# Ordering on the weight lattice

When given a finite dimensional complex Lie algebra $\mathfrak{g}$ that is also semisimple and a choice of Cartan subalgebra $\mathfrak{h}$ we may talk about its weight lattice $\Lambda_{W}$ in $\mathfrak{h}^{*}$. In order to study the representation theory of $\mathfrak{g}$ we introduce a total ordering on $\Lambda_{W}$ by means of defining a real linear map $l$ from $\Lambda_{W} \otimes_\mathbb{\mathbb{z}} {\mathbb{R}}$ to $\mathbb{R}$ and that is irrational with respect to $\Lambda_{W}$ (first question: could you please define what irrationality in this context means?)

Now my main problem is this:

In practice, say when $\text{dim}_\mathbb{C}\mathfrak{h}=2,$ what happens is that we draw a line in an illustration of $\mathfrak{h}^{*}_\mathbb{R}$ (the real span of the roots; $\mathfrak{h}^{*}_\mathbb{R}\subset\mathfrak{h}$ is a real inner product space with respect to the Killing form) on paper and in some way the line defines $l$ (second question: could you please explain the convention on how drawing a line defines $l$).

The most important/practical third question I want to ask is: given a drawn line could you use it to immediately tell which of two weights in $\Lambda_{W}$ is larger? A friend told me that it's just a matter of perpendicular distances (the plain definition for lines drawn on paper) from the line but I don't think this is true, especially since it seems to me that the arbitrariness of our illustration of $\mathfrak{h}^{*}$ should mean there ought to be no relation to a fixed notion of perpendicularity.

Suppose that $\Lambda\subset R^n$ is a lattice, i.e. a rank $n$ free abelian subgroup whose real span is the entire $R^n$. A linear function $l: R^n\to R$ is irrational with respect to the lattice $\Lambda$ if its restriction to $\Lambda$ is a monomorphism. In other words, if $v_1,...,v_n$ are generators of the lattice, their images are rationally independent real numbers. Given this, one defines an order on the lattice by: $$u< v \iff l(u)< l(v).$$ This also answers your second question: to compare two elements with respect to this order, just compute their images! Now, if you want to do this more geometrically, you can assume (by rescaling) that $l$ has unit norm. Therefore, it can be identified with the orthogonal projection to a certain 1-dimensional linear subspace $L$ (a line) in $R^n$. (Which, in turn, is identified with ${\mathbb R}$ by introducing a basis in this subspace.) Informally, you can think of the orthogonal projection to $L$ as the "height" above the hyperplane $H$ orthogonal to $L$. Then $l(v)$ can be understood as the "signed distance" to $H$ (not the ordinary distance your friend was using), just by the elementary linear algebra: The orthogonal projection of $v$ to $P$ is given by the formula $v\mapsto v - f(v)$, $f(v)\in L$. Then $d(v, P)= |v - (v-f(v))|=|f(v)|$. Thus, $f(v)=\pm d(v, P)$. Hope it helps.