Timeline for What is the Levi-Civita connection trying to describe?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Dec 13, 2020 at 19:25 | comment | added | Bence Racskó | @Qfwfq I doubt it. If $P:TM|_\Sigma\rightarrow T\Sigma$ is an idempotent vertical morphism (i.e. a tangential projection), $D$ is the projected connection and $\nabla$ is the ambient connection, we get $T^D(X,Y)=D_XY-D_YX-[X,Y]=P(\nabla_XY)-P(\nabla_YX)-[X,Y]=P(\nabla_XY)-P(\nabla_YX)-P[X,Y]=PT^\nabla(X,Y)=0$, where we have used the fact that for tangential vector fields $X,Y$ the bracket $[X,Y]$ is also tangential, therefore $[X,Y]=P[X,Y]$. | |
| Nov 16, 2020 at 13:50 | comment | added | Qfwfq | In the presence of torsion, can the $\nabla$ still be seen as euclidean covariant differentiation from the ambient followed by not necessarily orthogonal projection to the tangent space? | |
| Nov 15, 2020 at 22:03 | comment | added | Deane Yang | @AndrewNC, you don't actually need to extend $X$ and $Y$ to the ambient Euclidean space. Given $p \in M$, $XY(p)$ is the directional derivative of $Y$ in the direction $X$, which can be calculated by choosing any curve $c$ such that $c(0) = p$ and $c' = X$. In particular, $c$ can be chosen to lie in $M$. Then $$\left.XY = \frac{d}{dt}Y(c(t))\right|_{t=0}.$$ Then $\nabla_XY$ is the orthogonal projection of $XY$ onto $T_pM | |
| Nov 15, 2020 at 17:31 | comment | added | Andrew NC | Let me see if I get this straight: embed $M$ in some $\mathbb{R}^n$; extend $X$ and $Y$ in whatever way you want, and then project parallel transport coming from $\mathbb{R}^n$ onto the tangent space? And this is invariant not only of the ambient space, but also of the extensions of $X$ and $Y$? If I understand correctly, this is indeed a very intuitive of what Levi-Civita is! (Though the relationship between this intuition and the definition is a rift that I would still need to think about.) | |
| Nov 15, 2020 at 11:11 | comment | added | Ben McKay | @Gro-Tsen: oddly enough, that is exactly what Dirac does, embedding space time into a pseudo-Euclidean space (see page 10). He never worries about the physical interpretation of that pseudo-Euclidean space, since it is just a crutch to get the definitions going. | |
| Nov 15, 2020 at 10:54 | comment | added | Ben McKay | @Gro-Tsen: yes, we can. Roughly speaking, taking one derivative and then projecting doesn't ever use second derivatives, so we never feel curvature, so everything looks the same in Euclidean space and in any Riemannian manifold. | |
| Nov 15, 2020 at 10:15 | comment | added | Gro-Tsen | We can generalize this to higher dimensions, can't we? I assume that the Levi-Civita action can be defined on any Riemannian manifold $M$ by embedding the latter (at least locally) in a suitably large Euclidean space, performing parallel transport in the latter and then tangentially projecting to $M$. If so, this is (to my eyes) probably the best answer to OP's question. | |
| Nov 15, 2020 at 8:32 | history | answered | Ben McKay | CC BY-SA 4.0 |