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.
Best Answer
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.
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.