Tensor product of vector spaces?

I've been doing a lot of reading into multilinear algebra, and a good
deal of it is centered around the notion of the tensor product, which
is the apparently most general bilinear operation. I find the
definitions that I've seen of it to be rather confusing though. For
example, the article here defines it, for any two vector spaces V and
W, as the space V x W but with four equivalence relations

http://en.wikipedia.org/wiki/Tensor_product#Tensor_product_of_vector_spaces

But then it only lists 3 equivalence relations.

But even so, I still am not sure what a tensor product is, or what it
does, or most importantly what it "looks like." For instance, what
happens if we take the tensor product of the 2.3 subgroup and the 5.7
subgroup? How is it different from just the direct sum of those
spaces?

Even if there's no obvious, immediate application of this to interval
space, I'd still like to know what it is, because it comes up so often
in my pursuit to understand multilinear algebra that I find it a
serious impediment to my understanding to not know exactly what it is
and how it works. Can anyone share some insight?

Thanks,
Mike

Hi Mike,
in the wikipedia article, the third row in the list of the equivalence relations groups indeed two equivalences in one statement:
ce_(v,w)~ e(cv,w)
and
ce_(v,w)~ e(v,cw)

As to the "looks like" issue, in my opinion the most effective view is that of the "Kronecker product of matrices" (see again the wikipedia article; here "matrix" include the "row" and "column" cases).

Basically, "tensors" act by transforming vector-valued linear operators on vector spaces into vector-valued linear operators between spaces of different dimensions.

This includes the case of the underlying scalar fields considered as vector spaces over the same field, i.e. considering "linear functionals" too as "linear operators".

For instance, the tensor product of a column vector with 4 components and a row vector with 3 ones is a 4x3 matrix:
example:
1
2 tensor 0 1 2 =
3
4

=
0 1 2
0 2 4
0 3 6
0 4 8

I am afraid I cannot help you with the applications to music theory, since I still haven't been able to find the time to study your documentation :-(
All the best!
Claudi

--- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:

> But even so, I still am not sure what a tensor product is, or what it
> does, or most importantly what it "looks like." For instance, what
> happens if we take the tensor product of the 2.3 subgroup and the 5.7
> subgroup? How is it different from just the direct sum of those
> spaces?

You get something with a basis consisting of 2b
5, 2b
7, 3b
5 and 3b
7, whereas the direct sum would be 2,3,5, and 7. Since 2x2 = 2+2 = 4, the example might be clearer with more basis vectors.

> Even if there's no obvious, immediate application of this to interval
> space, I'd still like to know what it is, because it comes up so often
> in my pursuit to understand multilinear algebra that I find it a
> serious impediment to my understanding to not know exactly what it is
> and how it works. Can anyone share some insight?

It's like a wedge product, except vb
v isn't zero by definition, and ub
v and vb
u are just different, instead of like the wedge product, where ub'v = -vb'u. If you wedge together two elements of an n-dimensional space, you get something in an n(n-1)/2 dimensional space because of this relationship, whereas you get an n^2 dimensional space for the tensor product. It can be thought of as the nxn square matrix you get by taking u transpose times v.

--- In tuning-math@yahoogroups.com, "genewardsmith" <genewardsmith@...> wrote:

I thought Yahoo displayed unicode these days. :(

Thanks to all for your responses. A question for gene:

On Sat, Jan 7, 2012 at 7:32 PM, genewardsmith
<genewardsmith@sbcglobal.net> wrote:
>
> --- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:
>
> > But even so, I still am not sure what a tensor product is, or what it
> > does, or most importantly what it "looks like." For instance, what
> > happens if we take the tensor product of the 2.3 subgroup and the 5.7
> > subgroup? How is it different from just the direct sum of those
> > spaces?
>
> You get something with a basis consisting of 2b
5, 2b
7, 3b
5 and 3b
7, whereas the direct sum would be 2,3,5, and 7. Since 2x2 = 2+2 = 4, the example might be clearer with more basis vectors.

What if we tensor multiply a space with itself? For example if we
tensor multiply the 2.3.5.7 space with itself, what do we get?

> > Even if there's no obvious, immediate application of this to interval
> > space, I'd still like to know what it is, because it comes up so often
> > in my pursuit to understand multilinear algebra that I find it a
> > serious impediment to my understanding to not know exactly what it is
> > and how it works. Can anyone share some insight?
>
> It's like a wedge product, except vb
v isn't zero by definition, and ub
v and vb
u are just different, instead of like the wedge product, where ub'v = -vb'u. If you wedge together two elements of an n-dimensional space, you get something in an n(n-1)/2 dimensional space because of this relationship, whereas you get an n^2 dimensional space for the tensor product. It can be thought of as the nxn square matrix you get by taking u transpose times v.

One thing I find confusing is that tensor products can be applied to
entire vector spaces, and not just vectors. But, they can be applied
to individual vectors too. But, as far as I know, wedge products are
just applicable to vectors, and not to spaces. Is that right?

-Mike

--- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:

> What if we tensor multiply a space with itself? For example if we
> tensor multiply the 2.3.5.7 space with itself, what do we get?

That's a tensor power, you could think of it as a square matrix. For the full tensor algebra, get a non-commuting algebra over your base field or ring, in this case presumably Z. Since unicode is screwed up, I'll represent the tensor product by "x". Then you get an algebra over Z, with basis elements products like 3x2x7x7, etc., and where pxq is not equal to qxp. You do have the distributive and associative laws, so that 3x(2x5+7x2) = 3x2x5+3x7x2, etc, and 5x(2x7) = (5x2)x7 = 5x2x7.

> One thing I find confusing is that tensor products can be applied to
> entire vector spaces, and not just vectors. But, they can be applied
> to individual vectors too. But, as far as I know, wedge products are
> just applicable to vectors, and not to spaces. Is that right?

Nope! You can take exterior powers as well as tensor powers.

On Mon, Jan 9, 2012 at 10:04 AM, genewardsmith
<genewardsmith@sbcglobal.net> wrote:
>
> --- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:
>
> > What if we tensor multiply a space with itself? For example if we
> > tensor multiply the 2.3.5.7 space with itself, what do we get?
>
> That's a tensor power, you could think of it as a square matrix. For the full tensor algebra, get a non-commuting algebra over your base field or ring, in this case presumably Z. Since unicode is screwed up, I'll represent the tensor product by "x". Then you get an algebra over Z, with basis elements products like 3x2x7x7, etc., and where pxq is not equal to qxp. You do have the distributive and associative laws, so that 3x(2x5+7x2) = 3x2x5+3x7x2, etc, and 5x(2x7) = (5x2)x7 = 5x2x7.

Very interesting! And I assume that things like 3x2x7x7 are considered
to be atomic expressions that don't simplify any further?

I have a question about abstract algebra in general. I've been
thinking about wedge products as these families of parallelograms that
are defined by vectors, but from a purely algebraic standpoint that's
not what they are at all, right? The wedge product is just a generic
product with a set of generic axioms, and there's no visual,
geometric, or otherwise applied "use" other than that. And then, if
you're working with applied mathematics like we are here, you just
consider the axioms of whatever problem in real life you're trying to
analyze, and then find an analogous product, right?

So there's nothing that a tensor product "is." It's just a generic
operation that behaves as you've defined it, and the resultant basis
vectors "mean" nothing and have no interpretation. But, sometimes in
real life, things behave the same way, and so then it becomes useful.

Right?

> > One thing I find confusing is that tensor products can be applied to
> > entire vector spaces, and not just vectors. But, they can be applied
> > to individual vectors too. But, as far as I know, wedge products are
> > just applicable to vectors, and not to spaces. Is that right?
>
> Nope! You can take exterior powers as well as tensor powers.

But can you take the exterior power of one space with another space? I
know you can take the exterior power of a space, so that /\^2(V) is
the space of all 2-vectors generated by v1 ^ v2 in V. But can you do
something like V ^ W, where V and W are both different vector spaces?

-Mike

Just a quick bump, would be helpful if anyone could weigh in on this... thanks

-Mike

On Mon, Jan 9, 2012 at 4:28 PM, Mike Battaglia <battaglia01@gmail.com> wrote:
> On Mon, Jan 9, 2012 at 10:04 AM, genewardsmith
> <genewardsmith@sbcglobal.net> wrote:
>>
>> --- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:
>>
>> > What if we tensor multiply a space with itself? For example if we
>> > tensor multiply the 2.3.5.7 space with itself, what do we get?
>>
>> That's a tensor power, you could think of it as a square matrix. For the full tensor algebra, get a non-commuting algebra over your base field or ring, in this case presumably Z. Since unicode is screwed up, I'll represent the tensor product by "x". Then you get an algebra over Z, with basis elements products like 3x2x7x7, etc., and where pxq is not equal to qxp. You do have the distributive and associative laws, so that 3x(2x5+7x2) = 3x2x5+3x7x2, etc, and 5x(2x7) = (5x2)x7 = 5x2x7.
>
> Very interesting! And I assume that things like 3x2x7x7 are considered
> to be atomic expressions that don't simplify any further?
>
> I have a question about abstract algebra in general. I've been
> thinking about wedge products as these families of parallelograms that
> are defined by vectors, but from a purely algebraic standpoint that's
> not what they are at all, right? The wedge product is just a generic
> product with a set of generic axioms, and there's no visual,
> geometric, or otherwise applied "use" other than that. And then, if
> you're working with applied mathematics like we are here, you just
> consider the axioms of whatever problem in real life you're trying to
> analyze, and then find an analogous product, right?
>
> So there's nothing that a tensor product "is." It's just a generic
> operation that behaves as you've defined it, and the resultant basis
> vectors "mean" nothing and have no interpretation. But, sometimes in
> real life, things behave the same way, and so then it becomes useful.
>
> Right?
>
>> > One thing I find confusing is that tensor products can be applied to
>> > entire vector spaces, and not just vectors. But, they can be applied
>> > to individual vectors too. But, as far as I know, wedge products are
>> > just applicable to vectors, and not to spaces. Is that right?
>>
>> Nope! You can take exterior powers as well as tensor powers.
>
> But can you take the exterior power of one space with another space? I
> know you can take the exterior power of a space, so that /\^2(V) is
> the space of all 2-vectors generated by v1 ^ v2 in V. But can you do
> something like V ^ W, where V and W are both different vector spaces?
>
> -Mike

--- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:
> Very interesting! And I assume that things like 3x2x7x7 are considered
> to be atomic expressions that don't simplify any further?

I guess the question is whether you can write some equation like "3x2x7x7 = ..." where the stuff on the right side is the result of some operation on other elements of the tensor product space. If so, then yeah, that's obviously impossible. The only thing you can do with the elements is take linear combinations of them, and 3x2x7x7 isn't a linear combination of other elements. It's uniquely expressed as (1)(3x2x7x7) + (0)(everything else).

> I have a question about abstract algebra in general. I've been
> thinking about wedge products as these families of parallelograms that
> are defined by vectors, but from a purely algebraic standpoint that's
> not what they are at all, right? The wedge product is just a generic
> product with a set of generic axioms, and there's no visual,
> geometric, or otherwise applied "use" other than that. And then, if
> you're working with applied mathematics like we are here, you just
> consider the axioms of whatever problem in real life you're trying to
> analyze, and then find an analogous product, right?
>
> So there's nothing that a tensor product "is." It's just a generic
> operation that behaves as you've defined it, and the resultant basis
> vectors "mean" nothing and have no interpretation. But, sometimes in
> real life, things behave the same way, and so then it becomes useful.
>
> Right?

This sounds correct.

But of course, in math, whenever two or more definitions are exactly equivalent, there's never any reason to single out one as "the real definition" and the others as mere alternate characterizations. You can treat any one of them as the definition, and mathematicians do this all the time.

For example, I read a book about smooth manifolds, and the definition started "a smooth manifold is a subset of R^n such that...", and then gave a bunch of properties that the set of points in R^n had to have for it to be a smooth manifold. Then later I took a course, also about smooth manifolds, and the definition used was totally different. It was like "a smooth manifold is a Hausdorff space together with a set of homeomorphisms, called charts...".

These definitions seem totally unrelated at first. In the first definition, each point of the manifold is a point of R^n, so you can actually write down a list of n real numbers that are the coordinates of that point. In the second definition the manifold is a totally abstract space, where a point x in the manifold is simply the point x; you can specify x by its coordinates in one of the charts, but that's not unique, and it's not what the point x *is*, it's just one way of specifying it.

But because of the Whitney embedding theorem, the definitions are equivalent, and the book and the course were perfectly consistent with each other, just in a non-obvious way.

> But can you take the exterior power of one space with another space? I
> know you can take the exterior power of a space, so that /\^2(V) is
> the space of all 2-vectors generated by v1 ^ v2 in V. But can you do
> something like V ^ W, where V and W are both different vector spaces?

This makes no sense. The exponent is a positive integer, not another space.

Keenan

--- In tuning-math@yahoogroups.com, "Keenan Pepper" <keenanpepper@...> wrote:

> But because of the Whitney embedding theorem...

Not to mention the Nash embedding theorem, for Riemannian manifolds (yes, that Nash.)

--- In tuning-math@yahoogroups.com, "Keenan Pepper" <keenanpepper@...> wrote:

> I guess the question is whether you can write some equation like "3x2x7x7 = ..." where the stuff on the right side is the result of some operation on other elements of the tensor product space. If so, then yeah, that's obviously impossible. The only thing you can do with the elements is take linear combinations of them, and 3x2x7x7 isn't a linear combination of other elements. It's uniquely expressed as (1)(3x2x7x7) + (0)(everything else).

Well, recall these do represent (in this case) vals, and in other cases we have vectors, and in all cases we have the scalar product to consider, which works in a multilinear way.

--- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:
>
> Just a quick bump, would be helpful if anyone could weigh in on this... thanks
>
> -Mike
I
> > know you can take the exterior power of a space, so that /\^2(V) is
> > the space of all 2-vectors generated by v1 ^ v2 in V. But can you do
> > something like V ^ W, where V and W are both different vector spaces?
> >
> > -Mike
>
I would say yes, provided that your vector spaces are over the same scalar field and hence belong to the same exterior algebra. For instance, {two-forms}/\{three-forms}= {five forms}.
You could take a look at the older versions of wikipedia article, starting e.g. from this one:
http://en.wikipedia.org/w/index.php?title=Exterior_algebra&direction=next&oldid=890306
Also, if you read French, please take a look at
http://fr.wikipedia.org/wiki/Alg%C3%A8bre_ext%C3%A9rieure
That's a shorter article than the English one, which - in its present version - seems to have had too many contributors.
Hope this helps.
Best wishes,
Claudi

Hi all,
did anyone happen to read Jan Haluska's book:
The Mathematical Theory of Tone Systems (Chapman & Hall/CRC Pure and Applied Mathematics)
http://www.amazon.com/Mathematical-Systems-Chapman-Applied-Mathematics/dp/0824747143
please?
Many thanks,
Claudi

--- In tuning-math@yahoogroups.com, "clamengh" <clamengh@...> wrote:

> Also, if you read French, please take a look at
> http://fr.wikipedia.org/wiki/Alg%C3%A8bre_ext%C3%A9rieure
> That's a shorter article than the English one, which - in its present version - seems to have had too many contributors.

Even so, it has a notice on top about how terrible the article is, which lets you know this is Wikipedia.