Let $G$ be a connected, reductive, complex algebraic group. It is well known that if $G$ is moreover semisimple, i.e. its Lie algebra $\mathfrak{g}$ is semisimple (or equivalently, its center $Z(G)$ is finite, and so its Lie algebra is trivial), then all finite dimensional representations of the Lie algebra can be classfied (up to iso's) by "highest weights": More concretely, the Theorem of the highest weight states that there is an bijection between dominant integral elements to the set of equivalence classes of finite irreducible representations of the Lie algebra $\mathfrak{g}$.
Question: Are there any caveats why this result cannot(?) be straight forwardly extended to connected, reductive, complex algebraic groups? ("straight forwardly" in the sense by literally $1$-to-$1$ mimicking the construction from semisimple case)
Essentially on level of Lie algebras the reductive Lie algebra $\mathfrak{g}$ differs from semisimple one by sub-Lie algebra $\mathfrak{z} \subset \mathfrak{g}$ corresponding to its center $Z(G) \subset G$, thus $\mathfrak{g}/\mathfrak{z}$ becomes a semisimple Lie-algebra of commutator group $[G,G] \subset G$ (recall, if $G$ reductive, then $G=[G,G]Z(G)^{\circ}$).
Considerations: Clearly $G$ (since complex) is split, so contains an up to conjugacy unique torus $H \cong \mathbb{G}_m^r$ which contains $Z(G)^{\circ}$ as subtorus, and we obtain the same root decomposition of $\mathfrak{g}$ under adjoint $H$-action, can form (dominant) weights and so on inside $\mathfrak{h}^*$ (the dual of Lie algebra of $H$, also called maximal Cartan algebra) as in semisimple case.
Formulated it equivalently, on level of Lie algebras $\mathfrak{g}$ is reductive iff $\mathfrak{g}= \mathfrak{s} \oplus \mathfrak{a}$ with $\mathfrak{s}$ semisimple and $\mathfrak{a}$ abelian and contained in Cartan algebra $\mathfrak{h}$.
So cannot we just "mimic" naively straight forwardly $1$-to-$1$ the approach for semisimple Lie group, without additional care for this abelian $\mathfrak{z}$ which we simply regard as subalgebra of Cartan algebra $\mathfrak{h}$, with only difference that $\mathfrak{z}$ is not more zero as in semisimplicial case?
If we do so naively, into which problems/obstructions we are running trying extend the theorem of highest weight from connected semisimple to reductive complex groups in this naive manner?
Seemingly the the same techniques work in reductive setting, but nevertheless the result is only stated only for semisimple $G$.
What precisely goes wrong with this correspondence if we allow the existence of this abelian summand $\mathfrak{a}$ in the direct sum decomposition of $\mathfrak{g}$ as above?
Remark #1: Corollary 8.1.6 in Springer's Linear Algebraic Groups describes explicitly how one can pass from root system of a reductive group $G$, to its "semisimplification" $[G,G]$; on level of Lie algebras it's essentially as stated above by killing the center $\mathfrak{g} \mapsto \mathfrak{g}/\mathfrak{z}$. The rank drops from $r$ to so called semisimple rank (=$r-$ center's rank), the roots are isomorphic.
But I not see so far the reason where the highest weight approach breaks down in reductive case if we just naively mimic the construction in semisimple setting.
Remark #2: This question is closely related to problems I discussed here in sense of extending techniques from semisimple to reductive setting.
Best Answer
For a split reductive group there is exactly the same kind of story. I will describe the setup as precisely as possible so that you can see the connection. This may not be an answer to the question but it should go some way to describing what is happening.
Fix a field $k$, and let $T \subseteq B \subseteq G$ be a "pinning" of the split reductive group $G$, i.e. $B$ is a Borel subgroup containing a maximal torus $T$. The torus $T$ determines a weight lattice $X$ and coweight lattice $X^\vee$, a pair of free $\mathbb{Z}$-modules together in a perfect pairing $$ \langle-, - \rangle: X^\vee \times X \to \mathbb{Z}.$$ These $\mathbb{Z}$-modules are defined as hom-spaces in the category of algebraic groups: $$ X^\vee = \operatorname{Hom}_{\mathsf{AlgGrp}}(\mathbb{G}_m, T), \quad X = \operatorname{Hom}_{\mathsf{AlgGrp}}(T, \mathbb{G}_m),$$ and the perfect pairing is simply composing the homs to get an endomorphism of $\mathbb{G}_m$, which are classified by integers.
From $G$ we get sets of roots $\Phi \subseteq X$ and coroots $\Phi^\vee \subseteq X^\vee$, equipped with a bijection $\alpha \mapsto \alpha^\vee$. The Borel subgroup $B$ determines the set $\Phi^+$ of positive roots and the set $S \subseteq \Phi^+$ of simple roots.
Important note: The combinatorial data of $(X^\vee, X, \langle -, - \rangle, \Phi, \Phi^\vee, (-)^\vee: \Phi \to \Phi^\vee)$ completely determines the split reductive group $G$ up to isomorphism. This is called the root datum of the group.
The dominant weights $X_+ \subseteq X$ are those weights which pair with the simple coroots positively: $$ X_+ = \{\lambda \in X \mid \langle \alpha^\vee, \lambda \rangle \geq 0 \}.$$ Then the theorem is that the irreducible representations of $G$ are classified by the dominant weights $X_+$.
Here are some examples of root data.
So how is the case of $G$ any different to that of $\mathfrak{g} = \operatorname{Lie}(G)$? The key point is integrality: for reductive groups it is automatic and canonical, for reductive Lie algebras it is non-canonical and needs to be imposed.
Passing to $\mathfrak{g}$ loses some data even for semisimple algebraic groups, for instance $SL_n$ and $PGL_n$ are nonisomorphic as algebraic groups, but their Lie algebras are isomorphic. However, the root data sees the difference, which has to do with exactly how the roots and coroots sit inside their lattices.
Passing to $\mathfrak{g}$ for reductive groups which are not semisimple has a larger problem: we do not know what to call an "integral" weight for the centre, since we have thrown away the integrality that used to be there. What's worse is that there is no canonical way of getting it back.
And in fact, if you simply start with the reductive Lie algebra $\mathfrak{g}$, then anything goes for the centre! Consider for example the one-dimensional reductive Lie algebra $\mathfrak{g} = \mathfrak{gl}_1$: then irreducible finite-dimensional representations of this are classified by $\mathbb{C}$, not by any $\mathbb{Z}$-lattice. Similarly, irreducible finite-dimensional representations of $\mathfrak{gl}_n$ are classified by an irrep of $\mathfrak{sl}_n$ and a complex number $\mathbb{C}$.
To wrap this rambling post up, I would say that there is a connection between the classification of irreps in a reductive group $G$ and its reductive Lie algebra $\mathfrak{g} = \operatorname{Lie}(G)$, but the best way to uniformly state it is for reductive groups, and then extend it to the Lie algebra.
From the pinning $T \subseteq B \subseteq G$ of the group you get the Lie algebra $\mathfrak{g} = \operatorname{Lie}(G)$, a choice of Borel subalgebra $\mathfrak{b} = \operatorname{Lie}(B)$, and Cartan subalgebra $\mathfrak{h} = \operatorname{Lie}(T)$. Furthermore, every weight $\lambda \in X$ is a group homomorphism $\lambda: T \to \mathbb{G}_m$ which can be differentiated to a Lie algebra homomorphism $d\lambda: \mathfrak{h} \to \mathbb{C}$, i.e. an element of the dual $\mathfrak{h}^*$. So you get a canonical set of integral weights $X \subseteq \mathfrak{h}^*$ containing a root system and so on, and now you can define what it means to be an "integral" representation of $\mathfrak{g}$. And from here the classification will go through.