[SOLVED] Finding the Value of a Tensor

[Sudharaka]

Hi everyone,

Here's a problem that I recently encountered and want to get an hint on how to solve.

Problem:

Find the value \(F(v,\,f)\) of the tensor \(F=e^1\otimes e_2+e^2\otimes (e_1+3e_3)\in T_{1}^{1}(V)\), where \(v=e_1+5e_2+4e_3\), \(f=e^1+e^2+e^3\).

 

[Deveno]

Oh gee, it's been forever since I did this stuff, but I think if:

$v = v^1e_1 + v^2e_2 + v^3e_3$ (more compactly, $v = v^je_j$) and

$f = f_1e^1 + f_2e^2 + f_3e^3$ then:

$e^i \otimes e_j (v,f) = v^if_j$

so in this case:

$e^1 \otimes e_2 (v,f) = 1$

$e^2 \otimes (e_1 + 3e_3) (v,f) = (5)(1 + 3) = 20$

so that $F(v,f) = 21$ (assuming I have my "ups and downs" correct...it's been awhile).

 

[Sudharaka]

Oh gee, it's been forever since I did this stuff, but I think if:

$v = v^1e_1 + v^2e_2 + v^3e_3$ (more compactly, $v = v^je_j$) and

$f = f_1e^1 + f_2e^2 + f_3e^3$ then:

$e^i \otimes e_j (v,f) = v^if_j$

so in this case:

$e^1 \otimes e_2 (v,f) = 1$

$e^2 \otimes (e_1 + 3e_3) (v,f) = (5)(1 + 3) = 20$

so that $F(v,f) = 21$ (assuming I have my "ups and downs" correct...it's been awhile).

Thanks very much for the response. But can you please explain to me how you wrote,

\[e^i \otimes e_j (v,f) = v^if_j\]

I am new to tensors, so sorry if this is an obvious question.

 

[Deveno]

The way I learned it, a tensor is a multilinear map:

$T:\underbrace{V \times \cdots \times V}_{n\ copies} \times \underbrace{V^{\ast} \times \cdots \times V^{\ast}}_{m\ copies} \to F$

We can, by the universality of the tensor product, regard $T$ as an element of:

$\underbrace{V^{\ast} \otimes \cdots \otimes V^{\ast}}_{n\ copies} \otimes \underbrace{V \otimes \cdots \otimes V}_{m\ copies}$

(here, we are implicitly identifying $V$ with $V^{\ast\ast}$).

In this case, $m = n = 1$, which makes our tensor particularly easy to understand. In this case, we have that any such tensor is a linear combination of elementary tensors of basis covectors and vectors. So it suffices to determine what:

$e^j \otimes e_i (v,u^{\ast})$ is, for any two vectors $u,v \in V$.

Typically, $\{e_1,\dots,e_n\}$ is the standard (Euclidean) basis for $F^n$, and $\{e^1,\dots,e^n\}$ is the dual basis; that is, the linear functionals:

$\displaystyle e^j(v) = e^j\left(\sum_{i = 1}^n v^ie_i \right) = v^j$

(the linear functional that returns the $j$-th coordinate of $v$ in the standard basis).

It is typical to regard $e_i$ as an $n \times 1$ (column) matrix, and $e^j$ as a $1 \times n$ (row) matrix, in which case their tensor product is given by their Kronecker product, the $n \times n$ matrix (written below in block form):

$\begin{bmatrix}0e_i&\dots&1e_i&\dots&0e_n \end{bmatrix}$


$= E_{ij}$, which has a 1 in the $i,j$-th entry, and 0's elsewhere. That is:

$e^j \otimes e_i (v,u^{\ast}) = u^TE_{ij}v = (u^{\ast})_iv^j$

(I hope I have my indices correct....I get these backwards a lot).

In this example, we have $V = F^3$ (as near as I can surmise), with:

$v = (1,5,4)$ (in the basis of $\{e_1,e_2,e_3\}$ and

$f = (1,1,1)^{\ast}$ (in the dual basis), so that:

$e^1 \otimes e_2(v,f)$ returns the product of the first coordinate of $v$ with the second coordinate of $f$.

(Tensor products become very unwieldy to explictly compute once m and n start to get larger than 2, as the number of operations to do gets rather large).

 

[Sudharaka]

The way I learned it, a tensor is a multilinear map:

$T:\underbrace{V \times \cdots \times V}_{n\ copies} \times \underbrace{V^{\ast} \times \cdots \times V^{\ast}}_{m\ copies} \to F$

We can, by the universality of the tensor product, regard $T$ as an element of:

$\underbrace{V^{\ast} \otimes \cdots \otimes V^{\ast}}_{n\ copies} \otimes \underbrace{V \otimes \cdots \otimes V}_{m\ copies}$

(here, we are implicitly identifying $V$ with $V^{\ast\ast}$).

In this case, $m = n = 1$, which makes our tensor particularly easy to understand. In this case, we have that any such tensor is a linear combination of elementary tensors of basis covectors and vectors. So it suffices to determine what:

$e^j \otimes e_i (v,u^{\ast})$ is, for any two vectors $u,v \in V$.

Typically, $\{e_1,\dots,e_n\}$ is the standard (Euclidean) basis for $F^n$, and $\{e^1,\dots,e^n\}$ is the dual basis; that is, the linear functionals:

$\displaystyle e^j(v) = e^j\left(\sum_{i = 1}^n v^ie_i \right) = v^j$

(the linear functional that returns the $j$-th coordinate of $v$ in the standard basis).

It is typical to regard $e_i$ as an $n \times 1$ (column) matrix, and $e^j$ as a $1 \times n$ (row) matrix, in which case their tensor product is given by their Kronecker product, the $n \times n$ matrix (written below in block form):

$\begin{bmatrix}0e_i&\dots&1e_i&\dots&0e_n \end{bmatrix}$


$= E_{ij}$, which has a 1 in the $i,j$-th entry, and 0's elsewhere. That is:

$e^j \otimes e_i (v,u^{\ast}) = u^TE_{ij}v = (u^{\ast})_iv^j$

(I hope I have my indices correct....I get these backwards a lot).

In this example, we have $V = F^3$ (as near as I can surmise), with:

$v = (1,5,4)$ (in the basis of $\{e_1,e_2,e_3\}$ and

$f = (1,1,1)^{\ast}$ (in the dual basis), so that:

$e^1 \otimes e_2(v,f)$ returns the product of the first coordinate of $v$ with the second coordinate of $f$.

(Tensor products become very unwieldy to explictly compute once m and n start to get larger than 2, as the number of operations to do gets rather large).

Thanks so much for the detailed explanation. It would take some time to sink in all the details you provided but I am learning slowly.