# on a property of minuscules in weight lattice

Let $\Phi$ be a root system in inner product space $E=\mathbb{R}^l$ (inner product $(\cdot,\cdot)$ and $\Delta=\{\alpha,_1,\cdots,\alpha_l\}$ be a fundamental root system.

Consider the root lattice $\Lambda_r=\mathbb{Z}\alpha_1 + \cdots + \mathbb{Z}\alpha_l$ in $E$ and weight lattice $\Lambda$ $$\Lambda =\{\lambda\in E : \frac{2(\lambda,\alpha)}{(\alpha,\alpha)} \in\mathbb{Z} \forall \alpha\in \Delta\}.$$

Let $\Lambda^+ = \{ \lambda\in\Lambda : \frac{2(\lambda,\alpha)}{(\alpha,\alpha)} \geq 0 \forall \alpha\in\Delta\}$.

The miniscules are specific elements of $\Lambda^+$; they are minimal elements of $\Lambda^+$ w.r.t ordering defined as follows: $\lambda_1\leq \lambda_2$ if $\lambda_2-\lambda_1=\sum_{i=1}^l m_i\alpha_i$ with $m_i\geq 0$.

I wanted to see the proof of following fact (seen here this, page 2, top line)

For different minuscules $\lambda,\nu$, the cosets $\lambda+\Lambda_r$ and $\nu+\Lambda_r$ are distinct.

Is this easy to prove? How should we proceed?

I tried a simple case: suppose $\lambda,\nu$ are minuscules with $\lambda=\nu+\alpha-\beta$ where $\alpha,\beta$ are distinct fundamental (i.e. simple) roots. I couldn't get any direction to get contradiction from this.

The actual general problem is that if $\lambda+\Lambda_r=\nu+\Lambda_r$, then $\lambda-\nu$ is integral combination of fundamental roots, so y $\lambda=\nu+m_1\alpha_1 + \cdots + m_l\alpha_l$, so some coefficients could be positive and some negative. The simplest non-trivial case is then exactly one $m_i$ is $+1$, exactly one $m_j$ is $-1$ and rest are zero; I don't get any idea to proceed for contradiction.

• See Math.StackExchange to ask general questions in mathematics. – user21574 Feb 4 '17 at 15:29
• I agree Hassan, but in mathstack, I posted questions on some Lie algebra just "to check answer", and still I didn't get any response. – Learn_Math Feb 5 '17 at 8:42

To comment on the question here (in community-wiki mode), I should point out first that $\S13$ of my now-ancient book was meant to develop some properties of weights just in the framework of abstract root systems without having developed the representation theory where the ideas first arose. In this experimental spirit, my exercises were a bit speculative (and in some cases not worked out by me in detail until much later, since I knew the conclusions were correct, e.g., Exercise 13.10).

That being said, my Exercise 13.13 is intended to be done using just the tools of this chapter. However, the exercise itself comes from one in Bourbaki, VI, $\S2$, which I've now added (reluctantly) to the informal notes linked in the question. The Bourbaki exercise relies on their development in Chapter V of ideas about reflection groups, applied to affine Weyl groups. This gets complicated, since it uses the dual root system in an essential way. I didn't go into affine Weyl groups in my 1972 book but did include an account in Chapter 4 of my 1990 book on reflection groups. An advantage of that setting is to visualize things better: the affine Weyl group leads to a euclidean simplex (fundamental alcove) whose nonzero vertices are of the form $\varpi/c$ for the fundamental weights $\varpi$ and corresponding coefficients $c$ in the highest root of the (dual) root system. Only for $c=1$ do you get minuscule weights (none in some types of irreducible root system).

Anyway, one is led to the list in my Exercise 13.13 (or in Bourbaki's Chapter VIII). A subtle point is the dualization, which assigns (in Bourbaki's numbering which I adopted) $\lambda_\ell$ to type $B_\ell$ and $\lambda_1$ to type $C_\ell$. Bourbaki originally reversed these, but that seems to be corrected in the English translation. (Note too that they used E. Cartan's symbol $\varpi$, a version of the handwritten $\pi$ for poids, whereas I used $\lambda$. Their notation is more commonly used but is easy to confuse with $\omega$.)

Finally, a comment on the spelling of minuscule. This is the original version, emphasizing MINUS (I guess in contrast to majuscule), but often rendered as miniscule with emphasis on MINI. The question here mixes both spellings, which is undesirable.

• Thanks for comments; which part of Bourbaki's series the book is? Also can you suggest some points for the problem above? The solution posted is also partially not clear which I already commented there. – Learn_Math Feb 6 '17 at 6:38
• The references I've made are to Bourbaki's treatise Groupes et algebres de Lie, published in English by Springer under the title Lie Groups and Lie Algebras. The relevant chapters are mainly 6 and 8, though as I remarked there are essential references to 5 in some exercises. Since my 1972 book didn't include affine Weyl groups, it's probably best to follow the case-by-case treatment based on the classification (as in Table 1 of $\S13$) even though this is not so elegant. – Jim Humphreys Feb 6 '17 at 14:52
• I'm a bit sad to hear you think of your book as now-ancient. I lugged it with me in the 80s. Even had with me on the Finnish army reserve forces bootcamp in the summer of '87. I did my conscript duty in coastal artillery. So when we needed any math done, your book gave me enough street cred, and the officers would not question my math. – Jyrki Lahtonen Apr 20 '18 at 6:51

Possibly this argument works; I am also new in Lie groups and Lie algebras, so I am happy for experts comments.

Consider $\lambda$ a minuscule i.e. a minimal dominant weight (according to Humphreys).

Let $\{\lambda_1,\cdots,\lambda_l\}$ be dual basis of $\{\check{\alpha_1},\cdots,\check{\alpha_l}\}$ w.r.t. $(\cdot,\cdot)$.

1. $\lambda + m_1\alpha_1 + \cdots m_l\alpha_l$, for $m_i\geq 0$, is not minimal dominant weight unless all $m_i=0$.

Here all $m_i\geq 0$. If some $m_i>0$ then $\lambda'=\lambda + m_1\alpha_1 + \cdots m_l\alpha_l$ is bigger than $\lambda$ (i.e. by def.,$\lambda'-\lambda$ is sum of fundamental roots). Then minimality of both $\lambda$ and $\lambda'$ (as dominant weights) implies that they should be equal so $m_1\alpha_1 + \cdots m_l\alpha_l =0$; but $\alpha_i$'s are independent and one $m_i$ is non-zero, so contradiction.

2. Every dominant weight can be written as $\sum q_i\alpha_i$ where $q_i$ are positive rationals.

If $\lambda$ is dominant weight then this means $\lambda=c_1\lambda_1 + \cdots + c_l\lambda_l$ where $c_i\in\mathbb{Z}^+$. Further, each $\lambda_i$ can be written as $a_{i1}\alpha_1 + \cdot + a_{il}\alpha_l$ where $a_{ij}$ are positive rationals (see Exercise 13.8 in Humphreys Lie algebra). This proves the assertion (2).

3 (Main fact:) Every minimal dominant weight $\lambda$ can be written as $\sum q_i\alpha_i$ where $q_i$'s are positive rationals, less than $1$.

Write a dominant weight $\lambda$ as $\sum q_i'\alpha_i$ with $q_i'$ positive rationals. If some $q_j'>1$ then we can subtract $\alpha_j$ from $\lambda$ to get a smaller dominant weight.

4. If $\lambda,\nu$ are minimal dominant weights in same coset of $\Lambda_r$, then $\lambda-\mu$ is integral combination of $\alpha_i$'s; so $$\lambda=\nu + c_1\alpha_1 + \cdots + c_l\alpha_l, (c_i\in\mathbb{Z}).$$

If all $c_i's$ are $\geq 0$, apply $1$; if all $c_i\leq 0$, then take them to left side and apply 1.

If some $c_i>0$ then in RHS, coefficient of $\alpha_i$ will be rational $>1$, but in LHS, it is less than $1$, contradiction to (3).

• Proof of 3 is not clear to me. – Learn_Math Feb 5 '17 at 8:42
• A small comment on terminology: my use of "minimal" is straightforward relative to the usual partial ordering of dominant weights, but this allows the 0 weight to qualify as minimal (in the coset of the weight lattice corresponding to the root lattice itself). To exclude this rather trivial possibility, the less transparent term "minuscule" is used. – Jim Humphreys Feb 8 '17 at 23:23
• @Soluble: As suggested by Learn_Math's comment, your "Main fact" 3 isn't true (even with "minimal" replaced by "minuscule"). For example, look at the fundamental weights $\lambda_1, \lambda_6$ for type $E_6$. It may be safest to use case-by-case checking for the coset representatives, though I'd still prefer a uniform proof not relying on affine Weyl groups. – Jim Humphreys Apr 7 '17 at 17:52