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Hi,
The complex simple algebraic group $Sp_{m,\mathbb{C}}$ of $2m$-dimensional space $V$ has, for $m≥2$, an irreducible representation of dimension $m(2m−1)−1$ in a subspace of codimension $1$ of the space $\Lambda^2V$. Is it the irreducible representation of smallest dimension after $V$ itself?

Thank you.

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It follows from the Weyl dimension formula that the fundamental representations have minimal dimensions. So you only have to check the dimensions of these. –  Vít Tuček Jan 10 '13 at 17:52
    
Well, minimal was not the right word. What I meant to say that the representation with minimal dimension among all representations is the same as the representation which has the smallest dimension amongst fundamental representations. –  Vít Tuček Jan 10 '13 at 17:55
    
@robot: OP wants to know the second smallest dimension of a nontrivial irreducible representation of a group of type $C_m$. –  Mikhail Borovoi Jan 10 '13 at 18:59
    
The question itself just involves classical ideas, so it might be answered in the literature(?); anyway it's really about the Lie algebra, which may be a tag to add. As robot observes, Weyl's dimension polynomial gives bigger values for non-fundamental weights. The fundamental irreducibles are close to the exterior powers, with dimensions given as a difference of two binomial coefficients. A quick calculation for $m=3$ gives (I hope)` dimensions 6, 14, 14, but after that the later ones grow faster. Presumably the answer to your question is yes, but it needs an argument. –  Jim Humphreys Jan 10 '13 at 21:08
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The answer is yes, see my answer to mathoverflow.net/questions/118472/…. –  Mikhail Borovoi Jan 10 '13 at 22:27

2 Answers 2

up vote 3 down vote accepted

The irreducible complex representations of the simply connected simple group $G=Sp_{r,{\mathbb C}}$ of type $C_r$, for $r>1$, of dimension $n<{\rm dim}\ G$ are listed in the paper of Andreev, Vinberg, and Elashvili, Table 1 (see also the Russian version). They are the fundamental irreducible representations $R(\pi_1)$ of dimension $2r$, $R(\pi_2)$ of dimension $2r^2-r-1$, and, for $r=3$, $R(\pi_3)$ of dimension 14. For all $r\ge 2$, $r\neq 3$, we have ${\rm dim}\ R(\pi_1)=2r<2r^2-r-1={\rm dim}\ R(\pi_2)$, hence $R(\pi_2)$ is the nontrivial irreducible representation of second smallest dimension. For $r=3$, as Jim Humphreys noted, the dimensions are $6,14,14$, so ${\rm dim}\ R(\pi_2)={\rm dim}\ R(\pi_3)>{\rm dim}\ R(\pi_1)$, and $R(\pi_2)$ is a nontrivial irreducible representation of second smallest dimension.

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@Gabriel-Kj: Since you have accepted my answer and have thanked Jim, you can also vote up both answers... –  Mikhail Borovoi Jan 15 '13 at 19:01

It may be useful to expand my comments. The question involves Lie type $C_m$ with $m \geq 2$. Without developing Lie group or algebraic group language, it's enough to work with a simple Lie algebra over $\mathbb{C}$ of this type. Using the standard numbering of vertices in the Dynkin diagram, let $E_i$ be the fundamental representation of highest weight $\varpi_i$ for $i= 1, \dots, m$. Here $E_1$ is the standard module of dimension $2m$. For the others, there are numerous classical references. There is a thorough discussion of the construction in Bourbaki Groupes et algebres de Lie (also in English translation), Chap. VIII, $\S13$, no. 3, (IV). In particular, the well-known dimension formula is made explicit:

$$\dim E_i = \binom{2m}{i} - \binom{2m}{i-2} \text{ for } i \geq 2$$.

Clearly $\dim E_1 > \dim E_2$. The claim is that $\dim E_2 \geq \dim E_j$ for all $j >2$. This should require an elementary combinatorial comparison, not involving any Lie theory, though it would be interesting to see a conceptual argument.

Granted this inequality, Weyl's dimension formula (as already noted) will complete the desired argument for $ E_2$ being the second smallest nontrivial irreducible representation. The formula involves a fraction, whose denominator can be ignored. The numerator is an integral polynomial in the highest weight, which obviously grows larger as the coordinates of that weight increase relative to the $\varpi_i$.

P.S. I don't want to leave the impression that I've written down a formal proof. It's only a proof-scheme, but should be fairly easy to complete using straightforward methods. For the comparison between fundamental and non-fundamental weights, you'd need to look at the root system $C_m$ (say at the end of Chapters IV-VI of Bourbaki): a rough comparison of how often $\alpha_1, \alpha_2$ occur in each positive root shows for instance how the Weyl dimension for $2\varpi_1$ exceeds the dimension for $\varpi_2$, etc. I don't recall seeing all of this written down anywhere, but if there is motivation to do so it should be elementary to complete.

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Thanks to your great comments. –  purelymath Jan 14 '13 at 8:54
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@Gabriel-Kj: I don't have ready access to the paper cited by Mikhail, so was giving a more hands-on approach to the question. In any case, it might be appropriate to upvote at least one of our answers ;-) (Though I did resolve quite recently to avoid all future use of smileys.) –  Jim Humphreys Jan 15 '13 at 16:56
    
@Jim: You can find the Russian version for free in mathnet.ru/links/419af2a1d9d33839a49ff8898866b056/faa2839.pdf . There is just a table, no details of calculations. –  Mikhail Borovoi Jan 15 '13 at 18:53
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@Mikhail: The question and the two answers was very helpful for me. Thanks to all of you. As you said there is no details to the table in the cited article. Could you tell me where I can find some details to the table. I need to check possibility of some embeddings like the question. –  Nrd-Math Jan 15 '13 at 22:47
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@Nerd-Math: For split groups (and I guess compact groups), it's easy to decide via the highest weight, with triviality on the center iff the weight is in the root lattice. In your example, usually not. –  Jim Humphreys Jan 26 '13 at 14:49

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