When to use $\times$ and $\otimes$ Im wondering when to use $\times$ and when to use $\otimes$.
In some cases it seems very straightforward, for example $\times$ can be used when combining two elements into an n-tupel (for a product with n elements), for example in the case of the scalar product:
$$(.,.):V\times V\rightarrow\mathbb{C}:(u,v)\mapsto u\cdot v.$$
For the $\otimes$ i've seen it been used when for example one constructs for example the basis  for a $L^2$-Hilbert space where the x,y and z-coordinates can be split up. This yields:
$$f_{nml}(x,y,z)=f_n(x)\otimes f_m(y)\otimes f_l(z).$$
I was wondering if there was a general rule om when to use $\times$ and when to use $\otimes$ ? 
 A: The symbol $\times$ has a very strict definition and is almost never used outside it's definition: It is used to calculate the Cartesian product of two sets, meaning $$A\times B = \{(a,b)| a\in A, b\in B\}$$
The symbol $\otimes,$ on the other hand is a more algebraic symbol. It is used to denote the tensor product of two algebraic objects. The general definition (at least the one I know) uses notation from category theory to strictly define what a tensor product of two spaces is.
In the case when the objects are matrix spaces, the $\otimes$ means the Kronecker product of the two matrices, meaning that $$\mathbb R^{m,n}\otimes \mathbb R^{p,q}\subseteq \mathbb R^{mp,qn}.$$ (the cartesian product of the two sets would be $\mathbb R^{mn+pq}$).
Also, another annoying thing about notation is the fact that the elements of $A\otimes B$ are also written as $a\otimes b$ (although, unlike in $A\times B$, not all elements of $A\otimes B$ can be written as $a\otimes b$ for some pair $(a,b)\in A\times B$).
A: You can find vastly information about the Cartesian product, the second one, on the other hand is more sophisticated.
The tensor product, also called the Kronecker product $(\otimes)$, is defined as follows:
Let
$$X_1 = \begin{bmatrix}a_{1,1} &a_{1,2}\\a_{2,1}&a_{2,2}\end{bmatrix},\ \ X_2 = \begin{bmatrix}b_{1,1} &b_{1,2}\\b_{2,1}&b_{2,2}\end{bmatrix}$$ 
then:
$$\begin{bmatrix}a_{1,1} &a_{1,2}\\a_{2,1}&a_{2,2}\end{bmatrix}\ \otimes \begin{bmatrix}b_{1,1} &b_{1,2}\\b_{2,1}&b_{2,2}\end{bmatrix} = \begin{bmatrix} a_{1,1}\begin{bmatrix}b_{1,1} &b_{1,2}\\b_{2,1}&b_{2,2}\end{bmatrix} & a_{1,2}\begin{bmatrix}b_{1,1} &b_{1,2}\\b_{2,1}&b_{2,2}\end{bmatrix}\\ \\ a_{2,1} \begin{bmatrix}b_{1,1} &b_{1,2}\\b_{2,1}&b_{2,2}\end{bmatrix} & a_{2,2}\begin{bmatrix}b_{1,1} &b_{1,2}\\b_{2,1}&b_{2,2}\end{bmatrix} \end{bmatrix}$$ 
or:
$$X_1 \otimes X_2 =\begin{bmatrix}
a_{1,1}b_{1,1} & a_{1,1}b_{1,2} & a_{1,2}b_{1,1} & a_{1,2}b_{1,2}\\ 
a_{1,1}b_{2,1} & a_{1,1}b_{2,2} & a_{1,2}b_{2,1} & a_{1,2}b_{2,2}\\
a_{2,1}b_{1,1} & a_{2,1}b_{1,2} & a_{2,2}b_{1,1} & a_{2,2}b_{1,2}\\
a_{2,1}b_{2,1} & a_{2,1}b_{2,2} & a_{2,2}b_{2,1} & a_{2,2}b_{2,2}\end{bmatrix} $$
In general if:
$X_1 \in \mathbb{R}^{(m \times n)},\  X_2 \in \mathbb{R}^{(p \times q)}$ then $X_1 \otimes X_2 \in \mathbb{R}^{(mp \times nq)}$
Note that $X_1 \otimes X_2 \neq X_2 \otimes X_1$
Which one you should use depends completely on the task you want to solve.
PS. You have received two other fantastic answers concerning which of the two methods you should apply. Both answers are very thoughtful. 
Edit
One of many application the tensor product has (this one is not so popular as other applications, but is indeed a very important one if one is working with linear-algebra):
Let $X_1 \in \mathbb{R}^{(m \times n)}$ have a singular value decomposition ($U_{X_1} \Sigma_{X_1} V_{X_1}^t$) and let $X_2 \in \mathbb{R}^{(p \times q)}$ have a singular value decomposition ($U_{X_2} \Sigma_{X_2} V_{X_2}^t$). Then:
$$
(U_{X_1} \otimes\ U_{X_2})(\Sigma_{X_1} \otimes\ \Sigma_{X_2})(V_{X_1}^t \otimes\ V_{X_2}^t)
$$
yields a singular value decomposition of $X_1 \otimes X_2$, after a simple reordering of the diagonal elements of $(\Sigma_{X_1} \otimes \Sigma_{X_2})$ and the corresponding right and left singular vectors
A: This is by no means a full answer, but you can think of $\otimes$ as an operation on "functions" and $\times$ an operation on "spaces" which are in some sense dual to each other.  Vector spaces are often considered both ways (e.g. a vector bundle is a space, but linear functions on an algebraic variety are not -- this leads to endless confusion at least for me), so it admits both operations.
For instance, $\otimes$ is an operation which must consider some sort of linear structure: a common mistake in proving things about tensor products is to not consider sums.  On the other hand, $\times$ happens on the level of sets.
Also, $\otimes$ and $\times$ do correspond in a way: if $X$ and $Y$ are algebraic varieties, then functions on $X \times Y$ are given by
$$\mathbb{C}[X \times Y] = \mathbb{C}[X] \otimes \mathbb{C}[Y]$$
(Conversely, if $A$ and $B$ are rings, then $Spec(A \otimes B) = Spec(A) \times Spec(B)$)
Edit: Here's an addendum.  Often I find the direct product and direct sum are used interchangably for vector spaces when people really mean the direct sum (even though at least for finite dimensional vector spaces they are teh same, we should think about them differently).
In your example on scalar products, if you want the scalar product map:
$$\langle,\rangle: V ? V \rightarrow V$$
to be a morphism in the category of vector spaces, you MUST use $\otimes$, because we do not have that $\langle x, y \rangle + \langle w, z \rangle = \langle x + w, y + z \rangle$ which should hold if we used a direct sum (or direct product).
However, if you only want to think of it as a map in the category of sets, you can use $\times$ (and in your head you are extending it linearly)
