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Let $AG_k(\mathbb{R}^N)$ be the "affine Grassmannian" consisting of $k$-dimensional hyperplanes (i.e. affine subspaces) in $\mathbb{R}^N$. Is there any relation between $AG_k(\mathbb{R}^N)$ and the usual Grassmannian? Is $AG_k(\mathbb{R}^N)$ a classifying space of any Lie groups?

Let $AG_k(\mathbb{C}^N)$ be the "affine Grassmannian" consisting of $k$-dimensional complex hyperplanes in $\mathbb{C}^N$. Is there any relation between $AG_k(\mathbb{C}^N)$ and the usual complex Grassmannian? Is $AG_k(\mathbb{C}^N)$ a classifying space of any Lie groups?

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    $\begingroup$ It seems to me the answer is yes to all questions : The natural map $AG_k(\mathbf R^n)\to G_k(\mathbf R^n)$ has a contractible fiber (homeomorphic to $\mathbf R^{n-k}$) and is a homotopy equivalence. $\endgroup$
    – few_reps
    Commented Oct 21, 2015 at 9:03
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    $\begingroup$ Note that the term "affine Grassmanian" is usually reserved for a certain infinite dimensional analogs of the Grassmanians, associated to the so-called affine Lie algebras (which are themselves infinite dimensional analogs of the finite dimensional simple Lie algebras). $\endgroup$
    – André Henriques
    Commented Oct 21, 2015 at 9:20

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I assume you mean $k$-dimensional planes, not just hyperplanes. Then the space $AG_k(\Bbbk^n)$ deformation retracts onto the usual Grassmannian by sending each $k$-plane $E$ to $rE$ at time $r\in[0,1]$. More precisely, if $E=a+V$ with $a\in\Bbbk^n$ and $V\subset\Bbbk^n$ a linear $k$-dimensional subspace, then $rE$ denotes the affin plane $ra+V$.

In particular, these space approximate the classifying space of $Gl(k,\Bbbk)$ as $n\to\infty$. Because the deformation retractions above are compatible with the natural inclusions $AG_k(\Bbbk^n)\to AG_k(\Bbbk^{n+1})$, their colimit $AG_k(\Bbbk^\infty):=\lim_\to AG_k(\Bbbk^n)$ still has $\lim_\to G_k(\Bbbk^n)$ as a deformation retract. Therefore, $AG_k(\Bbbk^\infty)$ is a model for $BGl(k,\Bbbk)$.

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  • $\begingroup$ Is that really the deformation retraction? $\endgroup$
    – Michael Albanese
    Commented Oct 21, 2015 at 14:41
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    $\begingroup$ Yes, I think so, but regard $E\subset\Bbbk^n$ as a plane, not as a set of points. Then the limit of $rE$ as $r\to 0$ is a plane through the origin, and not just $\{0\}$. Does that answer your question? $\endgroup$
    – Sebastian Goette
    Commented Oct 21, 2015 at 14:53
  • $\begingroup$ I'm not sure what you mean by $rE$. You can write $E$ as $a + P$ where $P$ is a plane through the origin parallel to $E$ and $a \in E$, then the deformation retraction should be $ra + P$. Is that what you mean? $\endgroup$
    – Michael Albanese
    Commented Oct 21, 2015 at 16:56
  • $\begingroup$ @Michael Albanese: yes, and I edited the answer. $\endgroup$
    – Sebastian Goette
    Commented Oct 21, 2015 at 17:51

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