MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Littelmann path is a combinatorial tool to compute multiplicity. I have some questions about the definition of Littelmann path. It is said that a Littelmann path is a piecewise-linear mapping $$\pi:[0,1]\cap \mathbf{Q} \rightarrow P\otimes_{\mathbf{Z}}\mathbf{Q} \quad\quad (1)$$ such that $\pi(0) = 0$ and $\pi(1)$ is a weight. Here $\mathbf{Q}$ is the field of rational numbers, $P$ is the weight lattice of a Lie algebra $\mathfrak{g}$, $\mathbf{Z}$ is the ring of integers. I think that if we draw a Littelmann path in type $A_2$ case, then the graph is in a $3$-dimensional space. But in the example, on the top of page 43, the path is drawn in a plane. How to draw a Littelmann path using definition (1)? Does a Littelmann path represent an element in the basis of a representation of a Lie algebra? Thank you very much.

share|cite|improve this question
up vote 2 down vote accepted

Type $A_2$ means the Lie algebra is that of $SL_3$, which has a 2-dimensional maximal torus, so the weight lattice for that algebra is also of rank 2. Did you think dimension 3 because of $GL_3$? It has a maximal torus of dimension 3, but the group is not semisimple: it has a 1-dimensional central torus (scalar multiples of the identity). The center of the Lie algebra is not of great interest to Littelmann paths, so even if you are doing paths for the Lie algebra of $GL_3$, you wouldn't lose much information in projecting the paths along the "central" direction onto a 2-dimensional plane. Points so identified are weights that, viewed as linear functions on matrices living in the Lie algebra, differ by a multiple of the trace function (the trace annihilates the maximal torus of the $SL_3$ algebra, since it consists of traceless matrices). If $\varepsilon_i$ is the function on diagonal matrices that takes the $i$-th diagonal coefficient ($i=1,2,3$), then $\varepsilon_1+\varepsilon_2+\varepsilon_3$ is the trace, and this is 0 as a weight for $SL_3$; the weight lattice for this group is usually drawn with the $\varepsilon_i$ as equal length vectors at 120 degree angles, so that their sum is 0, and this gives the triangular lattice of the picture you referred to.

share|cite|improve this answer
@Marc, thank you very much. I think that $P$ can be drawn on a dimension 2 space (a plane) and $P\otimes_{\mathbf{Z}}\mathbf{Q}$ is of dimension 3. So I am confused. – Jianrong Li Nov 12 '11 at 19:22
If $P$ is a 2 dimensional lattice, then $P\otimes_{\mathbf Z}\mathbf Q$ is a vector space (over $\mathbf Q$), obtained by "filling in" that lattice by rational multiples. It then clearly has dimension 2, not 3. I don't understand your confusion. – Marc van Leeuwen Nov 15 '11 at 13:40

Your Answer


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.