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Yes, there is a standard procedure to analyze such systems, essentially, it is Cartan's method of prolongation combined with his theory of involutive systems. There are other approaches as well, but …

I may be misreading the sources that you list for the definition of Wick-rotatable, but, I believe that the following construction does fit that definition: According to the Wikipedia page that the O …

Here's how one can construct a specific example to illustrate what can happen: First, recall from my answer to this question that, if you have a smooth map $f:D\to\mathbb{R}^2$ with constant positive …

As I understand it, the equation you are imposing on the function $\theta(x,y)$, defined on a domain $D\subset\mathbb{R}^2$ in the $xy$-plane is that, for some positive constants $\lambda_1\not=\lambd …

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 …

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 …

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), …

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, …

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 …

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 …

In the OP's particular case, the situation is somehwat simpler than the general case that José discusses. That's because the family of left-invariant metrics on $\mathrm{SU}(4)$ that the OP wants to …

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 …

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 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, this only would make sense when $n$ is even, but there are two problems: First, except when $n=1$, the $2n$-sphere does not carry any symplectic structure. Second, the tangent bundle of the $2 …