No, it doesn't. A type (k, 1) tensor can be contracted with k dual vectors and l vectors to produce a scalar. That does not mean the k dual vectors and l vectors are "part of" the tensor.In both Wald and Carroll, a type (k,l) tensor has k dual vectors and l vectors
Yes, because a (1, 0) tensor can be contracted with 1 dual vector to produce a scalar, so it's a vector; and a (0, 1) tensor can be contracted with 1 vector to produce a scalar, so it's a dual vector.a (1,0) tensor is a vector and a (0,1) tensor is a dual vector.
A tensor, as they define it, is a multilinear map, it has k dual vectors and l vectors as input. So a (0,1) tensor will be a linear map that has a vector as an input i.e. it will be a dual vector.Summary:: Type (k,l) tensors, dual vectors and vectors.
In both Wald and Carroll, a type (k,l) tensor has k dual vectors and l vectors, yet a (1,0) tensor is a vector and a (0,1) tensor is a dual vector. I must be missing something simple. Please explain.
You are evidently misinterpreting statements from these sources, but unless you give specific quotes and where they are from, it's impossible to tell exactly what you are misinterpreting.In both Wald and Carroll
What on this page led you to believe what you said in the OP?General Relativity by Robert M. Wald, Page 20.
Thanks. Now I understand.No, it doesn't. A type (k, 1) tensor can be contracted with k dual vectors and l vectors to produce a scalar. That does not mean the k dual vectors and l vectors are "part of" the tensor.
Yes, because a (1, 0) tensor can be contracted with 1 dual vector to produce a scalar, so it's a vector; and a (0, 1) tensor can be contracted with 1 vector to produce a scalar, so it's a dual vector.
I was confused by thinking that type (0, 1) tensor meant (k, l) tensor, not understanding about the mapping to R.What on this page led you to believe what you said in the OP?