# Geometric interpretation of tensor product

I am not sure if this question is better suited for math SE, but given that I am asking for a geometrical (ideally physical) interpretation, I figured it would be best asked here.

I am looking for a way to understand tensor product spaces intuitively. I understand both the basis-dependent and the algebraic basis-free definitions. What I am struggling with is interpreting what something of the form $$v \otimes w, \quad v \in V, w \in W$$ means geometrically, where $$V$$ and $$W$$ are vector spaces.

For example, in $$\mathbb{R}^3$$ it is very easy to explain and visualize what any vector or basis looks like. For an arbitrary $$x \in \mathbb{R}^3$$ we can decompose it into $$x = a_1 e_1 + a_2 e_2 + a_3e_3$$ and this geometrically makes clear sense, we are simply breaking down the vector into its component parts. In other words, if $$x$$ denotes the position of some particle we can say it is $$a_i$$ distance in the $$i$$th direction ($$e_i)$$.

However, in the case of tensor products, does a similar geometric interpretation hold? How does one exactly interpret an arbitrary vector $$v \otimes w$$? To make the question simpler, take $$V = \mathbb{R}^2$$ (with basis $$\{e_i^1\}$$) and $$W = \mathbb{R}^3$$ (with basis $$\{e_i^2\})$$. Then \begin{align}v \otimes w = &a_1 e^1_1\otimes e_1^2 + a_2 e^1_2\otimes e_1^2 + a_3 e^1_1\otimes e_2^2 + \\&a_4 e^1_2\otimes e_2^2 + a_5 e^1_1\otimes e_3^2 + a_6 e^1_2\otimes e_3^2 \end{align}

The only picture I am able to have in mind is in the context of quantum mechanics. If the two vector spaces represent different observable quantities, then I interpret the tensor product as taking into account two things at once. But this is unsatisfying to me.

• Exterior algebra, which is related to tensor algebra, has an intuitive geometric interpretation. In my opinion, tensor algebra itself doesn’t. But there is no reason that it should. It seems to me to be natural (in the vague, undefined sense of that word) algebraically but not geometrically. Commented Sep 22, 2022 at 20:31
• @Ghoster Thank you I will take a look at exterior algebras, I'm not familiar with them at the moment. I suppose I will just have to consider tensor algebras are purely algebraic with no physical or geometric interpretation. Commented Sep 22, 2022 at 20:54
• Although I don’t know of a geometric interpretation, tensor algebra does have a physical interpretation in terms of the quantum state of multiple non-interacting quantum objects. If object $A$ has state $|A\rangle$ and similarly for B, then if they don’t interact the quantum state of the system is $|A\rangle\otimes|B\rangle$. Commented Sep 22, 2022 at 22:13

Well, in your example, you have a 3-vector w , each of whose components is multiplied by the same 2-vector v, to yield a 6-vector 𝑣⊗𝑤, which you may also think of as a 2×3 matrix, a sort of a dyadic with 6 components, all related to each other. That is, The first row is a 3-vector $$v_1$$ w, while the second row is proportional to it, $$v_2$$ w ; and likewise for the transpose, interchanging the roles of the vectors v and w.

Whether this is sufficiently avoidant of algebra for you in favor of geometry, I could not tell... If somebody gives me an isodoublet which is a color triplet at the same time, this is how I picture it in the eyes of my mind....

• Based on your answer and others it seems what I am asking for simply does not exist and the tensor project is inherently algebraic (and thus should be only seen as such). Commented Sep 22, 2022 at 21:06
• I always avoided drawing rigid lines between geometry and algebra... possibly a feature rather than a bug... Commented Sep 22, 2022 at 21:08
• Thank you for all the help! Commented Sep 22, 2022 at 21:09