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Feb 23, 2021 at 14:36 comment added Igor Khavkine @Harish The two generalizations that you proposed are not equivalent. The second one reduces to $D\Phi = B'$, where $B' = U B U^{-1}$. Now it is $B'$ that needs to satisfy the integrability condition, which becomes a condition on $U$. Unfortunately, the condition on $U$ is again non-linear and I also don't see how to approach it. Sorry!
Feb 23, 2021 at 13:59 comment added Harish I saw thhe problem bit differently, if the solution $\Phi$ is such that $D\Phi$ only need to preserve distances, then $D\Phi(x)= U(x) B (x) U(x)^{-1} $, for some distance preserving matric $U(x)$. where $B(x)= |det (A(x))|^{1/N+2} A(x)$. Now that we have more flexibility in solving the problem by choosing any $U$, does it not make the problem easier? (though I still not know the answer)
Feb 23, 2021 at 10:10 comment added Igor Khavkine @Harish By "this integrability condition" I referred to "\partial_k B_{ji} - \partial_j B_{ki} = 0". Since the $\xi$ in your new condition is arbitrary, equivalently $(D\Phi)^T (D\Phi) = B$ (collect the coefficients of $\xi_i\xi_j$) with $B$ a symmetric matrix proportional to $A^T A$ with the corresponding determinant normalization factor. This equation is no longer linear in $\Phi$ (it is now "fully nonlinear") and honestly I don't know how to approach it.
Feb 23, 2021 at 0:50 comment added Harish thank you for your answer, since I am not very familiar with differential forms, if yiu could clarify what "integrability condition" on $B_{ij}$ you are referring to? moreover, is the claim true if instead of equality, we have the following condition :$ |D\Phi (x) \xi| = \frac {|A(x)\xi|}{|det(A(x))|^{1/N+2}}$? this condition is more relaxed than one in question, instead of matrices being equal we only need to preserve distanced from otigin.
Feb 22, 2021 at 18:16 history answered Igor Khavkine CC BY-SA 4.0