# Tits algebra of the quasi-split semisimple algebraic groups

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?

## 1 Answer

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.