Does the "building of parabolics" of a semisimple group have a simplex corresponding to the entire group?

Question

Let $G$ be a semisimple (not just reductive) group over a field $k$. I believe that the question I am asking is what was meant in the second paragraph of Tits building of a linear algebraic group.

I reference the papers Ti: Tits - Buildings of spherical type and finite BN pairs; So: Solomon - The Steinberg character of a finite group with BN pair; CLT: Curtis, Lehrer, and Tits - Spherical buildings and the character of the Steinberg representation.

There are two sorts of objects that seem to go under the name of spherical building of $G$ over $k$, namely, in the terminology of [CLT], the "combinatorial building" that is an abstract simplicial complex, and the building defined in [CLT, Section 2] that is obtained by actually gluing together spheres.

A short form of my question, that reveals the essential difficulty, and where (I think) there is no difference between the combinatorial and spherical buildings: is the building of $\operatorname{SL}_2$ (either sort) the same as its flag variety, or is it the flag variety together with an extra point?

The simplices of the combinatorial building are parabolic subgroups of $G$, with an apartment consisting of the parabolic subgroups containing a fixed maximal split torus of $G$. The apartments of the CLT spherical building are again parameterised by maximal split tori, and a point of the apartment corresponding to $S$ is, by definition, a ray in the real vector space $X_*(S) \otimes_{\mathbb Z} \mathbb R$, where $X_*(S)$ is the lattice of cocharacters of $S$. In order for the apartment to be a sphere, as Section 1 of [CLT] makes clear that it is meant to be, a half-line must be viewed as spanned by a non-$0$ cocharacter. I'll comment on why this is important in a moment.

To each such ray $b$ one can associate a parabolic subgroup $P(b)$ of $G$ containing $S$, and Proposition 6.1 of CLT shows that this allows the spherical building to be viewed as a geometric realisation of the combinatorial building. However, since we insist that our rays are spanned by non-$0$ cocharacters, $P(b)$ can never equal $G$ (because $G$ is semisimple).

According to [So, Section 2], this seems to be fine; Solomon defines (in not quite this language) the vertices of the combinatorial building to be maximal parabolic subgroups (here clearly meaning maximal proper parabolic subgroups—else there would always be only one!), so that, for Solomon, the simplices in the spherical building are proper parabolic subgroups. On the other hand, [Ti, Theorem 5.2] says that the combinatorial building should have as simplices all parabolic subgroups; and maybe Tits implicitly means to say ‘proper’ here, but (setting aside whether such imprecision is likely from Tits) [Ti, Theorem 5.2(ii)] explicitly says that this notion of combinatorial building is the same as the one obtained from a BN-pair in [Ti, Theorem 3.2.6], and the definition there also allows all subgroups, not just proper subgroups, containing (a conjugate of) the $B$ of a BN-pair as simplices.

How does one reconcile these various definitions?

Answer 1

@UriyaFirst points out that every abstract simplicial complex must have an empty face of dimension $-1$, and it would make sense for this to correspond to the entire group. (This is consistent with the general phenomenon where projectivisation (passing from cocharacters to rays) drops the dimension of a facet by $1$, so that the dimension of the facet corresponding to a parabolic $P$ with Levi component $M$ is $\operatorname{rank}(\operatorname Z(M)/\operatorname Z(G)) - 1$.)

I was uncomfortable with the entire group corresponding to a facet because, for a semisimple group $G$, there is no non-$0$ cocharacter $b$ of $G$ such that $G = P(b)$; but that's OK, because CLT only claim that the map $b \mapsto P(b)$ from the spherical building to the combinatorial building can be lifted to a geometric realisation of the combinatorial building—and, of course, the empty facet does not show up in the geometric realisation!

(I am converting the comment to an answer because @UriyaFirst, and @GeoffRobinson who left a related but now-deleted comment, preferred not to do so, but it does answer the question.)