I'm trying to learn some Lie algebra without much knowledge of representation theory. While being asked to prove some things about an sl(2)-module, why can one assume that the module is irreducible and has dimension 2?
$\begingroup$
$\endgroup$
4
-
1$\begingroup$ @alyscia: When you ask a question because you're confused about some part of a proof, please include more information. What are you trying to prove? What is the basic argument? Which step are you confused about? Not only will this make it easier (or possible) for others to help you out, but the process of organizing that information will often help put the real problem into focus. (and sometimes even solve it!) $\endgroup$– Anton GeraschenkoCommented Nov 14, 2009 at 15:40
-
3$\begingroup$ I've downvoted this because the statement is simply false. Without much much more context, there is no way one can ``assume'' that an otherwise-unqualified sl(2)-module is irreducible and dimension 2. $\endgroup$– alekzanderCommented Nov 14, 2009 at 19:10
-
1$\begingroup$ also, i would encourage you to write more careful titles. If people can't figure out roughly what your question is from reading the title, they are less likely to click through. Aim for a title which is a complete sentence, stating your question (even if it is long). $\endgroup$– Ben Webster ♦Commented Feb 13, 2010 at 21:26
-
$\begingroup$ If you do not tell us what "things" you are being asked to prove, it is impossible to answer the question. $\endgroup$– Mariano Suárez-ÁlvarezCommented Feb 13, 2010 at 22:16
Add a comment
|
3 Answers
$\begingroup$
$\endgroup$
The representation theory of sl(2) is the basis for understanding the representation theory of any Lie algebra, and it is not difficult at all. An extremely readable account of it is given in chapter 11 of Fulton and Harris' "Representation Theory, a First Course".
answered Nov 14, 2009 at 18:31
$\begingroup$
$\endgroup$
2
You can't assume the module is irreducible, since many aren't!
However, if you want to learn something about modules of $sl_2$ it helps to make the following observations:
- Each module is a sum of irreducible modules
- It's easy to describe all irreducible modules.
In particular, when you'll be solving 2, you'll notice that there is exactly one irreducible modle of dimension 2. In fact, there is one of dimension 3, one of dimension 4, etc.
answered Nov 14, 2009 at 10:11
-
1$\begingroup$ In fact, for n > 2, the unique n-dimensional sl(2)-module is the (n-1):st symmetric power of the 2-dimensional one. So if in alyscia's situation one is asked to prove a property P for all sl(2)-modules, such that: 1) if M and N have property P , then M+N have property P; 2) if M has property P, then sym^n(M) has property P; then it suffices to prove it for the irreducible 2-dimensional module $\endgroup$ Commented Nov 14, 2009 at 12:35
-
$\begingroup$ 1. is a general fact about semisimple Lie algebras. 2. can be solved for semisimple Lie algebras in general (indeed, one can give an explicit description as the quotient of the enveloping algebra by a certain ideal), but the solution heavily uses the theory of sl2. $\endgroup$ Commented Feb 14, 2010 at 0:48
$\begingroup$
$\endgroup$
The irreducible modules of sl2 are of the form Sym^r(\C^2), i.e homogeneus polynomials of degree r and 2 variables x,y. So you have irreducible modules of dimension 1,2,3,4,.... etc.
Even more, sl2 acts on these spaces by x\diff y, y\diff x and the operator (x\diff y)(y\diff x)-(y\diff x)(x\diff y). From this it's easy to calculate the vectors with maximal weights in each irreducible representation.
