(1,1) tensor field maps vector to vector

Question

I am studying general relativity by Schutz. I am in curvature chapter, and the following concept I encountered.

Say $\vec V$ is vector field, and $x^\alpha$ be a set of coordinates, e.g. in polar case $x^1=r$ and $x^2=\theta$. No we are defining partial derivative on the vector field as following: $$\frac{\partial \vec V}{\partial x^\beta}=(\frac{\partial V^\alpha}{\partial x^\beta}+V^\mu\Gamma^\alpha_{\mu\beta})\vec e_\alpha$$

We can call $\frac{\partial V^\alpha}{\partial x^\beta}+V^\mu\Gamma^\alpha_{\mu\beta}$ as $V^\alpha_{;\beta}$

So $\frac{\partial \vec V}{\partial x^\beta}=V^\alpha_{;\beta}\vec e_\alpha$

And here comes the statement in the book, which says "we can regard $\frac{\partial \vec V}{\partial x^\beta}$ as associated with (1,1) tensor field which maps vector $\vec e_\alpha$ to vector $\frac{\partial \vec V}{\partial x^\beta}$." I can't understand the above statement. As I understand about (1,1) tensor field, they are defined as following:

$T:V^*\times \vec V \rightarrow\Bbb R$, so here in the above statement $\vec e_\alpha$ is member of vector space $\vec V$ , and from here I am lost ($V^*$ is the dual space of the vector space).

Q How vector maps to another vector

In the above case, once I supply a vector, then we have to supply a one form (member of dual space - correct?), but one-forms and vectors are considered in same footing, hence we can say something like this:

$T( , \vec e_\alpha)=\tilde { e_\alpha}()$, as one-forms and vectors are on same footing, then $\tilde { e_\alpha}()$ is a vector.

Is the above argument correct?!! It would greatly help if someone can actually show me a construction!

Help greatly appreciated.

ANSWER:

I think I got the answer.

(1,1) tensor can be written as $T:\tilde \omega \times \vec y \rightarrow \Bbb R$, where $\vec y$ is a vector. So once we supply a vector, the one form takes this vector and produces a scalar - so supplying the vector kills the one form. Hence we are remained with a vector. Hence we can say this tensor maps a vector to another vector.

Answer 1

In differential geometry, we define a vector $V$ as a mapping of functions $f \in C^\infty (M)$ to the Reals. Represented in a chart this concept looks like a merge of the concepts vector and derivative. \begin{equation} V: C^\infty (M) \to \mathbb{R}\quad V f = V^i \frac{\partial}{\partial x^i}f \end{equation}

When you look at the covariant derivative of a vector, as you want to do, you are trying to express the change in the same chart as a new vector. In the case of the covariant derivative you need to define a vector along which you differentiate. This is expressed $\nabla_X$, which takes vectors to vectors and convectors to convectors.

\begin{equation} \nabla_X V = \nabla_{X^i\frac{\partial}{\partial x^i} }V^j \frac{\partial}{\partial x^j}=X^i(\frac{\partial V^j}{\partial x^i}\frac{\partial}{\partial x^j}+V^j (\frac{\partial}{\partial x^i}\frac{\partial}{\partial x^j})) \end{equation} You can bring this equation into the shape of a (1,1) tensor acting on a vector. \begin{equation} X^i(\frac{\partial V^q}{\partial x^i}+V^j \Gamma_{ij}^q)\frac{\partial}{\partial x^q} \end{equation} Edit If you want to write down a linear map which maps a given chart basis $\vec{e}_\alpha=\frac{\partial}{\partial x^\alpha}$ to it's covariant derivative, the thing inside the brackets is your map. It's what's called an endomorphism.

\begin{equation} X^i(\frac{\partial V^q}{\partial x^i}+V^j \Gamma_{ij}^q)\frac{\partial}{\partial x^q} \end{equation}