Timeline for Quotients of Grassmannians
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Oct 17, 2017 at 10:39 | history | edited | Martin Sleziak | CC BY-SA 3.0 |
added (grassmanians) tag; corrected typo - smistable
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| Oct 11, 2017 at 0:23 | vote | accept | icmes imrf | ||
| Oct 11, 2017 at 0:15 | vote | accept | icmes imrf | ||
| Oct 11, 2017 at 0:23 | |||||
| Oct 10, 2017 at 21:24 | answer | added | Jason Starr | timeline score: 11 | |
| Oct 10, 2017 at 19:27 | comment | added | Sasha | @JasonStarr: In down-to-earth terms, an isomorphism $Gr(r,V) \cong Gr(n-r,V)$ is given by a nondegenerate bilinear form (a subspace $U \subset V$ goes to its orthogonal $U^\perp \subset V$ with respect to the form), and it is easy to see that there is no $T$-invariant bilinear form (unless $n = 2$). Just the space $V^\vee \otimes V^\vee$ of bilinear forms doesn't have zero weight vectors. | |
| Oct 10, 2017 at 19:23 | comment | added | Jason Starr | @Sasha. I agree that $P_r$ and $P_{n-r}$ are only exchanged by an outer automorphism $\phi$. But outer automorphisms still intertwine inner automorphisms $i_g$, $\phi(i_g(h))=\phi(ghg^{-1}) = \phi(g)\phi(h)\phi(g)^{-1} = i_{\phi(g)}(\phi(h))$. So the group of automorphisms of $P_r$ by conjugation by $T$ should get intertwined via $\phi$ with the group of automorphisms of $P_{n-r}$ by conjugation by $T'=\phi(T)$. Now the quotient of $P_{n-r}$ by $T'$ should be (abstractly) isomorphic to the quotient of $P_{n-r}$ by $T$ . . . | |
| Oct 10, 2017 at 19:15 | comment | added | Sasha | @JasonStarr: Yes, but I don't think $P_r$ and $P_{n-r}$ are conjugate. I guess they are only interchanged by an outer automorphism of $GL(V)$. | |
| Oct 10, 2017 at 18:56 | comment | added | Jason Starr | @Sasha. I do not get it. Fix any reductive group $G$ over an algebraically closed field. Fix any conjugacy class $c$ of parabolic subgroups. Let $X_c(G)$ be the scheme (SGA 3) of parabolic subgroups of $G$ that are in the conjugacy class $c$. There is a conjugation action of $G$ on $X_c(G)$. Any two maximal tori $T$ and $T'$ are conjugate, say $T'=gTg^{-1}$. Thus, the conjugation action of $T'$ on $X_c(G)$ is intertwined with the conjugation action of $T$ on $X_c(G)$ via conjugation by $g$. So the quotients of $X_c(G)$ by these two maximal tori are isomorphic as projective schemes. | |
| Oct 10, 2017 at 18:52 | comment | added | Sasha | @JasonStarr: Of course, but the question can be understand in two ways --- either about $Gr(r,V)$ and $Gr(n-r,V^\vee)$ --- in this case, of course, there is even an $GL(V)$-equivariant isomorphism --- or about $Gr(r,V)$ and $Gr(n-r,V)$ --- in this case, as far as I can see, there is no $T$-equivariant isomorphism (unless $n = 2r$). | |
| Oct 10, 2017 at 18:51 | comment | added | Jason Starr | Typo correction: the isomorphism $\textbf{PGL}_n(V)\to \textbf{PGL}_n(V^\vee)$ sends $[T]$ to $[S]$, where $\langle S(\chi),v\rangle = \langle \chi, T^{-1}(v) \rangle$, i.e., there is a $T^{-1}$ in there as well. Otherwise we get an isomorphism between "opposite" groups (reverse order of multiplication). | |
| Oct 10, 2017 at 18:40 | comment | added | Jason Starr | @Sasha. There is a $T$-equivariant isomorphism. Let $V$ be a vector space. There is a canonical isomorphism of $\textbf{PGL}_n(V)$ with $\textbf{PGL}_n(V^\vee)$ sending every linear transformation to its "transpose" or "dual" linear transformation. Choose a torus $T$ in $\textbf{PGL}_n(V)$ and identify $T$ with its image in $\textbf{PGL}_n(V^\vee)$ via the natural isomorphism. Now identify $\text{Grass}(r,n)$, resp. $\text{Grass}(n-r,r)$, with the set of parabolic subgroups in $\textbf{PGL}_n(V)$, resp. in $\textbf{PGL}_n(V^\vee)$, in the appropriate (respective) conjugacy classes. | |
| Oct 10, 2017 at 16:05 | comment | added | Sasha | @JasonStarr: To start with you need an isomorphism $Gr(r,n) \cong Gr(n-r,n)$ to be $T$-equivariant, which I doubt to be true. | |
| Oct 10, 2017 at 10:53 | comment | added | Jason Starr | The Picard group of the Grassmannian is free of rank $1$, i.e., $\text{Pic}\cong \mathbb{Z}$. So there is only one choice of linearization of any action of a reductive group on the Grassmannian, up to taking positive tensor powers (which does not affect the GIT quotient as a scheme). Thus, the semistable loci are isomorphic as schemes with a $T$-action. Finally, the geometric quotients by $T$ are isomorphic as proper schemes. At least when $r$ equals $2$, the GIT quotient appears to be a compactification of the Deligne-Mumford-(Grothendieck-Knudsen-Mayer) moduli space $M_{0,n}$. | |
| Oct 10, 2017 at 2:31 | history | asked | icmes imrf | CC BY-SA 3.0 |