Question about decomposable multivectors and addition

I've been trying to understand adding decomposable multivectors
together, and figuring out when the result will also be decomposable.
This is relevant to Gene's current work on Fokker blocks, for those
interested.

Gene pointed out in my last thread about this that the addition of any
two totally decomposable bivectors will yield another decomposable
bivector if they share a common vector as a "factor." Therefore, if
you add the wedgies for 7-limit porcupine and 7-limit pajara, you'll
get another wedgie that obeys the conditions of the Plucker embedding.

This is clear with a little bit of math done out on paper - if A can
be written as v^u and B can be written as v^w, A+B = v^u + v^w =
v^(u+w), and hence it's decomposable. Therefore, porcupine + pajara =
22p^15p + 22p^12p = 22p^(15p+12p) = 22p^27p = superpyth.

This still leaves me with a few questions, however:
1) It's clear that, for any two bivectors, the condition that they
share a common factor is -sufficient- to show that their sum will be
decomposable. However, is this condition -necessary-?
2) Is this condition sufficient for trivectors and multivectors of
grade>2 in general? For instance, a^b^c + a^d^e = a^(b^c+d^e), but
there's no guarantee here that b^c + d^e will be decomposable.

Any insights? Would much appreciate it.

-Mike

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

> 1) It's clear that, for any two bivectors, the condition that they
> share a common factor is -sufficient- to show that their sum will be
> decomposable. However, is this condition -necessary-?

Yes. If X V Y is the zero 4-val, then X and Y represent intersecting subspaces. You can find the common (projective) val either by solving directly that the wedge product with each of X and Y is zero, or taking the union of the commas and finding the corresponding val.

> 2) Is this condition sufficient for trivectors and multivectors of
> grade>2 in general? For instance, a^b^c + a^d^e = a^(b^c+d^e), but
> there's no guarantee here that b^c + d^e will be decomposable.

You can produce something just like the Fokker group for a wedgie of any order. That is, trivals which are all wedge products of vals with a given rank two temperament wedgie form a group, etc.

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

> Yes. If X V Y is the zero 4-val

X ^ Y, sorry.

On Sun, May 6, 2012 at 11:35 PM, genewardsmith <genewardsmith@sbcglobal.net>
wrote:
>
> --- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...>
> wrote:
>
> > 1) It's clear that, for any two bivectors, the condition that they
> > share a common factor is -sufficient- to show that their sum will be
> > decomposable. However, is this condition -necessary-?
>
> Yes. If X V Y is the zero 4-val, then X and Y represent intersecting
> subspaces. You can find the common (projective) val either by solving
> directly that the wedge product with each of X and Y is zero, or taking the
> union of the commas and finding the corresponding val.

How does that show, that for decomposable bivectors satisfying the
equation a^b + c^d = e^f, that all three of those terms can be
rewritten in terms of a common vector v?

-Mike

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

> How does that show, that for decomposable bivectors satisfying the
> equation a^b + c^d = e^f, that all three of those terms can be
> rewritten in terms of a common vector v?

One way to get this explicitly is to use the interior product, for instance with 2. If you have wedgies X and Y with X^Y = 0, take P =
(XV2)/N, Q = (YV2)/N, where the N is chosen so that P^V = X, Q^V = Y, which should mean it is V[1], the edo of our common val V.

On Mon, May 7, 2012 at 10:01 PM, genewardsmith <genewardsmith@sbcglobal.net>
wrote:
> --- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...>
> wrote:
>
> > How does that show, that for decomposable bivectors satisfying the
> > equation a^b + c^d = e^f, that all three of those terms can be
> > rewritten in terms of a common vector v?
>
> One way to get this explicitly is to use the interior product, for
> instance with 2. If you have wedgies X and Y with X^Y = 0, take P =
> (XV2)/N, Q = (YV2)/N, where the N is chosen so that P^V = X, Q^V = Y,
> which should mean it is V[1], the edo of our common val V.

This assumes that X^Y = 0 a priori. I'm asking, if you have two
wedgies X and Y, and X + Y = Z, and Z is also a wedgie, does X^Y
-HAVE- to be zero?

-Mike

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

> This assumes that X^Y = 0 a priori. I'm asking, if you have two
> wedgies X and Y, and X + Y = Z, and Z is also a wedgie, does X^Y
> -HAVE- to be zero?

It would if we are talking bivals, but you've given me a question I can't answer off the bat when it comes to trivals. For bivals, A^B = B^A, so (X + Y)^(X + Y) = X^X + 2 X^Y + Y^Y = 2 X^Y. Hence for Z^Z=0, we need that X^Y = 0, except in characteristic 2 which we don't care about. Could either try to find a proof or search for a counterexample among 13 or 17 limit trivals.

On Tue, May 8, 2012 at 1:34 PM, genewardsmith <genewardsmith@sbcglobal.net>
wrote:
>
> --- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...>
> wrote:
>
> > This assumes that X^Y = 0 a priori. I'm asking, if you have two
> > wedgies X and Y, and X + Y = Z, and Z is also a wedgie, does X^Y
> > -HAVE- to be zero?
>
> It would if we are talking bivals, but you've given me a question I can't
> answer off the bat when it comes to trivals. For bivals, A^B = B^A, so (X +
> Y)^(X + Y) = X^X + 2 X^Y + Y^Y = 2 X^Y.

This seems off to me. How are you defining ^ such that A^B = B^A?

I know that Wedgie(A^B) = Wedgie(B^A). However, if you actually define
^ like that, with the Wedgie(...) implicit, then ^ is no longer
bilinear and wouldn't obey the usual axioms of distributivity, so I'm
not sure how you're deriving the identity above.

Also, <12 19 28| = <7 11 16| + <5 8 12|. By the above reasoning, <12
19 28| ^ <12 19 28| = (<7 11 16| + <5 8 12|)^(<7 11 16| + <5 8 12|) =
2*(<7 11 16|^<5 8 12|) = <<2 8 8||.

> Hence for Z^Z=0, we need that X^Y =
> 0, except in characteristic 2 which we don't care about. Could either try to
> find a proof or search for a counterexample among 13 or 17 limit trivals.

For trivals, it doesn't seem as though sharing a single val would be
sufficient. For instance, consider a 5-dimensional vector space with
basis given by a, b, c, d, e. Then a^b^c + a^d^e = a(b^c+d^e), where
(b^c + d^e) is of order 2 by definition. As far as this trick with
distributivity is concerned, it seems as though trivals would have to
share a -bival- for it all to work out, so that a^b^c + a^b^d =
a^b^(c+d).

-Mike

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

> This seems off to me. How are you defining ^ such that A^B = B^A?

Standard way:

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

Look at the section "Graded structure".

> For trivals, it doesn't seem as though sharing a single val would be
> sufficient. For instance, consider a 5-dimensional vector space with
> basis given by a, b, c, d, e. Then a^b^c + a^d^e = a(b^c+d^e), where
> (b^c + d^e) is of order 2 by definition. As far as this trick with
> distributivity is concerned, it seems as though trivals would have to
> share a -bival- for it all to work out, so that a^b^c + a^b^d =
> a^b^(c+d).

Yes, this is exactly what I was telling you a few posts back.

On Tue, May 8, 2012 at 3:36 PM, genewardsmith <genewardsmith@sbcglobal.net>
wrote:
>
> --- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...>
> wrote:
>
> > This seems off to me. How are you defining ^ such that A^B = B^A?
>
> Standard way:
>
>
> http://en.wikipedia.org/wiki/Exterior_algebra#Formal_definitions_and_algebraic_properties
>
> Look at the section "Graded structure".

Ah. I thought you were saying that A^B and B^A were bivals, meaning
that A and B were vals. But you mean that A and B are bivals
themselves, so A^B and B^A are the wedge products of bivals, which
would be tetravals or something like that. OK thanks, now it all makes
sense. Very neat proof.

So if trival multiplication is anticommutative, that means that for
any trival T at all, T^T = 0 then?

-Mike

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

> So if trival multiplication is anticommutative, that means that for
> any trival T at all, T^T = 0 then?

Correct, which is why it is useless as a test to see if a trival is a wedgie. Instead, you use the Plueker relations: take the interior product with each prime, wedgie it with the trival, and see if it is the zero 6-val.

--- In tuning-math@yahoogroups.com, "genewardsmith" <genewardsmith@...> wrote:
Instead, you use the Plueker relations: take the interior product with each prime, wedgie it with the trival, and see if it is the zero 6-val.

zero 5-val.