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23h
comment Killing vector field along geodesic
... I mean $\nabla_X (\gamma')=\nabla_{\gamma'}(X)$ ...
1d
comment Killing vector field along geodesic
To compute $[X,\gamma']$ (which includes $\nabla_X \gamma'$) you have to define (extend) the vector field $\gamma'$ to some neighborhood of the geodesic $\gamma$. How you do this? Note, that it is possible to do so that $\nabla_X \gamma' = \nabla_X{\gamma'}$ along $\gamma$.
Apr
14
comment Jets and vertical differential
i.e., $V_u \pi$ is the vector space tangent to the fiber of the bundle $\pi: E\to M$. In the book local coordinates $(x,u)$ provide $T_u E$ with the splitting $V_u\pi + H_uE$, where (horizontal) subspace is identified with $T_uM$. so that taking vertical differential of a section equals the projection of the ordinary differential $d\phi$ to $V_uE$ along this $H_u$. Jet manifold $J^1_\pi$ has projections on $E$ and $M$ which makes it bundle, it is called associated since its structure group is defined by the structure group of the initial bundle $\pi$, see (3.5).
Apr
14
comment Jets and vertical differential
Janet the Physicist. Looks like your bundle is endowed with the connection, i.e., family of "horizontal" subspaces, while the vertical differential is the projection of $T_uE$ to vertical fibers $V_\pi$ of the bundle. How else you can define projection $d^V$?
Mar
8
comment Equivalence classes of mappings from $T^{2}$ to an arbitrary space $X$
$\pi_2(T^2)=0$ is not?
Feb
18
awarded  Supporter
Feb
2
comment Why does non-commutativity in quantum mechanics require us to use Hilbert spaces?
KCd - it should be a vector space (to have linearity) with Fourier transform defined, i.e., with scalar product. Does it mean - Hilbert space, right?
Feb
1
answered Why does non-commutativity in quantum mechanics require us to use Hilbert spaces?
Jan
16
comment Controlled-measurement of a quantum register
hope, you may find the answer in: physics.stackexchange.com/questions/26869/…