Do not understand adding wedgies

I keep trying to figure out what adding wedgies does to the null space
of a comma. Even better, I want to figure out what it does to the
(saturated) lattice of vals supporting that wedgie. It's difficult for
me to visualize.

So the meantone wedgie is <<1 4 10 4 13 12||, and the porcupine wedgie
is <<3 5 -6 1 -18 -28||. And if you add them, you get this wedgie <<4
9 4 5 -5 -16||. Which, assuming my "dewedgie" routine is right, leads
to this temperament here:

http://x31eq.com/cgi-bin/rt.cgi?ets=20cd_8bccc&limit=7

So I notice some kind of pattern in the null spaces of these
temperaments. Here's the LLL-reduced basis for the null space of
meantone

[-4, 4, -1, 0> (81:80)
[1, 2, -3, 1> (126:125)

Here's the LLL-reduced basis for the null space of porcupine:

[6, -2, 0, -1> (64:63)
[1, -5, 3, 0> (250:243)

And here's the null space of our goofy, combined, diminished-esque
temperament is

[5, 4, 0, -4> (2592:2401)
[5, -9, 4, 0> (20000:19683)

So I note that in our combined-wedgie temperament, you get (250/243) /
(81/80) = 20000/19683 in there. And I note that the same applies for
(64/63) / (126/125) = 2592/2401. But somehow, adding the wedgies for
meantone and porcupine has given us a new null space that's actually
subtracting meantone and porcupine.

I assume that wedgie normalization has something to do with this. And
if that were all that it has to do with it, then I'd be happy. But, I
note that subtracting random selections of other commas doesn't work.
For instance, (64/63) / (81/80) = 5120/5103 doesn't appear in our new
combined temperament.

How does this work?

-Mike

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

> So the meantone wedgie is <<1 4 10 4 13 12||, and the porcupine wedgie
> is <<3 5 -6 1 -18 -28||. And if you add them, you get this wedgie <<4
> 9 4 5 -5 -16||.

Actually, you don't, as that isn't a wedgie. Then problem is that there isn't a val supporting both potcupine and meantone. You have pajara^meantone = <<<<0|||| and pajara^porcupine=<<<<0||||, so they have a common val and you can add or subtract them. But meantone^porcupine = <<<<1||||, which means both that you can't add them, but also (since it is +-1) that they cover the whole 7-limit. If you know what both meantone and porcupine think a 7-limit interval is, you can recover it.

On Mon, Mar 26, 2012 at 11:02 PM, genewardsmith
<genewardsmith@sbcglobal.net> wrote:
>
> Actually, you don't, as that isn't a wedgie. Then problem is that there
> isn't a val supporting both potcupine and meantone.

Aha!

This brings me back to something that was confusing a little while
ago. On the page about wedgies, when you discuss Plucker embeddings,
you write this paragraph

"However, this is no longer the case in higher limits. There, not
everything which looks like a wedgie will be one; for instance the
wedgies must also satisfy the condition, for any wedgie W, that W∧W =
0, where the "0" means the multival of rank 2r obtained by wedging W
with W."

So I never understood this, because

1) I thought that W∧W = 0 is always true, by the definition of ∧. I
thought that that follows directly from anticommutativity. Or is that
not how the exterior product is defined?
2) It also then defines "0" as W∧W, which appears to be a circular definition.

What does this mean then?

-Mike

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

> 1) I thought that W∧W = 0 is always true, by the definition of ∧.

No, because W might be a linear combination of simple multivectors, that is, instead of for instance being A^B, it might be A^B + C^D. A wedgie is never like that, but in general you can't take any six integers which look like a 7-limit rank-two wedgie and associate it to a temperament.

--- In tuning-math@yahoogroups.com, "genewardsmith" <genewardsmith@...> wrote:
>
>
>
> --- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@> wrote:
>
> > 1) I thought that W∧W = 0 is always true, by the definition of ∧.
>
> No, because W might be a linear combination of simple multivectors, that is, instead of for instance being A^B, it might be A^B + C^D.

To expand on this, (A^B + C^D)^(A^B + C^D) = A^B^A^B + A^B^C^D + C^D^A^B + C^D^C^D. You can permute any two members of each product while reversing sign, so the result is -A^A^B^B + A^B^C^D + A^B^C^D -C^C^D^D = 2 A^B^C^D, and that works out to <<<<2 det||||, where "det is the determinant of the [A B C D] matrix.

On Tue, Mar 27, 2012 at 2:09 PM, genewardsmith <genewardsmith@sbcglobal.net>
wrote:
>
> >
> > No, because W might be a linear combination of simple multivectors, that
> > is, instead of for instance being A^B, it might be A^B + C^D.
>
> To expand on this, (A^B + C^D)^(A^B + C^D) = A^B^A^B + A^B^C^D + C^D^A^B +
> C^D^C^D. You can permute any two members of each product while reversing
> sign, so the result is -A^A^B^B + A^B^C^D + A^B^C^D -C^C^D^D = 2 A^B^C^D,
> and that works out to <<<<2 det||||, where "det is the determinant of the [A
> B C D] matrix.

Huh. So the thing about A^A = 0 just isn't always true then? It's only
true for basis multivectors?

Something about that doesn't make sense to me. Is a linear combination
of multivectors not also a multivector? What sortof multivectors have
the property that A^A = 0 then? Just ones that are in the basis?

-Mike

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

> Something about that doesn't make sense to me. Is a linear combination
> of multivectors not also a multivector? What sortof multivectors have
> the property that A^A = 0 then? Just ones that are in the basis?

A sufficient condition for A^A = 0 is that A be simple/decomposible/blade (and we should probably pick which terminology we want to use.) This covers all wedgies.

On Fri, Mar 30, 2012 at 12:22 PM, genewardsmith
<genewardsmith@sbcglobal.net> wrote:
>
> --- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...>
> wrote:
>
> > Something about that doesn't make sense to me. Is a linear combination
> > of multivectors not also a multivector? What sortof multivectors have
> > the property that A^A = 0 then? Just ones that are in the basis?
>
> A sufficient condition for A^A = 0 is that A be simple/decomposible/blade
> (and we should probably pick which terminology we want to use.) This covers
> all wedgies.

I come from the future to say that "decomposable" is the best term,
because it's the most descriptive. I think "blade" and "simple" will
go over a bit too much like Feynman's "wakalixes," but "decomposable"
actually gives an idea of what the condition is.

So I'm clear myself, an n-multivector is decomposable if it can be
expressed as a wedge product of n 1-vectors, right?

-Mike

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

> So I'm clear myself, an n-multivector is decomposable if it can be
> expressed as a wedge product of n 1-vectors, right?

Right!