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.