bio | website | |
---|---|---|
location | Salerno, Italy | |
age | 36 | |
visits | member for | 3 years, 5 months |
seen | 10 hours ago | |
stats | profile views | 833 |
Reasonably expert in Differential Geometry, I work mostly with jet bundles and their natural structures trying to exploit them in the areas of nonlinear PDEs and Calculus of Variations.
Jul
31 |
comment |
What does it mean that the Hessian is proportional to the metric?
Very illuminating - thanks! The statement "The level sets of your function formed by manifolds with constant normal curvature" is the most interesting one for me: do you have a reference for it? |
Jul
31 |
accepted | What does it mean that the Hessian is proportional to the metric? |
Jul
31 |
asked | What does it mean that the Hessian is proportional to the metric? |
Jul
30 |
comment |
Is there a well-known tautological bundle over $\mathbb{P}(\Lambda^nV)$?
Ok, let me try. If $[\omega]\in \mathrm{G}(n,V)$, then $\omega=v_1\wedge\cdots\wedge v_n$, and $\pi^{-1}([\omega])=\langle v_1,\ldots,v_n\rangle$, which is the easy implication. Conversely, knowing that $\dim \pi^{-1}([\omega])=n$, you choose a basis of $V$ such that $\{v_1,\ldots,v_n\}$ generates $\pi^{-1}([\omega])$: now you suggest to write down $\omega$ as a linear combination of $v_1\wedge\cdots\wedge v_n$ and other terms, and show that the coefficients in front of the latter must necessarily be zero. Yes, it should work, thanks! Do you have any clue about question #2? |
Jul
30 |
revised |
Is there a well-known tautological bundle over $\mathbb{P}(\Lambda^nV)$?
Removed unnecessary "$\times V$" from the title |
Jul
29 |
revised |
Is there a well-known tautological bundle over $\mathbb{P}(\Lambda^nV)$?
I have completely rewritten the question "Is the tautological bundle over the Grassmannian inherited from a bundle over the Plucker embedding space?" since from the users' feedback I realised it was lousily laid down. The content however is unaltered. |
Jul
25 |
comment |
Triangles in rigid Riemann surfaces
Now I understand. You are absolutely right. |
Jul
25 |
comment |
Triangles in rigid Riemann surfaces
Help me to understand your reasoning. First, by transitivity, you reduce the problem to the case when the two triangles have a vertex in common, say $x_0$: then you are basically asking whether the isotropy subgroup $G_{x_0}$ is 2-transitive (any triangle with one vertex $x_0$ is uniquely determined by two other points). And you say "NO" by using finiteness of $G_{x_0}$: is this enough? What if the surface is the hyperbolic plane: does this finiteness argument works? |
Jul
21 |
answered | Transformations that leave the Plucker embedding of G(2,4) invariant |
Jul
14 |
revised |
Is there a well-known tautological bundle over $\mathbb{P}(\Lambda^nV)$?
Changed "inherited by" in "inherited from" |
Jul
13 |
asked | Is there a well-known tautological bundle over $\mathbb{P}(\Lambda^nV)$? |
Jul
7 |
comment |
Stability of the Solar System
It also depends on the notion of "Solar System": if it comprises "parabolic" objects like comets (and I believe it should), then it is not stable at all. From my point of view, "stable" and "periodic" are interchangeable terms. Besides, how can you predict when/why/where a distant iceberg will decide to dive into the sun? |
Jul
7 |
revised |
Differential operators between modules, $\mathcal{D}_A(M, M)$ necessarily a filtered, almost commutative ring?
added 145 characters in body |
Jul
7 |
revised |
Equivalence of “Weyl Algebra” and “Crystalline” definitions of rings of differential operators between modules?
I forgot to add a comment on the kind of modules my reasoning works with |
Jul
7 |
answered | Differential operators between modules, $\mathcal{D}_A(M, M)$ necessarily a filtered, almost commutative ring? |
Jul
7 |
answered | Equivalence of “Weyl Algebra” and “Crystalline” definitions of rings of differential operators between modules? |
Jun
17 |
awarded | Good Question |
Jun
17 |
revised |
Why there is a relation among the second-order minors of a symmetric $4\times 4$ matrix?
Added a comment to Allen Knutson answer |
Jun
16 |
awarded | Self-Learner |
Jun
16 |
awarded | Yearling |