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(I did not find a solution of my problem in any forum so far. Sorry if it exists...)

I am new to Lie-Algebras and representations and actually do not need the mathematical background... I need only to apply the Weyl-Charakter-Formula on $so(5)$ ($B_2$-Type, $\text{rank }(so(5))=2$). In Fulton and Harris 'Representation Theory' they give on p.400 what I think is the same as https://en.wikipedia.org/wiki/Weyl_character_formula: \begin{align} \text{ch}_{\lambda} = \frac{\sum_{w\in\mathscr{W}} (-1)^{\ell(w)}e^{w(\lambda+\rho)}}{e^\rho \prod_{\alpha\in \Phi^+}(1-e^{-\alpha})} \end{align} So far I know/think the following:

$\Phi^+$ is the set of positive roots; I choose the simple roots to be $(-1,1)$ and $(1,0)$, so that the Weyl-Vector $\rho$ is $(0,1/2)$. The Weyl-Group $\mathscr{W}$ is isomorphic to the dihedral group of order $8$. $\ell(w)$ is the lenght of $w$ defined as the minimal number of secessary reflexions of simples roots.

Applying this on $su(2)$ was no problem, but I don't understand the meaning of $e^{\rho}$ and $e^{w(\lambda+\rho)}$. I was told once, that this is defined by $e^\rho:=e^{\rho_1}\cdot e^{\rho_2}$. Is this correct? Also this should still make sence if $e^{\beta\cdot\rho}$, where $\beta$ is a non-integral number.

Also I am confused by the fact, that the irreducible representation does only appear in $\lambda$, the highest weight in the fundamental Weyl-Chamber (Humphreys' Definition) which in turn is part of the Root-Lattice. Is that correct?

As I said, I just started learning about this, so please keep it as simple as possible. Thanks for any help! I somehoy cannot give usefull tags, so I will saty very general here...

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  • $\begingroup$ §16.4.6.4 in Soergel's notes ( home.mathematik.uni-freiburg.de/soergel/Skripten/XXALLES.pdf ) might help clarify the meaning of the $e^\alpha$ (provided that their meaning on the Wikipedia is the same as in Soergel's notes). Essentially, $e^\alpha$ is just a "wrapper" for $\alpha \in \mathfrak{h}^\ast$ whose purpose is to write the additive group $\mathfrak{h}^\ast$ multiplicatively (i.e., to find a multiplicative group isomorphic to the additive group $\mathfrak{h}^\ast$). Namely, we let $e^{\mathfrak{h}^\ast}$ denote the ... $\endgroup$ – darij grinberg Jul 28 '15 at 18:20
  • $\begingroup$ ... set of all $e^{\alpha}$ with $\alpha \in \mathfrak{h}^\ast$. Then, $e^{\mathfrak{h}^\ast}$ becomes a group, as follows: $e^{\alpha} e^{\beta}$ is defined to be $e^{\alpha+\beta}$, and $\left(e^{\alpha}\right)^u = e^{u\alpha}$ for $\alpha, \beta \in \mathfrak{h}^\ast$ and $u \in \mathbb{C}$ (this defines inverses, in particular). Thus, $\left(e^{\mathfrak{h}^\ast},\cdot\right) \cong \left(\mathfrak{h}^\ast, +\right)$ as groups. The reason why we are doing this is that we want to take the group ring of $\mathfrak{h}^\ast$, but working with group rings of additive groups is ... $\endgroup$ – darij grinberg Jul 28 '15 at 18:21
  • $\begingroup$ ... extremely painful (you need the symbol $+$ for two different things), so we prefer to make that additive group multiplicative. $\endgroup$ – darij grinberg Jul 28 '15 at 18:21
  • $\begingroup$ Related. $\endgroup$ – Jyrki Lahtonen Jul 28 '15 at 19:10
  • $\begingroup$ You can think of those $e^\lambda$:s as follows. If $\lambda=n_1\lambda_1+n_2\lambda_2$ (with $\lambda_1,\lambda_2$ the fundamental weights), and we declare $X_1=e^\lambda_1$, $X_2=e^\lambda_2$, then $$e^\lambda=X_1^{n_1}X_2^{n_2}.$$ In other words, those $e^\lambda$:s are just a bunch of placeholdes for compactly describing the character of the module. $\chi_V=\sum_\lambda m(\lambda)e^\lambda$ where $m(\lambda)$ is the multiplicity of the weight in $V$. $\endgroup$ – Jyrki Lahtonen Jul 28 '15 at 19:15

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