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Sep 16, 2012 at 10:52 comment added Tobias Ohrmann This is exactly what i wanted to know, sorry for covering it in a bunch of weird equations :-) And thank you for the answer!
Sep 16, 2012 at 9:37 comment added Michael Bächtold That seems all right. Are you asking if sections in $E\oplus (E\otimes T^*M)$ of the special form $(\phi,\nabla\phi)$ where $\phi$ is a section $E$ are in one to one correspondence with holonomic sections of $J^1 E$? The answer is yes.
Sep 16, 2012 at 7:29 comment added Tobias Ohrmann What I've done: 1. Expressing $j_x:T_x→T_{ϕ(x)}E$ in local coordinates: $j_x(∂_{x_i})=∂_{x_i}+∑^n_{j=1}y^j_i ∂_{y_j}$ 2. Splitting $j_x$ in horizontal and vertical part, according to ∇: $j_x=$ (horizontal part) $+ ∑^n_{j=1}(y^j_i−\sum^n_{k=1}Γ^j_{ik}y_k)∂_{y_j}$ 3. Project $j_x$ on the vertical part 4. Expressing $∇ϕ:T_xX→V_yY$ in local coordinates: $∇_∂_{x_i}\phi=∑^n_{j=1}(∂_{x_i}\phi_j−∑^n_{k=1}Γ^j_{ik}\phi_k)∂_{y_j} $5. Compare both expressions
Sep 15, 2012 at 17:37 comment added Michael Bächtold I'm not sure I understand the question. The identification we were talking about is bijective: to every section of $J^1 E$ corresponds a section of $E\oplus(E\otimes T^*M)$. By holonomic section in $J^1 E$ you mean a section which is the prolongation of a section in $E$?
Sep 15, 2012 at 14:07 comment added Tobias Ohrmann I still have one question: Does such an element $(\phi,\nabla \phi)$ exist for every $j_x=(x_i,y_j,y^j_i) \in (J^1E)_x$? As the condition, which have to be fulfilled for such an $\phi$, i get: $y_i^j-\sum_{k=1}^n \Gamma_{ik}^j y_k = \partial_{x_i}\phi^j - \sum_{k=1}^n\Gamma_{ik}^j \phi^k$ So, if $j$ is holonomic, one can choose the prolonged section of $j$ and the condition is fulfilled. Is holonomy also a necessary condition for the existence of such a section $\phi$?
Sep 14, 2012 at 14:29 vote accept Tobias Ohrmann
Sep 14, 2012 at 14:29 comment added Tobias Ohrmann Thank you very much, this helped me a lot, unfortunately i'm not autherized to click you answer up :)
Sep 13, 2012 at 6:37 history edited Michael Bächtold CC BY-SA 3.0
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Sep 12, 2012 at 22:59 comment added Tobias Ohrmann Thank you! What i still don't understand is that an element $j\in J^1E$ builds an subspace $R \subset T_\phi E$. With a given local trivialization $\psi$, $j$ could be represented by a point in the corresponding horizontal space $HE^\psi_\phi \subset T_\phi E$ of $\psi$. Why does the set of those points in $T_\phi E$ according to the set of all trivializations build a subspace? I tried to prove that, but what I need is that the trivializations build a vector space themselves, which is obviously wrong (take [$M=\mathbb R^n, E=TM, \alpha = id =-\beta$ local trivialisations] as a counterexample).
Sep 12, 2012 at 17:11 history answered Michael Bächtold CC BY-SA 3.0