Yahoo Tuning Groups Ultimate Backup tuning-math Improving on the wedge product for identifying contorted temperaments

I was talking about this to Gene on Facebook. He told me to post here about it.

Keenan and I were doing a lot of exploratory work on how to set up
some basic linear algebra stuff to handle melodic/scalar/categorical
structures. One way to go is to consider that all such "melodic"
structures define at least one interval - the 2/1, or whatever
equivalence interval you want - and as such can be considered
perversely contorted 2-limit temperaments, which happen to have
"mapping matrices" with more rows than columns. Of course, if such a
construction proves useful then these entities won't be "perverse" at
all, but rather elegant and simple extensions of what we're already
doing to consider such "mapping" matrices. But we'll see.

Either way, whether it works out or not, this and other related
approaches I've tossed around always end up coming back to the same
idea, which is that I wish we had some sort of way to formalize the
thing that Graham is calling "&" on his temperament finder as a
product which defines an "algebra" over some vector space. Whatever it
is, it is NOT the wedge product, because its behavior differs from the
wedge product where contorted temperaments are concerned: 12p&14p,
7p&24p, and 5p&19p all have +/- the same wedge product, which is <<2 8
8||, but they define uniquely contorted melodic structures. This means
that & distinguishes between contorted temperaments in a way that the
wedge product doesn't and in a way that I think might be particularly
useful.

One obvious (and inelegant) way to define & formally is to just treat
it as an operator which takes in n vectors and returns a tensor which
is the Hermite-reduced concatenation of those vectors into a matrix.
That satisfies all of the properties above. However, I can't help but
feel that there must be a more elegant solution to the problem. If two
matrices have the same Hermite normal form, it's surely an implication
of a much more basic property that they both share - in this case that
they have the same row space, as Keenan's pointed out. There's got to
be something simpler here.

Whatever "it" is, it can probably be applied to monzos too: <7 11 16]
& <12 19 28] defines a sublattice or subgroup of val space which
corresponds to meantone temperament, and likewise [-4 4 -1> & [-3 -1
2> can define a sublattice or subgroup of monzo space which
corresponds to that periodicity block we all know and love. I don't
think the nice dual properties will hold for contorted temperaments
(well, maybe), but it would be useful nonetheless.

Any ideas?

-Mike

🔗genewardsmith <genewardsmith@sbcglobal.net>

--- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:
>Of course, if such a
> construction proves useful then these entities won't be "perverse" at
> all, but rather elegant and simple extensions of what we're already
> doing to consider such "mapping" matrices. But we'll see.

This is the part I don't get, and wanted explained.

> Either way, whether it works out or not, this and other related
> approaches I've tossed around always end up coming back to the same
> idea, which is that I wish we had some sort of way to formalize the
> thing that Graham is calling "&" on his temperament finder as a
> product which defines an "algebra" over some vector space.

It should be noted that various people have been using the ampersand, I don't recall who started it. I've always taken it to mean the regular temperament you get from conjoining patent vals, or these days, including nonpatent vals with warts.

Whatever it
> is, it is NOT the wedge product, because its behavior differs from the
> wedge product where contorted temperaments are concerned: 12p&14p,
> 7p&24p, and 5p&19p all have +/- the same wedge product, which is <<2 8
> 8||, but they define uniquely contorted melodic structures.

I took it to mean you get <<2 8 8||, then reduce that to the wedgie, which is <<1 4 4||, ie meantone.

> One obvious (and inelegant) way to define & formally is to just treat
> it as an operator which takes in n vectors and returns a tensor which
> is the Hermite-reduced concatenation of those vectors into a matrix.

I don't know how tensors got into it, but you can certainly define something in this way.

> Any ideas?

One idea is that we agree on what "&" means. It's possible I started it, but if so I'm willing to let that go, but I'm worried I would have to check everything on the optimal patent val page now.

🔗Mike Battaglia <battaglia01@gmail.com>

On Dec 17, 2011, at 11:50 PM, "genewardsmith" <genewardsmith@sbcglobal.net>
wrote:

This is the part I don't get, and wanted explained.

Consider the MOS 5a7b with an equivalence interval we'll call 2/1. This is
equivalent to the statement that 5a + 7b = 2/1. Hence, 2/1 can be written
as [5 7> in this space. This can be written as the 1x2 "mapping matrix"
[<5] <7]>.

Now consider that we want to use a different interval of equivalence, and
call it 3/1. We might decide that 3/1 is reachable by 8a + 11b, so that it
can be expressed as [8 11>. If we concatenate this with the mapping for
2/1, we get the matrix [<5 8] <7 11]>. This also implies that those 8 and
11 steps can be arranged into an MOS, if you'd like, so that we can now
claim that the same generator creates a 5a7b MOS with respect to the 2/1
and an 8a11b MOS with respect to the 3/1. We'll consider negative steps as
pathological for now, but allow them to exist.

We can clearly continue this process for every prime until we've fully
defined our temperament. Hence, the following information uniquely
specifies meantone temperament:

5a7b with 2/1 as period
8a11b with 3/1 as period
12a16b with 5/1 as period

Let's say we want schismatic instead: then the 12a16b becomes 11a17b with
5/1 as period.

This is one potential way to lay things out. If the rows are interpreted as
step sizes, then each matrix implies a hexagon on the lattice, which I
-believe-, but am not sure, is the same as a hobbit in which you use
unweighted Kees expressibility as the norm instead of the Tenney-weighted
L2 norm. It may be more desirable instead to interpret matrices as
corresponding to Fokker blocks; Keenan has an algorithm for doing so and
I'll leave the specifics for him to explain.

I took it to mean you get <<2 8 8||, then reduce that to the wedgie, which
is <<1 4 4||, ie meantone.

Right. But if you WANT to explore contorted structures, they all fall flat.

I don't know how tensors got into it, but you can certainly define
something in this way.

Is this not a tensor?

> Any ideas?

One idea is that we agree on what "&" means. It's possible I started it,
but if so I'm willing to let that go, but I'm worried I would have to check
everything on the optimal patent val page now.

FWIW, Graham's temperament finder calls 7&31 Vicentino, and hence
distinguishes between contorted and non-contorted temperaments. If,
however, you want the & glyph specifically to refer to something which
filters out contorsion, I won't argue it. At that point it's already pretty
solidly defined in terms of the wedge product.

If that's the case, then the thing I am talking about is different, but
still useful. It might be nice to use the # sign instead, because it looks
like both a parallelogram and periodicity block and a sublattice, which in
some sense is what the thing I'm talking about reflects. 7p#12p can be
taken then to refer to the lattice generated by those vectors even if it
doesn't form a saturated subgroup of val space, and the same operator might
be applied to combinations of monzos as well. But how to formalize that
algebraically is where I need help.

-Mike

🔗Graham Breed <gbreed@gmail.com>

Mike Battaglia <battaglia01@gmail.com> wrote:

> FWIW, Graham's temperament finder calls 7&31 Vicentino,
> and hence distinguishes between contorted and
> non-contorted temperaments. If, however, you want the &
> glyph specifically to refer to something which filters
> out contorsion, I won't argue it. At that point it's
> already pretty solidly defined in terms of the wedge
> product.

My temperament finder uses the & operator to distinguish
contorsion because I've been using it that way before
anybody was doing anything else. Whether or not I argue
about it again I'll keep using it that way.

Originally, & was an operator for MoS classes. There, it's
clear that 7&31 is not the same as 12&31. Temperament
classes are a generalization of MoS classes from octaves to
prime intervals. I chose & because it's a Python operator
and I was using + to add EqualTemperament objects to get
another EqualTemperament object.

I was also arguing way back about not specifying ambiguous
equal temperaments by their octave division.

If we want an operator for contorsion-free combination of
equal temperaments, there's always ^. It happens that both
& and ^ have some sense of intersection. I prefer to think
of temperament classes as unions of equal temperaments, but
intersection works in terms of unison vectors.

Graham

🔗Mike Battaglia <battaglia01@gmail.com>

On Sun, Dec 18, 2011 at 5:00 AM, Graham Breed <gbreed@gmail.com> wrote:
>
> My temperament finder uses the & operator to distinguish
> contorsion because I've been using it that way before
> anybody was doing anything else. Whether or not I argue
> about it again I'll keep using it that way.

Is there some kind of background issue over the validity of contorsion
that I'm not aware of? The only person who I know who's complained
about it is Paul - but mainly because he didn't want contorted
temperaments to be called "temperaments" but rather something else.

What I have to say about contorsion is better said by this knowsur album:

http://split-notes.com/004/

This is an album in 14-EDO in which he never ever goes outside of the
"5-limit," if you want to call it that. More importantly, he never
plays a chord which utilizes notes on both "chains" of 7-EDO at the
same time. Instead he sticks to 7-EDO, and uses 14-EDO as an ambient
"chromatic scale" to have nice things like "passing notes" and other
purely melodic effects. This means that he's using 14-EDO as a
contorted 5-limit temperament (or a 2.3.11/9 temperament or however
you want to think of it) in which the second chain is quite on purpose
NOT harmonically related to the first. Good show.

On the last tune of the album, he gets creative and starts messing
around with creating what he calls a "mode shift" by modulating from
one 7-EDO chain to the other. He doesn't do so in nacho cheese fashion
by forcing a modulation by some "consonant" 7-limit interval which
connects the two chains. He just moves up by a half step, which he
apparently doesn't feel pressured to relate back to the tonic by any
sort of consonant harmonic relationship. It sounds kind of neat
(although the production is kind of whack to my spoiled western
aestheticized ears).

The only person who has a problem with any of this that I know of is
Paul. AFAIK he thinks they're great, but he just doesn't want to use
the word "temperament" for them. OK, fine. Is "wakalix" taken? Moving
on...

> Originally, & was an operator for MoS classes. There, it's
> clear that 7&31 is not the same as 12&31. Temperament
> classes are a generalization of MoS classes from octaves to
> prime intervals.

Do you mean by this the same thing I wrote in my email? As in, MOS's
are defined with respect to an equivalence interval (usually 2/1), and
that by defining how the same step pattern generates 3/1 and 5/1 and
other primes, you get a temperament.

> I was also arguing way back about not specifying ambiguous
> equal temperaments by their octave division.

What do you mean by this?

> If we want an operator for contorsion-free combination of
> equal temperaments, there's always ^. It happens that both
> & and ^ have some sense of intersection. I prefer to think
> of temperament classes as unions of equal temperaments, but
> intersection works in terms of unison vectors.

The non-contorted version of & I think would be defined perfectly as
v1&v2 = abs(v1 ^ v2)/gcd(v1^v2), where gcd returns the GCD of the
coefficients of the elements of the exterior algebra.

The contorsion-permissive version of & I don't think can be defined in
terms of the exterior product, because information is lost in the
latter that's not lost in the former. But, I think that it might be
possible to define the exterior product in terms of the & product.

I still wonder if there's a simpler way to define it though than just
that it takes two vectors as input, and returns a tensor which is the
Hermite-reduced concatenation of both. There's got to be some simple
way to do it, something which works equally well to describe
combinations of monzos as it does to describe combinations of vals,
and something which can be axiomatized nicely such that one can work
out by hand the basic operations of regular temperament theory on the
back of a napkin...

-Mike

🔗genewardsmith <genewardsmith@sbcglobal.net>

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

> The only person who has a problem with any of this that I know of is
> Paul. AFAIK he thinks they're great, but he just doesn't want to use
> the word "temperament" for them. OK, fine. Is "wakalix" taken?

I've objected for years to calling contorted scales temperaments, because they aren't a result of tempering.

> The non-contorted version of & I think would be defined perfectly as
> v1&v2 = abs(v1 ^ v2)/gcd(v1^v2), where gcd returns the GCD of the
> coefficients of the elements of the exterior algebra.

Or in other words, Wedgie(v1 ^ v2)

> I still wonder if there's a simpler way to define it though than just
> that it takes two vectors as input, and returns a tensor which is the
> Hermite-reduced concatenation of both.

I wish I knew why you are calling this a tensor and not just a matrix.

🔗genewardsmith <genewardsmith@sbcglobal.net>

🔗Mike Battaglia <battaglia01@gmail.com>

On Sun, Dec 18, 2011 at 10:17 AM, genewardsmith
<genewardsmith@sbcglobal.net> wrote:
>
> --- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:
>
> > The only person who has a problem with any of this that I know of is
> > Paul. AFAIK he thinks they're great, but he just doesn't want to use
> > the word "temperament" for them. OK, fine. Is "wakalix" taken?
>
> I've objected for years to calling contorted scales temperaments, because they aren't a result of tempering.

Fine. How about "contemperament?"

> > I still wonder if there's a simpler way to define it though than just
> > that it takes two vectors as input, and returns a tensor which is the
> > Hermite-reduced concatenation of both.
>
> I wish I knew why you are calling this a tensor and not just a matrix.

Are matrices not tensors?

-Mike

🔗Carl Lumma <carl@lumma.org>

Mike wrote:
>Is there some kind of background issue over the validity of contorsion
>that I'm not aware of? The only person who I know who's complained
>about it is Paul - but mainly because he didn't want contorted
>temperaments to be called "temperaments" but rather something else.

Neither Gene nor I want contorted things to be called
temperaments.

>What I have to say about contorsion is better said by this knowsur album:
> http://split-notes.com/004/
>This is an album in 14-EDO in which he never ever goes outside of the
>"5-limit," if you want to call it that. More importantly, he never
>plays a chord which utilizes notes on both "chains" of 7-EDO at the
>same time. Instead he sticks to 7-EDO, and uses 14-EDO as an ambient
>"chromatic scale" to have nice things like "passing notes" and other
>purely melodic effects. This means that he's using 14-EDO as a
>contorted 5-limit temperament (or a 2.3.11/9 temperament or however
>you want to think of it) in which the second chain is quite on purpose
>NOT harmonically related to the first. Good show.

Eh? I think it means he's using 7 as a non-contorted 5-limit
temperament with 14-chromaticism for passing notes and other
purely melodic effects.

-Carl

🔗Mike Battaglia <battaglia01@gmail.com>

On Dec 18, 2011, at 10:26 AM, genewardsmith <genewardsmith@sbcglobal.net>
wrote:

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

> We can clearly continue this process for every prime until we've fully
> defined our temperament. Hence, the following information uniquely
> specifies meantone temperament:
>
> 5a7b with 2/1 as period
> 8a11b with 3/1 as period
> 12a16b with 5/1 as period

You are looking at [<5 8 12|, <7 11 16|] in terms of scales, but how does
this relate to the idea that you want to define your product to be [<1 0
-4|, <0 1 4|] not [<5 8 12|, <7 11 16|]?

The notion of interpreting matrices as scales isn't as immediately related
to the specific thing I asked about here, which is why I glossed over it in
my first post. But you asked specifically about how I thought to look at
purely melodic structures as contorted temperaments, so I went into more
depth about some of my thoughts on it.

I'm trying to approach the same problem from multiple angles. One of those
angles requires finding some way to explore contorted temperaments with the
same sort of power that the exterior algebra gives us for non-contorted
ones. It's just another aspect of the same problem I keep running into.

-Mike

🔗Mike Battaglia <battaglia01@gmail.com>

On Dec 18, 2011, at 1:43 PM, Carl Lumma <carl@lumma.org> wrote:

Eh? I think it means he's using 7 as a non-contorted 5-limit
temperament with 14-chromaticism for passing notes and other
purely melodic effects.

What's the difference?

-Mike

🔗genewardsmith <genewardsmith@sbcglobal.net>

🔗Carl Lumma <carl@lumma.org>

🔗Mike Battaglia <battaglia01@gmail.com>

On Sun, Dec 18, 2011 at 3:48 PM, genewardsmith
<genewardsmith@sbcglobal.net> wrote:
>
> > > I wish I knew why you are calling this a tensor and not just a matrix.
> >
> > Are matrices not tensors?
>
> It hardly adds to clarity to use the term when "matrix", more precise and known to anyone who's done his sophomore linear algebra, will do.

I used the term tensor because I thought something that already exists
might be defined as a product of the tensors acting on a vector space.
But we can talk about matrices instead.

There seems to be some confusion over whether the & sign refers to the
thing I'm talking about or not, so I'm going to use # instead for now.
# is what Graham calls & on his website, but is not what Gene is
calling &. If at the end we want to change # back to &, then we can do
so.

The entire point of #, and the reason it diverges from ^, is that # is
NOT bilinear. For example, we don't want (2*5p) # 12p to equal 5p #
(2*12p). So since this product is not bilinear, then any vector space
endowed with this product could not be said to constitute an
"algebra." I don't know what it would be called.

Here's what I know so far, which seems to be implicit in the way we're using it:

1) # is commutative, ideally, but could be anticommutative so that we
simply look at the absolute value of the result.
2) # does -not- satisfy the axiom of bilinearity.
3) # is associative, so that (a # b) # c = a # (b # c).

#3 means that the way I proposed earlier - just defining a # b as the
matrix which is the Hermite-reduced concatenation of the two vectors -
isn't going to work. If you take two vectors, stuff them into a
matrix, and Hermite reduce it -- it is NOT the same as if you take
three vectors, stuff them into matrix, and Hermite reduce the same
thing all at once.

I'm reading about something called "matroids" now which seems to be
very relevant.

-Mike

🔗genewardsmith <genewardsmith@sbcglobal.net>

🔗Keenan Pepper <keenanpepper@gmail.com>

🔗Mike Battaglia <battaglia01@gmail.com>

On Tue, Dec 20, 2011 at 12:33 PM, Keenan Pepper <keenanpepper@gmail.com> wrote:
>
> --- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:
> > #3 means that the way I proposed earlier - just defining a # b as the
> > matrix which is the Hermite-reduced concatenation of the two vectors -
> > isn't going to work. If you take two vectors, stuff them into a
> > matrix, and Hermite reduce it -- it is NOT the same as if you take
> > three vectors, stuff them into matrix, and Hermite reduce the same
> > thing all at once.
>
> Are you sure about this? I thought that hermitereduce(stack(hermitereduce(stack(A,B)), C)) == hermitereduce(stack(A,B,C)). Can you give an example where this fails?

Sorry, you're right. I thought I had a counterexample to this last
night, but it was an error on my part. I'm still not sure if this
holds for all cases though.

But another question is, is it ideal to define # or & as an operator
that returns the Hermite reduced matrix you arrive at by concatenating
the vectors? For example, why not make it a lattice operator, where v1
# v2 is defined as the lattice spanned by v1 and v2?

Clearly the matrix represents such a lattice, but why not just define
it as the lattice itself? A lattice of vals seems like exactly what a
temperament is to me. Since every val is a linear functional on a
monzo, there will be a nice corresponding dual lattice of monzos which
returns 0 for every linear functional in the lattice acting on it.

-Mike

🔗Keenan Pepper <keenanpepper@gmail.com>

--- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:
> > Are you sure about this? I thought that hermitereduce(stack(hermitereduce(stack(A,B)), C)) == hermitereduce(stack(A,B,C)). Can you give an example where this fails?
>
> Sorry, you're right. I thought I had a counterexample to this last
> night, but it was an error on my part. I'm still not sure if this
> holds for all cases though.

It does, because just as row reduction of a matrix over a field preserves the row space (subspace spanned by the row vectors), row reduction of matrix over the integers preserves the lattice spanned by the row vectors. Also, any two matrices whose rows span the same lattice have the same HNF. Therefore an HMF matrix is a good, unique representation of a lattice.

> But another question is, is it ideal to define # or & as an operator
> that returns the Hermite reduced matrix you arrive at by concatenating
> the vectors? For example, why not make it a lattice operator, where v1
> # v2 is defined as the lattice spanned by v1 and v2?
>
> Clearly the matrix represents such a lattice, but why not just define
> it as the lattice itself? A lattice of vals seems like exactly what a
> temperament is to me.

Sounds like a fine idea.

> Since every val is a linear functional on a
> monzo, there will be a nice corresponding dual lattice of monzos which
> returns 0 for every linear functional in the lattice acting on it.

...but this part doesn't work like you're expecting it to work. It's possible for two matrices to have the same kernel (aka nullspace) even though their rows span different lattices. This is exactly the case with contorted temperaments. So if you define a possibly-contorted temperament as a lattice of vals, that makes perfect sense and the resulting objects are actually realizable as musical scales, but if you try to define it as a lattice of monzos (that represent commas), then the contorted ones make no sense. The duality is only perfect if nothing is contorted.

Allow me to explain in more detail. Consider 5-limit meantone, 7p & 12p. It has three different contorted versions of index 2: 14c & 12p (5-limit injera), 7p & 24p (5-limit maqamic, and 5p & 19p (5-limit godzilla). The lattices spanned by these four pairs of vals are all distinct (specifically the contorted ones are index-2 sublattices of the meantone lattice), and accordingly the HNF matrices are all distinct. However, all four matrices have the same kernel: the 1D lattice generated by 81/80. So it is not possible to tell these contorted temperaments apart by their lattices of commas.

What happens if you do take a sublattice of the comma lattice? Well, in this case there is only *one* index-2 contorted version, rather than three different versions. It's the lattice generated by (81/80)^2 rather than 81/80. Is there any way to actually implement this as a musical scale? I think not. For each of the three different val sublattices, you know what's being split in half: 14c&12 divides the period, 7&24 divides the 3/2, and 5&19 divides the 4/3. But in the case of the single monzo sublattice, it makes no sense. 81/80 is now a different pitch from 1/1, because 81/80 is not tempered out, only its square is. But how can 81/80 be any nontrivial interval if its square is tempered to 1/1? We're forced to dismiss monzo sublattices as nonsense.

So, in conclusion, your idea of defining possibly-contorted temperaments as lattices is fine and dandy, but they *must* be lattices of vals, not lattices of commas. If you want to embrace contorsion, it comes with the price of ruining the perfect duality.

BTW, do you know how many different contorted versions of a rank-2 temperament there are with index N? For N=2 there are always 3 versions, as above, but what is the general formula? Answer in rot13 in case you want to figure it out yourself: Vg'f gur fhz bs qvivfbef bs a.

Keenan

🔗Mike Battaglia <battaglia01@gmail.com>

On Wed, Dec 21, 2011 at 11:28 AM, Keenan Pepper <keenanpepper@gmail.com> wrote:
>
> It does, because just as row reduction of a matrix over a field preserves the row space (subspace spanned by the row vectors), row reduction of matrix over the integers preserves the lattice spanned by the row vectors. Also, any two matrices whose rows span the same lattice have the same HNF. Therefore an HMF matrix is a good, unique representation of a lattice.

Aha. Very nice.

> > Since every val is a linear functional on a
> > monzo, there will be a nice corresponding dual lattice of monzos which
> > returns 0 for every linear functional in the lattice acting on it.
>
> ...but this part doesn't work like you're expecting it to work. It's possible for two matrices to have the same kernel (aka nullspace) even though their rows span different lattices. This is exactly the case with contorted temperaments. So if you define a possibly-contorted temperament as a lattice of vals, that makes perfect sense and the resulting objects are actually realizable as musical scales, but if you try to define it as a lattice of monzos (that represent commas), then the contorted ones make no sense. The duality is only perfect if nothing is contorted.

I probably abused the word "dual" here. Does "dual" have to mean
unique? I was aware of this but thought it desirable that all of these
contorted meantone temperaments should be "dual" to the 81/80 lattice.

> What happens if you do take a sublattice of the comma lattice? Well, in this case there is only *one* index-2 contorted version, rather than three different versions. It's the lattice generated by (81/80)^2 rather than 81/80. Is there any way to actually implement this as a musical scale? I think not. For each of the three different val sublattices, you know what's being split in half: 14c&12 divides the period, 7&24 divides the 3/2, and 5&19 divides the 4/3. But in the case of the single monzo sublattice, it makes no sense. 81/80 is now a different pitch from 1/1, because 81/80 is not tempered out, only its square is. But how can 81/80 be any nontrivial interval if its square is tempered to 1/1? We're forced to dismiss monzo sublattices as nonsense.

And that's fine with me, because that's basically how things already
work in real life. Contorted temperaments (which I guess we're calling
"contemperaments" now) actually do make some sort of sense: a
contorted val does accurately represent how many steps along some
generator chain some ratio maps to. There is a sensible way to
interpret them, whether you want to treat them as "true temperaments"
or not. But I don't think anyone has any similarly sensible way to
handle torsion.

> So, in conclusion, your idea of defining possibly-contorted temperaments as lattices is fine and dandy, but they *must* be lattices of vals, not lattices of commas. If you want to embrace contorsion, it comes with the price of ruining the perfect duality.

I assume this means I was right in guessing that I messed up with the
word "duality." Oh well. But we can still create lattices of monzos
and work out what lattices exist "in the dual space" which are not
actually themselves "dual" to the original lattices for which the
original lattice is a null space. Much better!

> BTW, do you know how many different contorted versions of a rank-2 temperament there are with index N? For N=2 there are always 3 versions, as above, but what is the general formula? Answer in rot13 in case you want to figure it out yourself: Vg'f gur fhz bs qvivfbef bs a.

Must... sleep... gah.

-Mike

🔗Mike Battaglia <battaglia01@gmail.com>

On Wed, Dec 21, 2011 at 11:28 AM, Keenan Pepper <keenanpepper@gmail.com> wrote:
>
> --- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:
> > > Are you sure about this? I thought that hermitereduce(stack(hermitereduce(stack(A,B)), C)) == hermitereduce(stack(A,B,C)). Can you give an example where this fails?
> >
> > Sorry, you're right. I thought I had a counterexample to this last
> > night, but it was an error on my part. I'm still not sure if this
> > holds for all cases though.
>
> It does, because just as row reduction of a matrix over a field preserves the row space (subspace spanned by the row vectors), row reduction of matrix over the integers preserves the lattice spanned by the row vectors. Also, any two matrices whose rows span the same lattice have the same HNF. Therefore an HMF matrix is a good, unique representation of a lattice.

Coming back to this again off of a new wave of linear algebra studies:
is this correct? Isn't the "row space" supposed to be a continuous
thing, so that the row space of the matrix [1 0 0] also includes the
row vector [0.5 0 0]? The row space is supposed to be the space of
"all linear combinations" of the rows, and my best interpretation of
what I've seen on this is that this also includes what you get if you
multiply each row by a real number and not just an integer.

If this is true, then all of the different contorted versions of
meantone would have the same row space, although they'd have different
HNF's. So row spaces would then represent "true" temperaments, and
something like a wedgie would then be the unique representation of it.
And also, it would mean that two matrices which have the same HNF
share some different property than having the same row space. In
group-theoretic terms I guess it would mean that their rows bases for
the same subgroup, or perhaps that the free abelian groups defined
with their rows as bases are "automorphic" to one another. But I feel
that there has to be some far more simple version of this property...

-Mike

🔗Graham Breed <gbreed@gmail.com>

🔗Keenan Pepper <keenanpepper@gmail.com>

--- In tuning-math@yahoogroups.com, Graham Breed <gbreed@...> wrote:
>
> Mike Battaglia <battaglia01@...> wrote:
>
> > Coming back to this again off of a new wave of linear
> > algebra studies: is this correct? Isn't the "row space"
> > supposed to be a continuous thing, so that the row space
> > of the matrix [1 0 0] also includes the row vector [0.5 0
> > 0]? The row space is supposed to be the space of "all
> > linear combinations" of the rows, and my best
> > interpretation of what I've seen on this is that this
> > also includes what you get if you multiply each row by a
> > real number and not just an integer.
>
> A space is continuous but a lattice isn't. If you apply
> the definition to a lattice, you'll see it makes sense of
> (and distinguishes) contorted temperaments.

Exactly. This is why I intentionally avoided using the phrase "row space" and instead said "lattice spanned by the row vectors".

Keenan