Let $\mathfrak{g}$ a complex semisimple Lie algebra. It is well known that by the theorem of the highest weight, all finite-dimensional complex irreps of $\mathfrak{g}$ are (up to iso) classified by dominant integral weights $\alpha \in \mathfrak{h}^*$ (ie those integral weights lying in the cone spanned by fundamental weights). That's classical, concrete constructions can be found for instance in Humphreys' or Hall's introductions to representation theory & (semisimple)Lie algebras.
Question: Where precisely the dominant integral weights construction described in quoted sources for semisimple Lie groups (... and leading to an integral cone where all weights representing irreps sit) would break down "irreparably" (...and probably causes "nonsense" results) if we would start naively with a reductive Lie group instead (ie having nontrivial center $\mathfrak{z}$) and would mimic stubbornly the construction for semisimple case?
I would like to note that it is familar to me that an educated approach would be based on considering decomposing
$\mathfrak{g}=\mathfrak{s} \oplus \mathfrak{a}$ into abelian and semisimple part and treat reps of both parts separately.
But that's not what I want to discuss in this question. I would like to chew through the construction in order to understand where (ie at which intermediate steps) exactly the dominant weights construction (and it's connection to irreps) for semisimple Lie algebra badly fail (and to which extend? Maybe this would give a hint to a partial result) if we apply would "with head trough the wall" perform the semisimple construction to reductive algebra.
Let's see: Firstly, we have in reductive setting still the root decomposition of the Lie algebra as $\mathfrak{g}=\mathfrak{h}\oplus \bigoplus_{\alpha \in R}\mathfrak{g}_{\alpha}$ under adjoint representation by maximal Cartan subalgebra $\mathfrak{h}$ which contains the center $\mathfrak{z}$ where $R$ are the roots.
We can still fix positive roots and obtain inclusion $R_s \subset R_+ \subset R \subset \mathfrak{h}^*$ of simple and positive roots.
Here appear first differences in contrast to semisimple case: $R$ not span (as vector space) the whole $\mathfrak{h}^*$ but a subspace of dimension which equals to cardinality of the set of simple elements $|R_s|$ which in turn equals to the semisimple rank defined as $\dim \mathfrak{h}/\mathfrak{z}$.
Note that the Killing form $K(,)$on $\mathfrak{h}$ which induces nondegenerate form on $\mathfrak{h}/\mathfrak{z}$ (indeed only the center makes it degenerated) and thus induces a bilinear form on the dual spaces $K^*(,)$ gives us still a way to to declare dominant weights (whose elements in $h \in \mathfrak{h}^*$ with $K^*(h,r) \ge 0$ for all positive roots $r \in R_+$.
In contrast in the construction of fundamental weights there would be more involved in terms of non-canonical choices:
Indeed, recall in case with $\mathfrak{g}$ semisimple, let $r_1,..., r_d \in R_s$ be the simple roots, there exist for each $r_i$ a unique $b_i \in \mathfrak{h}^*$ - a fundamental weight - satisfying $K^*(b_i,r_j)=\delta_{ij}$ (maybe the Killing form should have been normed somewhere in a previous step).
What changes if $\mathfrak{g}$ is not semisimple but reductive in the constructio ? If I'm not missing something, only that the fundamental weights are not not more uniquely determined but uniquely modulo $\mathfrak{z}^*$ as $\mathfrak{z}^*$ forms exactly the radical of induced Killing form on the dual of the Cartan space, right? And that's it, or is the story a bit longer?
Having once made these choices for the fundamental weights $z_i$, the story should continue similarly as in semisimple case: They span a cone and intersecting this cone with space of dominant weights gives us dominant integral weights which should witness as highest weights the irreps (...at least that's what the construction in semisimple case would be about).
The only problem I see so far is that these span not a full cone, ie it's topological dimension is due to rank defect smaller then dimension of $\mathfrak{h}^*$, it's a kind of "degenerated" cone.
What can we do at this stage? Can we for example complement these chosen fundamental weights $b_i$ by a basis from $\mathfrak{z}^*$ and then build (now nondegenerate cone)? If that latter step not works, we have still our degenerated cone spanned by "chosen modulo $\mathfrak{z}$ unique" fundamental weights intersected with dominant weights.
Still, to which irreps of $\mathfrak{g}$ these would correspond, even if due to these obstructions we probably cannot expect a one-to-one correspondence between isomorphism classes of irreps and highest integral weights living in this cone?
Firstly, are there even more problems in this naive attempt to mimic the (first part of the) highest weight construction to reductive Lie algebras?
And still if we stay at the end of the day with this "degenerated cone": To which irreps are integral weights there correspond?
