Given a simisimple Lie algebra $\mathfrak{g}$ and a weight $\lambda$. Let $W$ be the Weyl group of $\mathfrak{g}$. Is there an algorithm (reference or software) to compute the set $W \cdot \lambda$ (the set obtained by the action of $W$)? Many thanks.
[Math] Algorithms to compute the orbit of the action of the Weyl group of a semisimple Lie algebra on a given weight
lie-algebrasrepresentation-theory
Related Solutions
It depends on the Lie algebra. We have that $-w_0$ induces some permutation of the simple roots since it sends the positive Weyl chamber to itself. Since $-w_0$ respects inner products, it must induce an automorphism of the Dynkin diagram of $\mathfrak g$. This immediately tell us that $-w_0$ is the identity for the simple Lie algebras $B_n$, $C_n$, $E_7$, $E_8$, $F_4$, and $G_2$.
For $A_n$, the root system can be taken to have simple roots $$ \boxed{e_1-e_2}-\boxed{e_2-e_3}-\cdots -\boxed{e_n-e_{n+1}} $$ where the $e_i$ are standard basis vectors, and the inner product is the standard one. The Weyl group is generated by reflections through $e_i-e_{i+1}$, which switch $e_i$ and $e_{i+1}$, so the Weyl group is $S_{n+1}$, acting by permuting the $e_i$. If I can find an element of the Weyl group which takes the set of simple roots to the set of negatives of the simple roots, it must be $w_0$. The permutation sending $e_1,\dots, e_{n+1}$ to $e_{n+1},\dots, e_1$ does the trick, so we see that $-w_0$ reverses the order of the simple roots. That is, it sends the above Dynkin diagram to $$ \boxed{-e_{n+1}+e_n}-\boxed{-e_n+e_{n-1}}-\cdots -\boxed{-e_2+e_1} $$
For $D_n$, we can take the simple roots to be $$ \boxed{e_1-e_2}-\cdots -\boxed{e_{n-2}-e_{n-1}}<{\boxed{e_{n-1}-e_n}\atop\boxed{e_{n-1}+e_n}} $$ Here, the Weyl group again acts by permuting the $e_i$, but we also have reflection through $e_n+e_{n+1}$, which sends $e_n$ and $e_{n+1}$ to $-e_{n+1}$ and $-e_n$, respectively. Thus, the Weyl group can arbitrarily permute the $e_i$, and it can negate an even number of them. If $n$ is even, we can take $w_0$ to negate all the $e_i$, in which case $-w_0$ is the identity. If $n$ is odd, we can take $w_0$ to negate all the $e_i$ except $e_n$, in which case, $-w_0$ switches the two "horns" of the diagram.
For $E_6$, we can take the root system to be $$ \begin{align*} \boxed{e_1-e_2}-\boxed{e_2-e_3}&-\boxed{e_3-e_4}-\boxed{e_4-e_5}-\boxed{e_5-e_6}\\ &\qquad\qquad|\\ &\boxed{e_4+e_5+e_6} \end{align*} $$ Again, we can permute $e_1,\dots, e_6$ arbitrarily, and this time we have the additional operation of reflecting through $e_4+e_5+e_6$. I don't see an easy way to negate the set of simple roots using these operations. If you do, please add a comment. I feel like we might be able to embed the root system into $E_7$ (by adding a $\boxed{e_6-e_7}$ to the end), then say that the longest word negates all the simple roots (since there are no diagram automorphisms), then argue that the longest word for $E_6$ should "do the same thing" (this part is not clear to me), so $-w_0$ should be the identity.
Incidentally, since the dual of $V_\lambda$ is $V_{-w_0(\lambda)}$, this answers the following question: Which simple $\mathfrak g$ have the property that every finite-dimensional representation is isomorphic to its dual?
For question 2, whenever a group $G$ acts on a vector space $V$, it also acts on the dual space $V^*$ as follows. If $g\in G$ and $\alpha\in V^*$ then $g\alpha$ is the element of $V^*$ that (regarded as a function from $V$ to scalars) sends each $v\in V$ to $\alpha(g^{-1}v)$.
Perhaps a nicer way to remember the formula is the equivalent form $(g\alpha)(gv)=\alpha(v)$.
Best Answer
A word of warning: in this context the notation $w\cdot\lambda$ often means the so called dot action, where you conjugate the usual action of the Weyl group by a fixed translation, IOW $$ w\cdot\lambda=w(\lambda+\rho)-\rho, $$ where $\rho$ equals one half of the sum of the positive roots. I assume that you mean the usual action, because a solution to that gives you a solution to the problem w.r.t to the dot action as well.
I don't know of an implementation, so this may not be quite what you want, but (depending on the size of the Weyl group that interests you) the comments may still be useful.
The common theme in my comments is length reduction. The length $\ell(w)$ of an element $w$ of the Weyl group is the number of those positive roots that $w$ maps to a negative root or equivalently: $\ell(w)=|\Phi^+\setminus w(\Phi^+)|$. This is also equal to the number of reflections $s_i=s_{\alpha_i}$, $\alpha_i$ a simple root, in the shortest presentation of $w$ as a product of the generators $s_i$, $i=1,\ldots,n$, $n$ = the rank of the root system. We always have $\ell(ws_i)=\ell(w)\pm1$ and $\ell(s_iw)=\ell(w)\pm1$ all depending on whether $w(\alpha_i)$ (resp. $w^{-1}(\alpha_i)$) is a positive root or not in the former (resp. latter) case.
The first observation is that without loss of generality we can assume that $\lambda$ is a dominant weight. Each orbit has a unique dominant weight, and given any weight it is easy to map it back to the (closure of the) dominant Weyl chamber by a sequence of simple reflections: if $(\lambda,\alpha_i)<0$, then $s_i(\lambda)$ is 'closer to being dominant than $\lambda$ is. Rinse. Repeat.
Assuming that $\lambda$ is dominant, I would then build the orbit recursively as a union of lists $R_i$ of weights $i=0,1,\ldots,|\Phi^+|$. The list $R_i$ consists of those weights $w(\lambda)$ such that A) $\ell(w)=i$, and B) $w(\lambda)$ does not appear on any of the earlier lists $R_j$, $j<i$. A way of generating the lists goes as follows.
In the end the union of the lists is your answer. The proof that this works uses standard properties of the Coxeter groups, and is not too difficult. The double loop in step 3 takes a little time, because you also need to remove the duplicates. The list $R_i$ is at its longest when $i$ is approximately $|\Phi^+|/2$.
Note: You can calculate the size of the orbit in advance. The stabilizer of a dominant weight $\lambda$ in $W$ is generated by those simple reflections $s_i$ that are in it (i.e. $s_i(\lambda)=\lambda$). The cardinality of that group can easily be computed using the Dynkin diagram by removing the nodes corresponding to those generators that don't belong to the stabilizer - the stabilizer is the direct product of the Weyl groups of the components of whatever remains of the diagram. Then we can use the identity $|W(\lambda)|=|W|/|Stab_W(\lambda)|.$
Alternatively you could try to find the coset representatives of the stabilizer in the Weyl group, and use those. That would make the removal of duplicates automatic. Because the word problem in a Coxeter group has a solution, the necessary algorithms may have been implemented in a package like GAP. I don't know for sure. Sorry.