Suggested operatonal definition for a tensor [duplicate]

Question

The two tensor definitions I'm (newly) familiar with, by transformation rules, and as a map from a tensor product space to the reals, don't tell me what a tensor does, and to the best of my knowledge they don't make it apparent. So, I'm looking for an operational definition, and suggesting the following one -

This question was marked as a duplicate and here I try to explain why I don't think it is. See paragraph 1. Moreover - this definition is different because it defines tensors of rank 2 and above as a means of defining maps from tensor to tensor, and defines the operation of one tensor mapping another as an extension of the angle bracket operator for covectors and vectors (which I denote by >cv, v< as typing them correctly causes them not to print). So, this definition tells you explicitly what a tensor does and how it does it. I think it is much easier to understand if you're starting with no knowledge of tensors.

Scalars are rank 0 tensors.

For tensors of higher rank we start with a n-dimensional vector space V with basis e1, e2,... en and dual covector space V* with basis e1*, e2*,... en*. Vectors and covectors are rank 1 tensors.

A tensor of rank r>1 is a means of defining a linear map from tensors of rank m to tensors of rank n, where m and n are generally < r, but not necessarily, that is, the definition doesn't require it.

A tensor of rank r>1 is a linear combination of dyadic basis tensors of rank r, where a dyadic basis tensor product of rank 1 basis tensors a1 x a2 x a3 x....x ar where each ai is either a basis vector ej of V or a basis covector ej* of V*.

The result of applying one dyadic basis tensor a1 x a2 x a3 x....x ar to another b1 x b2 x b3 x....x bs is computed by evaluating >ar,b1< and if that is not 0, evaluating >ar-1,b2<, and if that is not 0 continuing till one of the dyadic basis tensors (normally b1 x b2 x b3 x....x bs.) is used up, if no 0 was produced the dyadic basis tensor that remains is the result, it's either a 1 or a dyadic basis tensor. Note when evaluating >ar-(k-1),bk< one must be a vector and one a covector, else it's an error.

Note: I'm using non-standard brackets >cv,v< to indicate applying a covector to a vector, the normal angle brackets don't print correctly.

The a dyadic basis tensor eats the b dyadic basis tensor till one of the nibbles is a 0 and the result is 0, or till one is used up and the result is 1 or the part of the a dyadic basis tensor that didn't get a bite (or the b dyadic basis tensor that didn't get bitten).

A tensor A maps tensor B by applying each term (coefficient times dyadic basis tensor) in A to each term in B and multiplying their coefficients when the result is not 0, and summing the resulting terms to get the result of A applied to B.

I think this is how tensors work, and I think this definition does explicitly spell it out and make it clear. But I'd like to have it verified or corrected if necessary.

Answer 1

So, I'm looking for an operational definition ...

I've kept the following definition of a general tensor handy since I first read it in "Gravitation" (top of page 75):

one defines the most general tensor $\mathbf H$ and its rank $\left( \begin{array}{c} n\\ m\\ \end{array} \right)$ as follows: $\mathbf H$ is a linear machine with $n$ input slots for $n$ 1-forms and $m$ input slots for $m$ vectors; given the requested input, it puts out a real number denoted

$$\mathbf{H\left(\sigma, \lambda, ... ,\beta, u,v,...,w \right)}$$

(In the above, the greek arguments denote the $n$ 1-forms and the latin arguments denote the $m$ vectors.)

From the above, it is evident that a vector is a rank $\left( \begin{array}{c} 1\\ 0\\ \end{array} \right)$ tensor, and a 1-form is a rank $\left( \begin{array}{c} 0\\ 1\\ \end{array} \right)$ tensor.

Lastly, a scalar can be thought of as a $\left( \begin{array}{c} 0\\ 0\\ \end{array} \right)$ tensor, accepting zero 1-forms and zero vectors as 'inputs' and 'putting out' a real number.

Answer 2

I can't follow your definition, but it does sound very very similar to the version where it a tensor is a multilinear map (and from that we can get the transformation rule).

So you have your scalars (rank 0), and you vectors and your covectors (rank 1). and there are no other rank zero or rank one tensors. Let's no think of them asx functions. A scalar is a function that has no arguments, give it nothing and it gives you a real. A vector is like a function with just one hungry mouth, hungry for covectors, feed it a covector and it gives you a real. A covector is like a function with just one hungry mouth, hungry for vectors, feed it a vector and it gives you a real.

So now our rank 0 and rank 1 tensors are functions of some number (possibly zero) of vectors and/or covectors and that gives you a real. Next we notice that these functions are linear.

If you imagine a beast with two mouths, each hungry for either a covector or a vector, then you can feed either mouth, and the result is a beast with only one hungry mouth, so feeding a rank two tensor gives you a rank one tensor. Since the rank one tensors are linear functions, and you could have fed either mouth, it makes sense to insist that in some way all the mouths are linear. The proper way to do this is to say that the function is multilinear. This means that if if feed just one mouth then you get a function and you can add functions and scale them, so we can insist that feeding any mouth is done in a linear way.

So yes, now you can work it out with a basis if you so desire, and then the transformation rules for basis transformation follow since the tensor needs to adapt to the changing basis in such a way as to make the operation on a vector (which doesn't depend on basis) act as a function (so doesn't depend on the basis, just on the vector fed it).

The issue I had following your post was about products. There are multiple ways to make multiple products. For instance if you have two beasts with various hungers you could do an outer product where you just chain them together, so that if they all get fed you just multiply the numbers together. If you had 3 mouths and 4 mouths you can have a 7 mouthed beast and we know what happens if you fed each of the seven mouths.

But there are other products in tensor algebra (and tensor analysis). There are contractions or inner products. These are just as important, but are not so easily imagined as a beast with various hungers, but I'll try. If you fed all the hungers except one hunger for a vector and one hunger for a covector you basically have a linear transformation (something that hungers for a vector but if fed gives you another vector), so since it is linear it is like a matrix, so instead of feeding it you can take the trace, that's a way to get a number. So you could single out that pair of hungers from the beginning and say that after everything else is fed, you will take the trace of the matrix, so now it's like you have two fewer hungers because you now have a rule to assign numbers to the remaining hungers in a multilinear manner.