Here $F$ is a locally compact non-archimedean non-discrete field.

Let $X$ be the reduced (affine) Bruhat-Tits building of ${\rm GL}(n,F)$. Fix a maximal split torus $T$. Let $B$ be a Borel subgroup containing $T$ and write $U$ for the unipotent radical of $B$. Let $A$ be the unique apartment of $X$ stabilized by the normalizer $N_G (T)$ of $T$ in $G$. Finally fix a vertex $x$ of $X$.

It is not difficult to see that there exists a unique point $x_U$ of $A$ such that $x_U =u.x\in A$ for some $u\in U$ (use Iwasawa decomposition). Making $B$ vary among the Borel subgroups containing $T$, one gets $n!$ points $x_U$ in $A$.

My first question is:

What is the link between the points $x_U$ and the projection $x_A$ of $x$ onto the apartment $A$ ($A$ is a closed convex subset of the CAT$(0)$ space $X$ and this projection is well defined) ? (For $n=2$, $x_A$ is the isobarycenter.)

Now fix $B$ and consider $x$ and $u$ as above. Assume that $x\not\in A$. Then $u$ fixes a sector $C$ of $A$. Let $c$ be a point of $C$. Consider the geodesic segments $[c,x]$ and $[c, x_U ]$ (so that $[c,x_U ]=u.[c,x]$). Let $c_0$ be the unique point of $[c,x]$ such that $[c,c_0 ]\subset A$ and $(c_0 ,x]\cap A =\emptyset$.

My second question is:

Is the point $c_0$ close to $x_A$ ? Can one choose $c$ such that $c_0 =x_A$ ?

My third question is :

Do we have $d(x,x_A )=\frac{1}{2}d(x,x_U )$ ?

  • $\begingroup$ These questions seem well formulated though they may be beyond the technical range of most Math Overflow users. Since there are several questions, it would help to number them and also highlight them using the command > $\endgroup$ Nov 19, 2010 at 23:45
  • $\begingroup$ Romeo, the connection of the last two questions that you tagged nt-number.theory to number theory is very tangential. Maybe it's just me, but I think that your re-tagging is not helping to keep the tag-space on MO clean. $\endgroup$
    – Alex B.
    Nov 20, 2010 at 6:00
  • 2
    $\begingroup$ @Paul- I'm not sure Jim was trying to say the question was too technical (and shame on him if he was). No question is too technical for MO, though lots of questions are too technical to have a good chance of getting an answer; it may well be you are actually the most knowledgeable current MO user on the subject of affine buildings. I guess we'll find out soon enough. $\endgroup$
    – Ben Webster
    Nov 20, 2010 at 12:54
  • 4
    $\begingroup$ I approve the idea of creating a tag for building. I think that will encourage questions to be asked. This question seems difficult since it is not a pure geometric question on building: the $x_U$ are defined group theoretically. $\endgroup$
    – Joël
    Nov 20, 2010 at 13:20
  • 1
    $\begingroup$ This is a bit of a specialized research question, I think. I certainly don't know the answer off the top of my head. But perhaps the paper "Kostant Convexity for Affine Buildings" (preprint on ArXiv, now in Forum. Math.?), by P. Kitzelberger will help. I think the result on p.4 comes close to an answer to the first question, and the rest of the questions are in the same spirit as Kitzelberger's paper too. Good luck! $\endgroup$
    – Marty
    Nov 21, 2010 at 4:57

1 Answer 1


I think that the answers to your three questions are negative. Here's an example for $n=3$.

Choose first $x_A$ to be a vertex of $A$. In the link of $x_A$, it is possible to choose a chamber $d$ which is at distance $2$ from $2$ chambers in $A$, and at distance $3$ from the $4$ others. Choose $x$ in the alcove $d$ so that it projects on $x_A$. To fix the ideas, you can take the middle of the side opposite to $x_A$.

Then the geodesic ray from $x$ to $x_A$ can be continuated in only four ways, namely via one of the four alcoves opposite $d$ in the link of $x_A$. This answers negatively to your second question. The retraction from the 4 points towards which these geodesics are going are easily calculated: you just have to retract the geodesic segment from $x$ to $x_A$ from somewhere further on your geodesic ray; the four points you get are the four points in the alcove opposite to the four alcoves which were opposite to $d$ (and, of course, on the middle of the side opposite to $x_A$).

From this, it is possible also to answer negatively to the third question: there is some $x_U$ which is on in one of the two alcoves which are at distance $2$ from $d$. Then the distance from $x$ to this $x_U$ is the distance between the middle of two sides of a regular hexagon which are at distance $2$, and the distance from $x$ to $x_A$ is the distance to the center of this hexagon, so we do not have $d(x,x_A)=\frac 1 2 d(x,x_U)$.

The calculation of the retractions centered at the two other points is not so easy. Let $\xi_1$ be one of them. The sector from $x_A$ to $\xi_1$ starts with some alcove $c$ which is at distance $2$ from $d$. Let $d'$ be the alcove adjacent to both $c$ and $d$, and let $H$ be the wall of $A$ which separates $c$ and $d'$. $H$ separates $A$ in two half-planes, say $\alpha$ and $-\alpha$, with $\xi_1$ in $\alpha$. Let $\beta$ be another half-plane bounded by $H$, such that $d'\in \beta$. Then $\beta\cup\alpha$ is another apartment $A_1$.

The retraction on $A_1$ centered at $\xi_1$ sends $d$ to a chamber which is at distance $2$ from $c$ and adjacent to $d'$, so it is the alcove in $\beta$ which is adjacent to $d'$. The retraction of $d'$ on $A$ is then the retraction of this alcove is the alcove which is at distance 2 from $c$ and in $-\alpha$. Since this alcove is already opposite to an alcove which is opposite to $d'$, we get the same point as some other $x_U$.

Of course, the retraction centered at the last point is treated in a similar way. So, in conclusion, there are only 4 different retractions of $x$. The two "double" points form a segment whose middle is $x_A$. The two other are symmetric with respect to $H$, but are not in the same sector. So the barycenter of our 6 points is some point of $H$ which is not $x_A$.


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