Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $V, W$)


share|cite|improve this question
@WillieWong : Thank you very much. Yes I have. However, the links above just mentioned the algebraic properties/structures. What I could not imagine is the concrete picture of tensor product. – Arsenaler Mar 2 '12 at 13:29
The tensor product is an algebraic convenience and does not in general admit a geometric interpretation. (On the other hand, it is easy to give a geometric interpretation of the wedge product.) – Zhen Lin Mar 2 '12 at 17:18
Do you have a geometric appreciation for bilinear forms? – Dylan Moreland Apr 1 '12 at 16:57
@eduardo: Related:… – user23238 Feb 23 '13 at 9:04

You want to stay concrete, so let's let $V$ be a two-dimensional real vector space and $W = \operatorname{Hom}(W,\mathbb{R})$. 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 $w = w_1e^1 + w_2e^2$, where $e_1,e_2$ 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.

share|cite|improve this answer
Double multi-indices in a question regarding "geometric intuition"! – Martin Brandenburg Nov 25 '15 at 9:21

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 $v\otimes w$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.