Pushing, pulling? Co- or contravraint?

[Qfwfq]

I caught it and ate it.

I knew you would! :hihi:

 

The following fact may be mildly disconcerting: the field [math]C[/math] of complex numbers, when considered as a vector space, is a REAL vector space, since the scalar coefficients will always be in [math]R[/math]. Wow!

 

So, by your note, we will have that [math]R^2 \simeq C[/math] as real vector spaces.

Arrrrrgh nooooooo! :shocked:

 

[math]\mathbb{C}[/math] is a 1 dimensional vector space over itself! Otherwise it isn't a field! If you consider it 2D over [math]\mathbb{R}[/math] then it isn't a field, so it isn't [math]\mathbb{C}[/math] at all but [math]\mathbb{R}^2[/math] instead! :Guns:

[Ben]

Arrrrrgh nooooooo!

 

[math]\mathbb{C}[/math] is a 1 dimensional vector space over itself! Otherwise it isn't a field! If you consider it 2D over [math]\mathbb{R}[/math] then it isn't a field, so it isn't [math]\mathbb{C}[/math] at all but [math]\mathbb{R}^2[/math] instead! :Guns:

Oooo goody, a gunfight!

 

While I admit to a slight abuse of English as we know it (later), I don't agree; first, by considering [math]\mathbb{C}[/math] as a vector space, we are "forgetting" the field axiom that every element has a multiplicative inverse, so we no longer require it to be a field.

 

Bang!

 

Second, if I offer you, say [math]1+i[/math] as a basis for [math]C[/math] when considered as a vector space, how would you interpret the coefficients in, say [math]z = a +bi[/math] as an element in this vector space? Real or complex? Are they, or are they not, by definition, elements in the field over which this vector space is defined?

 

Bang, bang.

 

However, I admit to this: when I call any vector space real, I mean this to refer to the field over which it is defined. This usage does occur, but I concede it is non-standard.

 

You winged me there (just a flesh wound)

[Qfwfq]

Actually, I agree with the terminology of real or complex vector space, as a way of saying over which field it is.

 

Those bangs are aimed wrong and don't even reach my kevlar!

While I admit to a slight abuse of English as we know it (later), I don't agree; first, by considering [math]\mathbb{C}[/math] as a vector space, we are "forgetting" the field axiom that every element has a multiplicative inverse, so we no longer require it to be a field.

 

Bang!

That would go for considering any field as a 1D vector space over itself, by the canonical isomorphism. It doesn't make us "forget" the field axioms nearly the way in which regarding it as [imath]\mathbb{R}^2[/imath] makes us forget there's any such a thing as multiplication at all. That's why, the moment you construct the product in the 2D space, you go and turn it right back into a field again and hence 1D over itself!!! :D

 

Second, if I offer you, say [math]1+i[/math] as a basis for [math]C[/math] when considered as a vector space, how would you interpret the coefficients in, say [math]z = a +bi[/math] as an element in this vector space? Real or complex? Are they, or are they not, by definition, elements in the field over which this vector space is defined?

 

Bang, bang.

Errrr, can you find a proper subfield of [imath]\mathbb{C}[/imath] to which that element belongs (and yet shares the same additive and multiplicative neutrals)? Given [imath]\mathbb{C}[/imath] for Christmas instead of having constructed it, after all those other numbers starting from Peano, one could define [imath]\mathbb{R}[/imath] as being its only uncountable proper subfield, unless you can find another one.

 

This does bring me to a quiz I've sometimes pondered (but alas, not enough to argue conclusively): Could one abstractly define [imath]\mathbb{C}[/imath] as being THE uncountable field which is also algebraically complete?

There might be theorems out there by which one could say so...

[Ben]

OK. As long as we looking at the same object from a different perspective, we are never going to agree. I think, in fact I know, we are both right from our own perspective, I suggest we leave it be. Agreed?

This does bring me to a quiz I've sometimes pondered (but alas, not enough to argue conclusively): Could one abstractly define [imath]\mathbb{C}[/imath] as being THE uncountable field which is also algebraically complete?

There might be theorems out there by which one could say so...

Yes there is. It goes something like this. First, a field [math]F[/math] is algebraically complete iff every polynomial with coefficients in [math]F[/math] has all roots in [math]F[/math].

 

Suppose otherwise, that is [math] F[/math] is not algebraically complete in this sense.

 

Theorem: Any field [math]F[/math] has an algebraically complete extension.

 

A little thought shows this is a nice generalization of the fundamental theorem of algebra. And since [math]R[/math] and [math]C[/math] are the only uncountable fields in sight, and since [math]C[/math] extends [math]R[/math], then [math]C[/math] is the uncountable algebraic completion of the (uncountable) reals.

 

(This is without any checking, so let me get back on this when I have)

 

Anyhoo, to continue my diatribe

 

Let's clean up our notation. Recall I wrote a type (2. 0) tensor as [math]v \otimes w[/math], which not what anyone was expecting to see, right?

 

So let's look at this guy in grisly detail. Using the usual definition of a vector as the sum of scalar products of some chosen basis, let's have a set of basis vectors as [math]\{e_i\}[/math]. Then

 

[math]v \otimes w =\sum_i a^ie_i \otimes \sum_j b^j e_j = \sum_{i,j}a^ib^j e_i \otimes e_j = \sum_{i.j}a^{ij}e_i \otimes e_j[/math].

 

The sets [math]\{a^i\},\,\{b^j\}[/math] are called the "components" of the vectors [math]v,\,w[/math], and the set [math]\{a^{ij}\}[/math] is the component of the tensor product of these vectors. Notice that if [math]v,\,w[/math] depend on the [math]n[/math] quantities [math]a^i,\,b^j[/math], then their tensor product depends on on the [math]n^2[/math] quantities [math]a^{ij}[/math]

 

Now for reasons that are slightly obscure, it is customary to refer to tensors by their components, thus we will have that [math]v \otimes w \equiv a^{ij}[/math]. Moreover, it is usual to capitalize, so one writes [math]A^{ij}[/math] is a type (2. 0) tensor.

 

And a type (0, 2) tensor will look like [math]B_{hk}[/math], a type (1, 1) tensor as [math]C^p_q[/math].

 

Ah... that looks more familiar, right? Phew!

[Qfwfq]

And since [math]R[/math] and [math]C[/math] are the only uncountable fields in sight

What I've been looking for, actually, are theorems that show there aren't other ones.

 

Another bun:

Notice that if [math]v,\,w[/math] depend on the [math]n[/math] quantities [math]a^i,\,b^j[/math], then their tensor product depends on on the [math]n^2[/math] quantities [math]a^{ij}[/math]

Strictly, you are considering a dyad i. e. result of a tensor product which can't give all possible tensors of the same ranking. therefore you don't actually have [imath]n^2[/imath] independent quantities. The simplest way to construct them all is to take as a basis all the tensor products of the basis [imath]\{e_i\}[/imath].

 

Now for reasons that are slightly obscure, it is customary to refer to tensors by their components

Yep, sure. That's why the "tensor" is so customarily defined as "something that transforms", instead of saying that its components transform. ;)

[Ben]

Note: I have edited out some utter gibberish.

 

Another bun:Strictly, you are considering a dyad i. e. result of a tensor product which can't give all possible tensors of the same ranking. therefore you don't actually have [imath]n^2[/imath] independent quantities. The simplest way to construct them all is to take as a basis all the tensor products of the basis [imath]\{e_i\}[/imath].

I am not completely sure what you are saying here, but I suspect you are pre-empting my next piece of self-aggrandizement.

 

Tensors have a superficially simple arithmetic. Let's start with "multiplication".

 

Suppose that [imath]A^{ij} \equiv \sum_{i,j} a^{ij}e \otimes e_j[/imath] is a type (2, 0) tensor and that [imath]B^{hkl} \equiv \sum_{h,k,l}b^{hkl} e_h \otimes e_k \otimes e_l[/imath] is a type (3, 0) tensor.

 

Then their product, assuming associativity (a bold assumption!!) will be, omitting the tensor product symbol (customary)

 

[math]A^{ij}B^{hkl} =(\sum_{i,j} a^{ij}e_i \otimes e_j) \otimes (\sum_{h,k,l} b^{hkl} e_h \otimes e_k \otimes e_l) [/math]

 

[math]= \sum_{i,j,h,k,l} a^{ijhkl}e_i \otimes e_j \otimes e_h \otimes e_k \otimes e_l[/math],

 

a type (5, 0) tensor called (by its friends) [math]C^{ijhkl}[/math].

 

Multiplying type (0, r) tensors by type (s, 0) tensors is far more interesting.

 

Later (though nobody else cares - why do I bother?)