2
$\begingroup$

(I'm not sure that I even phrased the question correctly. I will explain more about this below.)

Given a k-tensor $T$, we can define an alternating k-tensor $Alt(T)$ in the following way:

where $\epsilon$ is the sign function.

My first question is, what do we even call the $Alt$? In my title, I called it the "Alternating Operator". I would like to know its formal name.

And here's my main question. Given alternating k- and l- tensors $\theta, \eta$, we have:

enter image description here

It seems that the "Alternating Operator" distributes to the two tensors in the first equality. Why is this?

I am new to tensors (just starting studying this month). I am aware that the tensor product distributes, but is this in any way related to the reason why the Alternating Operator also distributes? If so, how?

$\endgroup$
4
$\begingroup$

I think if you call it the "Alternating operator" people will generally understand what you mean. But at least in my circles it is more frequently called "the (total) antisymmetrization of $T$".

For your main question: the operator $\mathscr{A}$ is a linear mapping from $\mathscr{T}^k$ to itself. For any linear mapping you have the distribution law $L(x - y) = L(x) - L(y)$.

(It is in fact not just a linear mapping but a projection, meaning that $\mathscr{A}^2 = \mathscr{A}$.)

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks. Just curious, what is your circle? $\endgroup$ – Sally G May 20 at 1:22
  • $\begingroup$ For the purpose of this answer: math physics / differential geometry $\endgroup$ – Willie Wong May 20 at 3:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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