# Why are vectors considered to be rank (0,1) tensors and dual vectors considered to be rank (1,0) tensors?

Sean Carrol in his book of general relativity, he defines a tensor to be a multilinear map from a collection of dual vectors and vectors to $\mathbb{R}$: $T:T^*_p \times...\times T^*_p \times T_p \times...\times T_p \to \mathbb{R}$.

He then goes on to say that a scalar is a type $(0,0)$ tensor, a vector is a type $(1,0)$ tensor and a dual vector is a type $(0,1)$. However, I thought that dual vectors are elements of a dual space, $T^*_p$, so shouldn't it be that dual vectors are rank $(1,0)$ tensors and vectors rank $(0,1)$ tensors?

• A vector takes a dual vector to give a scalar. A dual vector takes a vector to give a scalar. It sounds like the notation is that the first number is the number of dual vector arguments and the second number is the number of vector arguments (which makes sense given the way that you have written the domain of $T$).
– Ian
Feb 2, 2016 at 21:48

Given a vector $v$ and a dual vector $f$, you can produce a scalar $f(v)$. This can be viewed as a map $v \mapsto f(v)$ or $f \mapsto f(v)$.
So a vector determines a (linear) map from the space of dual vectors to scalars (i.e. a $(1,0)$-tensor since we have 1 dual vector input and no vector inputs). Likewise a dual vector determines a (linear) map from the space of vectors to scalars (i.e. a $(0,1)$-tensor since we have no dual vector inputs and 1 vector input).
We say T is a $(a,b)$ tensor $T:T^*_p \times...\times T^*_p \times T_p \times...\times T_p \to \mathbb{R}$ where $a$ is the number of $T^*_p$s and $b$ is the number of $T_p$s.
A dual vector is an element of $T_p^*$, which means that it gives a map $T_p\rightarrow \mathbb R$, and so it's a $(0,1)$ tensor.
A vector is an element of $T_p$, which means that it gives a map $T_p^*\rightarrow \mathbb R$, and so it's a $(1,0)$ tensor. (The map is the "evaluation map" mentioned by Bill Cook.)