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: 



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> 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.)
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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: 



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> Is the point $c\_0$ close to $x\_A$ ? Can one choose $c$ such that $c\_0 =x\_A$ ?
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My third question is : 



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> Do we have $d(x,x\_A )=\frac{1}{2}d(x,x\_U )$ ? 
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Distance to an apartment of the affine building of GL(N)
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 $x\_A$.


