It is possible to formulate elasticity on Riemann manifolds. However, the Cauchy decomposition theorem is preferably used in rheology in conjuction with irreversible thermodynamics. Therefore, it would be a great loss not to have the corresponding decomposition in Riemann spaces, which I do not beleive. In case it still does not exist, what is the crucial property of Euclidian spaces which allows for the Cauchy decomposition theorem as opposed to Riemann spaces?
The polar decomposition applies to any nonsingular square matrix. Hence, I believe you can apply it to any matrix that represents locally a non-singular (invertible) transformation between smooth Riemanninan manifolds of the same dimension.
Mr. Goddard, thank you very much for your suggestion!
What are the neccessary conditions for the existence of a global map such that it is made up of an isometry and another map that provides the same strains as the original one?
The need for a global decomposition is certainly justified because the deformation of bodies are maps defined globally.
Next, having a decomposition of the deformation gradient, can we reproduce the original global map and its decomposition into maps whose tangent maps are the above components? And, in addition, uniquely?
I feel the difficulty in Euclidean spaces that tensors and tangent maps can be easily exchanged. For, one of the components of the Cauchy polar decomposition is a tangent map, the other is a tensor. (I restrict here the definition of a tensor to a particular tangent space, as is usual.) In Riemann spaces this distinction is made clearly.
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