# Seeing Tensor Products

• Mandelbroth
In summary, one way to better understand the tensor product is to think about its properties in lower dimensional spaces and use visualization techniques. It can also be helpful to think of it as a machine for processing vectors, where the resulting vector is in the direction of the first vector and has a magnitude based on the dot product with the second vector. This perspective may be more relatable for those with an engineering background.

#### Mandelbroth

"Seeing" Tensor Products

Is there a way to "visualize" the tensor product of two (or ##n##) vectors/tensors/algebras/etc.?

I'm having a lot of trouble making the tensor product feel intuitive. I know its properties, and I can usually apply it without too much of a problem, but it does not feel "easy." Any ideas?

For tensor actions in general, I always started by thinking about the properties in lower dimensional spaces, where they can make more physical sense or can be mapped out/drawn explicitly on paper. It tends to bolster your intuition for what happens with higher rank objects.

Then, you can just think of a general action as something that you can already visualize, but in a projected sense (ie, you are imagining only a part of the true behavior, as more exists "behind the scenes," but it is enough to know what's going on).

I'm probably coming at this from a different perspective from you, but maybe this can help. My background is in engineering. I like to visualize that tensor product as a machine for processing vectors. You feed a vector to the machine, and the tensor product maps the vector into a new vector. If $\vec{L}\otimes \vec{R}$ is the vector product of the vectors $\vec{L}$ and $\vec{R}$, then if I dot the vector product on the right by a vector $\vec{V}$, I get:

$$(\vec{L}\otimes \vec{R})\centerdot\vec{V}=\vec{L}(\vec{R}\centerdot\vec{V})$$
This is a vector in the direction of $\vec{L}$, with a magnitude equal to the magnitude of $\vec{L}$ times $(\vec{R}\centerdot\vec{V})$. If I dot the vector product on the left by a vector $\vec{V}$, I get an analogous result.

As a physicist or mathematician, I don't know if this makes any sense to you. But to me as an engineer it makes perfect sense, and I have used it many time in practice when working with the engineering stress tensor to map a unit normal vector to a surface into the traction vector.

Chet