Wedgies and multivals

I'm trying to get a more intuitive understanding the properties of
exterior algebra, and I have a few questions.

On wedgies, Gene's article states: "an alternating multilinear map
which is a multilinear function taking a certain number n of monzos as
arguments and returning an integer as a value we may call an n-map."
So if we're wedging together two vals u and v, I assume that u^v is a
bilinear map which takes 2 monzos as arguments and returns a scalar.

Wikipedia's definition of a bilinear map is a function f: V1 x V2 ->
W, where V1, V2, and W are all different vector spaces or modules. So
let's call unweighted interval space M, and call its dual space M* = V
for tuning space, the space of vals. Then I assume the following
statement is true for u, v in V:

u^v: M x M -> R

That is, u^v is a bilinear map that takes two monzos and returns a real.

So here are my questions then, and I'd appreciate some clarity on these issues:

1) Let's say I want to actually apply this map to two monzos and see
what scalar results. Is the simplest way to do so by first calculating
their bimonzo, and then for <<a b c d e f||g h i j k l>> I simply
compute ag + bh + ci + dj + ek + fl?
2) What music-theoretic interpretation does the resulting scalar have?
For example, for any val v and monzo m, <v|m> tells you what position
m maps to on the chain of generators specified by v. What does
<<u^v||m^n>> tell you? What position the m^n maps to on the chain of
"unit parallelograms" specified by u^v?

Thanks,
Mike

Mike Battaglia <battaglia01@gmail.com> wrote:

> 1) Let's say I want to actually apply this map to two
> monzos and see what scalar results. Is the simplest way
> to do so by first calculating their bimonzo, and then for
> <<a b c d e f||g h i j k l>> I simply compute ag + bh +
> ci + dj + ek + fl?

No. Whatever you mean. If <a b c d e f] and [g h i j k l>
are vectors, you can form a bracket like that, and call it
an inner product. Once you move onto bivectors, it gets
more complicated. But the modulus of a multivector -- the
inner product of the multivector with itself -- does work
out as the sum of the squares of the elements.

> 2) What music-theoretic interpretation
> does the resulting scalar have? For example, for any val
> v and monzo m, <v|m> tells you what position m maps to on
> the chain of generators specified by v. What does
> <<u^v||m^n>> tell you? What position the m^n maps to on
> the chain of "unit parallelograms" specified by u^v?

It must be something like that. <v | m> is generally the
size of the interval m where v defines the prime intervals
you use to measure it. <v | v> is the square of the TE
complexity of a temperament that follows the mapping v,
where [v> can be of any rank and so not necessarily a val
and <v] is the weighted complement of [v>. You can define
<m | m> much the same way where m describes the unison
vectors.

Graham

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

> 1) Let's say I want to actually apply this map to two monzos and see
> what scalar results. Is the simplest way to do so by first calculating
> their bimonzo, and then for <<a b c d e f||g h i j k l>> I simply
> compute ag + bh + ci + dj + ek + fl?

Suppose your bival is <<1 4 10 4 13 12|| (meantone.) Then the alternating bilinear map is, in terms of monzos a=|a1 a2 a3 a3> and b=|b1 b2 b3 b4>

Meantone(a, b) = (a1b2-b1a2) + 4(a1b3-b1a3) + 10(a1b4-b1a4) + 4(a2b3-b2a3) + 13(a2b4-b2a4) + 12(a3b4-b3a4)

You can consider this to be a matrix product a.M.transpose(b), where M is antisymmetric.

> 2) What music-theoretic interpretation does the resulting scalar have?

Meantone(a, b) is 0 if a or b is a comma; it tells how many generator steps it takes to get to b if a is 2. If Meantone(a, b) = +-1 then you can use a and b as your generators; the matrix of monzos from a and b together with 81/80 and 126/125 is unimodular.

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

> Meantone(a, b) is 0 if a or b is a comma; it tells how many generator steps it takes to get to b if a is 2. If Meantone(a, b) = +-1 then you can use a and b as your generators; the matrix of monzos from a and b together with 81/80 and 126/125 is unimodular.

I should add that the general rule is that if W is the multiliear map corresponding to the wedgie of a temperament of rank r, and if a1, a2, ..., ar are r monzos, then |W(a1, a2, ..., ar)| = N means that the tempering of these r intervals defines a subgroup of index N of the entire group of tempered notes.

On Mon, Nov 21, 2011 at 7:45 AM, genewardsmith
<genewardsmith@sbcglobal.net> wrote:
>
> --- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:
>
> > 1) Let's say I want to actually apply this map to two monzos and see
> > what scalar results. Is the simplest way to do so by first calculating
> > their bimonzo, and then for <<a b c d e f||g h i j k l>> I simply
> > compute ag + bh + ci + dj + ek + fl?
>
> Suppose your bival is <<1 4 10 4 13 12|| (meantone.) Then the alternating bilinear map is, in terms of monzos a=|a1 a2 a3 a3> and b=|b1 b2 b3 b4>
>
> Meantone(a, b) = (a1b2-b1a2) + 4(a1b3-b1a3) + 10(a1b4-b1a4) + 4(a2b3-b2a3) + 13(a2b4-b2a4) + 12(a3b4-b3a4)

OK, that's what Keenan had worked out as well. So if the bival is in
the form <<v1 v2 v3 v4 v5 v6||, and if two monzos a and b have a wedge
product a^b = ||m1 m2 m3 m4 m5 m6>>, it's just Sum_i vi*mi? There was
also a more complicated formula for sending a single monzo into a
bival I never quite got my head wrapped around, but I got the gist
that if I send |-3 -1 2> into <<1 4 4||, I get the val <7 11 16| as a
result.

> You can consider this to be a matrix product a.M.transpose(b), where M is antisymmetric.

. is matrix multiplication?

> > 2) What music-theoretic interpretation does the resulting scalar have?
>
> Meantone(a, b) is 0 if a or b is a comma; it tells how many generator steps it takes to get to b if a is 2. If Meantone(a, b) = +-1 then you can use a and b as your generators; the matrix of monzos from a and b together with 81/80 and 126/125 is unimodular.
//snip
> I should add that the general rule is that if W is the multiliear map corresponding to the wedgie of a temperament of rank r, and if a1, a2, ..., ar are r
> monzos, then |W(a1, a2, ..., ar)| = N means that the tempering of these r intervals defines a subgroup of index N of the entire group of tempered notes.

OK, that makes perfect sense. But does saturation show up anyway in
this result? Will a saturated subgroup appear to have a different
result than an unsaturated one?

I also note that if you temper out 81/80, 25/24, and 3/2, you get the
rank-0 temperament "4," which is unfortunately contorted.

-Mike

On Mon, Nov 21, 2011 at 1:58 PM, Mike Battaglia <battaglia01@gmail.com> wrote:
> On Mon, Nov 21, 2011 at 7:45 AM, genewardsmith
> <genewardsmith@sbcglobal.net> wrote:
>>
>>
>> Meantone(a, b) = (a1b2-b1a2) + 4(a1b3-b1a3) + 10(a1b4-b1a4) + 4(a2b3-b2a3) + 13(a2b4-b2a4) + 12(a3b4-b3a4)
>
> OK, that's what Keenan had worked out as well. So if the bival is in
> the form <<v1 v2 v3 v4 v5 v6||, and if two monzos a and b have a wedge
> product a^b = ||m1 m2 m3 m4 m5 m6>>, it's just Sum_i vi*mi?

I say this because, if memory serves correctly, ||(a1b2-b1a2)
(a1b3-b1a3) (a1b4-b1a4) (a2b3-b2a3) (a2b4-b2a4) (a3b4-b3a4)>> is the
correct formula for the wedge product of a^b in a 4D space.

-Mike

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

> OK, that's what Keenan had worked out as well. So if the bival is in
> the form <<v1 v2 v3 v4 v5 v6||, and if two monzos a and b have a wedge
> product a^b = ||m1 m2 m3 m4 m5 m6>>, it's just Sum_i vi*mi?

That's true in general for multivals; that's the multilinear map.

There was
> also a more complicated formula for sending a single monzo into a
> bival I never quite got my head wrapped around, but I got the gist
> that if I send |-3 -1 2> into <<1 4 4||, I get the val <7 11 16| as a
> result.

That's the interior product.

> > You can consider this to be a matrix product a.M.transpose(b), where M is antisymmetric.
>
> . is matrix multiplication?

Right.

> OK, that makes perfect sense. But does saturation show up anyway in
> this result? Will a saturated subgroup appear to have a different
> result than an unsaturated one?

If a list of monzos is unsaturated the multivector you get by wedging them will have GCD > 1.

> I also note that if you temper out 81/80, 25/24, and 3/2, you get the
> rank-0 temperament "4," which is unfortunately contorted.

You note that, do you? Meantone(3/2, 25/24) = 4, at any rate.

On Mon, Nov 21, 2011 at 5:32 PM, genewardsmith
<genewardsmith@sbcglobal.net> wrote:
>
> There was
> > also a more complicated formula for sending a single monzo into a
> > bival I never quite got my head wrapped around, but I got the gist
> > that if I send |-3 -1 2> into <<1 4 4||, I get the val <7 11 16| as a
> > result.
>
> That's the interior product.

If you mean this

http://en.wikipedia.org/wiki/Interior_product

"In mathematics, the interior product is a degree -1 antiderivation on
the exterior algebra of differential forms on a smooth manifold."

I could never understand this page, and I still can't understand it.
How are wedgies and monzos differential forms? What's the smooth
manifold in this case?

It looks like ιXω is operating on a bunch of vector fields, and I
don't see what all the fields are in this case.

> > OK, that makes perfect sense. But does saturation show up anyway in
> > this result? Will a saturated subgroup appear to have a different
> > result than an unsaturated one?
>
> If a list of monzos is unsaturated the multivector you get by wedging them will have GCD > 1.

Oh right, I get it. So the key is that if the GCD of the multimonzo is
1, but <<V||M>> is > 1, it's a saturated subgroup

> > I also note that if you temper out 81/80, 25/24, and 3/2, you get the
> > rank-0 temperament "4," which is unfortunately contorted.
>
> You note that, do you?

Yes!

> Meantone(3/2, 25/24) = 4, at any rate.

Because it's the interior product of <<1 4 4]] and [-1 1 0>, which is
the rank-1 temperament <1 1 0], and then the interior product of that
with [-3 -1 2>, which is the rank-0 temperament "4". Or also, it's how
many steps that 3/2 maps to in the val <7 11 16], and how many steps
25/24 maps to in the val <1 1 0]. Or also, an index-4 subgroup of
meantone. That is, if I've made all these connections correctly.

-Mike

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

> If you mean this
>
> http://en.wikipedia.org/wiki/Interior_product

This is better:

http://en.wikipedia.org/wiki/Exterior_algebra#Interior_product

It still reads like it's written for mathematicians.

Then there's this:

http://mathworld.wolfram.com/InteriorProduct.html

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

> If you mean this
>
> http://en.wikipedia.org/wiki/Interior_product
>
> "In mathematics, the interior product is a degree -1 antiderivation on
> the exterior algebra of differential forms on a smooth manifold."
>
> I could never understand this page, and I still can't understand it.

It wasn't written by an algebraist, but by someone who thinks the only use for multilinhear algebra is in differential topology. To heck with it and him.

> How are wedgies and monzos differential forms? What's the smooth
> manifold in this case?

There is none. Forget it.

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

> > I could never understand this page, and I still can't understand it.
>
> It wasn't written by an algebraist, but by someone who thinks the only use for multilinhear algebra is in differential topology. To heck with it and him.

Maybe a page explaining the interior product on the Xenwiki would be good?