I was wondering if anyone knows where I can find a formula for the dimension of an irreducible module of highest weight $\Lambda$ expressed only in terms of the Young diagram corresponding to $\Lambda$.
Thank you.
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2$\begingroup$ I'm puzzled about the reason for wanting this sort of formula, since Weyl's dimension formula (for any Lie type) is straightforward to compute given a highest weight. Anyway, for general or special linear groups and their Lie algebras, there are a number of textbooks which show how to realize representations concretely in tensor spaces: Goodman-Wallach, Procesi, etc. Much of this is in the setting of Schur-Weyl duality, where the easier results about symmetric groups given in terms of Young diagrams and tableux imply results for the Lie groups. $\endgroup$– Jim HumphreysCommented Jan 31, 2012 at 1:36
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1$\begingroup$ More to the point, I think in order to ask this question, you are required to specify what aspect of the Weyl dimension formula you think is unsuitable. I can imagine reasonable answers to this, but I would need to know yours before I could answer the question. $\endgroup$– Ben Webster ♦Commented Jan 31, 2012 at 2:26
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1$\begingroup$ Hook-length formula? $\endgroup$– Vít TučekCommented Jan 31, 2012 at 14:14
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$\begingroup$ Im writing in a hurry so please forgive carelessness. Indeed the formula seems to be something like this: You take the diagram and fill in the squares with the numbers such that the $(i,j)$ square has $n+j-i$ where the $n$ is from $\mathfrak{sl}_n$. Then the formula should be the product of these numbers divided by the product of the hooklengths of each box. This is probably contained in the exercises of those texts that were recommended. $\endgroup$– Michael SunCommented Jan 31, 2012 at 21:35
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$\begingroup$ Something about this can be found in Fulton-Harris's book on rep theory Exercise 6.4 around page 78 and proposition 15.15 on page 223. Thank you for posting replies, I have found your answers and comments to be very useful. $\endgroup$– Michael SunCommented Feb 3, 2012 at 19:37
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3 Answers
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One influential older source to consult is the concise Springer Lecture Notes No. 682 by Gordon James The Representation Theory of the Symmetric Groups (1978). Section 26 applies the symmetric group theory to general linear groups, using the language of "Weyl modules" and partitions. See in particular his Theorem 26.5 for the dimension formula in terms of semistandard partitions. It takes some work, however, to get into his notational scheme, though his overall approach to the subject has the virtue of leading naturally into fields of prime characteristic while also treating the classical Schur-Weyl duality.
Note by the way that in the classical setting there is no essential difference between the group and Lie algebra theories here, while the transition from general linear to special linear is quite easy.
ADDED: I'm still unsure what is really needed, but I don't see the partition formulation of weights for general (or special) linear groups or Lie algebras to be helpful in considering duality. In the notes by James, or in many textbooks like Fulton-Harris (with varying notation), one studies the irreducible representations of a fixed $GL_d$ on the $n$-th tensor powers of the natural $d$-dimensional module for variable $n$ by using the representation theory of the symmetric groups $S_n$. Here a partition of $n$ into at most $d$ parts corresponds nicely to a highest weight. But it's tricky to bring the transpose (conjugate) partition into this picture, since its number of parts might exceed $d$.
The role of the transpose partition for $S_n$ itself is more straightforward: Partitions of $n$ naturally index irreducible representations of $S_n$. Then the transpose partition indexes the product of the given representation and the sign representation, with no change in dimension. But the relationship of this machinery (via Schur functors) to representations of each fixed $GL_d$ is indirect.
answered Jan 31, 2012 at 14:46
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It is quite possible formulas exist or can be found. I don't know of any off the top of my head, but I know that it amounts to counting semistandard tableaux. I do not know of a formula for this number based on the shape of the diagram, but perhaps someone else can elaborate.
This fact comes from the theory of quantum groups; that is, a certain $q$-deformation of the enveloping algebra. This deformation breaks some of the symmetry of the usual enveloping algebra and allows us to see that all modules have certain bases called crystal bases (one of many remarkable facts in quantum groups) with very nice properties.
In this theory one can show that all modules of the quantum group are deformations of modules of the usual enveloping algebra, and that the Weyl character formulas are the same, so we just need to count the size of the crystal bases. Using properties of tensor products of these bases, you can show that each simple module's crystal base can be parametrized by certain Young tableaux in the shape of the Young diagram corresponding to the weight; in the $A_n$ case, it is just semistandard tableaux in the letters $1,2,\ldots, n+1$, but for other types the constraints get a bit more complicated.
Some good introductory references I know for this theory are Jantzen's Lectures on Quantum Groups and Hong and Kang's Introduction to Quantum Groups and Crystal Bases
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Thank you both for taking the time to reply. The main reason I want such a formula is because I want to show that dimensions are preserved when taking the transposed diagram.
It is my hope that the formula will make this obvious and will also be easy to generalise to the q-dimension.
It might be the case that Weyl's formula is perfectly adequate to achieve this, but I just thought it would be interesting to see this other formula.
My overall goal is to say something about the level rank duality of representations of affine $\mathfrak{sl}_n$ at level $m$ (where the duality is between the $n$ and the $m$).
P.S Hi Ben, hope you are well in your new position.
Prof. Humphreys, I really enojoyed your book!
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$\begingroup$ You should add this to the question, so as to reserve the answer box to, well, answers! (You cannot yet leave comments, but you can edit your own questions) $\endgroup$ Commented Jan 31, 2012 at 6:59
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$\begingroup$ Ah yes, thank you I should do that. I am not yet quite sure how to register an account, I will try to be more on top of these things in the future. $\endgroup$ Commented Jan 31, 2012 at 7:29
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$\begingroup$ The dimension is not preserved when taking the transposed diagram. The dimension is preserved for the symmetric group but not $\mathfrak{sl}_n$. On the other hand, the $\mathfrak{sl}_n$ dimension is preserved by taking the complement in a rectangle with $n$ rows (and rotating 180 degrees). $\endgroup$ Commented Feb 1, 2012 at 1:33
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$\begingroup$ Ah yes, but the quantum dimension is preserved right? Is it true that it was you who proved the formula for the quantum dimension? $\endgroup$ Commented Feb 3, 2012 at 19:38