Iwasawa Decomposition Does anyone know where I can find a proof of the Iwasawa decomposition for reductive groups? I know that there are a couple of related results that are called the Iwasawa decomposition, but I am interested in the following statement:
Let G be a complex reductive group, let O be Taylor series with complex coefficients, and let K be Laurent series with complex coefficients. The G(K) = G(O) * T(K) * U(K), where T is a maximal torus and U is a maximal unipotent subgroup of G. 
I am interesting in finding the proof because this is how one shows that the semi-infinite cells in the affine Grassmannian cover the entire space. I have been using this fact for quite a while now and am becoming uncomfortable about not knowing where to find the proof. A proof in the case where K is a p-adic field and O is its ring of integers would also be great since I am sure a proof would carry over to the above case.
 A: The original reference for this is the paper of Bruhat-Tits (available on NUMDAM), see Prop. 4.4.3. Another reference, probably easier to read, is the book of Macdonald, "Spherical functions on a group of p-adic type", Theorem 2.6.11.
There is a nice proof of this fact using the geometry of buildings, which goes as follows. You can think only about trees (eg for G=SL(2)) to get the main ideas.
Consider the affine building X associated to G(K). The buildings at infinity of X (which is also the space of flags) can be seen as equivalence classes of sectors in X. Then U(K) is the union of fixators of sectors in such a class \xi. The group T(K) is the group of translation in some apartment A, and B:=T(K)U(K) is the stabiliser of the equivalence class of sector. G(O) is the stabiliser of some vertex o.
The main point is that the building is the union of all apartments containing a sector pointing towards \xi. It follows that, for every x in X, there is an element u of U(K) such that u.x is in A.
Let g in G. Applying this to the element x=g.o, we see that the vertex ug.o is in A. By transitivity of the action of T(K) on vertices of A, there is an element t in T such that tug.o=o. Thus tug is in G(O), which gives the decomposition of g.
A: Is it possible to use the fibration B -> G -> G/B for a chosen Borel subgroup B in G to prove the covering statement? As H^1(K,B) = *, this fibration should say that, G(K)/B(K) = G/B(K). By properness of G/B, G/B(K) = G/B(O). By the previous argument on O, we have G/B(O) = G(O)/B(O). In particular, we get that G(K) = G(O) B(K). Since H^1(K,U) = *, the exact sequence U -> B -> T gives B(K) = T(K) U(K), etc. I have no idea if this actually works! 
A: This reverses your logic a bit, but perhaps you can prove directly that the semi-infinite orbits cover the Grassmannian and then deduce the Iwasawa decomposition.
Here is a possible approach to doing this.
First, show that the semi-infinite orbits are the attracting sets for a C^* action on Gr.  This C^* action comes from a generic sub-C^* of the maximal torus T(C) of G(C) (for example from the coweight \rho).
The fixed points of this C^* action will be the same as the fixed points of the T(C) action action, namely the points t^\mu, where \mu is a coweight.  The semiinfinite orbits are the attracting sets for this C^* action.  
To show that the semi-infinite orbits cover Gr, we just need to show that every point in Gr has a limit under this C^* (ie is in a C^* orbit).  This would be automatic if Gr was projective.  However, the C^* action preserves the \overline{Gr^\lambda}, and every point lives in some \overline{Gr^\lambda}, so every point has a limit under the C^* action.
