Tensor Rank: Scalar, Vector, Matrix, & More

[KFC]

Hi there,
I have a question about tensor rank. As we know, zero-rank tensor is scalar, rank one tensor is a vector and rank two tensor is a 3x3 matrix. Moreover, scalar and vector can also be written in the form of matrix. However, for higher rank tensor, says rank 4, according to the definition, there are 3^4 = 81 entries. And many textbook wrote rank 4 tensor, in index form, as

$$T_{\alpha, \beta, \gamma, \delta}$$

i.e., there are four indices. So can we also write higher rank tensor (rank 4 or above) with a square matrix? If so, what does each index mean?

 

[ice109]

4 indices means a means a hypercube, not a really big square matrix. think about it, each element has 4 "coordinates" in the matrix - obviously you need 4 dimensions to display em all.

 

[Fredrik]

(I won't be writing any summation sigmas in this post, because we always sum over those indices that appear twice, and only those. This is the Einstein summation convention).

If V is a vector space, you can define the dual space V* as the set of all linear functions from V into the real numbers. A tensor of type (n,m) is a multilinear (linear in all variables) function

$$T:\underbrace{V^*\times\cdots\times V^*}_{\mbox{n factors}}\times\underbrace{V\times\cdots\times V}_{\mbox{m factors}}\rightarrow\mathbb R$$

What you call a tensor of rank 4 is a tensor of type (0,4).

Given a basis $\{\vec e_i\}$ of V, you can define a basis $\{\tilde e^i\}$ of V* by

$$\footnotesize\tilde e^i(\vec e_j)=\delta^i_j$$

where the right-hand side is the Kronecker delta (i.e. it's =1 when i=j and zero otherwise). This basis is called the dual basis of $\{\vec e_i\}$.

The set of all tensors of type (0,4) also has a natural vector space structure, and we can use any basis of V* to construct a basis for it. For example, the one constructed from $\{\tilde e^i\}$ is $\{\tilde e^i\otimes\tilde e^j\otimes\tilde e^k\otimes\tilde e^l\}$. The $\otimes$ symbol has a simple definition. I'll just give an example: If $\tilde\alpha$ and $\tilde\beta$ are members of V*, we have

$$\tilde\alpha\otimes\tilde\beta(\vec u,\vec v)=\tilde\alpha(\vec u)\tilde\beta(\vec v)$$

for all $\vec u$ and $\vec v$ in V.

OK, here's the definition of your T with 4 indices. It's the components of T when we express it using a basis:

$$T=T_{ijkl}\tilde e^i\otimes\tilde e^j\otimes\tilde e^k\otimes\tilde e^l$$

It's easy to show that

$$T_{ijkl}=T(\vec e_i,\vec e_j,\vec e_k,\vec e_l)$$

Note that the tensor itself is something that's completely independent of all bases. It's the components of the tensor that changes when you decide to use another basis.

If you're wondering what any of this has to do with changing coordinate systems, the answer is that the vector space V is usually a tangent space of a manifold (there's one at each point), and you can use a coordinate system to construct a basis for the tangent space at any point where the coordinate system is defined.

 

[nicksauce]

Note that a rank 2 tensor is not necessarily a 3x3 matrix. For example the kronocker delta is represented by a 2x2 matrix, and the Minkowski metric is represented by a 4x4 matrix.

A rank 3 matrix could be represented by a matrix that is being stretched out into a third dimension. See http://en.wikipedia.org/wiki/Levi-Civita_symbol for such an image.

 

[Fredrik]

For example the kronocker delta is represented by a 2x2 matrix,

Actually it's an nxn matrix where n is the dimension of the manifold.