A Need some clarifications on tensor calculus please

[pitaly]

TL;DR Summary

Need some clarifications on tensor calculus

I've started reading up on tensors. Since this lies well outside my usual area, I need some clarifications on some tensor calculus issues.

Let $A$ be a tensor of order $j > 1$. Suppose that the tensor is cubical, i.e., every mode is of the same size. So for example, if $A$ is of order 3 then $A \in R^{n \times n \times n}$.

Further assume that $A$ is symmetric (sometimes called supersymmetric), i.e., it is invariant under permutation of indices. For example, if the order is 3 then $A_{ijk} = A_{ikj} = A_{jik} = A_{jki} = A_{kij} = A_{kij}$.

We say that a tensor is positive semidefinite if $x \cdot A \cdot x \geq 0$ for all vectors $x \neq 0$. So for example, if $A$ is of order 2, we have a standard quadratic form and if $A$ is positive semidefinite then $x \cdot A \cdot x \geq 0$.

Now, to my two questions:

1) Is there any clean notation for tensor differentiation? If $A$ is of order 2 then the gradient of $x \cdot A \cdot x$ with respect to $x$ is $2 x \cdot A$. How do I write the gradient of $x \cdot A \cdot x$ with respect to $x$ when $A$ is a tensor of any order $j>2$?

2) Consider a tensor $A$ of order 2 and the quadratic form $(x-y) \cdot A \cdot (x-y)$ for any vectors $x,y$. If $A$ is positive semidefinite then $x \cdot A \cdot x - y \cdot A \cdot y \geq 2y \cdot A \cdot (x-y)$. Does there exist any analogous inequality for tensors of any order $j>2$, where the tensor $A$ is positive semidefinite?

Any help on this is highly appreciated!

 

[jedishrfu]

Wikipedia has some examples on this page:

https://en.wikipedia.org/wiki/Tensor_derivative_(continuum_mechanics)

 

[fresh_42]

Summary:: Need some clarifications on tensor calculus

I've started reading up on tensors. Since this lies well outside my usual area, I need some clarifications on some tensor calculus issues.

Let $A$ be a tensor of order $j > 1$. Suppose that the tensor is cubical, i.e., every mode is of the same size. So for example, if $A$ is of order 3 then $A \in R^{n \times n \times n}$.

Further assume that $A$ is symmetric (sometimes called supersymmetric), i.e., it is invariant under permutation of indices. For example, if the order is 3 then $A_{ijk} = A_{ikj} = A_{jik} = A_{jki} = A_{kij} = A_{kij}$.

We say that a tensor is positive semidefinite if $x \cdot A \cdot x \geq 0$ for all vectors $x \neq 0$. So for example, if $A$ is of order 2, we have a standard quadratic form and if $A$ is positive semidefinite then $x \cdot A \cdot x \geq 0$.

What is $x \cdot u \otimes v \otimes w \cdot x\;$?

Now, to my two questions:

1) Is there any clean notation for tensor differentiation? If $A$ is of order 2 then the gradient of $x \cdot A \cdot x$ with respect to $x$ is $2 x \cdot A$. How do I write the gradient of $x \cdot A \cdot x$ with respect to $x$ when $A$ is a tensor of any order $j>2$?

The first that comes to my mind is the boundary operator in cohomology theory.

2) Consider a tensor $A$ of order 2 and the quadratic form $(x-y) \cdot A \cdot (x-y)$ for any vectors $x,y$. If $A$ is positive semidefinite then $x \cdot A \cdot x - y \cdot A \cdot y \geq 2y \cdot A \cdot (x-y)$. Does there exist any analogous inequality for tensors of any order $j>2$, where the tensor $A$ is positive semidefinite?

Any help on this is highly appreciated!

Good question. I had the same (see above).

Let's see what you have actually done. You fed a tensor with two vectors and computed a scalar. Hence we need $u^*,v^*\in V^*$ to get $(u^*\otimes v^*)(x,y)= u^*(x)v^*(y)\in \mathbb{R}.$ We certainly can fill up this equation with arbitrary many $w^{(k)}\in V$ to get
$$
(u^*\otimes v^*\otimes w^{(1)}\otimes \ldots\otimes w^{(m)})(x,y)= u^*(x)v^*(y) \cdot w^{(1)}\otimes \ldots\otimes w^{(m)}
$$
but then we lose the possibility to compare it with the scalar $0.$

I think you are approaching this from the wrong side. What do you want to achieve is the question you should ask, not how can I generalize something.

Since you asked in a mathematics forum, you may want to have a read:
https://www.physicsforums.com/insights/what-is-a-tensor/
https://www.physicsforums.com/insights/pantheon-derivatives-part-iii/
https://www.physicsforums.com/insights/pantheon-derivatives-part-iv/

 

[suremarc]

Summary:: Need some clarifications on tensor calculus

1) Is there any clean notation for tensor differentiation? If A is of order 2 then the gradient of x⋅A⋅x with respect to x is 2x⋅A. How do I write the gradient of x⋅A⋅x with respect to x when A is a tensor of any order j>2?

Not an expert, but I know my way around some basic tensor math. Normally I would the tensor product in Einstein notation, ie $A_{ijk}x^ix^j$. Then I would write the derivative as a tensor-like expression: $\partial_i = \frac{\partial}{\partial x^i}$. And then we can apply it:
$$\partial_l A_{ijk}x^ix^j = A_{ijk}\partial_l(x^ix^j)$$ Knowing the Leibniz rule and that $\frac{\partial x^i}{\partial x^j}=\delta^i_j$, we can expand out the derivative: $$ \begin{align} \partial_l A_{ijk}x^ix^j & = A_{ijk}((\partial_l x^i)x^j + x^i\partial_l(x^j)) \\ & = A_{ijk}(\delta_l^ix^j+x^i\delta_l^j) \\ & = A_{ljk}x^j+A_{ilk}x^i\end{align}$$ In the case that $A$ is symmetric, we can freely swap the first two indices to get the expression $A_{jlk}x^j+A_{ilk}x^i$, which are really two sums with the same form, so we can change the index to be the same:
$$A_{ilk}x^i+A_{ilk}x^i=2A_{ilk}x^i$$ I guess it looks kinda hand-wavey, but it’s really just a compact way of writing sums along indices. Add as many indices beyond $l$ to get the same identity for order >3.

 

[haushofer]

Further assume that $A$ is symmetric (sometimes called supersymmetric), i.e., it is invariant under permutation of indices. For example, if the order is 3 then $A_{ijk} = A_{ikj} = A_{jik} = A_{jki} = A_{kij} = A_{kij}$.

I've never seen the expression "supersymmetric" for this, and seems very confusing; supersymmetry is something completely different. But maybe it's common in engineering texts.

 

[jedishrfu]

Summary:: Need some clarifications on tensor calculus

For example, if the order is 3 then Aijk=Aikj=Ajik=Ajki=Akij=Akij.

I think the last A term should be k j i .