Mysterious morphism Until I get a clearer understanding of what follows, and it is the aim of this post, I am sorry I cannot make the question title more explicit.
In the setting of algebraic groups I found the following notations written many times, and I don't figure out what they are, both intuitively and formally. $G$ is a reductive group. $X(G)$ is the additive group of homomorphisms from $G$ to $GL(1)$, it is a free abelian group. Then we consider
$$\mathfrak{a}_G = Hom_{\mathbb Z} (X(G), \mathbb R)$$
which notation makes me think of a Lie algebra. Is there a natural interpretation as such? As e.g. the Lie algebra of the connected component of the central subgroup (usually denoted $A_G$)?
Most importantly, we define an homomorphism
$$H_G : G(\mathbb A) \to \mathfrak{a}_G$$
by
$$\langle H_G(x), \chi \rangle = |\log (x^\chi)|$$
How to make this definition more explicit, e.g. in the simple cases of $GL_2$ or $GL_n$? Is it a kind of norm?
 A: Even though your tags imply this is in the realm of algebraic groups proper, it's really an object which appears more naturally in the context of automorphic representations.
The point is that in the theory of automorphic representations of $G$ (a reductive group over $\mathbb{Q}$) one wants to do harmonic analysis on the locally compact Hausdorff group $G(\mathbb{A})$ or, more precisely, to study the harmonic analysis of the measure space $G(\mathbb{A})/G(\mathbb{Q})$ (the measure coming from the quotient measure of the Haar measure on $G(\mathbb{A})$). In particular, as is usually the case when doing harmonic analysis, one wants to study something like the space of square-integrable complex valued functions on $G(\mathbb{A})/G(\mathbb{Q})$. This is slightly complicated though by the fact that $G(\mathbb{A})/G(\mathbb{Q})$ doesn't have finite volume (with the Haar measure). That said, there is a way of harmlessly modifying $G(\mathbb{A})/G(\mathbb{Q})$ to have finite volume.
To do this we need, in some sense, the set of points of 'bounded norm' in $G(\mathbb{A})$. Of course, we need to explain what we mean by the 'norm' here. Note that for any character $\chi$ in $X^\ast(G)$ one gets an induced map
$$|\cdot|_\chi:G(\mathbb{A})\to \mathbb{R}:g\mapsto |\chi(g)|$$
where, as per usual, for an element $(a_v)\in \mathbb{A}^\times$ we set
$$|(a_v)|:=\prod_v |a_v|$$
which makes sense since $a_v\in\mathcal{O}_v^\times$ for all but finitely many $v$ and so, consequently, $|a_v|=1$ for all but finitely many $v$.
One can then form the subgroup $G(\mathbb{A})^1$ as follows:
$$G(\mathbb{A})^1:=\left\{g\in G(\mathbb{A}):|g|_\chi =1\text{ for all }\chi\in X^\ast(G)\right\}$$
Note that $G(F)\subseteq G(\mathbb{A})^1$ which is precisely the product theorem from basic valuation theory.
One then has the following spectacular theorem:

Theorem(Borel, Conrad, Oesterle, Harder, [1, §5.3]): The quotient $G(\mathbb{A})^1/G(\mathbb{Q})$ has finite volume.

Now to explain why this modification is 'harmless' it's useful to understand $G(\mathbb{A})^1$ in terms of the objects you described. So, let us preliminarily take your definitions
$$\mathfrak{a}_G:=\mathrm{Hom}_\mathbb{Q}(X^\ast(G),\mathbb{R})$$
and (I think you have a typo here)
$$H_G:G(\mathbb{A})\to \mathfrak{a}_G$$
defined by letting $H_G(g)$ be the $\mathbb{Q}$-linear map $X^\ast(G)\to\mathbb{R}$ which associates to $\chi$ the element $\log|\chi(g)|\in\mathbb{R}$.
We then note that
$$\begin{aligned}\ker(H_G)&=\left\{g\in G(\mathbb{A}): H_G(g)=0\right\}\\ &= \left\{g\in G(\mathbb{A}):H_G(\chi)=0\text{ for all }\chi\in X^\ast(G)\right\}\\ &= \left\{g\in G(\mathbb{A}):\log|\chi(g)|=0\text{ for all }\chi\in X^\ast(G)\right\}\\ &= \left\{g\in G(\mathbb{Q}):\chi(g)=1\text{ for all }\chi\in X^\ast(G)\right\}\\ &= G(\mathbb{A})^1\end{aligned}$$
To continue it is useful to address your other question. Namely, you astutely noted that the notation $\mathfrak{a}_G$ looked like the notation for a Lie algebra. The reason for that is as follows. Let $A_G$ denote the largest split subtorus of $Z(G)$. Then, we have that
$$\mathrm{Lie}(A_G)\otimes_\mathbb{Q}\mathbb{R}\cong \mathrm{Hom}_\mathbb{Q}(X^\ast(G),\mathbb{R})$$
Indeed, note that
$$X^\ast(G)=X^\ast(G^\mathrm{ab})\cong X^\ast(Z(G))\cong X^\ast(A_G)$$
where the first isomorphism is obvious, the second is from the fact that the natural map $Z(G)\to G^\mathrm{ab}$ is an isogeny, and the final is since $Z(G)$ is isogenous to $A_G\times T'$ where $T'$ is an anisotropic subtorus of $Z(G)$. But, we then see that
$$\mathrm{Hom}_\mathbb{Q}(X^\ast(G),\mathbb{R})=\mathrm{Hom}_\mathbb{Q}(X^\ast(A_G),\mathbb{R})=X_\ast(A_G)\otimes_\mathbb{Q}\mathbb{R}\cong \mathrm{Lie}(A_G)\otimes_\mathbb{Q}\mathbb{R}$$
where the last isomorphism associates to a cocharacter $\mu:\mathbb{G}_m\to A_G$ the element $(d\mu)(1)\in \mathrm{Lie}(A_G)$.
In fact, using this one can show the following:

Proposition: The map $$G(\mathbb{A})^1\times A_G(\mathbb{R})\to G(\mathbb{A}):(g,a)\mapsto ga$$ is an isomorphism.

In particular, the reason that $G(\mathbb{A})^1/G(\mathbb{Q})$ is functionally not much different than $G(\mathbb{A})/G(\mathbb{Q})$ is that they differ by the term $A_G(\mathbb{R})$ which, since this is central and a direct factor of $G(\mathbb{A})$, can be twisted away in practice.
References:
[1] Platonov, V., Rapinchuk, A. and Rowen, R., 1993. Algebraic groups and number theory. Academic press.
