# Cartan decomposition for upper triangular matrices

Due to the comments, I have the impression that I have to be more precise.

Consider $G= GL_n(F)$ for a non-Archimedean field $F$ with ring of integers $o$. Let $K= GL_n(o)$ and let $I$ the Iwahori subroup of $K$ enlarged by the centrum. Let $M$ be the Levisubgroup (=diagonal matrices) of the standard Borel (=upper triangular matrices) in $G$ and $N$ the Normalizer (diagonal matrices * permutations) of $M$ in $G$.

Then $(I,N)$ is a BN pair for $G$.

The Cartan decomposition is given as $$G = K M K,$$ and the affine Bruhat decomposition as $$G = INI.$$

Is it easier to read off the Cartan or alternatively the affine Bruhat decomposition, if the matrix is an upper triangular matrix? With read off, I mean is there a way to see the decomposition from the norms of the entries.

The usual algorithmic proof of the Cartan decomposition in the non-Archimedean case goes as follows:

1. Multiply by permutation matrices from the right and left such that we have at position $(1,1)$ the entry with maximal norm.

2. Kill the entries in the first column and first row by elementary matrices from $GL(n,o)$.

3. iterate the procedure for the matrix with the first column and row removed...

What is an algorithmic proof of the affine Bruhat decomposition, which works on $GL_n(F)$ rather than one the building?

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Presumable $o$ denotes the ring of integers of $F$ ... ? – Max Horn Jan 8 '12 at 14:15
Like Benjamin, I also wonder what exactly you mean... If by affine Bruhat decomposition, you mean the Bruhat decomposition $G=BWB$ with respect to an affine (type $\tilde A_{n-1}$) $BN$-pair of $G$, then with the standard choice for $B$ (namely the subgroup of upper triangular matrices), you trivially get the decomposition of an upper triangular matrix. Maybe you had something else in mind, though, such as forcing $B$ to be the lower triangular matrices, or arbitrary? – Max Horn Jan 8 '12 at 14:20
Also, what is the Cartan decomposition of a group for you? The Iwasawa decomposition? I.e. $G=KB$, where $K$ is a maximal compact subgroup (or the centralizer of an involution, or...) and $B$ is a Borel subgroup? Then again, you either get a trivial decomposition of an upper triangular matrix (assuming $B$ is again the subgroup of upper triangular matrices); or you need to specify more clearly what you want... – Max Horn Jan 8 '12 at 14:25
Thanks for the clarifications. So with afine Bruhat decomposition, you mean exactly what I thought you meant (only that I screwed up in thinking that $B$ was upper triangular, which of course it isn't -- that would be the a Borel for a spherical Bruhat decomposition). However, I don't think $(K,M)$ is a BN-pair. If it was, then its Weyl group would $W$be trivial, as $M\subset K$ and $W=M/(M\cap K)=1$. So we would have to have $G=K$, which is false. However, I think it is a Gelfand pair and as such $K\backslash H / K$ is an abelian Hecke algebra... – Max Horn Jan 9 '12 at 12:27
And neither is $(I,M)$ a BN-pair, but $(I,N)$ is. Since you wrote the Bruhat decomposition for $(I,N)$, I assume this is a typo? – Max Horn Jan 9 '12 at 12:34