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Probably, the best thing to do would be to write $x = f(u)$ and then use $$ \int_{u^{-1}(a)}^{u^{-1}(b)} L(x,u,u') dx = \int_a^b L\left(f(u),u,\frac{1}{f'(u)}\right)f'(u)\ du = \int_a^b M\left(u,f(u), …

The answer depends on the values of the constants $J^x$, $J^y$, and $J^z$. Here is what direct computation yields: If $J^x=J^y=J^z=0$, so that $A=0$, then $B_1$ and $B_2$ span a $2$-dimensional abel …

If you let $I$ denote multiplication by $\sqrt{-1}$, then the two operators $I$ and $M$ on your vector space (say, $V$) satisfy $$ I^2 = -1,\qquad M^2 = 1,\qquad\text{and}\qquad IM=-MI. $$ (The former …

If you have a nondegenerate Lagrangian $L:TM\to\mathbb{R}$ (such as the energy Lagrangian of a pseudo-Riemannian metric or the square of a Finsler metric, though these are not the only cases) with the …

Well, I don't have the complete answer, but then I don't think that a 'complete' answer is going to be simple. For example, you haven't ruled out the case $L=0$, which amounts to giving a `normal for …

The answer to your question is 'yes', that is the general solution. This is one of the basic results in the theory of the variational bicomplex. It is a statement of the vanishing of a certain cohom …

Actually, there is a fair amount known in the first nontrivial case: $(k,n) = (2,4)$. For example, see Calibrations on $R^8$ by J. Dadok, R. Harvey and F. Morgan Transactions of the American Mathemat …

Another couple of examples are the sphere rolling on the plane (or any surface, for that matter) without twisting or slipping, which is described by a connection on a principal SO(3)-bundle over the s …

Note: I'm updating my answer to give a better (i.e., simpler) example plus a little more information about how to derive the example from Douglas' results (which may not be entirely clear upon first …

To understand the local geometry of this equation, I think one should apply the Calabi resolution of the Killing equation. (See E. Calabi, On compact, Riemannian manifolds with constant curvature. I, …

New answer: I now have an answer for the subgroup case that the OP originally asked about. In fact, one has the following result: Let $G$ be a connected compact Lie group and let $p = (p_1,\ldots,p …

In any case, the proof is very simple. Consider the $3$-form $$ \phi_0 = dx^1\wedge dx^2\wedge dx^3 + dx^4\wedge dx^5\wedge dx^6. $$ I claim that the subgroup $G\subset\mathrm{GL}(6,\mathbb{R})$ that …

Cartan showed that the lowest dimensional (nontrivial) $E_8$-module is ${\frak{e}}_8$ itself, i.e., the adjoint representation, which has dimension $248$. The next smallest nontrivial irreducible mod …

Yes, it is well-known that a $6$-manifold has an $\mathrm{SU}(3)$-structure if and only if it is orientable and spinnable (i.e., it has a spin structure). The necessity of these two conditions is c …

The $E_6$ and $E_7$ decompositions you list are explained in Cartan's 1894 thesis (see pages 89–92 for these formulae). For $E_8$, Cartan instead gives a decomposition (a $\mathbb{Z}_3$-grading) of t …