# Tensor fields and multiplication

by Kontilera
Tags: fields, multiplication, tensor
 Sci Advisor HW Helper PF Gold P: 4,771 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.