Linear symmetric spaces are spaces with ''orthogonal complements''?

The collection of marked unimodular lattices in the euclidean $n$-space corresponds to the symmetric space $S_n:=SO_n(\mathbb{R}) \backslash SL_n(\mathbb{R})$.

I have only recently been made aware that $S_n$ is a so-called $linear$ symmetric space. Apparently this means the following. The collection of (marked) nondegenerate lattices in $n$-space corresponds to the collection $P_n$ of positive definite $n\times n$ symmetric real matrices. The space $P_n$ forms an open convex cone in $\mathbb{R}^N$, $N=n(n-1)/2$. The unimodular elements in $P_n$ consists precisely of $S_n$. Thus $S_n$ is a hypersurface in the cone $P_n$. More specifically $S_n$ is a 'section' of the cone, ie. $S_n$ is homeomorphic to an open ball $B$. The significance is this: taking the closure $\bar{B}$ of the ball gives a compactification. To extend the homeomorphism (if we can) now to the closure gives a compactification of $S_n$.

But still, I do not know what the ''linearity'' of $S_n$ $really$ means.

I would like to know how it relates to the following very remarkable property of unimodular lattices:

Take an unmarked rank $n$ unimodular lattice $\Gamma$. (ie. $\Gamma$ is an element in $S_n / SL_n(\mathbb{Z})$). The remarkable fact is this: for a given rational subspace $W$ in $\Gamma \otimes \mathbb{R}$ there is a CANONICAL complementary rational subspace $W^o$ --( of course, $W^o$ is just the orthogonal complement of $W$ in $\Gamma \otimes \mathbb{R}$).

My question is then how do we see the LINEARITY of $S_n$ as responsible for the fact that rational subspaces have CANONICAL rational complements? Any remarks on what a ''linear'' symmetric space really $is$ would be appreciated.

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I haven't heard the terminology "linear symmetric spaces" before, could you give a reference where the term is defined, I'm just curious about it. –  Thomas Richard Mar 29 '12 at 7:04
@jmart: I'm a little puzzled by your omission of the unimodularity condition. The space $S_n$ is actually embedded as a hypersurface in the convex cone of positive definite quadratic forms on $\mathbb{R}^n$, a hypersurface that is asymptotic to the boundary of the convex cone. Are you asking for noncompact Riemannian symmetric spaces $M = G/K$ for which there is a $G$-module $V$ such that there is an open, proper convex cone $C\subset V$ that is $G$-invariant and such that $K$ is the stabilizer of a point inside $C$ and $\dim V = 1+\dim M$? –  Robert Bryant Mar 29 '12 at 15:28
I cannot provide a reference for a proper definition of ''linearity'' because I only know the term as described in MacPherson & McConnell's Inventiones paper "Explicit reduction theory for Siegel modular threefolds". –  J. Martel Mar 29 '12 at 17:13
@Bryant: As originally phrased, my description of $S_n$ as linear space iswrong--i shall correct it. From MacPherson/McConnell it seems that ''linearity'' describes how the symmetric space $S_n/$ compactifies. Roughly (and this is what i'd like to refine for myself) a symmetric space is linear if it compactifies like the open ball. But this is a very amateur description. The business of these various compactifications (Satake, Borel-Serre, toroidal, ...) I find quite difficult to understand. –  J. Martel Mar 29 '12 at 17:30
Even the question "how do we compactify a disk" I don't see as so clear. More unclear, is of course,"how do we compactify a disk modulo $SL_n(\mathbb{Z})$". So my question above, on interpreting the linearity of $S_n$ as the existence of canonical rational complements, is really a twisted way of trying to learn more about these various compactifications. From what I've been learning, flags of rational subspaces are almost the boundary components of the (Borel-Serre or Satake) compactifications. On these points I am over my head--but not too over. I am trying to work these questions out. –  J. Martel Mar 29 '12 at 17:37

Definition of "linear symmetric space" appears in the book of Borel and Ji "Compactifications of symmetric and locally symmetric spaces", p. 286. Their definition is not very precise, but can be restated as follows: A symmetric space $X$ is called linear if there exists a convex domain $D$ in the projective space such that $(X, Isom(X))$ is equivariantly diffeomorphic to $(D, Proj(D))$, where $Isom(X)$ is the group of isometries of $X$ and $Proj(D)$ is the group of projective transformations of $D$. With this definition, there are very few irreducible linear symmetric spaces, see e.g. http://www.math.umn.edu/~garrett/m/v/classical_domains.pdf Whatever the property of lattices you need (I still do not understand which one since it is stated only for $SL(n, {\mathbb R})$) you can probably check case-by-case. I think, the main advantage of such spaces is that fundamental domains of lattices (used in the reduction theory) could be described by linear inequalities in this case.
Update: Below is my interpretation of what "complementary" lattice is in the general setting. First, consider $SL(n)$. Then every rational linear subspace $V$ in ${\mathbb Q}^n$ has a "canonical" orthogonal complement $V^\perp$ (also defined over rationals) once we have fixed a rational positive-definite quadratic form. (The complementary lattice is then obtained by taking the integer points in $V^\perp$.) This amounts to choosing a rational point $x$ in the symmetric space $X$ of $SL(n, {\mathbb R})$, i.e., points which is fixed by rational Cartan involution $\theta=\theta_x$. From the viewpoint of the symmetric space $X$, the space $V$ is a point $p_V$ on the ideal boundary of $X$, more precisely, rational Tits building $Y$ sitting inside of the Tits boundary of $X$. Applying $\theta$ to $p_V$ we obtain an antipodal (rational) point $q\in Y$, which is nothing but $p_{V^\perp}$. This interpretation goes through verbatim for every semisimple algebraic group $G$ defined over ${\mathbb Q}$ provided that $G$ contains a rational Cartain involution (one that preserves $G({\mathbb Q})$ and, hence, the corresponding rational Tits building). Existence and uniqueness of such involutions (up to conjugation in $G({\mathbb Q})$) was studied by several people, like I.Satake "On the rational structures of symmetric domains", (parts I and II which appeared in 1989, 1991), A.Helmnick "On the classification of k-involutions", http://www4.ncsu.edu/~loek/research/class.pdf and, I think, many others. Google "rational Cartan involution" and "rational points of symmetric space" to find more references. Jim Humphries who comments frequently on MO can surely provide much more information than I do. The bottom line is that linearity of the symmetric space is irrelevant for the construction of the "canonical" complementary lattice.
In a euclidean vector space $V$, for given subspace $W$ we can identify a canonical (coordinate free) complement $W^o$. This complement affords us an embedding, say of $GL(W) \times GL(W^o) \to SL(V)$. Given a nondegenerate symplectic vector space $(V, \omega)$ we have no obvious (as far as I can tell) means of assigning canonically (ie. coordinate free-ly) an analagous complement for a given totally isotropic subspace $W$. That is, there is no coordinate free means of embedding $GL(W)$ into $Sp(V)$. Now supposedly $U_g(\mathbb{C}) \backslash Sp_{2g}(\mathbb{R})$ is non-linear symmetric space. –  J. Martel Mar 30 '12 at 23:52
Moreover, in the Borel-Serre compactifications of such symmetric spaces (really, of the associated locally symmetric space $U_g \backslash Sp_{2g} / Sp_{2g}(\mathbb{Z})$) there is a large role played by flags of totally isotropic subspaces. I wanted to know to which extent non-linearity is the cause for this non-canonicity. NOTE: In above comment, I should have said "the complement $W^o$ affords us an embedding of $GL(W)$ into \$SL(V)". –  J. Martel Mar 30 '12 at 23:56