My question arose from the proof of Proposition 31.7 of "The book of involutions."
It says "… is the Tits algebra of the quasisplit group $(G\_{\nu\_G})\_{F\_{\chi}}$, hence it is trivial." I understood every part of the proof but this sentence. I guess this sentence is true due to the positive answer to the following question.


Let $G$ be a quasi-split simple connected semisimple algebraic group over a field $F$ and $\Gamma=\operatorname{Gal}(F\_\text{sep}/F)$. If $\chi$ is a $\Gamma$-invariant character of the center of $G$, is the minimal Tits algebra $A\_{\chi}$ of $G$ split?


I didn't look closely at the construction of Tits algebra but I'm just taking the notion of Tits algebra as a Galois descent of $\operatorname{End}\_L(V)$, where $L$ is a splitting field of $G$ and $V$ is the minimal representation of $G\_L$ corresponding to $\chi$. From this, however, I cannot see why the Tits algebra of a quasisplit group is split. Is it true? If so, how can I show this?



Tits algebra of the quasi-split semisimple algebraic groups
I found and read the original paper ***<Représentations linéaires irréductibles d'un groupe réductif sur un corps
quelconque>*** by Jacques Tits.
In Theorem 3.3 of this paper, it is shown that the Tits algebra of a quasi-split semisimple algebraic group $G$ is split.


