# Group of characters for parabolic subgroups

Let us fix a linear algebraic group $$G$$ (in my case $$G$$ is also reductive).

It is a well-known fact that for an algebraic torus $$T \cong (\mathbb{C}^*)^n$$ in $$G$$ the group of characters $$M(T)$$ is isomorphic to $$\mathbb{Z}^n$$ via the map $$diag(t_1,\ldots ,t_n) \mapsto t_1^{m_1}\cdot \ldots \cdot t_n^{m_n}$$ for some $$m_1,\ldots,m_n \in \mathbb{Z}$$.

I have also found that for a Borel subgroup $$B \subset G$$ it holds the same, so $$M(B) \cong M(T) \cong \mathbb{Z}^n$$ (but if someone gives me a rigourous reference, I will be happier).

At this point, consider a parabolic subgroup $$P \subset G$$ (that is a subgroup of $$G$$ containing $$B$$), what is $$M(B)$$? Is there a nice description as for tori and Borel subgroups?

Let $$H$$ be any algebraic group (I don't know what framework you are using, but affine and finite type, and smooth (although this is automatic over $$\mathbb{C}$$) is what I care about). Then any homomorphism $$H\to \mathbb{G}_m$$ sends $$R_u(H)$$, the unipotent radical of $$H$$, to a unipotent subgroup of $$\mathbb{G}_m$$, of which there is only the trivial group. So, in fact, you see that $$M(H)=M(H/R_u(H))$$.
In particular, if now $$H=P$$ is a parabolic with Levi factor $$L$$ then $$M(P)=M(L)$$. Note that $$L$$ is a reductive group, and it's easy to see that $$M(L)=M(L^\mathrm{ab})$$ (where $$L^\mathrm{ab}$$ is the abelianization of $$L$$). Note, moreover, that since $$L$$ is reductive that $$Z(L)^\circ\to L^\mathrm{ab}$$ (where $$\circ$$ denotes the connected component) is a surjection with finite kernel (an isogeny). So, if you only care about the rank of $$M(L)$$ then you can replace it with $$M(Z(L)^\circ)$$ (note that $$Z(L)^\circ$$ is a torus).
Example: If $$P=B$$ a Borel of $$G$$ then the above says that $$M(B)=M(B/R_u(B))$$ but if $$T\subseteq R_u(B)$$ is a maximal torus then this is (essentially) a Levi subgroup of $$B$$--in other words $$B/R_u(B)\cong T$$. Thus, $$M(B)=M(T)\cong \mathbb{Z}^n$$.
Example: If $$P$$ is the maximal parabolic of $$G$$, namely $$P=G$$, then $$M(P)=M(G)=M(G^\mathrm{ab})$$ and this has the same rank as $$M(Z(G)^\circ)$$. For example, if $$G=\mathrm{GL}_n$$ then $$M(G)=M(G^\mathrm{ab})=M(\mathbb{G}_n)=\mathbb{Z}$$ (which is the same rank as $$M(Z(G)^\circ)=M(\mathbb{G}_n)=\mathbb{Z})$$. But, as an extreme example, note that if $$G$$ is semisimple (e.g. $$G=\mathrm{SL}_n,\mathrm{Sp}_{2n},...$$) then $$Z(G)^\circ$$ is trivial and so the above shows that $$M(G)=\{0\}$$
Example: If $$P$$ is the parabolic of $$\mathrm{GL}_n$$ consisting of block upper diagonal matrices with sizes $$\ell+k=n$$ then $$M(P)=M(L)$$ where $$L$$ is the Levi factor of $$P$$ which is just $$\mathrm{GL}_k\times\mathrm{GL}_\ell$$. I leave it to you to then compute that $$M(L)=M(L^\mathrm{ab})=\mathbb{Z}^2$$.