Why is a dual vector a type (0,1) tensor?

Question

I am reading "An introduction to Spacetime and Geometry" by Sean Carroll, and on page 21 section 1.6, it is mentioned that the dual of a vector is a type (0,1) tensor.

The definition given of the page for a tensor is as follows (1):

\begin{equation}\tag{1} T:T^*_p\underbrace{\times...\times}_{k \text{ times}}T^*_p\times T_p\underbrace{\times...\times}_{l \text{ times}}T_p\mapsto\mathbb{R} \end{equation}

In (1), $T_p$ stands for the tangent space at a point of a manifold $M$, and $T^*_p$ the dual to the same space. $\times$ denotes the Cartesian product.

By this definition (1), a $(k,l)$ tensor would have $k$ dual vectors and $l$ regular vectors.

However, the textbook claims that a dual vector can be thought of as a type (0,1) tensor and a regular one a type (1,0). This seems to be in contradiction of (1), where dual vectors appear $k$-times and regular ones $l$-times.

Which leads to my question:

Have I misinterperated the definition? If not, why are the indices of dual and regular vectors switched for a type $(k,l)$ tensor?

Answer 1

A dual vector is an element of the dual space $T_p^*$. This means it is a map from $T_p$ to $\mathbb{R}$. In general, a $(k,l)$ tensor is a map which takes $k$ dual vectors and $l$ vectors as inputs, and outputs a real number. A dual vector is hence a $(0,1)$ tensor. Your statement that a $(k,l)$ tensor has $k$ dual vectors and $l$ vectors is imprecise, and is the source of the confusion.