1
$\begingroup$

Hi I am trying to find weights for representation of the lie algebra $sl(n,c)$ .

I understand how to find weights for representation of $sl(3,c)$ on $C^3$. i.e the fundamental representations. I can do this by first finding roots via adjoint representation. and then by hand find weights.

however i am totally stuck on finding weights of representation in different vector spaces.

for example how do i get weights of $sl(3,c)$ on vector space $C^5$ or $C^4$. really stuck on even associating a 5X5 matrix to the basis of cartan subalgebra of $sl(3,c)$.

I am self studying this topic. this question is just my idea cant find any examples or anything related to this in the textbooks or internet.

Note: I know how to find roots, fundamental weights however i am stuck on how to find weights of a particular representations that is the reason for asking this question.

Thanks for any help.

$\endgroup$
  • $\begingroup$ This is discussed in Fulton & Harris book Representation Theory (Graduate Texts in Mathematics, Volume 129). $\endgroup$ – Spenser Sep 1 '16 at 7:11
  • $\begingroup$ @Spenser Thanks for comment. I checked Fulton & Harris for $sl(3,C)$ he does lot of examples but all in the standard $V=C^3$ or its dual or their tensor product spaces. for $sl(4,C)$ he does the same again as in he works on vector space $C^4$ , its dual and their several inner product spaces. My question is if i have $sl(3,C)$ and take representation on $C^5$ that's not answered in the book. If i have missed and if its actually in the book can you give me perhaps page number i looked everywhere. Thanks again $\endgroup$ – MRK Sep 1 '16 at 8:10
0
$\begingroup$

Every finite-dimensional complex representation of $sl_n(\mathbb C)$ (or every other semisimple complex Lie algebra) is completely reducible, i.e. decomposes as sum of irreducible representations. To classify those irreducible representations ("irreps") is one of the main points of any book, article, blog post or set of lecture notes on this, of which there are plenty.

Note that one always has the $1$-dimensional trivial irrep.

The results are kind of easy for $sl_2$, where for each dimension $k$, there is exactly one irrep; for $k=1$, this is the trivial one, for $k=2$ the defining ("standard") and fundamental one, for $k=3$ the adjoint one, then it goes higher.

For $sl_3$, it's already more involved, as the irreps can be classified by pairs $(p,q) \in \mathbb Z^{\ge 0}$, where the irrep indexed by $(p,q)$ has dimension $\frac12 (p+1)(q+1)(p+q+2)$. So we have

  • one $1$-dimensional irrep, the trivial one, index $(0,0)$
  • two $3$-dimensional irreps, both fundamental (in math terminology), one of them the defining ("standard") one and the other its dual, indices $(0,1)$ and $(1,0)$
  • one $8$-dimensional irrep, the adjoint one, index $(1,1)$
  • then it goes higher.

So if you want to see a representation of $sl_3$ on $\mathbb C^k$ with dimension $k=4$ or $5$ (or $6$ or $7$), you cannot have an irreducible one. For $k=4$ you can have a sum of the trivial one and either of the $3$-dimensional ones; or four trivial ones. For $k=5$ you have a sum of either of the $3$-dimensional ones and two copies of the trivial one; or five trivial ones. For $k=6$, you could now add the two $3$-dimensional ones. Or choose one $3$-dimensional one and add it to itself. Or choose one $3$-dimensional one and add three trivial ones. I think you can see how it goes from there. Starting at $k=8$ you'll have another building block, so plenty of options for $k=9, 10, ...$ etc.

For $sl_4$, the first irreps occur in dimension $k=1$ (trivial), $k=4$ (two: the standard one, and its dual), $k=6$ (another fundamental one, in math terminology), and it goes higher.

One of the greatest results is that all these building blocks occur in some tensor product of the fundamental ones. So when you complain in your comment

"for $sl(3,C)$ he does lot of examples but all in the standard $V=C^3$ or its dual or their tensor product spaces. for $sl(4,C)$ he does the same again as in he works on vector space $C^4$ , its dual and their several inner product spaces."

you are touching on (but slightly missing) the coolest result: Namely, that by doing that, "he" (the authors) are actually talking about all representations. Namely, all irreps are contained as summands in what they do there. They leave the easy part of putting together those irreps to match some dimension $k$ you like to you (or me, as outlined above).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.