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visits | member for | 2 years, 7 months |
seen | 2 days ago | |
stats | profile views | 127 |
May 7 |
comment |
Making the identification $\tau M\approx TM\oplus (TM\odot TM)$
@MichaelBächtold I'm curious about this too. Lazaro Cami says that $\tau M$ is naturally a vector bundle over $M$ (which seems to make sense). Meanwhile, it looks like the arXiv paper you sent needs additional structure to give $T^2M$ a vector bundle structure. |
May 6 |
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Making the identification $\tau M\approx TM\oplus (TM\odot TM)$
@MichaelBächtold Sure. I first saw this use of this terminology in this PHD thesis by Joan Andreu Lazaro Cami gmcnet.webs.ull.es/files/thesis/alazaro.pdf. See p. 33. The book "Global and Stochastic Analysis with Applications to Mathematical Physics" by Yuri E. Gliklikh also uses it. On page 66 of the book, there are a number of other references. |
May 4 |
awarded | Yearling |
May 3 |
answered | Lagrangian flow preserves symplectic form |
May 3 |
comment |
Lagrangian flow preserves symplectic form
you can read about this topic starting on page 6 of arxiv.org/pdf/math/9807080v1.pdf in the section titled "The Variational Approach." |
Apr 29 |
answered | Which sections of $T^*M\odot T^*M$ have reproducing kernel “primitives”? |
Apr 29 |
revised |
Which sections of $T^*M\odot T^*M$ have reproducing kernel “primitives”?
updated posting to include a (very) partial solution. |
Apr 20 |
revised |
Which sections of $T^*M\odot T^*M$ have reproducing kernel “primitives”?
Explained my motivation for asking the question |
Apr 19 |
asked | Which sections of $T^*M\odot T^*M$ have reproducing kernel “primitives”? |
Apr 18 |
comment |
Making the identification $\tau M\approx TM\oplus (TM\odot TM)$
$\pi$ doesn't take values in $\tau M$; I think you might want $S\circ \pi$ instead of $\pi\circ S$? |
Apr 18 |
awarded | Teacher |
Apr 17 |
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Are sections of $\tau M$ differential operators on the exterior algebra?
This is exactly what I was looking for. |
Apr 17 |
accepted | Are sections of $\tau M$ differential operators on the exterior algebra? |
Apr 17 |
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Momentum a cotangent vector
I should also say that $\mathbf{F}L(v_q)=(DL_q)(v_q)$. |
Apr 17 |
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Momentum a cotangent vector
@JoséFigueroa-O'Farrill I guess it depends on who you ask. In Abraham and Marsden on p. 219 they say "The transformation $\mathbf{F}L:TQ\rightarrow T^*Q$ thus maps the Lagrange equations into the Hamilton equations. In the literature $\mathbf{F}L$ itself is sometimes called the Legendre transformation (e.g. Sternberg [1964]), while classically the name is usually reserved for the map that takes...[$L$ to $H$]." The Sternberg reference is this I think: amazon.com/Lectures-Differential-Geometry-Chelsea-Publishing/dp/…. |
Apr 17 |
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Momentum a cotangent vector
Did you mean to write $p\in T^*_qM$ instead of $p\in T^*_pM$? |
Apr 17 |
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Momentum a cotangent vector
What is $u$? Also, I thought the momentum of a particle at $q\in M$ is a linear functional on the tangent space $T_qM$, as the question suggests, and not a linear functional on $T_{(q,\dot{q})}TM$. Note that the dimension of $T_{(q,\dot{q})}TM$ is twice the dimension of $M$, whereas the classical formula $p_i=\partial L/\partial q^i$ suggests that the momentum should have only $\text{dim}(M)$ components. |
Apr 17 |
answered | Momentum a cotangent vector |
Apr 15 |
asked | Are sections of $\tau M$ differential operators on the exterior algebra? |
Apr 15 |
comment |
Making the identification $\tau M\approx TM\oplus (TM\odot TM)$
I think we want a $\Gamma:\tau M\rightarrow TM$, right? Should it be $\Gamma=\iota^{-1}\circ(\text{id}-S\circ\pi)$, where $\iota:TM\rightarrow \iota(TM)\subset \tau M$ is the inclusion? |