Let $G$ be a connected reductive group defined over a field $F$. Let $T$ be a maximal torus of $G$ which is defined over $F$, $A\_0$ the maximal $F$-split subtorus of $T$, and $X\_0$ the cotorsion free subgroup of $X = X(T)$ corresponding to $A\_0$. If $L$ is a finite Galois extension of $F$ over which $T$ splits, then $X\_0$ can be described as 


$$X\_0 = \{ \chi \in X : \sum\limits\_{\sigma \in \textrm{Gal}(L/F)} \chi^{\sigma} = 0 \}$$


Now, let $\Phi$ be the set of roots of $T$ in $G$, and $\Delta$ a base for the root system $(\langle \Phi \rangle\_{\mathbb{Q}}, \Phi)$, where $\langle \Phi \rangle\_{\mathbb{Q}}$ denotes the vector subspace of $\mathbb{Q} \otimes\_{\mathbb{Z}} X$ spanned by $\Phi$ .


The base $\Delta$ comes from a total ordering $<$ of $\langle \Phi \rangle\_{\mathbb{Q}}$, or even of $X$. Suppose that $<$ is compatible with $X\_0$ in the sense that if $\chi, \chi' \in X$ with $\chi + X\_0 = \chi' + X\_0$ with $\chi \not\in X\_0$ and positive, then $\chi'$ is also positive. Then $<$ induces a total ordering on $X/X\_0$.


Suppose $\Delta$ is coming from such a total ordering. Let $\overline{\Delta}$ be the image of $\Delta$ in $X/X\_0$, if necessary removing the zero element. 


It may not be the case that the reduction of $\Delta - \Delta \cap X\_0$ modulo $X\_0$ is injective. Nevertheless, let $\overline{\Delta} = \{\gamma\_1, ... , \gamma\_t \}$ for distinct $\gamma\_i \in X/X\_0$. These are positive elements in the total ordering on $X/X\_0$. **Is it the case that the $\gamma\_i$ are linearly independent?**


I have tried to prove this without success. I was thinking that $\overline{\Delta}$ should be a base for a possibly nonreduced root system. The elements of the image of $\Phi$ in $X/X\_0$, removing zero, are integer-linear combinations of $\overline{\Delta}$ which are either all positive or all negative, but the linear independence does not seem obvious.



Is $\overline{\Delta}$ a linearly independent set?
Indeed.


As explained in my answer to your previous question (<https://mathoverflow.net/q/255159>), the map $X\to X/X\_0$ is the restriction $X(T)\to X(A\_0)$. In your present question, you gave those character groups compatible orders as in "Borel Tits, Groupes réductifs". Denote $\_{F}\Delta$ (respectively $\Delta$) the base of the root system $\Phi (G,A\_0)$ (respectively $\Phi$) corresponding to this order. Then $\overline{\Delta} =~ \_{F}\Delta $ (by "Borel Tits, Groupes réductifs, 6.8"). Being a base of a root system, they are in particular linearly independent.


In summary, as you guessed, the image of $\Phi$ in $X/X\_0$ (removing zero) is itself a root system (namely the root system $\Phi (G,A\_0)$, also called the relative root system), and $\overline{\Delta} =~ \_{F}\Delta $ is a base of it.


