Visual intuition for direct sum vs. tensor product of vector spaces I completely understand the formal mathematical distinction between the direct sum and the tensor product of two vector spaces. I also understand that the direct sum has a nice visual interpretation (especially the direction sum of two 1D vector spaces, or of a 2D and a 1D vector space), where you simply think of attaching the vector spaces together at their respective origins in orthogonal directions inside some higher-dimensional ambient Euclidean space. Question 1: Is there any such simple geometrical picture for the tensor product? (Unfortunately, the simplest nontrivial tensor product - that of two 2D vector spaces - is four-dimensional, so perhaps not.)
Now assume that the vector spaces $V$ and $W$ have (finite) dimensions $N_V$ and $N_W$ respectively. Then there is a natural map from $V \oplus W$ to $V \otimes W$, where we map $(v \oplus w) \in (V \oplus W)$ to $(v \otimes w) \in (V \otimes W)$. The image of this map is a $(N_V + N_W - 1)$-dimensional set $M \subset
(V \otimes W)$. (The "$-1$" is because this map is not one-to-one: the preimage of $v \otimes w \neq 0$ is the one-dimensional manifold
$$\{ c v \oplus (1/c) w\ |\ c \in (\mathbb{F} \setminus \{0\}) \}.$$
) Note that $M$ is not closed under vector addition, so it does not form a subspace of $V \otimes W$ (indeed, this fact is exactly the reason why the direct sum and the tensor product of vector spaces are different!). In terms of visual intuition, the fact that the manifold $M$ contains the origin but is not a vector space means that it is "curved" rather than "flat".
For example, if $V = \mathbb{R}^3$ and $W = \mathbb{R}^2$, and we choose bases for $V$ and $W$ in which $(v \in V) \to (v_1, v_2, v_3)$ and $(w \in W) \to (w_1, w_2)$, then we have that
$$(v_1, v_2, v_3, w_1, w_2) \in V \oplus W$$
is naturally identified with
$$(v_1 w_1, v_1 w_2, v_2 w_1, v_2 w_2, v_3 w_1, v_3 w_2) \in V \otimes W,$$ and that the expression on the LHS parameterizes the four-dimensional smooth manifold $M \subset \mathbb{R}^6$. Question 2: If there is a way to visualize the tensor product of vector spaces, is there an intuitive way to see within that visualization why the natural embedding of $V \oplus W$ into $V \otimes W$ is "curved"?
 A: This is probably not what you are looking for, but my "geometric" picture for tensor product comes from function spaces. Consider $V$ to be the space of functions $f:A\to K$ and $W$ the space of functions $g:B\to K$ for two sets $A, B$ and a field $K$. Then the tensor product of $V$ and $W$ consists of the functions $A\times B \to K$.
If it comes to bases and such, the simplest case is the case of Hilbert spaces. If $(f_i)$ is an orthonormal basis of $V$ and $(g_j)$ is an orthonormal basis of $W$, then an orthonormal basis of $V\otimes W$ is given by the functions $(f_i\otimes g_j)(x,y) = f_i(x)g_j(y)$. If find this picture pretty intuitive and geometric. Even more concrete is the example of $A=B=[0,1]$ where the orthonormal basis in both spaces is the (real or complex) Fourier basis.
As a concrete example: The space of square integrable measurable functions $L^2(\Omega)$ it holds that
$$
L^2(\Omega_1)\otimes L^2(\Omega_2) = L^2(\Omega_1\times\Omega_2)
$$
and if $(\phi_k)$ and $(\psi_k)$ are orthonormal bases of $L^2(\Omega_1)$ and $L^2(\Omega_2)$, respectively, then the functions
$$\Phi_{j,k}(x,y) = \phi_j(x)\psi_k(y)$$
form an orthonormal basis of $L^2(\Omega_1\times\Omega_2)$.
