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comment Symplectic orthogonality and projective duality: how do they work together?
@Libli: nice reference, thanks! As I have explained in my last comments to Robert Bryant's answer, I'm mostly interested in curves in $\mathbb{P}^3$, whereas, for higher values of $n$, the space $\mathbb{P}^{2n-1}$ has to be replaced by ``something more general''. Nevertheless, after your remark, it seems that the only curves in $\mathbb{P}^3$ having one-dimensional duals, are $\mathbb{P}^1$'s (as just observed by Bryant), which happen to be $\omega$-self-dual iff the $\mathbb{P}^1$'s are the projectivizations of Lagrangian planes (see below).
Feb
9
comment Symplectic orthogonality and projective duality: how do they work together?
... and now I'm sure that the problem of characterising, e.g., the 5-folds $\mathcal{E}$ in the 6-dimensional Lagrangian Grassmannian (i.e., with $n=3$), which can be brought to the form $\mathcal{E}_X$, with $\dim X=2$, can be dealt with in terms of $GL_2$-structures, as Abraham D. Smith did in his work (arxiv.org/abs/0912.2789).
Feb
9
comment Symplectic orthogonality and projective duality: how do they work together?
With $n=3$ and beyond, the situation becomes more intricate, and I barely touched it. It goes also beyond the scope of the OP, as $\mathbb{P}V$ needs to be replaced with $Gr_\omega(V,n-1)$, the Grassmannian of $(n-1)$-dimensional $\omega$-isotropic subspaces of $V$ (carrying a contact form $\theta_\omega$ inherited by $\omega)$. As before, $Gr_\omega(V,n-1)$ contains very special $n-1$-dimensional varieties $X$, namely the (standard) Grassmannians $Gr(L,n-1)$, with $L$ Lagrangian subspace of $V$, and the corresponding $\mathcal{E}_X$ is once again an $n$-dim parabolic Monge-Ampere equation...
Feb
9
comment Symplectic orthogonality and projective duality: how do they work together?
[these non-editable comments are really annoying: above I meant $n=2$, of course!]
Feb
9
comment Symplectic orthogonality and projective duality: how do they work together?
... and I guess that the answer is somehow buried in this work of Dennis The (arxiv.org/abs/1009.1364), where he cooks out a differential invariant whose vanishing is a necessary condition for $\mathcal{E}$ to be - as he puts it - singly ruled. In turn, this "singly-ruledness" should guarantee the existence of a curve $X$ such that $\mathcal{E}=\mathcal{E}_X$, but I could not yet work out the details (and Dennis himself could not help much). [The main reason I planned this workshop was to answers questions like these (and, as I explained in my email, there can be other chances to meet)]
Feb
9
comment Symplectic orthogonality and projective duality: how do they work together?
in the case $n=4$ (i.e., when $V$ is the contact plane in $J^1(2,1)$), I'm interested in curves $X$ in $\mathbb{P}V$. Indeed, such a curve determines the 2-fold (surface) $\mathcal{E}_X:=\{ L\subset V\mid \omega|_L\equiv 0, \mathbb{P}L\cap X\neq\emptyset\}$ in the Lagrangian Grassmannian $L(V)$ of $V$ (which is a 3-fold). Basically, I'm trying to understand to what extent such surfaces in $L(V)$ can be studied through the corresponding curves $X$ in $\mathbb{P}V$. And the first question is: how to test an arbitrary surface $\mathcal{E}$ to be of the form $\mathcal{E}_X$? ...
Feb
8
comment Symplectic orthogonality and projective duality: how do they work together?
... Then, you're right: the "non-symplectic" part of my question is purely algebraic-geometric, and requires different tools (the 2nd observation of yours, though correct, is not useful when it comes to PDEs, since, if $X$ is "too big", then the corresponding PDE is trivial). Speaking of which, as you like so much discussing this stuff, why don't you join us into a dedicated workshop (impan.pl/en/sites/gmoreno/home/workshop-2016) we are organising in Warsaw? Pawel Nurowski and Gianni Manno have both sent you an invitation, a couple of weeks ago.
Feb
8
comment Symplectic orthogonality and projective duality: how do they work together?
Now the Force is back into balance! I like, in particular, the 1st observation: me and my friends long suspected it was so, but none of us was able to find such a simple argument to prove it! As you may guess, we are interested into applications to 2nd order nonlinear PDEs: if $V$ is the contact plane in $J^1(2,1)$, and $X$ is 1-dimensional, then $X$ determines in a natural way an hypersurface in $J^2(2,1)$, and you have just proved that $X$ is $\omega$-self-dual if and only if the hypersurface is a parabolic Monge-Ampere equation...
Feb
6
comment Symplectic orthogonality and projective duality: how do they work together?
@Libli : I also think it's a pity that the first Robert Bryant's comment has disappeared. After all, the incidence correspondence in $\mathbb{P}V\times \mathbb{P}V^*$ covers both $X$ and $X^*$, so that the distinction between 'conormal' and 'dual' is very soft. Besides, he made some comments about developables (related to my updated question above) which I found interesting, but now I cannot recall them... It reminds me of the Siths erasing the whereabouts of planet Kamino from the Jedi archives: now all surrounding comments are still pulled towards it, but the central comment has gone!
Feb
6
revised Symplectic orthogonality and projective duality: how do they work together?
added a comment after Libli's reply
Feb
6
comment Symplectic orthogonality and projective duality: how do they work together?
@Libli : I kind of feel ashamed now, for I had all the time the Tevelev's paper in my e-library, and I never thought of looking at it! Of course, you're right: the - how he calls it - "reflexivity theorem" allows one to 'solve the equation' $(1)$ by starring both members: $X_\bullet:=(X^\perp)^*$. So, the answer to my first question is: "for all $X$" - as you said. Thanks.
Feb
5
comment Symplectic orthogonality and projective duality: how do they work together?
Of course that $X^\perp$ is just $X$, but without this passage I cannot compare $X$ with its dual, or - if you prefer - conormal, which is essential for my purposes. So, if you want to bypass the identification $V^*=V$, then I'm just asking if there is some obvious differential invariant associated with a variety $Y$, for instance, in $\mathbb{P}^3$, whose vanishing is a necessary (or even sufficient) condition for $Y$ to be the the conormal variety of another variety $X$. And possibly a method to recover $X$ out of $Y$... What is the standard reference for this kind of business?
Feb
5
asked Symplectic orthogonality and projective duality: how do they work together?
Feb
1
comment Does a $G$-structure on $M$ automatically descend to a contact $G$-structure on $\mathbb{P}T^*M$?
Concerning the $\hat{G}$-reduction, do you have any reference? It's enough one title: I'll track back the citations by myself!
Feb
1
comment Does a $G$-structure on $M$ automatically descend to a contact $G$-structure on $\mathbb{P}T^*M$?
Very illuminating explanation: thanks! However, I still detect a certain conflict between @Robert Bryant's first comment above and the first sentence of your answer. You claim that the $CSp(2n)$-structure is 'equivalent' to the contact structure $H$: I can easily understand that given the latter, you obtain the former, but what about the other way around? As Bryant said, we don't have a genuine $CSp(2n)$-structure, but just a $CSp(2n)$-reduction of the frames of $H$: in particular, we already have $H$! In this sense, I fail to appreciate the 'equivalence'.
Feb
1
revised Does a $G$-structure on $M$ automatically descend to a contact $G$-structure on $\mathbb{P}T^*M$?
Added a comment after Bryant's reply
Feb
1
comment Does a $G$-structure on $M$ automatically descend to a contact $G$-structure on $\mathbb{P}T^*M$?
@RobertBryant: thanks for spotting my rough mistake! The sense of the main question, however, does not suffer from it. Can you please point out a good departing point, in the literature, to start learning about this prolongation process? (You said by yourself that, "in some cases, it exists").
Feb
1
asked Does a $G$-structure on $M$ automatically descend to a contact $G$-structure on $\mathbb{P}T^*M$?
Jan
19
awarded  Popular Question
Dec
14
accepted Are all bidimensional second-order PDE at most quadratic in the top derivatives of Monge-Ampère type?