Geometric intuition of tensor product

Question

Let $V$ and $W$ be two algebraic structures, $v\in V$, $w\in W$ be two arbitrary elements.

Then, what is the geometric intuition of $v\otimes w$, and more complex $V\otimes W$ ? Please explain for me in the most concrete way (for example, $v, w$ are two vectors in 2 dimensional vector spaces $V, W$)

Thanks

Answer 1

You want to stay concrete, so let's let $V$ be a two-dimensional real vector space and $W = \operatorname{Hom}(V,\mathbb{R})$. Then $V = T^1_0(V)$ and $W = T^0_1(V)$, so for any $v\in V$ and $w\in W$, $v\otimes w\in T^1_1(V)$.

Each of $v,w$ has two components; $v = v^1e_1 + v^2e_2$ and $w = w_1e^1 + w_2e^2$. Here $e_1,e_2$ is a basis for $V$ and $e^1,e^2$ is the dual basis in $V^*$.

The components of $v\otimes w$ are all the components of $v$ times all the components of $w$: $$(v\otimes w)^i_j = v^iw_j.$$

To see this, observe that $(v\otimes w)(\theta, x)=v(\theta)w(x),$ so that $(v\otimes w)(e^i, e_j) = v(e^i)w(e_j).$

More generally, if $A = A^{i_1\cdots i_p}_{j_1\cdots j_q}\in T^p_q$ and $B = B^{k_1\cdots k_r}_{l_1\cdots l_s}\in T^r_s$, then

$$(A\otimes B)^{i_1\cdots i_pk_1\cdots k_r}_{j_1\cdots j_q l_1\cdots l_s} = A^{i_1\cdots i_p}_{j_1\cdots j_q}B^{k_1\cdots k_r}_{l_1\cdots l_s}.$$

Note that our example is from the derived tensor algebra over a two-dimensional vector space, $T^p_q(V) = (V^*)^{\otimes p}V^{\otimes q}.$ Hopefully this helps build your intuition about the case where $V$ and $W$ are two vector spaces of potentially different dimension.

Answer 2

The difference between the ordered pair $(v,w)$ of vectors and the tensor product $v\otimes w$ of vectors is that for a scalar $c\not\in\{0,1\}$, the pair $(cv,\;w/c)$ is different from the pair $(v,w)$, but the tensor product $(cv)\otimes(w/c)$ is the same as the tensor product $v\otimes w$.

Answer 3

For a totally different angle to the subject, we can get a taste of what the tensor product means by considering Dehn's invariant. I'll try to shortly explain what it is, if we have a polyhedron with edge lengths $l_i$ and dihedral angles $\theta_i$, we can define the Dehn's invariant through the quotient:

$$ R \otimes_{\mathbb{Z} } ( R / 2 \pi \mathbb{Z})$$

Now, the point is, if we have a polyhedron, then upon cutting it into two pieces, the Dehn's invariant is fixed!

For a beautiful exposition of this topic, see this Numberphile video