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Let $\mathcal K = k((t))$ be the field of formal Laurent series over $k$, and by $\mathcal O = k[[t]]$ the ring of formal power series over $k$.

Let $G$ be an algebraic group over $k$. The affine Grassmannian $Gr_G$ is the functor that associates to a $k$-algebra $A$ the set of isomorphism classes of pairs $(E, \varphi)$, where $E$ is a principal homogeneous space for $G$ over $Spec A[[t]]$ and $\varphi$ is an isomorphism, defined over $Spec A((t))$, of $E$ with the trivial $G$-bundle $G \times Spec A((t))$.

By choosing a trivialization of $E$ over all of $Spec \mathcal O$, the set of $k$-points of $Gr_G$ is identified with the coset space $G(\mathcal K)/G(\mathcal O)$.

Let $V$ be an $n$-dimensional vector space. Then $GL(V)$ acts transitively on the set of all $r$-dimensional subspaces of $V$. Let $H$ be the stabilizer of this action. Then the usual Grassmannian is $GL(V)/H$.

What are the relations between affine Grassmannian and Grassmannian? Why it is important to study affine Grassmannian? Thank you very much.

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  • $\begingroup$ It seems that you have forgotten part of your text: what is $G$? What is $E$? What is $Gr_G$? $\endgroup$
    – abx
    Commented Dec 30, 2013 at 10:40
  • $\begingroup$ @abx, thank you very much. I will edit. $\endgroup$
    – Jianrong Li
    Commented Dec 30, 2013 at 11:16
  • $\begingroup$ I seem to recall having read about this somewhere in Lusztig's paper called Singularities, Character Formulas, and a q-Analog of Weight Multiplicities. You might find your answer there. $\endgroup$
    – Peter Crooks
    Commented Dec 30, 2013 at 12:02

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I am not an expert, but the affine Grassmannian is intimately related to the representation theory of the Langlands dual group $G^{\vee}$. Assume that $G$ is complex semisimple and simply-connected (for simplicity, so to speak). Recall that the dominant weights of $G^{\vee}$ index the finite-dimensional irreducible complex $G^{\vee}$-modules. The dominant weights of $G^{\vee}$ are the dominant coweights of $G$.

Note also that the dominant coweights of $G$ index the strata in a stratification of $Gr_G$. Given a dominant coweight $\lambda$ of $G$, let $Gr^{\lambda}$ denote the corresponding stratum. It turns out that the intersection homology $IH_*(\overline{Gr^{\lambda}})$ is naturally a $G^{\vee}$-module, and is isomorphic to the irrep of $G^{\vee}$ of highest coweight $\lambda$. This description also yields some nice bases for irreps of $G^{\vee}$, called MV cycles. You might read some papers by Kamnitzer and Mirkovic-Vilonen on this subject.

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    $\begingroup$ How does this relate to the usual Grassmannian? $\endgroup$
    – Dori Bejleri
    Commented Dec 30, 2013 at 17:27
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    $\begingroup$ I think the idea is to first realize $Gr_G$ as the affine Grassmannian discussed in Pressley-Segal. (The equivalence of these versions is the subject of mathoverflow.net/questions/150171/…) The Pressley-Segal version of the affine Grassmannian is defined in terms of subspaces of some Hilbert space satisfying some technical analytic conditions. $\endgroup$
    – Peter Crooks
    Commented Dec 30, 2013 at 17:53
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    $\begingroup$ If $G=PGL_n$, then $Gr$ has $n$ components, each of which contains a unique minimal $Gr^\lambda$. The minimal $Gr^\lambda$ in the $k$th component of $Gr$ is the ordinary Grassmannian $Gr(k,n)$. $\endgroup$
    – Allen Knutson
    Commented Feb 12, 2014 at 6:04
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The relation is merely an analogy : for $G=GL(r)$, $Gr_G$ parametrizes certain lattices ($\cong \mathcal{O}^r$) in $\mathcal{K}^r$, see Proposition 2.3 here. The same paper may give you an idea of what the affine Grassmannian is good for.

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  • $\begingroup$ I think the relationship may be more extensive, but I am not sure. I believe that Pressley and Segal introduced a differential-geometric notion of the affine Grassmannian. This version of the affine Grassmannian is more explicitly related to the finite-dimensional case. The trick is then to relate the Pressley-Segal version of the affine Grassmannian to $Gr_G$ when $G$ is a complex reductive group. $\endgroup$
    – Peter Crooks
    Commented Dec 30, 2013 at 13:32

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