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[from tuning] Re: webpage update: multimonzo (math terminology warning)

🔗monz <monz@attglobal.net>

8/5/2004 8:37:42 PM

hi Robert,

i decided that this discussion was gettin too math-technical
for the tuning list, so i moved it here.

--- In tuning@yahoogroups.com, "Robert Walker" <robertwalker@n...>
wrote:

> Hi Gene,
>
> > It's the monzo for 1, which is the unison, so why isn't
> > it the unison monzo?
>
> Just established convention that's all. The
> periodicity block vectors have been called unison intervals,
> so unison in this context seems to mean equivalence under
> the mapping induced by the vectors. Either that or redefine
> unison vector, or make it so unison vector and unison monzo
> use different ideas of unison. But it is a bit late to
> redefine unison vector and seems okay to me understood as
> an equivalence relation.

actually, as i just told Gene, i did just rewrite the
definition of "unison-vector", because the old one was
not entirely correct.

> > It's an element of a group which is free and finitely
> > generated is what else it is; the generators being
> > the primes in some p-limit. The coefficients of a monzo,
> > in other words, should be integers. We can extend
> > this to a vector space over Q but I don't think that
> > should be the primary meaning. This way the vals are
> > homomorphisms to Z, not linear functionals to Q, which
> > means they are composed of Q-valuations and not
> > valuations on a field of algebaic numbers. This is,
> > after all, what we want if we are going to interpret
> > temperaments as approximations to just intervals.
>
> Yes - but it is still a vector. The group is a free
> finitely generated group indeed (non techy note - free
> here just means that there are no additional equivalence
> type relationships between the elements added to reduce
> the number of elements on the group), but its elements
> are vectors. Just as the elements of a symmetry group
> are symmetries.
>
> Seems strange on first sight to extend it to
> rational powers of a prime.
> But I suppose any number that can be written in the
> form 2^a*3^b*5^c*... for a b c rational has a unique
> representation in that form (if two were identical
> then you would have some expression of that form
> equal to 1 which is easily seen to be impossible
> by multiplying out and unique factorisation),
>
> So it is consistent to do that and well formed.
> Obviously relevant for ets as e.g. 2^1/12 etc.
>
> Well it makes sense mathematically anyway.
> Though having a dense field instead of a
> discrete lattice makes it rather different
> from what one is used to. I wonder what
> it would be useful for.
>
> I see what you mean though - calling them vectors
> implies to a mathematician that they
> need to be elements of a vector space
> which is fair enough
> - and the lattice isn't a vector space
> because the scalar multiplication
> is by integers and they don't form a field.
>
> Putting that in plain english, for them
> to be vectors, it must make sense to talk
> about a vector of half the length of a lattice
> unit vector such as 2^1, 3^1 or whatever as well
> as double the length, and a third of the length.
> of any vector in the lattice and so on.
>
> However, I think that does make sense.
>
> One way to think of it is that
> the lattice is embedded in
> the pitch space of not just rational
> multiples but even of real multiples
> of the generating vectors.
>
> 2^pi * 3^e *5^rho makes
> sense as a way of defining an interva;
> - one could define an interval that
> way if one wanted to - doesn't have to be a unique
> representation of an interval.
> Just means that the map from
> the intervals to the lattice is
> now one to many - and from the
> lattice back to intervals remains
> many to one. That's okay.
>
> It seems to be the underlying
> implied vector space that the
> lattice is embedded in and the
> reason why it is mathematically
> correct to refer to them as vectors.
>
> Though the rational exponents type
> vectors will also do as a
> vector space to embed it in as well.
>
> So I see no problem calling them
> vectors and making that part of the
> definition.
>
> Robert

i've been arguing for a long time in favor
of having rationals in the monzo. (before anyone
was calling it a monzo)

i like this because it enables me to plot
"linear temperaments" *as* a line on the lattice.

-monz