Tensor product of $\partial_i$ and $dx^j$ In general the tensor product is not commutative. But I have a doubt:
in defining the absolute derivative (or covariant derivative) of a vector field $X$, as $\nabla X\in \cal{T_{1}^{1}(\cal{M})}$ (tensor field of type $(1,1)$) I have read that it is defined by:
$$\nabla X=\xi_{;j}^{i}dx^j\otimes \partial_i$$ with $X=\xi^i\partial_i$.
But in defining  a vector field of type $(1,1)$ in coordinates I know that it can be written as
$$t=t_i^j\partial_i\otimes dx^j$$
So it seems that $\partial_i$ and $dx^j$ are shifted in the order in defining $\nabla X$ with respect to the general definition of a tensor field of type $(1,1)$.
How it is possible? Maybe it is true that $\partial_i\otimes dx^j=dx^j\otimes \partial_i$?
 A: It really depends on your conventions. Some books define a $(1,1)$ tensor to be a section of $T^{*}M \otimes TM$ (the cotangent bundle appears first) while others define a $(1,1)$ tensor to be a section of $TM \otimes T^{*}M$ (the tangent bundle appears first). The bundles are of course isomorphic but still one makes a certain choice.
Let's say your convention is $TM \otimes T^{*}M$. Then the expression $dx^i \otimes \partial_j$ is not a "legal" $(1,1)$ tensor since it is not a section of $TM \otimes T^{*}M$. Once a choice is made, one uses certain isomorphisms to identify $(1,1)$-tensors with bilinear or linear maps. This choice is also somewhat arbitrary but most people try to be consistent. In the case of $TM \otimes T^{*}M$, one can identify sections of this bundle with smoothly varying bilinear maps $T_p^{*}M \times T_pM \rightarrow \mathbb{R}$ by declaring that
$$ (\alpha \otimes X)(Y, \omega) = \alpha(Y) \cdot \omega(X) $$
(and extending linearly). Note that the first argument $Y$ is a vector field and it "acts" on the first factor $\alpha$ (which is a one-form) and similarly the second argument $\omega$ is a one-form and it acts on the second factor $X$ (which is a vector field). One could do it the other way around but this is not common.
So in your case, the notation $(\nabla X)(Y,\omega)$ suggests that $(1,1)$-tensors are considered as sections of $T^{*}M \otimes TM$ and the formula $\nabla X = \xi^i_{;j} dx^j \otimes \partial_i$ is consistent with this guess.
A: Some authors define a tensor of type $(1,1)$ to be, in local coordinates:
$$ \label{1}
T = \sum_{i,j} T^i_j \partial_i \otimes \mathrm{d}x^j, \tag{1}
$$
while others prefer to define it the other way
$$ \label{2}
T = \sum_{i,j} T^i_j \mathrm{d}x^j\otimes \partial_i \tag{2}
$$
I personally prefer \eqref{2} because if given a vector field $X$, one can feel that in the writing $\left(\mathrm{d}x^j\otimes \partial_i\right)(X)$, it becomes $\mathrm{d}x^j(X) \partial_i$ and the left part is a coefficient $\times $ the vector field $\partial_i$. In this case, a $(1,1)$-tensor is a section of $T^*M\otimes TM$.
In case \eqref{1}, $(1,1)$-tensors are sections of $TM\otimes T^*M$. This looks a bit unnatural to me as I learned during my studies that we have a natural isomorphism $\mathrm{End}(E) \simeq E^*\otimes E$ (coefficient $\times$ vector), not the other way. Again, this is a personal taste.
One a choice is made, you have to try being consistent. Your first equation $\nabla X = \xi^i_j \mathrm{d}x^j \otimes \partial_i$ is suggesting that the convention is \eqref{2}.
Edit I did not notice that there was another answer (I took my time writing this up), and it seems we are basically saying the same thing. The other answer is a bit more accurate as it explains how to understand the notation $\alpha \otimes X \left(Y,\omega \right)$.
