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Let $G$ be a connected, simply connected, complex, semi-simple group with affine Grassmannian $\mathcal Gr \cong G(\mathbb C[t,t^{-1}])/G(\mathbb C[t])$. Fix a choice of maximal torus $T \subset G$ and positive roots, so that $B, N^\pm$ are the corresponding Borel and connected unipotent subgroups.

There is an evaluation map $\text{ev}: G(\mathbb C[t]) \to G$ mapping $\gamma \to \gamma(0)$, and we define the Iwahori subgroup to be $I=\text{ev}^{-1}(B)$. Additionally, let $J = \text{ev}^{-1}(N^+)$.

Let $X^*(T)$ be the correspondings set of coweights and $t^\lambda$ be the class of $\lambda \in X^*(T)$ in $\mathcal Gr$. The literature is abound with the fact that $\mathcal Gr$ admits a Bruhat decomposition $$ \mathcal Gr = \bigsqcup_{\lambda \in X^*(T)}e_\lambda,$$ where the $e_\lambda$ are the Schubert cells. The main proofs I have seen of this fact are usually Mitchell's Building's paper or Pressley & Segal's book Loop Groups.

Unfortunately, there seems to be some discrepencies in the statement of the Bruhat decomposition. Overwhelmingly, the literature seems to say that the Schubert cells are the orbits of the Iwahori subgroup: $e_\lambda = It^\lambda$. However, Pressley and Segal (which is rather well cited!) uses instead cells which are the orbits $e_\lambda = Jt^\lambda$. My question:

Are these decompositions topologically equivalent? Or are they instead similar notions which have been endowed with the same name. If they are the same, are there any advantages using one framework over the other. If they are different, why has the community overwhelmingly chosen the Iwahori orbit viewpoint?

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    $\begingroup$ Yes I would guess these orbits are the same (not only topologically equivalent). As each $t^\lambda$ is a $T$-fixed point the $J$ and $I$ orbit through it are the same. The finite dimensional version of your question is whether one considers $N^+$ orbits or $B$-orbits. This seems to be a matter of taste (as long as you are not worrying about equivariant cohomology / sheaves, where the difference can matter.) $\endgroup$ Oct 21, 2014 at 21:25
  • $\begingroup$ @Tyler: I don't recognize at first your definition of the affine Grassmannian, since I'm used to seeing a field of formal power series there. (Also, the name is "Pressley".) $\endgroup$ Oct 21, 2014 at 23:02
  • $\begingroup$ @GeordieWilliamson I may actually be working in the equivariant case. I'll be sure to be careful there. Thank you for pointing that out. $\endgroup$ Oct 22, 2014 at 17:30
  • $\begingroup$ @JimHumphreys Sorry about the spelling of Pressley, it was just a typo (I spelt it correctly earlier). As for the affine Grassmannian, I'm just taking my cue here from Ginzburg (Perverse Sheaves ...). The use of the $\mathbb C[t]([t^{-1}])$ is more convenient (to me) for the identification of $\mathcal Gr$ with the algebraic based loop group of the maximal compact subgroup $\Omega_{\text{alg}} K$. $\endgroup$ Oct 22, 2014 at 17:31

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They are definitely the same sets. If we embed $T$ as the constant functions in $G((t))$, then $I=JT$, so the orbit of the $I$ and $J$ through any $T$-fixed point is the same.

As for why: if you want to consider the $I$-action on them (quite common) you'll want to use $I$ orbits. On the other hand, for seeing their topological structure, it can be cleaner to think of them as $J$-orbits. Of course, there's no reason to choose.

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