Basis of tensor product of subspaces Consider two vector spaces $S$ and $S\otimes S$, both of which  are subspaces of $H\otimes H$, where $H$ is of $d$ dimension and so $H\otimes H$ is of $d^2$ dimension.
We assume that $S$ is of $n$-dimension and thus $S\otimes S$ is of $n^2$ dimension. 
A basis of $S$ is given as $\{e_1, \cdots, e_n\}$, where each $e_i$ is a $d^2$-dimension vector. According to standard textbooks, $\{e_i\otimes e_j\mid 1\leq i,j\leq n\}$ gives a basis of $S\otimes S$. However, now $e_i\otimes e_j$ is a $d^4$-dimension vector. Clearly this is not right. 
My question is, what is the correct formulation of the basis of $S\otimes S$ in this case?
 A: Say $\{h_1,\cdots,h_d\}$ is a basis for $H$, so that $\{h_i\otimes h_j\}$ is a basis for $H\otimes H$. This is a collection of $d^2$ different vectors, but if we arbitrarily choose an ordering of these $d^2$ elements (there is no canonical ordering, so we'd have to pick a way of listing them out arbitrarily) we could if we wanted to call them for example $\{g_1,\cdots,g_{d^2}\}$.
If $S$ is a subspace of $H$, then of course every element of $S$ can be written as a linear combination of the basis vectors $h_1,\cdots,h_d$. But $S$ doesn't necessarily have a basis consisting of these $h_i$ vectors, for instance $\{(1,0),(0,1)\}$ is a basis for $\mathbb{R}^2$ but the subspace defined by $x=y$ does not have a basis using these vectors. Let's say $S$ has a basis $\{s_1,\cdots,s_n\}$ (in particular $\dim S=n$).
Then $S\otimes S$ will be a $n^2$-dimensional subspace of $H\otimes H$, and $\{s_i\otimes s_j\}$ will be a basis. If we want, yet again we could arbitrarily choose a way of listing these elements out in some order and call the basis $\{r_1,\cdots,r_{n^2}\}$.
Now, if $H$ is a coordinate vector space so its elements look like $(x_1,\cdots,x_d)$, then that means the subspace $S$ is also comprised of $d$-tuples even though $S$ is $n$-dimensional not $d$-dimensional, so there is no contradiction between a subspace having lesser dimension than the number of entries in its tuples. For instance, the diagonal line $y=x$ in $\mathbb{R}^2$ contains $(1,1)$, even though the diagonal line is $1$-dimensional and $(1,1)$ has not one but two coordinates.
The same applies to $S\otimes S$ and $H\otimes H$.
Keep in mind, not every vector is a coordinate vector, and even when it is it's not necessary for the number of coordinates to match the dimension of the vector space we're talking about. In general, at least to mathematicians, a vector space is a set with some operations satisfying some axioms (definitionally, at least), and there's no restrictions on what it's elements look like. Its elements could be polynomials or functions or formal sums of things or whatever. 
But every vector space $V$ has a basis $\{v_1,\cdots,v_k\}$ (say $V$ is finite-dimensional) so that every $v\in V$ is uniquely expressible as $a_1v_1+\cdots+a_kv_k$ for some scalars $a_1,\cdots,a_k$. Such a basis induces a choice of coordinates, i.e. a linear function $V\to\mathbb{R}^k$ (say the scalar field is $\mathbb{R}$) defined by $v\mapsto (a_1,\cdots,a_k)$ (where $v$ is as before expressible as $v=a_1v_1+\cdots+a_kv_k$). Keep in mind that in general there may be more than one choice of coordinates $V\to\mathbb{R}^k$, so that a fixed vector $v\in V$ could have different coordinate representations. If we want to create a $k$-dimensional vector space from scratch we might as well use $\mathbb{R}^k$ though.

Let's get concrete. Consider $V=\mathrm{span}\{e_1,e_2\}$. Then $\{e_1\otimes e_1,e_1\otimes e_2,e_2\otimes e_1,e_2\otimes e_2\}$ is a (not "the" but "a") basis for $V$. If we choose to write


*

*$f_1:=e_1\otimes e_1$

*$f_2:=e_1\otimes e_2$

*$f_3:=e_2\otimes e_1$

*$f_4:=e_2\otimes e_2$


then we can say $\{f_1,f_2,f_3,f_4\}$ is a basis for $V\otimes V$. Notice I didn't reuse the letter $e$ for these new basis vectors. If had instead labelled them $e_1,e_2,e_3,e_4$, then I wouldn't know whether $e_1,e_2$ refer to the original basis vectors for $V$ or the new basis vectors for $V\otimes V$! I could end up accidentally interpreting $e_1\otimes e_1$ as an element of $(V\otimes V)\otimes(V\otimes V)$ for example. That's what you've done.
