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.