We know this fact that the first jet Bundle $J^1M$ is diffeomorphic with $T^*M×\mathbb{R}$.i.e, ($J^1M=T^*M×\mathbb{R}$)
Is there something like this identity for higher jet bundle $J^kM$?
I editted my question after comment of Michael Murray
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$\begingroup$ Your diffeomorphism is true when you take the jet bundle of the trivial line bundle $M \times \mathbb{R} \to M$, but false in general. $\endgroup$– S. Carnahan ♦Nov 2, 2013 at 0:13
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$\begingroup$ You can see this fact in Example 1.3. in part Cotangent bundles and 1-Jet Bundles. here math.ias.edu/files/wam/Traynorlecture.pdf $\endgroup$– user21574Nov 2, 2013 at 1:09
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$\begingroup$ Second reference , see Example 6.7 of faculty.tcu.edu/richardson/Seminars/symplectic.pdf $\endgroup$– user21574Nov 2, 2013 at 1:17
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1$\begingroup$ Okay, it seems some mathematicians choose a definition that only covers this special case. Thank you for clearing that up. $\endgroup$– S. Carnahan ♦Nov 2, 2013 at 1:44
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1$\begingroup$ You don't need $M$ to be compact. This is a local fact about the exterior derivative splitting the jet exact sequence. $\endgroup$– Michael MurrayNov 2, 2013 at 7:10
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1 Answer
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First, note that there is a natural exact sequence of bundle maps $$ Sym^kT^*M \rightarrow J^kM \rightarrow J^{k-1}M $$ but there is no natural splitting of this sequence. You can, however, do it by choosing a Riemannian metric and defining a map $J^{k-1}M \rightarrow J^kM$ by extending a $(k-1)$-jet to the $k$-jet whose symmetrized $k$-th order covariant derivatives all vanish at the point. From this you can conclude that there exists a bundle isomorphism $$ J^kM = \mathbb{R} \oplus T^*M \oplus Sym^2T^*M \oplus\cdots \oplus Sym^kT^*M. $$ This is all explained in the algebraic geometric context in section 2.4 of these notes. My guess is that it is all explained in the book "The Geometry of Jet Bundles" by Saunders.
answered Nov 2, 2013 at 18:29
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$\begingroup$ Dear Deane Yang@ what is the third part ,i.e,$B$ of your equality $$ J^kM = \mathbb{R} \oplus T^*M \oplus B\cdots \oplus Sym^kT^*M. $$ $\endgroup$– user21574Nov 2, 2013 at 18:47
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$\begingroup$ It is $B=Sym^2T^*M$? $\endgroup$– user21574Nov 2, 2013 at 18:54
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$\begingroup$ Deane Yang@ Can you clarify you notation of $Sym^kT^*M$ ? $\endgroup$– user21574Nov 4, 2013 at 0:37
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