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I am reading the book Introduction to lie algebras and representation theory. I have some difficulty in understanding some parts of the book for Freudenthal's formula.

  1. Page 120, line 6, why $m(\mu-i\alpha)=n_0+\cdots+n_i$? Here $m=\langle\mu,\alpha\rangle$ is an integer and $\mu,\alpha$ are weights. So $m(\mu-i\alpha)$ is an element in $H^{*}$ where $H$ is the Cartan subalgebra. But $n_0+\cdots+n_i$ is an integer. How to prove this formula? What does figure 1 on this page mean?

  2. Page 121, section 22.3, line 7, why $(\mu,\mu)=\sum_{i,j}a_ia_j\kappa(h_j,h_i)$? Here $\mu(h_i)=\sum_{j}a_j\kappa(h_j,h_i)$.

  3. Page 123, example 22.4, I cannot compute some of the data in table 1. For example, $(\mu+\delta, \mu+\delta)$. When $\mu=\lambda=\lambda_1+3\lambda_2$, $\delta=(\alpha_1+\alpha_2)/2$, $(\mu+\delta, \mu+\delta)=(13\alpha_1/6+17\alpha_2/6, 13\alpha_1/6+17\alpha_2/6)$. But $(\alpha_1,\alpha_1)=(\alpha_2,\alpha_2)=2$, $(\alpha_1,\alpha_2)=\sqrt{2}\times \sqrt{2}\cos (2\pi/3) = -1$. I cannot get $(\mu+\delta, \mu+\delta)=28/3$.

Thank you very much.

1 Answers 1

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[Edit: Massive editing coming up. I now have my copy of Humphreys with me]

1) Here $m(\mu)$ denotes the multiplicity of a weight $\mu$ in the representation under discussion. It is perhaps a bit unfortunate that here $m$ is both a constant $m=\langle\mu,\alpha\rangle$ and a mapping $m=m_V:\Lambda\rightarrow\mathbf{Z}$. You have to look at the context to figure out which it is at each point.

2) This goes back to §8. You are adviced to review it. I had to, too! It is easy to forget the origin of the inner product (the Killing form), after you have absorbed the geometric viewpoint, where we identify the weights as a lattice in a Euclidean space. Here $(\mu,\mu)= \kappa(t_\mu,t_\mu)$, so by plugging in the expansion $t_\mu=\sum_ia_ih_i$ we get $$ (\mu,\mu)=\sum_{i,j}a_ia_j\kappa(h_j,h_i)= \sum_i a_i\left(\sum_ja_j\kappa(h_j,h_i)\right)=\sum_ia_i\mu(h_i)=\sum_i\mu(h_i)\mu(k_i).$$ [/Edit]

3) Surely $\delta=\lambda_1+\lambda_2=(\alpha_1+\alpha_2)$. Also $\cos2\pi/3=-1/2$. But why don't you use the basis of the fundamental weights? $||\lambda_1||^2=||\lambda_2||^2=1/3$ and the angle between them is $\pi/3$, so $||a\lambda_1+b\lambda_2||^2=(a^2+ab+b^2)/3.$ Here $\mu+\delta=2\lambda_1+4\lambda_2$, so the squared norm is $(2^2+2\cdot4+4^2)/3=28/3$.

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    thank you very much. But by definition (page 122, the equation below (12)), $\delta=(\alpha_1+\alpha_2)/2=(\lambda_1+\lambda_2)/2$.2011-07-08
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    No. In all these character formulas $\delta$ equals the sum of the fundamental weights, and also half the sum of all positive roots. This example is about $A_2$, so the positive roots are $\alpha_1,\alpha_2$ and $\alpha_1+\alpha_2$. Their sum is $2\alpha_1+2\alpha_2$, and half of that is $\alpha_1+\alpha_2$.2011-07-08
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    What Humphreys denotes $\delta$ is often also denoted $\rho$. But whatever the notation, it *is* an element of the weight lattice, IOW is an integer linear combination of the fundamental weights. Either there is a typo in that formula, or you misread "simple root" in place of "positive root". IIRC Jantzen uses $\rho$. That is also used in e.g. http://en.wikipedia.org/wiki/Weyl_character_formula .2011-07-08
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    BTW. I don't mean to sound negative. It is good for you to work your way thru these examples. I recall needing to backpedal several times in this chapter. I produced several sheets of papers with the weight lattices of $A_2$, $B_2$ and $G_2$ drawn on them. It really helps when working on things like Weyl's character formula. Courage and patience!2011-07-08
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    @user9791: Hopefully the edited answer helps you. May be you have figured all this out on your own by now?2011-07-11
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    thank you. I still cannot deduce that $m(\mu-i\alpha)=n_0+\ldots+n_i$. By definition, we have to find a basis of the weight space $V_{\mu-i\alpha}(\lambda)$. It seems that we can use the result that each weight space of a highest weight module of $sl_2$ is of dimension $1$. But why they know $m(\mu-i\alpha)=n_0+\ldots+n_i$ without finding a basis of $V_{\mu-i\alpha}(\lambda)$?2011-07-11
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    @user9791: That fact follows from the study of $W=V_\mu+V_{\mu-\alpha}+\cdots+V_{\mu-m\alpha}$ as a representation of $S_\alpha$. This is really $sl_2$-theory, because $S_\alpha$ is just a copy of $sl_2$. As such an $sl_2$-module $W$ has weights $m,m-2,m-4,\ldots,-m$, so it splits into a direct sum $$ W\cong \bigoplus\sum_{i=0}^{[m/2]} V(m-2i)^{n_i}.$$ When we keep trac of the action of all of $H$, we need to replace $m-2i$ with $\mu-i\alpha$ here.2011-07-12
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    ... and the weight $\mu-j\alpha$ (or $m-2j$, if we restrict ourselves to $sl_2$) has multiplicity 1 in $V(m-2i)$, if $i\le j$ (again by the known $sl_2$-theory) and 0 otherwise.2011-07-12