The length of this question has got a little bit out of hand. I apologize.

Basically, this is a question about the relationship between the cohomology of Lie groups and Lie algebras, and maybe periods.

Let $G$ be a complex reductive (connected) Lie group and let $T$ be a maximal torus of $G$. Set $\mathfrak{g}=Lie(G)$ and $\mathfrak{t}=Lie(T)$. Notice that $\mathfrak{t}$ has a natural integral structure: $\mathfrak{t}=\mathfrak{t}(\mathbf{Z})\otimes_\mathbf{Z}\mathbf{C}$ where $\mathfrak{t}(\mathbf{Z})$ is formed by all $x$ such that $\exp(2\pi ix)$ is the unit $e$ of $G$.

All there is to know about $G$ can be extracted from the (covariant) root diagram of $G$, which is formed by $\mathfrak{t}(\mathbf{Z})$, the sublattice $M$ corresponding to the connected component of the center of $G$ (this is a direct summand) and the coroot system $R$ of $G$, which is included in some complementary sublattice of $\mathfrak{t}(\mathbf{Z})$. For example, $\pi_1(G)$ is the quotient of $\mathfrak{t}(\mathbf{Z})$ by the subgroup spanned by $R$. See e.g. Bourbaki, Groupes et alg`ebres de Lie IX, 4.8-4.9 (Bourbaki gives a classification in terms of compact groups, but this is equivalent).

The question is how to extract information on the cohomology of $G$ (as a topological space) from the above.

For the complex cohomology there are no problems whatsoever. We only need $\mathfrak{g}$: restricting the complex formed by the left invariant forms to the unit of $G$ we get the standard cochain complex of $\mathfrak{g}$.

The next step would be the rational cohomology. One possible guess on how to get it would be to notice that $\mathfrak{g}$ is in fact defined over $\mathbf{Q}$. So one can find an algebra $\mathfrak{g}(\mathbf{Q})$ such that $\mathfrak{g}=\mathfrak{g}(\mathbf{Q})\otimes_\mathbf{Q}\mathbf{C}$. We can identify $H^{\bullet}(\mathfrak{g},\mathbf{C})=H^{\bullet}(\mathfrak{g}(\mathbf{Q}),\mathbf{Q})\otimes\mathbf{C}$ and so we get two rational vector subspaces in the complex cohomology of $G$. One is the image of of $H^{\bullet}(G,\mathbf{Q})$ and the other is the image of $H^{\bullet}(\mathfrak{g}(\mathbf{Q}),\mathbf{Q})$ under $$H^{\bullet}(\mathfrak{g}(\mathbf{Q}),\mathbf{Q})\to H^{\bullet}(\mathfrak{g},\mathbf{C})\to H^\bullet(G,\mathbf{C})$$ where the last arrow is the comparison isomorphism mentioned above.

1). What, if any, is the relationship between these subspaces? More precisely, apriori the second subspace denends on the choice of $\mathfrak{g}(\mathbf{Q})$ (I don't see why it shouldn't, but if in fact it doesn't, I'd be very interested to know) and the question is if there is a $\mathfrak{g}(\mathbf{Q})$ such that the relationship between the above subspaces of $H^{\bullet}(G,\mathbf{C})$ is easy to describe.

Notice that this is somewhat similar to what happens when we compare the cohomology of the algebraic de Rham complex with the rational cohomology. Namely, suppose we have a smooth projective or affine algebraic variety defined over $\mathbf{Q}$; its algebraic de Rham cohomology (i.e. the (hyper)cohomology of the de Rham complex of sheaves) sits inside the complex cohomology, but this is not the same as the image of the topological rational cohomology. Roughly speaking, the difference between the two is measured by periods, e.g. as defined by Kontsevich and Zagier.

2). If question 1 has a reasonable answer, then what about the integral lattice in $H^{\bullet}(G,\mathbf{C})$? Again, a naive guess would be to take a $\mathfrak{g}(\mathbf{Z})$ such that $\mathfrak{g}(\mathbf{Z})\otimes\mathbf{C}=\mathfrak{g}$. At present, I'm not sure whether such an integral form exist for any reductive $G$ (and I'd be very interested to know that), but in any case it exists for $SL(n,\mathbf{C})$. By taking the standard complex of $\mathfrak{g}(\mathbf{Z})$ and extending the scalars we get a lattice in $H^{\bullet}(\mathfrak{g},\mathbf{C})\cong H^{\bullet}(G,\mathbf{C})$. Is there a choice of $\mathfrak{g}(\mathbf{Z})$ for which this lattice is related to the image of the integral cohomology in $H^\bullet(G,\mathbf{C})$ in some nice way?

3). If even question 2. has a reasonable answer, then what about the integral cohomology itself? Here, of course, the answer is interesting even up to isomorphism. A very naive guess would be to take an appropriate integral form $\mathfrak{g}(\mathbf{Z})$ as in question 2 and compute the integral cohomology of the resulting standard complex.