Comma(s) from 7-limit vectors

Hi,

Could someone please show me the formula for extracting comma(s) from
7-limit interval vectors? (I know how to do the cross-product to get
the comma from two 5-limit vectors.) Is this a wedge product? Does
it always generate two commas?

Thanks

Paul

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@u...> wrote:
> Hi,
>
> Could someone please show me the formula for extracting comma(s) from
> 7-limit interval vectors? (I know how to do the cross-product to get
> the comma from two 5-limit vectors.) Is this a wedge product? Does
> it always generate two commas?

It sounds almost like you are asking how to extract commas from commas.
I can show how to extract commas from wedgies. If l is a 7-limit
wedgie, then what I call the subgroup commas are these:

2^l[4]*3^(-l[2])*5^l[1], 2^(-l[5])*3^l[3]*7^(-l[1]),

2^l[6]*5^(-l[3])*7^l[2], 3^l[6]*5^(-l[5])*7^l[4]

Each of these uses only three of the possible four primes. These
commas are not linearly independent nor are they usually the simplest
available; Hermite reducing them is one way to simply this.

Do you have something available which handles linear algebra?

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul.hjelmstad@u...> wrote:
> > Hi,
> >
> > Could someone please show me the formula for extracting comma(s)
from
> > 7-limit interval vectors? (I know how to do the cross-product to
get
> > the comma from two 5-limit vectors.) Is this a wedge product? Does
> > it always generate two commas?
>
> It sounds almost like you are asking how to extract commas from
commas.

* Isn't a 7-limit interval vector (12,19,28,34) for example?
(That's what I meant, is this a val?)

> I can show how to extract commas from wedgies. If l is a 7-limit
> wedgie, then what I call the subgroup commas are these:
>
> 2^l[4]*3^(-l[2])*5^l[1], 2^(-l[5])*3^l[3]*7^(-l[1]),
>
> 2^l[6]*5^(-l[3])*7^l[2], 3^l[6]*5^(-l[5])*7^l[4]
>
>
> Each of these uses only three of the possible four primes. These
> commas are not linearly independent nor are they usually the
simplest
> available; Hermite reducing them is one way to simply this.
>
Thanks, I understand everything here except "Hermite reduction"

> Do you have something available which handles linear algebra?

Right now, just Excel. What do you recommend (without having to
spend too much?)

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@u...> wrote:

> > Do you have something available which handles linear algebra?
>
> Right now, just Excel. What do you recommend (without having to
> spend too much?)

I recommend you tell me what Excel can do for now. Can it row reduce a
matrix, for example? Can it find a nullspace?

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul.hjelmstad@u...> wrote:
>
> > > Do you have something available which handles linear algebra?
> >
> > Right now, just Excel. What do you recommend (without having to
> > spend too much?)
>
> I recommend you tell me what Excel can do for now. Can it row
reduce a
> matrix, for example? Can it find a nullspace?

No. Just finds Determinants, Inverses and Multiplies matrices.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
>
> I can show how to extract commas from wedgies. If l is a 7-limit
> wedgie, then what I call the subgroup commas are these:
>
> 2^l[4]*3^(-l[2])*5^l[1], 2^(-l[5])*3^l[3]*7^(-l[1]),
>
> 2^l[6]*5^(-l[3])*7^l[2], 3^l[6]*5^(-l[5])*7^l[4]
>
>
> Each of these uses only three of the possible four primes. These
> commas are not linearly independent nor are they usually the
simplest
> available; Hermite reducing them is one way to simply this.
>
> Do you have something available which handles linear algebra?
>
Question about this and an earlier post:

/tuning-math/messages/6615

I take it that it doesn't matter if the wedgie is the product
of vals or monzos. Why did you choose monzos in message 6615?

Also, does the formula above work for all wedgies? At the top
of message 6615 the subgroup commas are based on different
formulas:

2^u6 3^(-u2) 5^u1
2^u5 3^u3 7^(-u1)
2^u4 5^(-u3) 7^u2
3^u4 5^u5 7^u6

Wedgie of monzos:

[r[3]*s[4] - s[3]*r[4], r[4]*s[2] - r[2]*s[4], -r[3]*s[2] + r[2]*s
[3], r[1]*s[4] - s[1]*r[4], r[3]*s[1] - r[1]*s[3], -r[2]*s[1] + r[1]*s
[2]]

Wedgie of vals:

[-r[2]*s[1] + r[1]*s[2], r[1]*s[3] - r[3]*s[1], r[1]*s[4] - s[1]*r
[4], -r[3]*s[2] + r[2]*s[3], r[2]*s[4] - r[4]*s[2], r[3]*s[4] - s[3]*r
[4]]

If I use these formulas to calculate the wedgie, which formula(s) do
I use to obtain the subgroup commas?

Thanks

Paul Hj

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@a...> wrote:
>
> --- In tuning-math@yahoogroups.com, "Gene Ward Smith"
<gwsmith@s...>
> wrote:
> >
> > I can show how to extract commas from wedgies. If l is a 7-limit
> > wedgie, then what I call the subgroup commas are these:
> >
> > 2^l[4]*3^(-l[2])*5^l[1], 2^(-l[5])*3^l[3]*7^(-l[1]),
> >
> > 2^l[6]*5^(-l[3])*7^l[2], 3^l[6]*5^(-l[5])*7^l[4]
> >
> >
> > Each of these uses only three of the possible four primes. These
> > commas are not linearly independent nor are they usually the
> simplest
> > available; Hermite reducing them is one way to simply this.
> >
> > Do you have something available which handles linear algebra?
> >
> Question about this and an earlier post:
>
> /tuning-math/messages/6615
>
> I take it that it doesn't matter if the wedgie is the product
> of vals or monzos. Why did you choose monzos in message 6615?
>
> Also, does the formula above work for all wedgies? At the top
> of message 6615 the subgroup commas are based on different
> formulas:
>
> 2^u6 3^(-u2) 5^u1
> 2^u5 3^u3 7^(-u1)
> 2^u4 5^(-u3) 7^u2
> 3^u4 5^u5 7^u6
>
>
> Wedgie of monzos:
>
> [r[3]*s[4] - s[3]*r[4], r[4]*s[2] - r[2]*s[4], -r[3]*s[2] + r[2]*s
> [3], r[1]*s[4] - s[1]*r[4], r[3]*s[1] - r[1]*s[3], -r[2]*s[1] + r[1]
*s
> [2]]
>
> Wedgie of vals:
>
> [-r[2]*s[1] + r[1]*s[2], r[1]*s[3] - r[3]*s[1], r[1]*s[4] - s[1]*r
> [4], -r[3]*s[2] + r[2]*s[3], r[2]*s[4] - r[4]*s[2], r[3]*s[4] - s[3]
*r
> [4]]
>
> If I use these formulas to calculate the wedgie, which formula(s)
do
> I use to obtain the subgroup commas?
>
> Thanks
>
> Paul Hj
>

No need to answer either question. (The first was a dumb question,
and the second I figured out by using the subgroup comma formulas
in message 8501)