Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In the book of Humphreys, page 139, Weyl character formula is $$\left(\sum_{w\in W} \operatorname{sn}(w)\epsilon_{w\delta}\right) * \operatorname{ch}_{\lambda} = \sum_{w\in W} \operatorname{sn}(w) \epsilon_{w(\lambda + \delta)}\tag{1} $$ where $*$ is convolution but not usual multiplication of functions. In other books, Weyl formula is given by $$ \operatorname{ch}_{\lambda} = \frac{\sum_{w\in W} \operatorname{sn}(w) \epsilon_{w(\lambda + \delta)}}{\sum_{w\in W} \operatorname{sn}(w)\epsilon_{w\delta}}.\tag{2}$$ Are these two formulas (1) and (2) equivalent? In some other books, Weyl formula is given by $$ \operatorname{ch}_{\lambda} = \frac{\sum_{w\in W} \operatorname{sn}(w) \epsilon_{w(\lambda + \delta)-\delta}}{\sum_{w\in W} \operatorname{sn}(w)\epsilon_{w\delta - \delta}}.\tag{3}$$ Is this because in (2) $w \in W$ act as dot pruduct and (3) $w\in W$ act as usual Weyl group action? If we are given a highest weight $\lambda$ explicitly, say, $\lambda = 3\omega_1 + 2\omega_2$, where $\omega_1, \omega_2$ are fundamental weights of $\mathfrak{g} = \mathfrak{sl}_3$ (type $A_2$), how can we compute explicitly the character of the irreducible $\mathfrak{g}$-module $V_{\lambda}$ with highest weight $\lambda$ using formula (2)? I am confused with the form of the formula since the character should be of the form $\operatorname{ch}(V_{\lambda}) = \sum_\mu\dim(V_\mu)\epsilon_\mu$ but the Weyl formula is not of this form. Thank you very much.

share|improve this question

1 Answer 1

up vote 4 down vote accepted

Hmm. IMHO convolution is the natural product on a group ring. May be you are used to viewing them as functions as opposed to elements of the group ring $\mathbf{Z}[\Lambda]$, where $\Lambda$ is the additive group of weights (= a free abelian group generated by the fundamental dominant weights)? A function $f:\Lambda\rightarrow \mathbf{Z}$ with a finite support is turned into an element of the group ring naturally as $\sum_{\lambda\in\Lambda}f(\lambda)e_{\lambda}$. The quantities $e_\lambda$ are multiplied by the natural formula $e_{\lambda}*e_{\mu}=e_{\lambda+\mu}$, and the convolution is just the linear extension of this (according to the distributive law).

Weyl character formula is actually very useful for the rank two cases, where you can just stare at the diagram. As I don't have a useful tool for drawing a sequence of 2-D diagrams of weight lattices, let's do an example of the case $A_1$, where you should know the formal character anyway (so we can use that to test the formula). In that case we can identify the weight $m\lambda_1$ with the integer $m$. In the group ring we then have $e_{m\lambda_1}=e_{\lambda_1}^m=z^m$, where I use the abbreviation $z=e_{\lambda_1}$. With this identification $\delta=\lambda_1=1, e_\delta=z$. The only non-trivial element of the Weyl group in this case is the negation of weights, and that has length =1 and hence a negative sign. Therefore $$ \sum_{w\in W}sn(w)e_{w\delta}=e_\delta - e_{-\delta}= z -z^{-1} $$ and $$ \sum_{w\in W}sn(w)e_{w(\lambda+\delta)}=e_{\lambda+\delta} - e_{-\lambda-\delta}= z^{m+1} -z^{-m-1}, $$ where again $\lambda=m\lambda_1, m\ge0$. This time we know from $sl_2$-theory that $$ ch_\lambda=\sum_{\mu}\dim(V_\mu)e_\mu=z^m+z^{m-2}+z^{m-4}+\cdots+z^{4-m}+z^{2-m}+z^{-m}, $$ because for the irreducible module $V(m)$ of highest weight $m$ the dimensions of the weight space $V(m)_k$ is equal to 1, if $k$ is in the arithmetic progression $m,m-2,m-4,\ldots,-m+2,-m$ and 0 otherwise. So in this case the Weyl character formula says that $$ z^m+z^{m-2}+z^{m-4}+\cdots+z^{4-m}+z^{2-m}+z^{-m}=\frac{z^{m+1} -z^{-m-1}}{z -z^{-1}}. $$ This identity in the group ring $\cong \mathbf{Z}[z,z^{-1}]$ is easy to verify and/or derive. Either you can write the l.h.s. as a geometric sum, or you can multiply both sides with the denominator $z-1/z$, and the usual avalanche of termwise cancellations gives you the result. Note that the group ring is an integral domain, so your equations (1) and (2) are equivalent.

In the case of $A_2$ we can proceed similarly. For example, if we write $z_i=e_{\lambda_i}$ and $s_i=s_{\alpha_i}$, then $\delta=\lambda_1+\lambda_2$, $s_1(\delta)=\delta-\alpha_1=-\lambda_1+2\lambda_2=\alpha_2$, $s_2(\delta)=\delta-\alpha_2=2\lambda_1-\lambda_2$, $s_1s_2(\delta)=-2\lambda_1+\lambda_2$, $s_2s_1(\delta)=\lambda_1-2\lambda_2$ and finally $s_1s_2s_1(\delta)=-\lambda_1-\lambda_2$. Therefore (taking the parity of the length of the group element into account) $$ \sum_{w\in W}sn(w)e_{w\delta}=z_1z_2-z_1^{-1}z_2^2-z_1^2z_2^{-1}+z_1^{-2}z_2+z_1z_2^{-2}-z_1^{-1}z_2^{-1}, $$ where I wrote the terms (i.e. the elements of $W$) in the same order as above.

To do your example, you need to also compute the similar expansion for $\sum_{w\in W}sn(w)e_{w(\lambda+\delta)}$. Then the Weyl character formula tells you that $ch_{\lambda}$ is the element of the group ring $\mathbf{Z}[z_1,z_2,z_1^{-1},z_2^{-1}]$ that is the (unique) solution of the equation (1) or (of equation (2)).

This answer has just become unmanageably long, so I cannot continue. I recommend that you first work out the cases $\lambda=\lambda_1$, $\lambda=\lambda_2$ and $\lambda=\delta$ where you know the answer already: the first rep in this list is the natural 3-dimensional rep of $sl_3$. The second is its dual, and the last is the 8-dimensional adjoint representation as its highest weight is also the highest root. If your have problems with those, ask another question showing where you got stuck, and somebody here will take a look. Ping me, if necessary.

Edit: I just recalled that I already wrote an example on how to apply Weyl's formula in another answer.

share|improve this answer
    
thank you very much. In type $A_1$, why $\delta \neq \alpha_1/2$? Is the root system of type $A_1$ as follows: $\{\alpha_1, -\alpha_1\}$? –  LJR Aug 26 '11 at 19:48
    
I see. $\lambda_1 = \alpha_1/2$. –  LJR Aug 26 '11 at 19:52
    
@user9791: Correct. –  Jyrki Lahtonen Aug 30 '11 at 9:36

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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