# Non commutative tensor products / simple example?

In O'Neil's "Semi-Riemannian Geometry" it is stated that if $A$ is a $(r,0)$-tensor field and $B$ is $(0,s)$-tensor field then it is ""from the definition'' that $A \otimes B = B \otimes A$. I still do not understand tensor fields very well (I give my definition of tensor field below in case it is necessary). I have the following question:

Is there a simple example with physical or geometrical interpretation of a tensor product not commuting?

My doubts are actually more general since I would like to understand the physical and geometrical meaning of

a) tensor products themselves

b) commutation of tensor products.

But I will be happy with some example or bilbiographical references. Thank you in advance.

Definition (tensor field). If $\mathfrak{X}(M)$ denotes the class of vector fields in $M$, $\mathfrak{X}^*(M)$ denotes the set of one-forms in $M$ and $\mathfrak{F}(M)$ denotes the set of smooth real-valued functions in $M$, then a (r,s)-tensor field is a $\mathfrak{F}(M)$-linear function $A$ of the form $$A: \underbrace{\mathfrak{X}^*(M)\times \cdots \times \mathfrak{X}^*(M)}_{r\text{ times}} \times \underbrace{\mathfrak{X}(M)\times \cdots \times \mathfrak{X}(M)}_{s\text{ times}} \longrightarrow \mathfrak{F}(M).$$

That is because we always arrange the "form parameters" of your tensor function first. Examples: let $e_1=\frac{\partial}{\partial x}$ and $e_2=\frac{\partial}{\partial y}$, $dx\otimes dy \ne dy \otimes dx$ since $dx\otimes dy(e_1, e_2)=1$ and $dy\otimes dx(e_1, e_2)=0$. However if they're of different tensor types we always arrange the components so that "form parameters" come first: $(e_1 \otimes dx)(dx, e_1)=1$. You cannot apply something like $dx \otimes e_1$ on $(dx, e_1)$ like $(dx \otimes e_1)(dx, e_1)$ since it unwindes to $dx(dx)\cdot e_1(e_1)$ which is undefined.