A proof of Keenan's identity using p-adic numbers and Weil height

On Thu, Aug 16, 2012 at 3:49 PM, Mike Battaglia <battaglia01@gmail.com> wrote:
>
> log(max(3,1)) + log(max(1/5,1)) + log(max(5/3,1)) = log(3) + log(1) +
> log(5/3) = log(3) + log(5/3) = log(5)
>
> Right? I still don't know what a "place" is, but this seems to work out.

This is a precursor to the next part about how to define Weil norms on
monzos. There's another way to make sense of this definition for Weil
height above, which is to note that for n/d, you get

log(max(|n/d|_2, 1)) + log(max(|n/d|_3, 1)) + log(max(|n/d|_5, 1)) + …
+ log(max(|n/d|_r, 1))

where |n/d|_r is the usual Archimedean absolute value on reals. But
since max(|n/d|_p, 1) will always be 1 if the only factors of p are in
the numerator, then the above is equivalent to replacing all of the
n/d's above with 1/d's in all of the p-adic norm terms:

log(max(|1/d|_2, 1)) + log(max(|1/d|_3, 1)) + log(max(|1/d|_5, 1)) + …
+ log(max(|n/d|_r, 1))

Since |1/d|_p can never be < 1, we can now get rid of the max terms

log(|1/d|_2) + log(|1/d|_3) + log(|1/d|_5) + … + log(max(|n/d|_r, 1))

Multiply within the logs gives us

log(|1/d|_2 * |1/d|_3 * |1/d|_5 * ...) + log(max(|n/d|_r, 1))

But since each p-adic norm simply gives you the largest factor of the
denominator which is a prime power of p, then the log term on the left
is the product of all of these factors, and hence is just the product
of the prime factorization of d itself. Therefore, we're left with

log(d) + log(max(|n/d|_r, 1))

Or, if you like,

log(d) + max(log(n/d), 0)

And now it's obvious what's going on; you only replace log(d) with
log(n) if n > d. Conversely, it should be immediately obvious from the
above that the following identity will also hold:

log(n) - min(log(n/d), 0)

Now, as for Keenan's identity: it's clear at this point that both of
these expressions are equal to max(log(n,d)). Since both of these
expressions must be equal, their arithmetic mean must also be equal:

(log(d) + max(log(n/d), 0) + log(n) - min(log(n/d), 0))/2

Or, expressed more neatly

(log(n*d) + max(log(n/d), 0) - min(log(n/d), 0))/2

It's trivial to show that for any x, max(x,0) - min(x,0) = |x|_r,
therefore we're left with:

(log(n*d) + |log(n/d)|)/2

Note that, in musical terms, this is the arithmetic mean of Tenney
height and span for any rational n/d. In monzo form for a weighted
monzo |a b c ...>, this is

(|a| + |b| + |c| + ... + |a+b+c+...|)/2

This is Keenan's identity, proven for any rational number in any limit.

Note that the above is also the L1 norm of the augmented monzo

|a/2 b/2 c/2 ...; (a+b+c+...)/2>

where ; is there to represent that the coefficient following is the
augmented coefficient.

This means that the Weil height of any interval can be constructed by
mapping the interval onto a vector space that is one dimension larger
and then taking the L1 norm of the image of the monzo. This means that
the Weil norm is just the induced L1 norm on the subspace of this
larger space that corresponds to the image of interval space. This
will come in handy when we look at Weil optimal tunings.

-Mike

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

> This means that the Weil height of any interval can be constructed by
> mapping the interval onto a vector space that is one dimension larger
> and then taking the L1 norm of the image of the monzo.

So long as the interval is expressible as a fractional monzo. Weil height also covers such intervals as O.

On Fri, Aug 17, 2012 at 4:20 PM, genewardsmith <genewardsmith@sbcglobal.net>
wrote:
>
> --- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...>
> wrote:
>
> > This means that the Weil height of any interval can be constructed by
> > mapping the interval onto a vector space that is one dimension larger
> > and then taking the L1 norm of the image of the monzo.
>
> So long as the interval is expressible as a fractional monzo. Weil height
> also covers such intervals as φ.

I think that's supposed to say phi. How do you calculate the Weil
height of phi? What's the 2-adic, 3-adic, etc norm on phi?

Also, why just rational monzos? This ought to work for real monzos too, I hope.

-Mike

On Fri, Aug 17, 2012 at 4:56 PM, Mike Battaglia <battaglia01@gmail.com> wrote:
>
> I think that's supposed to say phi. How do you calculate the Weil
> height of phi? What's the 2-adic, 3-adic, etc norm on phi?

I guess that using your alternative definition of Weil height, it
turns out that phi has a Weil height of 1.

If you extend the relationship that WH(n/d) = (log(n*d) +
|log(n/d)|)/2 to algebraic numbers x, and generalize log(n*d) to a
function TH(x) on algebraic numbers, you get

WH(x) = (TH(x) + |log(x)|)/2

and hence

TH(x) = 2WH(x) - log(x)

which is a generalization of Tenney Height to algebraic numbers. So
using that identity, TH(phi) = log(4/phi). But then again, I dunno if
there's a better way to generalize Tenney Height, or if it's really
useful anyway.

-Mike

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

> I think that's supposed to say phi. How do you calculate the Weil
> height of phi? What's the 2-adic, 3-adic, etc norm on phi?

The minimal polynomial for phi is phi^2-phi-1, which is of degree two. Phi is a PV number, so the only conjugate greater than 1 in absolute value is phi. Hence, H(phi) = sqrt(phi), widely lauded and acclaimed as a regular temperament generator. Phi is a unit, with p-adic norm everywhere a unit, with absolute value 1.

> Also, why just rational monzos? This ought to work for real monzos too, I hope.

How, exactly, do real monzos uniquely represent real numbers?

On Fri, Aug 17, 2012 at 6:01 PM, Mike Battaglia <battaglia01@gmail.com> wrote:
> On Fri, Aug 17, 2012 at 4:56 PM, Mike Battaglia <battaglia01@gmail.com> wrote:
>>
>> I think that's supposed to say phi. How do you calculate the Weil
>> height of phi? What's the 2-adic, 3-adic, etc norm on phi?
>
> I guess that using your alternative definition of Weil height, it
> turns out that phi has a Weil height of 1.

No, I totally ruined this. I'm supposed to be looking at only the
zeros of the minimal polynomial with absolute value greater than 1,
and also the thing you gave was for arithmetic Weil height and not
logarithmic Weil height. Fixing that awful mess gives

WH(phi) = sqrt(phi)

and

TH(x) = 2WH(x) - log(x)

TH(phi) = 2log(sqrt(phi)) - log(phi) = log(phi) - log(phi) = 0.

So phi has a TH of 0. I fold.

-Mike

On Fri, Aug 17, 2012 at 6:12 PM, genewardsmith <genewardsmith@sbcglobal.net>
wrote:
>
> The minimal polynomial for phi is phi^2-phi-1, which is of degree two. Phi
> is a PV number, so the only conjugate greater than 1 in absolute value is
> phi. Hence, H(phi) = sqrt(phi), widely lauded and acclaimed as a regular
> temperament generator. Phi is a unit, with p-adic norm everywhere a unit,
> with absolute value 1.

Wait a sec, something here doesn't seem right. I can see how under
your alternative definition for Weil height, this works out, but this
doesn't seem to also agree with the original definition.

If the logarithmic Weil height of any number x is the sum of
log(max(|x|_v, 1)) over all absolute values v, and if phi has p-adic
norm of 1 for all p, doesn't that mean that log(max(|phi|_v, 1)) = 0
for all non-Archimedean absolute values in the summation (e.g. the
ones corresponding to p-adic norms), leaving only the ordinary
Archimedean absolute value?

So since those terms all cancel out to 0, then the only thing left is
log(max(|phi|_r, 1)), where | · |_r is the Archimedean absolute value.
So this works out to log(phi). Or, if you want the non-log version of
Weil height, that's just phi.

So why does this way give H(phi) = phi but your way give H(phi) =
sqrt(phi)? If I've made an arithmetical error somewhere above, I can't
see it.

-Mike

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

> So why does this way give H(phi) = phi but your way give H(phi) =
> sqrt(phi)? If I've made an arithmetical error somewhere above, I can't
> see it.

The relative height H_K(phi) = phi, where K = Q(phi) = Q(sqrt(5)). This is of degree 2, so the absolute Weil height is sqrt(phi).

Time to get back to all this now...

On Sat, Aug 18, 2012 at 3:34 AM, genewardsmith
<genewardsmith@sbcglobal.net> wrote:
>
> --- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:
>
> > So why does this way give H(phi) = phi but your way give H(phi) =
> > sqrt(phi)? If I've made an arithmetical error somewhere above, I can't
> > see it.
>
> The relative height H_K(phi) = phi, where K = Q(phi) = Q(sqrt(5)). This is of degree 2, so the absolute Weil height is sqrt(phi).

But isn't arithmetic Weil height the product of all the places of the
number? So if each p-adic norm on phi is 1, and the Archimedean norm
on phi is just phi, the product of everything is just phi. So why
isn't the Weil height of phi just equal to phi?

-Mike

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

> But isn't arithmetic Weil height the product of all the places of the
> number? So if each p-adic norm on phi is 1, and the Archimedean norm
> on phi is just phi, the product of everything is just phi. So why
> isn't the Weil height of phi just equal to phi?

There
s also the absolute and relative way of normalizing places.