/ticket/10052&gt;.&nbsp; I basically only had to define products, antip=
odes, and coproducts on basis elements, and the framework took care of ever=
ything else.&nbsp; (Note, by the way, that the Steenrod algebra is infinite=
-dimensional, and this is not a problem for working with CombinatorialFreeM=
odules.)<br><br>-- <br>John<br><br></div>
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