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I'm interested in the question in the title:

Does a spherical building $B$ always embeds in a building $\tilde B$ of type $A_n$ for some $n$?

By embedding I mean an isometric embedding with respect to their $CAT(1)$ metric. If the question has a positive answer, then how does this embedding work? For instance, does an automorphism of $B$ extend to an automorphism of $\tilde B$?

Edit: I forgot that there are strange polygons. I would like to restrict the question to the case of irreducible thick buildings of rank at least 3. In particular, they are Moufang.

Edit 2: As Dima Pasechnik pointed out, if the building $B$ has residues of type $A_2$ which are non-Desarguesian projective planes, then we cannot have such an embedding, since projective spaces of dimension $\geq 3$ must be Desarguesian. This is a problem if $B$ is of type $B_3$ or $F_4$. In the former case, it is known that there are non-embeddable buildings (see Tits' "Buildings of spherical type and BN-pairs", chapter 9). Hence the following question:

What about type $F_4$?

Buildings of type $B_n$ with $n\geq 4$ are embeddable (see Tits' "Buildings of spherical type and BN-pairs", chapter 8) into a projective space of possibly infinite rank. So, I assume that the answer to my quesion in this case is "No in general" (because of the infinite rank). Is this correct?

Buildings of type $D_n$ should be embeddable, since they correspond to the group $SO_{2n}(k)$ for some field $k$. Hence these buildings are embeddable in $A_{2n}$.

Type $E_n$, $n=6,7,8$, should not be a problem either, since they correspond to algebraic groups. Is this also correct?

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  • $\begingroup$ you probably need to restrict your $B$ somehow. There are very weird generalised quadrangles (and thus buildings of type $C_2$) known which aren't embeddable in projective spaces at all. $\endgroup$
    – Dima Pasechnik
    Commented Apr 2, 2014 at 19:51
  • $\begingroup$ @Dima Pasechnik: Yes, that is true. I was thinking more in higher rank (see edit). $\endgroup$
    – Luc
    Commented Apr 3, 2014 at 6:51

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If $B$ is of sufficiently high rank (at least 5) then it comes from a polar space, something that is well-known. If the singular subspaces are Desarguesian then it's known to be embeddable in the projective space in the usual way. See Chapter 8 of the book "Diagram geometry" by Buekenhout and Cohen. So in this case (as well as in slightly more general case of $B$ coming from a polar space with Desarguesian singular planes) the answer is yes.

I don't know about the non-Desarguesian case (as well as about small rank non-polar space case, i.e. $B=F_4$, or $B=E_k$ for $k=6,7,8$, case), perhaps there is more in loc.cit.

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  • $\begingroup$ Yes, non-Desarguesian should be a problem. If we forget about the rank 3 case, then all buildings with exception of type $F_4$ have residues of type $A_3$ so the corresponding spaces are Desarguesian $\endgroup$
    – Luc
    Commented Apr 3, 2014 at 15:24
  • $\begingroup$ there has been some work done for $F_4$ and $E_k$ along the same lines as for polar spaces, but I don't know how conclusive are the results. Certainly, when there is just one building, given a field (e.g. for finite fields), the answer is yes. $\endgroup$
    – Dima Pasechnik
    Commented Apr 3, 2014 at 18:41
  • $\begingroup$ I took again a look at Tits' book "Buildings of spherical type and BN-pairs". The embeddings of polar spaces in projective spaces is also explained there. But as I understand it, the projective spaces are sometimes of infinite rank. So we do not get an embedding to a building of type $A_n$. $\endgroup$
    – Luc
    Commented Apr 4, 2014 at 9:04
  • $\begingroup$ well, I won't be surprised if this has been improved (perhaps even by Tits himself) - that's why I referred to the book that is more or less the state of the art. $\endgroup$
    – Dima Pasechnik
    Commented Apr 4, 2014 at 19:03
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A comment on the metric aspect of the question.

In the case of the embedding of a polar space $X$ into a projective space $\tilde{X}$ described by Buekenhout and Cohen, hyperplanes in $X$ are sent to points in $\tilde{X}$, and points in $X$ are sent to subspaces in $\tilde{X}$ which are not hyperplanes (unless $X$ is already a projective space). In building terms, this translates to a (poset) embedding of $B$ into $\tilde{B}$ of type $A_n$, whose image contains vertices of type $1$, but no vertex of type $n$.

This cannot be an isometric embedding with respect to the CAT(1) metrics (even after rescaling), since opposite points in $B$ should be sent to opposite points in $\tilde{B}$, because of the non-uniqueness of geodesic segments between them. But the points in $\tilde{B}$ opposite to a vertex of type $1$ are vertices of type $n$.

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  • $\begingroup$ I'm not sure I fully understand the embedding in Buekenhout and Cohen. But I think, we just have to interpret it in the right way. (Again supposing that the projective space is of finite rank:) A point $x$ in $\tilde X$ in the image of the embedding is always accompanied by the orthogonal $x^\perp$ and $x\subset x^\perp$, so as metric spaces we should think of the image of an hyperplane in $X$ as the midpoint of the edge between $x$ and $x^\perp$. So in building terms, the image is me midpoint of an edge $1n$ and these have antipodes of the same type. $\endgroup$
    – Luc
    Commented Apr 4, 2014 at 16:08
  • $\begingroup$ The easiest example is the embedding of the building $B$ of $Sp(2 \times 2,\mathbb{K})$ into the building $\tilde{B}$ of $PGL(6,\mathbb{K})$: an isotropic plane $P$ in $\mathbb{K}^4$ is sent to the line $\Lambda^2 P$ in $\Lambda^2 \mathbb{K}^4 \simeq \mathbb{K}^6$, and an isotropic line $L$ in $\mathbb{K}^4$ is sent to the $3$-plane $\Lambda^2 L^\perp$ in $\Lambda^2 \mathbb{K}^4$. In particular, if $L \subset P$, the distances are not preserved: $d_B(L,P)=\frac{\pi}{4}$ and $d_{\tilde{B}}(\Lambda^2 L,\Lambda^2P)=arccos(\sqrt{1/3})$. $\endgroup$
    – Thomas Haettel
    Commented Apr 7, 2014 at 11:08
  • $\begingroup$ So this is what the embedding does? Why just not take $Sp(4)\hookrightarrow GL(4)$? $\endgroup$
    – Luc
    Commented Apr 7, 2014 at 12:50
  • $\begingroup$ If you take $Sp(4) \hookrightarrow GL(4)$, you will obtain a cellular embedding from the building of $Sp(4)$ into the building of $GL(4)$, which will not be locally isometric : the length of the edges are different, and the image of the geodesic bewteen two non-opposite isotropic planes will not be a geodesic in the building of $GL(4)$. However, if you take $SL(2) \times SL(2) \hookrightarrow SL(4)$ for instance, there you will get an isometric embedding of the buildings (not simplicial). I think a necessary condition is that the root system of one group is a subroot system of the other one. $\endgroup$
    – Thomas Haettel
    Commented Apr 8, 2014 at 5:48
  • $\begingroup$ Yes, that was my point in my first comment. Using $Sp(4)\hookrightarrow GL(4)$ the image of an isotropic plane $P\subset \mathbb{K}^4$ should be $P$ and the image of an isotropic line $L$ should be the midpoint of the edge $L\subset L^\perp$. This is an isometric embedding. $\endgroup$
    – Luc
    Commented Apr 8, 2014 at 6:52

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