Converting a dominant weight of $\mathfrak{gl}_n$ to $\mathfrak{sl}_n$

Question

The rank of $\mathfrak{sl}_n$ is $n-1$, so given a dominant $\mathfrak{gl}_n$ weight $(d_1, d_2, \cdots, d_n)$ such that $d_1 \geq d_2 \geq \dots \geq d_n$, the corresponding weight for $\mathfrak{sl}_n \subset \mathfrak{gl}_n$ should be $(d_1 - d_2, d_2-d_3, \dots, d_{n-1}-d_n)$. But this does not satisfy the same condition $d_1-d_2 \geq d_2-d_3 \geq \dots$, take for example the $\mathfrak{gl}_3$ weight $(1,1,0)$.

I ask this because in a theorem of Chriss-Ginzburg 4.2.3, we are given that a simple $\mathfrak{sl}_n$-module is given an associated highest weight $(d_1 \geq d_2 \cdots \geq d_n)$, which I am sure is a typo.

However, even if it is, we are not guaranteed the highest weight for $\mathfrak{sl}_n$. So what do the authors mean here?

Edit 1: To make clear my understanding, a basis of the weight space of $\mathfrak{gl}_n$ is given by the functionals $\epsilon_i : \mathfrak{h} \to \mathbb{C}, 1 \leq i \leq n$ which sends the diagonal matrix to its $i$-th entry. A basis of the weight space of $\mathfrak{sl}_n$ is given by the functionals $$ \varpi_i := \epsilon_i - \frac{1}{n}(\epsilon_1 + \dots + \epsilon_n), 1 \leq i \leq n-1. $$

It is not clear to me how the $\mathfrak{gl}_2$ weight $(\lambda_1, \lambda_2)$ can be converted to the $\mathfrak{sl}_2$ weight $(\lambda_1-\lambda_2)\varpi_1$.

Edit 2: Also, the CSA $\mathfrak{h} \subset \mathfrak{sl}_n$ is a subalgebra of $\mathfrak{h}_0 \subset \mathfrak{gl}_n$. Then $\mathfrak{h}^* = \mathfrak{h}_0^* / \mathfrak{h}^\perp$. However it is not clear to me that $\mathfrak{h}^\perp = \{\lambda \in \mathfrak{h}_0^* : \lambda(\mathfrak{h}) = 0\}$ is equal to $\mathbb{C}(\epsilon_1 + \cdots + \epsilon_n)$. Clearly the latter is included in the former, but showing the converse is not easy for me. I have that $$\mathfrak{h}^\perp = \{\sum_i \lambda_i \epsilon_i : \sum_i \lambda_i \epsilon_i(h) = 0 \text{ for all } h \in \mathfrak{h}\},$$ but all this says is that $$\lambda_1h_1 + \dots + \lambda_nh_n = 0,$$ and (even when trying to use that $h_1 + \dots + h_n = 0$ for all $h \in \mathfrak{h}$) there is no way to conclude that $\lambda_1 = \dots = \lambda_n$ from this condition.

Answer 1

The notation for $\mathfrak{gl}_n$ weights and $\mathfrak{sl}_n$ weights you are using is different. It is clearer if we write things in terms of a basis.

The weight space of $\mathfrak{gl}_n$ is $\mathbb{C}\{\epsilon_1, \ldots, \epsilon_n\}$, where $\epsilon_i \colon \mathfrak{gl}_n \to \mathbb{C}$ is the linear map taking a diagonal matrix $D$ to the entry $D_{ii}$. A weight $\lambda = \lambda_1 \epsilon_1 + \cdots + \lambda_n \epsilon_n$ is conventionally written $(\lambda_1, \ldots, \lambda_n)$. In this post I will write $\lambda = \epsilon(\lambda_1, \ldots, \lambda_n)$ so we don't get confused. The condition that $\lambda$ is dominant is equivalent to $\lambda_1 \geq \cdots \geq \lambda_n$.

The weight space of $\mathfrak{sl}_n$ is $\mathbb{C}\{\epsilon_1, \cdots, \epsilon_n\} / \mathbb{C}(\epsilon_1 + \cdots + \epsilon_n)$, a quotient space. So in terms of the $\epsilon$ labelling, there are many weights of $\mathfrak{gl}_n$ that descend to the same $\mathfrak{sl}_n$ weight, for example when $n = 3$, the weights $\epsilon(3, 2, 1)$ and $\epsilon(4, 3, 2)$ become the same $\mathfrak{sl}_3$ weight.

The $\mathfrak{sl}_n$ weight space is $(n - 1)$-dimensional, and it has a basis given by the fundamental weights $\varpi_1, \ldots, \varpi_{n-1}$ where $\varpi_i = \epsilon_1 + \cdots + \epsilon_i$ (I mean the image of this element inside the quotient). If you compute the change-of-basis inside the $\mathfrak{sl}_n$ weight space from the $\epsilon$ basis to the $\varpi$ basis, you will find that $$ \epsilon(\lambda_1, \ldots, \lambda_n) = \lambda_1 \epsilon_1 + \cdots + \lambda_n \epsilon_n \mapsto (\lambda_1 - \lambda_2) \varpi_1 + \cdots + (\lambda_{n-1} - \lambda_n) \varpi_{n-1} = \varpi(\lambda_1 - \lambda_2, \ldots, \lambda_{n-1} - \lambda_n).$$ So the $\mathfrak{gl}_3$ weight $\epsilon(3, 2, 1)$ becomes the $\mathfrak{sl}_3$ weight $\varpi(1, 1)$. If $\lambda = \varpi(d_1, \ldots, d_{n-1})$ is an $\mathfrak{sl}_n$ weight, the the condition for $\lambda$ to be dominant is simply $d_i \geq 0$.

So before writing a weight down as just a list of numbers, make sure you know what basis of the weight space you are working in.

Answer 2

I will give an answer specifically on the relationship between the weight spaces of $\mathfrak{gl}_n$ and $\mathfrak{sl}_n$. For this I will introduce some nonstandard notation: let $D_n$ be the vector space of diagonal $n \times n$ matrices (the Cartan subalgebra of $\mathfrak{gl}_n$), and $S_n \subseteq D_n$ be the vector space of traceless diagonal $n \times n$ matrices (the Cartan subalgebra of $\mathfrak{sl}_n$).

Weights of $\mathfrak{gl}_n$ restrict to weights of $\mathfrak{sl}_n$: A weight of $\mathfrak{gl}_n$ is simply a linear map $\lambda \colon D_n \to \mathbb{C}$. Since $S_n \subseteq D_n$, we get a restricted map $\overline{\lambda} \colon S_n \to \mathbb{C}$ by composing the inclusion $S_n \hookrightarrow D_n$ with $\lambda$. We do not need to choose bases or anything to do this, this is simply restricting a linear function to a subspace.

The weight space of $\mathfrak{sl}_n$ is a quotient of the weight space of $\mathfrak{gl}_n$: Using the restriction map above, we have a map of vector spaces $$ \operatorname{Hom}(D_n, \mathbb{C}) \to \operatorname{Hom}(S_n, \mathbb{C}), \quad \lambda \mapsto \overline{\lambda}. $$ It is easy to see this map is surjective and that the dimension of the space on the right is $n - 1$, and hence there is a one-dimensional kernel. The trace function $\operatorname{tr} \colon D_n \to \mathbb{C}$ is nonzero and belongs to the kernel (since the trace function restricted to traceless matrices is zero), hence by the first isomorphism theorem we have $$ \operatorname{Hom}(D_n, \mathbb{C}) / (\mathbb{C} \operatorname{tr}) \cong \operatorname{Hom}(S_n, \mathbb{C}). $$

A basis for the weight space of $\mathfrak{gl}_n$: Let $\epsilon_i \colon D_n \to \mathbb{C}$ be the linear map taking a diagonal matrix to its $(i, i)$ entry. Then $\epsilon_1, \ldots, \epsilon_n$ form a basis for the weight space of $\mathfrak{gl}_n$. Weights of $\mathfrak{gl}_n$ are often written in two ways. The first is $\lambda = \lambda_1 \epsilon_1 + \cdots + \lambda_n \epsilon_n$ where the dominance condition is $\lambda_1 \geq \cdots \geq \lambda_n$. The second way is by defining $\varpi_i = \epsilon_1 + \cdots + \epsilon_i$ and noting that $\varpi_1, \ldots, \varpi_n$ forms another basis for the weight space. The weight $\lambda = w_1 \varpi_1 + \cdots + w_n \varpi_n$ is dominant if and only if $w_i \geq 0$ for all $1 \leq i \leq n - 1$. (The last coefficient $w_n$ can be freely varied and has no effect on the dominance of the weight. Note that $\varpi_n = \operatorname{tr}$). Some people call this second basis the "fundamental weight" basis of $\mathfrak{gl}_n$ although that is an abuse of notation: it is a particular choice of basis which obeys the condition $\langle \varpi_i, \alpha_j^\vee \rangle = \delta_{ij}$ for all simple coroots $\alpha_j^\vee$ for $1 \leq j \leq n - 1$, but is in no way unique. It is very convenient though, as $\varpi_i$ is the highest weight of the $i$th wedge power of the standard representation.

A basis for the weight space of $\mathfrak{sl}_n$: For instance $\overline{\epsilon_1}, \ldots, \overline{\epsilon_{n-1}}$ would be a basis, as the weight space of $\mathfrak{sl}_n$ is $n-1$ dimensional, and these functions remain linearly independent when restricted to $S_n$. However this is a pretty arbitrary choice, rather than leaving out $\overline{\epsilon_n}$ I could have left out any one of the $\overline{\epsilon_i}$ from my list and obtained a basis of the weight space. What is convenient is that the weights $\overline{\varpi_i} = \overline{\epsilon_1} + \cdots + \overline{\epsilon_i}$ remain linearly independent (aside from $\varpi_n$ which gets sent to zero). These $n-1$ weights are truly the fundamental weight basis for $\mathfrak{sl}_n$.

An example: Consider the $\mathfrak{gl}_3$ weight $\lambda = 4 \epsilon_1 + \epsilon_2 + \epsilon_3$. This could also be expressed as $\lambda = 3 \varpi_1 + \varpi_3$. In the quotient, we have $$\overline{\lambda} = 4 \overline{\epsilon_1} + \overline{\epsilon_2} + \overline{\epsilon_3} = 3 \overline{\epsilon_1} + 0\overline{\epsilon_2} + 0\overline{\epsilon_3} = 2 \overline{\epsilon_1} - \overline{\epsilon_2} - \overline{\epsilon_3}.$$ (Note that many $\mathfrak{gl}_n$ weights map to $\overline{\lambda}$, such as $2 \epsilon_1 - \epsilon_2 - \epsilon_3$). Using the fact that $\lambda = 3 \varpi_1 + \varpi_3$ we can also see that $\overline{\lambda} = 3 \overline{\varpi_1} + 0 \overline{\varpi_2}$ in the fundamental weight basis.

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So that was quite long, and there is some fiddly stuff in dealing with weight spaces and different bases. (I think this is fine, in Lie theory there are a lot of related vector spaces and vectors and bases and it takes some time to get across them all). I think the basis you have written in your original post which involves $\frac{1}{n}$ is confusing the fundamental weights with the fundamental coweights, which is why it is unnatural: the weights of $\mathfrak{gl}_n$ naturally restrict to weights of $\mathfrak{sl}_n$, while the coweights of $\mathfrak{sl}_n$ naturally embed into the coweights of $\mathfrak{gl}_n$. The $\varpi_i$ you have written there clearly and naturally becomes a $\mathfrak{gl}_n$-thing, and this should only happen for coweights. There should not be a natural way of taking an $\mathfrak{sl}_n$ weight to a $\mathfrak{gl}_n$ weight.