# Tensor fields and multiplication

by Kontilera
Tags: fields, multiplication, tensor
 P: 98 Hello! I'm currently reading John Lee's books on different kinds of manifolds and three questions has appeared. 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
 Sci Advisor HW Helper PF Gold P: 4,756 Regarding the second question... What Lee is saying is that a connection ∇: $\Gamma(TM)\times \Gamma(TM)\rightarrow \Gamma(TM)$ looks like a (2,1) tensor (compare with Lemma 2.4), but it is not one as it is not $C^{\infty}(M)$-linear in its second argument. Later, he defines the covariant derivative of a tensor, and remarks that if you take a tensor T of type (k,l), and take its covariant derivative ∇T, you get a tensor of type (k+1,l). In particular, if you take a vector field Y (tensor of type (0,1)) and jam it up the second slot of the connection map like so: ∇Y, you get a tensor, because the problem was in the second argument of ∇ and you've now eliminated that problem.
 P: 98 Thanks for the answer! Nobody that could give some response to the idea of the new multiplication? Maybe its just not so useful so Lee doesnt mention it..