Simple identity to make the interior product easier to compute

I have this complicated interior product routine drafted up, and I
can't help but think I'm doing it wrong. For instance, say that V is a
multival, and M is a monzo. And I want to compute V v M, the interior
product of V and M. Can't I just do this?

If grade(V) > grade(M), then
V v M = (V° ^ M)°

If grade(V) < grade(M), then
V v M = (V ^ M°)°

where ° is the dual.

So in plain English, the first case says "add M to the kernel of V and
compute the resulting multival", and the second case says "add V to
the list of vals tempering out M and compute the resulting kernel."

-Mike

--- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:
>
> I have this complicated interior product routine drafted up, and I
> can't help but think I'm doing it wrong. For instance, say that V is a
> multival, and M is a monzo. And I want to compute V v M, the interior
> product of V and M. Can't I just do this?
>
> If grade(V) > grade(M), then
> V v M = (V° ^ M)°
>
> If grade(V) < grade(M), then
> V v M = (V ^ M°)°

Isn't M always grade 1?

On Thu, Nov 8, 2012 at 9:11 AM, genewardsmith <genewardsmith@sbcglobal.net>
wrote:
>
> > I have this complicated interior product routine drafted up, and I
> > can't help but think I'm doing it wrong. For instance, say that V is a
> > multival, and M is a monzo.
>
> Isn't M always grade 1?

Typo; should have said M is a multimonzo.

BTW, is there a name for the operation which takes two multimonzos, or
two multivals, and returns the multimonzo/multival representing the
subspace the two share in common? So for instance, if @ is the
operator, then <<1 4 4|| @ <<2 1 -3|| = <7 11 16|.

So in general, A @ B = A° v B, where v is the interior product.

Have I discovered the long-lost regressive product?

-Mike

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

> BTW, is there a name for the operation which takes two multimonzos, or
> two multivals, and returns the multimonzo/multival representing the
> subspace the two share in common? So for instance, if @ is the
> operator, then <<1 4 4|| @ <<2 1 -3|| = <7 11 16|.
>
> So in general, A @ B = A° v B, where v is the interior product.

If we do this with two grade k multivectors, we get a grade n-2k multivector. What is the interpretation in general? I just tried winston and 13-limit meantone, and got the zero multival. The two have commas in common, of course, and vals supporting them in common also.

On Fri, Nov 9, 2012 at 9:42 AM, genewardsmith <genewardsmith@sbcglobal.net>
wrote:
>
> If we do this with two grade k multivectors, we get a grade n-2k
> multivector. What is the interpretation in general? I just tried winston and
> 13-limit meantone, and got the zero multival. The two have commas in common,
> of course, and vals supporting them in common also.

Can you give me the wedgies you were using?

I think it's something like: if there are no commas in common, you get
the temperament which is the union of their kernels (and the
intersection of their row spaces, if you were to convert them into
mapping matrices). If there are commas in common, you get the zero
vector. For instance, try taking the interior product of 7-limit
meantone and |-4 4 -1 0> and you'll also get the zero vector. The same
applies here.

The wedge product doesn't like if you temper out the same comma twice;
you tend to get the zero vector.

-Mike

bump

At 12:27 AM 2012/11/11, Mike wrote:
>On Fri, Nov 9, 2012 at 9:42 AM, Gene wrote:
>>
>> If we do this with two grade k multivectors, we get a grade n-2k
>> multivector. What is the interpretation in general? I just tried
>> winston and 13-limit meantone, and got the zero multival. The two have
>> commas in common, of course, and vals supporting them in common also.
>
>Can you give me the wedgies you were using?
>
>I think it's something like: if there are no commas in common, you get
>the temperament which is the union of their kernels (and the
>intersection of their row spaces, if you were to convert them into
>mapping matrices). If there are commas in common, you get the zero
>vector. For instance, try taking the interior product of 7-limit
>meantone and |-4 4 -1 0> and you'll also get the zero vector. The same
>applies here.
>
>The wedge product doesn't like if you temper out the same comma twice;
>you tend to get the zero vector.
>
>-Mike