- #1

- 137

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In 'Introduction to Smooth Manifolds' Lee writes that a tensor of rank 2 always can be decomposed into a symmetric and an antisymmetric tensor:

A = Sym(A) + Alt(A).

We define a product which looks at the antisymmetric part of A \otimes B according to:

AB = Sym(A \otimes B),

while the wedge product describes the antisymmetric part:

A \wedge B = Alt(A \otimes B).

Now first of all the fact that a tensor of, lets say, rank 3 can not be decomposed in this way seems quite counter-intuitive, for me. How do you think of it? Is there any easy way to picture it?

Secondly: Can we define a product for this last term (that is neither symmetric or antisymmetric) of our tensors of rank higher than 2? In other words:

A * B = (A \otimes B) - Sym(A \otimes B) - Alt(A \otimes B) ?

The last question concerns the total covariant derivative that is definied in the book on Riemannian manifolds. Lee first sets out to claim:

'

*Although the definition of a linear connection resembles the characterization of (2,1)-tensor fields [...], a linear connection is not a tensor field because it is not linear over C^∞(M) in Y, but instead satisfy the product rule.*' (- 'Riemannian Manifolds: An Introduction to Curvature' by John Lee)

Later however he states that the total covariant derivative (the generalization of this linear connection) is a (k+1, l)-tensor field. This seems to be contradictive.. or am I mixing something up?

Thanks for all the help!

Kindly Regards

Kontilera