bio | website | |
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visits | member for | 2 years, 6 months |
seen | 4 hours ago | |
stats | profile views | 108 |
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 |
comment |
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 |
comment |
Momentum a cotangent vector
I should also say that $\mathbf{F}L(v_q)=(DL_q)(v_q)$. |
Apr 17 |
comment |
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 |
comment |
Momentum a cotangent vector
Did you mean to write $p\in T^*_qM$ instead of $p\in T^*_pM$? |
Apr 17 |
comment |
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? |
Apr 15 |
accepted | Making the identification $\tau M\approx TM\oplus (TM\odot TM)$ |
Apr 12 |
awarded | Commentator |
Apr 12 |
comment |
Making the identification $\tau M\approx TM\oplus (TM\odot TM)$
@MatthiasLudewig Thank you. I thought it seemed very similar in spirit to the notion of a so-called Ehresmann connection. However, I didn't see how the details worked out because, in the case of the Ehresmann connection, the exact sequence is $V_eE\rightarrow T_eE\rightarrow T_{\pi(e)}B$, where $\pi:E\rightarrow B$ is some fiber bundle and $VE\subset TE$ is the vertical subbundle. I couldn't see how $\tau_m M$ plays the role of $T_eE$. |
Apr 12 |
asked | Making the identification $\tau M\approx TM\oplus (TM\odot TM)$ |
Feb 26 |
comment |
Obstruction to the existence of global isometries on a constant-curvature Riemannian manifold
@ThiKu your point is well-taken. However I wonder if there is a terminology issue here. There are apparently two kinds of "Killing vectors", those with complete flows and those without. The ones with complete flows are special because they are elements of the Lie algebra of the isometry group. Are both types of vectors commonly referred to as "Killing vectors"? |
Feb 26 |
comment |
Obstruction to the existence of global isometries on a constant-curvature Riemannian manifold
Thank you. This answer together with the comments above have greatly cleared up a lot of my confusion. But how can we tell which, if any, of the pulled-back killing vectors are genuine Killing vectors on $M^m$? For instance, suppose $m=2$, $k=1$, and $dev(M^m)$ is the domain of the stereographic projection $S^2\rightarrow\mathbb{R}^2$. The killing vector that fixes the north pole will pull back to a genuine killing vector, but the others would not. |
Feb 26 |
asked | Obstruction to the existence of global isometries on a constant-curvature Riemannian manifold |