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Let $V$ be a finite dimensional vector space over some field (say, $\mathbb C$). Consider the set $\operatorname{GLI}(V)$ of all linear isomorphisms between subspaces of $V$. This is a monoid under natural multiplication (in fact an inverse monoid). Its elements can be represented by triples: two elements of the Grassmannian of $V$ of degree $k\le n$ representing the domain and the range, and a non-singular $k\times k$-matrix representing the map. I am interested in developing a theory of representations of finite inverse monoids (pseudogroups) in $\operatorname{GLI}(V)$. What is the structure of $\operatorname{GLI}(V)$ from the algebraic geometry or geometric topology point of view?

Edit: It looks like the question is not completely clear. For comparison, if somebody gives me a group and asks what can I say about it, I would try to decide whether the group is finite or infinite, solvable or not, hyperbolic or not, what is the derived subgroup and the lower central series, is it residually finite and what is the profinite completion, etc. I want a similar analysis of $\operatorname{GLI}$ (but from the algebraic geometry point of view). One of the goals is to study representation varieties of groupoids (=pseudogroups, inverse semigroups). These varieties are complicated even for easy finite groupoids. The starting point would be to understand $\operatorname{GLI}$ itself.

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  • $\begingroup$ I think you answered your own question -- it's a disjoint union of principal bundles over products of Grassmannians. What more do you want? $\endgroup$
    – Vivek Shende
    Commented Oct 1, 2010 at 11:32
  • $\begingroup$ I want to know standard information about it (say, what is the co-homology ring, singularities, etc.) If somebody considered such varieties before, I would like to see a reference. From the topological point of view, I would like to see if there exists a boundary similar to the boundary of Lie groups. $\endgroup$
    – user6976
    Commented Oct 1, 2010 at 13:31
  • $\begingroup$ I'm a little confused as to why you call it a monoid and then a groupoid. Do some people have more flexible notions of monoids -- that the binary operation need not be defined on the entire product? $\endgroup$
    – Ryan Budney
    Commented Oct 1, 2010 at 19:14
  • $\begingroup$ @Ryan: If you take an inverse monoid, and remove the 0, you get a groupoid. So the difference is cosmetic. $\endgroup$
    – user6976
    Commented Oct 1, 2010 at 19:21
  • $\begingroup$ @RyanBudney You don't have to remove zero. $M=\mathrm{GLI}(V)$ is canonically the set of arrows of some groupoid, whose set of objects is the set of subspaces of $V$, and whose set of arrows are the isomorphisms between those subspaces, defining $gf$ only if $codom(f)=dom(g)$. So viewing it as a groupoid is just losing some structure (as one restricts the composition): this groupoid is purely defined from the monoid structure: $codom(f)=dom(g)$ indeed means that for every idempotent $e$, we have $(eg=g)\Leftrightarrow (fe=f)$. Removing zero just artificially removes an isolated point. $\endgroup$
    – YCor
    Commented Apr 30, 2020 at 13:51

2 Answers 2

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Some small comments.

Let $n=dim(V)$, so I'll think of $V$ as $\mathbb R^n$, then as a space, $GLI(V)$ you could think of as

$$ V_{n,k} \times_{O_k} V_{n,k} $$

where $V_{n,k}$ is the Stiefel manifold of orthonormal $k$-frames in the vector space $V$. i.e. this is the space $V_{n,k}^2$ mod the diagonal action of $O_k$.

So you could view it as a bundle over $G_{n,k}^2$ with fiber $O_k$, or as a bundle over $G_{n,k}$ with fiber $V_{n,j}$. $G_{n,k}$ is the Grassmannian of $k$-dimensional subspaces of $V$.

The map $V_{n,k} \times_{O_k} V_{n,k}$ to $GLI(V)$ is given by sending a pair $(A,B) \in V_{n,k} \times V_{n,k}$ to:

The span of $A$, the span of $B$ and the corresponding linear isometry represented by $B\circ A^{-1}$ where we think of $A$ and $B$ as representing isometric embeddings $\mathbb R^k \to \mathbb R^n$.

So the homotopy-type of this space is at least fairly reasonable as $V_{n,k}$ is highly connected. I think this bundle likely has a lot of other nice properties lurking near the surface. Is this the kind of thing you're asking about? In particular as a bundle over $G_{n,k}^2$ you'd have some nice Schubert-cell type constructions. i.e. you could view $V_{n,k} \times_{O_k} V_{n,k}$ as the "diagonal" $V_{n,k}$ subspace union "Schubert cells".

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  • $\begingroup$ Ryan, thanks. Yes, this kind of information is most helpful. Also it would be nice to have references to places where all that is explained in details. I have heard about Schubert cells, but never read about it, for example. $\endgroup$
    – user6976
    Commented Oct 1, 2010 at 19:30
  • $\begingroup$ The basics appears in Milnor and Stasheff's book "Characteristic Classes". If you just want the idea of Schubert cells rather than any actual details, I wrote up a computation of the Euler Characteristic of $G_{n,k}$ using related ideas, here: en.wikipedia.org/wiki/Grassmannian#Schubert_cells $\endgroup$
    – Ryan Budney
    Commented Oct 1, 2010 at 21:05
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I'm no expert, but paging through "Linear Algebraic Monoids" by Lex Renner suggests to me that it has a lot of information you could use.

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  • $\begingroup$ No, the book is about submonoids of $M_n$ (the monoid of all $n\times n$-matrices). $\endgroup$
    – user6976
    Commented Sep 25, 2010 at 23:49
  • $\begingroup$ In any case, the question is not about representations of monoids or algebraic groups, it is about alg. geom and geom. topological properties of one particular object. $\endgroup$
    – user6976
    Commented Sep 27, 2010 at 12:02

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