I've been percolating on the posts in this thread and some online references and I'm making some progress. I just wanted to write down what I understand so far. I think I need to just keep plugging away at the symbology. Thanks to all for the discussion so far.

2. Tensors are essentially multilinear maps from the Cartesian product of vector spaces to the Reals

That's a very clarifying remark. Especially in light of this:

First suppose a vector space with . Then to any such space we may associate another vector space - called the dual space - which is the vector space of all linear mappings , that is .

Obviously then, for then .

So the tensor (or direct) product of two vector spaces is written as the bilinear mapping , where elements in are the ordered pairs (of vectors) , so that, for , by definition,

The object is called a TENSOR. In fact it is a rank 2, type (0,2) tensor

The above is extremely clear. By which I mean that it became extremely clear to me after I worked at it for a while

And I definitely got my money's worth out of this example. Let me see if I can say it back.

Given

and

as above, let

be functionals, and define a map

. In other words

is a function that inputs a pairs of elements of

, and outputs a real number. Specifically,

where the right hand side is just ordinary multiplication of real numbers. Note that

doesn't mean anything, it's notation we give to a particular function.

It's clear (and can be verified by computation) that

is a bilinear map.

In other words

is a function that inputs a pair of functionals, and outputs a

*function* that inputs pairs of vectors and outputs a real number.

So that's one definition of a tensor.

I'm not clear on why your example is of rank

, I'll get to that in a moment.

Another way to understand the tensor product of two vector spaces comes directly from the abstract approach I talked about earlier. In fact in the case of finite-dimensional vector spaces, it's especially simple. The tensor product

is simply the set of all finite linear combinations of the elementary gadgets

where the

's are a basis of

, subject to the usual bilinearity relationships.

Note that I didn't talk about duals; and the tensors are linear combinations of gadgets and not functions. In fact one definition I've seen of the rank of a tensor is that it's just the number of terms in the sum. So

is a tensor of rank

, and

is a tensor of rank two. Note that I have a seemingly different definition of rank than you do.

In general, a tensor is an expression of the form

. This is important because later on you derive this same expression by means I didn't completely follow.

By the way if someone asks, what does it mean mathematically to say that something is a "formal linear combination of these tensor gadgets," rest assured that there are technical constructions that make this legit.

Now if I could bridge the gap between these two definitions, I would be making progress. Why do the differential geometers care so much about the dual spaces? What

*meaning* do the duals represent? In differential geometry, physics, engineering, anything?

Likewise I understand that in general tensors are written as so many factors of the dual space and so many of the original space. What is the meaning of the duals? What purpose or meaning do they serve in differential geometry, physics, and engineering?

Now one more point of confusion. In your most recent post you wrote:

Physics and engineering would be unthinkable without a metric, although this causes no problems to a mathematician. Specifically, a vecto space is called a "metric space" if it has an inner product defined.{edit "with" to "without"}

I think that can't be right, since metric spaces are much weaker than inner product spaces. Every inner product gives rise to a metric but not vice versa. For example the Cartesian plane with the taxicab metric is not an inner product space. I'm assuming this is just casual writing on your part rather than some fundamentally different use of the word metric than I'm used to.

Now an inner product is defined as a bilinear, real-valued mapping (with certain obvious restrictions imposed), that is where .

In the case that our vector space is defined over the Reals, we have that

Agreed so far. Although in complex inner product spaces this identity doesn't hold, you need to take the complex conjugate on the right.

Turn to the dual space, with This means that for *any* and *any* that

Yes.

In the case of a metric space there always exists some *particular* for all .

The correspondence between the dual space and the inner product is not automatic, it needs proof. Just mentioning that.

And likewise by the symmetry above, there exists a . But writing as their product, we see this is just , so that .

Now here I got lost but I need to spend more time on it. You're relating tensors to the inner product and that must be important. I'll keep working at it.

And if we expand our dual vectors as, say and , then as before we may write then, dropping all reference to the basis vectors, we may have that .

Aha! The right side is exactly what I described above. It's a finite linear combination of elementry tensor gadgets. And somehow the functionals disappeared!

So I know all of this is the key to the kingdom, and that I'm probably just a few symbol manipulations away from enlightenment

Therefore the are called the components of a type (0,2) metric tensor.

Right, it's (0,2) because there are 0 copies of the dual and 2 copies of V. But where did the functionals go?

It is important in General Relativity (to say the least!!)

Should I be thinking gravity, photons, spacetime? Why are the duals important? And where did they go in your last calculation?

I'll go percolate some more. To sum up, the part where you define a tensor as a map from the Cartesian product to the reals makes sense. The part about the duals I didn't completely follow but you ended up with the same linear combinations I talked about earlier. So there must be a pony in here somewhere.

**Edited by wtf, 11 February 2017 - 04:40 AM.**