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.

  • $\begingroup$ It seems that you have forgotten part of your text: what is $G$? What is $E$? What is $Gr_G$? $\endgroup$
    – abx
    Dec 30, 2013 at 10:40
  • $\begingroup$ @abx, thank you very much. I will edit. $\endgroup$ 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. $\endgroup$ Dec 30, 2013 at 12:02

2 Answers 2


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.

  • 1
    $\begingroup$ How does this relate to the usual Grassmannian? $\endgroup$ Dec 30, 2013 at 17:27
  • 2
    $\begingroup$ 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. $\endgroup$ Dec 30, 2013 at 17:53
  • 4
    $\begingroup$ 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 $Gr(k,n)$. $\endgroup$ Feb 12, 2014 at 6:04

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.

  • $\begingroup$ 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. $\endgroup$ Dec 30, 2013 at 13:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.