question about rank 2 keyboard layouts

Paul E. and I have been discussing rank 2 keyboard layouts
over on facebook. If the buttons of a 2-D keyboard are given
x,y coordinates, Paul says we can only put a pitch on every
key if the determinant of the keyboard vectors representing
the period and generator is +/- 1.

Example: For pajara, suppose we use [2,-1] for the fourth
(generator) and [2, 1] for the half-octave (period). The
determinant is -4 (or 4) and it seems we can only map 1 of
every 4 buttons on the keyboard. The reasoning is, the
vectors define a parallelogram of volume 4 and can't address
points inside it. That seems to make sense.

I ran into trouble trying to think about it more abstractly.
To map every key, I think we need to cover Z2 with a linear
combination of the generators. I think that means they need
to be a basis for Z2 (or perhaps just a spanning set, which
is like a basis but without linear independence...).

Wikipedia claims that a matrix of the appropriate size (2x2
in this case) is a basis iff the determinant of its transpose
is nonzero. See here: http://bit.ly/eQpSG9

However I think they are talking about R2. On this page
http://en.wikipedia.org/wiki/Invertible_matrix#Properties
it says the transpose of a matrix is a basis over a *field*
if the determinant is nonzero. I would guess Z2 is a field.
But it adds,

"In general, a square matrix over a commutative ring is
invertible iff its determinant is a unit in that ring."

That's a bit different. Z2 must not be a field... is it
a ring with where the only unit is 1?

-Carl

--- In tuning-math@yahoogroups.com, Carl Lumma <carl@...> wrote:

> That's a bit different. Z2 must not be a field... is it
> a ring with where the only unit is 1?

The important point is that Z (the integers) are a ring, but not a field. Hence a matrix, to be invertible, must have determinant a unit, which in the case of Z means must be +-1. The integers mod 2 are a field, and hence also a ring. If the determinant of any matrix over a field is non-zero, it is invertible. But in this case, that means it is 1. The integers mod 3 are similar, in that if something is not zero,it is +-1.

>> That's a bit different. Z2 must not be a field... is it
>> a ring with where the only unit is 1?
>
>The important point is that Z (the integers) are a ring, but not a
>field. Hence a matrix, to be invertible, must have determinant a unit,
>which in the case of Z means must be +-1. The integers mod 2 are a
>field, and hence also a ring. If the determinant of any matrix over a
>field is non-zero, it is invertible. But in this case, that means it
>is 1. The integers mod 3 are similar, in that if something is not
>zero,it is +-1.

Thanks. That's as I suspected then. It looks like Q and R are
fields, which is why the standard advice is any nonzero determinant
works. Because, uh, there is always 1/n where n is nonzero.
But in Z the only identity is 1. Right?

-Carl

Carl Lumma <carl@lumma.org> wrote:
> Paul E. and I have been discussing rank 2 keyboard layouts
> over on facebook. If the buttons of a 2-D keyboard are
> given x,y coordinates, Paul says we can only put a pitch
> on every key if the determinant of the keyboard vectors
> representing the period and generator is +/- 1.

He's right!

> I ran into trouble trying to think about it more
> abstractly. To map every key, I think we need to cover Z2
> with a linear combination of the generators. I think
> that means they need to be a basis for Z2 (or perhaps
> just a spanning set, which is like a basis but without
> linear independence...).

Z2??? Should there be a superscript?

What we want is a torsion-free basis. I don't know the
mathematical term. Lattice basis?

> Wikipedia claims that a matrix of the appropriate size
> (2x2 in this case) is a basis iff the determinant of its
> transpose is nonzero. See here: http://bit.ly/eQpSG9
>
> However I think they are talking about R2. On this page
> http://en.wikipedia.org/wiki/Invertible_matrix#Properties
> it says the transpose of a matrix is a basis over a
> *field* if the determinant is nonzero. I would guess Z2
> is a field. But it adds,

Yes, they're talking about reals. With integers, what you
have is a group, or a lattice when you add the norm.

> "In general, a square matrix over a commutative ring is
> invertible iff its determinant is a unit in that ring."
>
> That's a bit different. Z2 must not be a field... is it
> a ring with where the only unit is 1?

A ring has two binary operations, corresponding to addition
and multiplication. A field allows for division as well.
So what we have is certainly not a field, but it looks
simpler than a ring. Still, theorems for rings
will likely apply to it.

Graham

Graham wrote:
>> I ran into trouble trying to think about it more
>> abstractly. To map every key, I think we need to cover Z2
>> with a linear combination of the generators. I think
>> that means they need to be a basis for Z2 (or perhaps
>> just a spanning set, which is like a basis but without
>> linear independence...).
>
>Z2??? Should there be a superscript?

Yes. Also boldface.

-Carl

--- In tuning-math@yahoogroups.com, Carl Lumma <carl@...> wrote:
>
> >> That's a bit different. Z2 must not be a field... is it
> >> a ring with where the only unit is 1?
> >
> >The important point is that Z (the integers) are a ring, but not a
> >field. Hence a matrix, to be invertible, must have determinant a
> >unit, which in the case of Z means must be +-1. The integers mod 2
> >are a field, and hence also a ring. If the determinant of any
> >matrix over a field is non-zero, it is invertible. But in this
> >case, that means it is 1. The integers mod 3 are similar, in that
> >if something is not zero,it is +-1.
>
> Thanks. That's as I suspected then. It looks like Q and R are
> fields, which is why the standard advice is any nonzero determinant
> works. Because, uh, there is always 1/n where n is nonzero.

Exactly so.

> But in Z the only identity is 1. Right?
>

No. It is 1 or -1.

I ran into a similar problem when I was experimenting with 2d layouts for 22edo, Z/22Z being also a ring, not a field. A 2d mapping for 22edo cannot, for example, cover all pitches if both generators are even, for the same reason as above.
--
Hans Straub

--- In tuning-math@yahoogroups.com, Carl Lumma <carl@...> wrote:

> But in Z the only identity is 1. Right?

In any ring you call the multiplicative identity 1. Units are invertible elements, which in this case would be +-1.

--- In tuning-math@yahoogroups.com, Graham Breed <gbreed@...> wrote:

> A ring has two binary operations, corresponding to addition
> and multiplication. A field allows for division as well.
> So what we have is certainly not a field, but it looks
> simpler than a ring. Still, theorems for rings
> will likely apply to it.

It's not anything unless you define it. Z^2 with coordinatewise operations is a ring, with additive identity [0 0] and multiplicative identity [1 1]. It has orthogonal idempotent elements [1 0] and [0 1].

>> Thanks. That's as I suspected then. It looks like Q and R are
>> fields, which is why the standard advice is any nonzero determinant
>> works. Because, uh, there is always 1/n where n is nonzero.
>
>Exactly so.

Thanks, Hans!

>It's not anything unless you define it. Z^2 with coordinatewise
>operations is a ring, with additive identity [0 0] and multiplicative
>identity [1 1]. It has orthogonal idempotent elements [1 0] and [0 1].

>In any ring you call the multiplicative identity 1. Units are
>invertible elements, which in this case would be +-1.

Wikipedia sez:
"uv = vu = 1R, where 1R is the multiplicative identity element"

In Z, 1R = 1. The units are supposed to be 1 and -1.

If it's ok that u = v, then 1 * 1 = 1 and -1 * -1 = 1.
Otherwise I don't see how to make it work.

In Z^2, 1R = [1 1]. How are the units still 1 and -1?
If the unit is an element of Z^2 it seems it must be a
pair of integers. ?

-Carl

--- In tuning-math@yahoogroups.com, Carl Lumma <carl@...> wrote:

> Wikipedia sez:
> "uv = vu = 1R, where 1R is the multiplicative identity element"
>
> In Z, 1R = 1. The units are supposed to be 1 and -1.
>
> If it's ok that u = v, then 1 * 1 = 1 and -1 * -1 = 1.
> Otherwise I don't see how to make it work.

That's fine. 1^(-1)=1, (-1)^(-1) = -1, and all are happy. By the way, normally "1R" is just called 1, so you can skip the R in most contexts.

> In Z^2, 1R = [1 1]. How are the units still 1 and -1?

[1 1]^(-1) = [1 1], [-1 -1]^(-1) = [-1 -1]. But these aren't the only units; there are also [1 -1] and [-1 1].

>> If it's ok that u = v, then 1 * 1 = 1 and -1 * -1 = 1.
>> Otherwise I don't see how to make it work.
>
>That's fine. 1^(-1)=1, (-1)^(-1) = -1, and all are happy. By the way,
>normally "1R" is just called 1, so you can skip the R in most contexts.
>
>> In Z^2, 1R = [1 1]. How are the units still 1 and -1?
>
>[1 1]^(-1) = [1 1], [-1 -1]^(-1) = [-1 -1]. But these aren't the only
>units; there are also [1 -1] and [-1 1].
>

Thanks but I'm afraid I'm clueless as usual. Looks like ^ is
exponentiation, but the negative exponents produce rationals,
which aren't allowed in Z. ?

Furthermore I thought the unit had to be a member of the ring.
So 1 and -1 can't be units of Z^2. [1 -1] and [-1 1] look like
more likely suspects.

I know 1R is often called 1. But obviously I couldn't use that
convenience above and I hope to god you're not either.

-Carl

--- In tuning-math@yahoogroups.com, Carl Lumma <carl@...> wrote:

> >> In Z^2, 1R = [1 1]. How are the units still 1 and -1?
> >
> >[1 1]^(-1) = [1 1], [-1 -1]^(-1) = [-1 -1]. But these aren't the only
> >units; there are also [1 -1] and [-1 1].
> >
>
> Thanks but I'm afraid I'm clueless as usual. Looks like ^ is
> exponentiation, but the negative exponents produce rationals,
> which aren't allowed in Z. ?

The exponent -1 is standardly used to denote the multiplicative inverse. There are worse abuses of terminology--I am thinking of its use denoting inverse functions.

> Furthermore I thought the unit had to be a member of the ring.
> So 1 and -1 can't be units of Z^2. [1 -1] and [-1 1] look like
> more likely suspects.

The four units of Z^2 are [1 1], [-1 -1], [1 -1] and [-1 1]; that was the point I was trying to make.

>The four units of Z^2 are [1 1], [-1 -1], [1 -1] and [-1 1]; that was
>the point I was trying to make.

OK! The determinant, though, produces a single number. No?

-Carl

--- In tuning-math@yahoogroups.com, Carl Lumma <carl@...> wrote:
>
> >The four units of Z^2 are [1 1], [-1 -1], [1 -1] and [-1 1]; that was
> >the point I was trying to make.
>
> OK! The determinant, though, produces a single number. No?

Yes, from a pair of elements of Z^2.

>> >The four units of Z^2 are [1 1], [-1 -1], [1 -1] and [-1 1]; that was
>> >the point I was trying to make.
>>
>> OK! The determinant, though, produces a single number. No?
>
>Yes, from a pair of elements of Z^2.

So they're telling a fib when they say,

"In general, a square matrix over a commutative ring is
invertible iff its determinant is a unit in that ring."

Some technicality is being left out...

-Carl

--- In tuning-math@yahoogroups.com, Carl Lumma <carl@...> wrote:

> >> OK! The determinant, though, produces a single number. No?
> >
> >Yes, from a pair of elements of Z^2.
>
> So they're telling a fib when they say,
>
> "In general, a square matrix over a commutative ring is
> invertible iff its determinant is a unit in that ring."

That's correct. What's the problem?

On Fri, Dec 10, 2010 at 2:21 AM, genewardsmith
<genewardsmith@sbcglobal.net> wrote:
>
> > "In general, a square matrix over a commutative ring is
> > invertible iff its determinant is a unit in that ring."
>
> That's correct. What's the problem?

I'm confused too. The units of Z^2 are [1 1], [1 -1], [-1 1], and [-1
-1]. The determinant of any square matrix is going to be a scalar, and
the units of Z^2 are all two-column matrices. How could the
determinant of a square matrix be a unit in Z^2 if the units aren't
scalars?

-Mike

Gene wrote:

>> >> OK! The determinant, though, produces a single number. No?
>> >
>> >Yes, from a pair of elements of Z^2.
>>
>> So they're telling a fib when they say,
>>
>> "In general, a square matrix over a commutative ring is
>> invertible iff its determinant is a unit in that ring."
>
>That's correct. What's the problem?
>

The determinant's not a member of the ring. -Carl

Mike wrote:

>> > "In general, a square matrix over a commutative ring is
>> > invertible iff its determinant is a unit in that ring."
>>
>> That's correct. What's the problem?
>
>I'm confused too. The units of Z^2 are [1 1], [1 -1], [-1 1], and [-1
>-1]. The determinant of any square matrix is going to be a scalar, and
>the units of Z^2 are all two-column matrices. How could the
>determinant of a square matrix be a unit in Z^2 if the units aren't
>scalars?

Yeah, that. -Carl

--- In tuning-math@yahoogroups.com, "genewardsmith" <genewardsmith@...> wrote:
>
>
> --- In tuning-math@yahoogroups.com, Graham Breed <gbreed@> wrote:
>
> > A ring has two binary operations, corresponding to addition
> > and multiplication. A field allows for division as well.
> > So what we have is certainly not a field, but it looks
> > simpler than a ring. Still, theorems for rings
> > will likely apply to it.
>
> It's not anything unless you define it. Z^2 with coordinatewise
> operations is a ring, with additive identity [0 0] and
> multiplicative identity [1 1]. It has orthogonal idempotent
> elements [1 0] and [0 1].
>

Well, yes, that is a ring - with multiplication defined as [a b] * [c d] = [a*b c*d]. But how exactly is this ring structure of Z^2 of importance for keyboard layouts? I have to say I am confused, too.
--
Hans Straub

--- In tuning-math@yahoogroups.com, Carl Lumma <carl@...> wrote:

> >> "In general, a square matrix over a commutative ring is
> >> invertible iff its determinant is a unit in that ring."
> >
> >That's correct. What's the problem?
> >
>
> The determinant's not a member of the ring. -Carl

It's a member of the ring Z, and that is the ring the matrix is over. As I said before, the important thing is that the units of Z are +-1.

--- In tuning-math@yahoogroups.com, "hstraub64" <straub@...> wrote:

> Well, yes, that is a ring - with multiplication defined as [a b] * [c d] = [a*b c*d]. But how exactly is this ring structure of Z^2 of importance for keyboard layouts? I have to say I am confused, too.

So far as I know it isn't, and the only important thing is the abelian group structure. I've been waiting for someone to tell us why they wanted to know about the ring structure of Z^2, but apparently that isn't going to happen.

Hans wrote:
> Well, yes, that is a ring - with multiplication defined as [a b] *
>[c d] = [a*b c*d]. But how exactly is this ring structure of Z^2 of
>importance for keyboard layouts? I have to say I am confused, too.

Wikipedia mentioned that the basis of a ring must have a unit
determinant.

-Carl

--- In tuning-math@yahoogroups.com, Carl Lumma <carl@...> wrote:

> Wikipedia mentioned that the basis of a ring must have a unit
> determinant.

Not sure what you mean. If R is the ring of integers of an algebraic number field (say, what you get by adding the golden ratio to Q), then it has an integral basis. An example would be the basis {1, phi} for the ring of integers of Q(phi), so that every element of the ring is a+b*phi for integers a and b. Then the square of the determinant of the matrix of the various complex embeddings of the ring of integers is a fundamental invariant of the ring (and so of the number field) called the discriminant. For example

det [1 phi] = -1/phi - phi = -sqrt(5)
[1 -1/phi]

so the discriminant is 5.

>> Wikipedia mentioned that the basis of a ring must have a unit
>> determinant.
>
>Not sure what you mean.

Basis of a vector space (keyboard) derived from the ring, I suspect.

I did provide links to my sources. See my original post:

/tuning-math/message/18421

-Carl

"genewardsmith" <genewardsmith@sbcglobal.net> wrote:

> So far as I know it isn't, and the only important thing
> is the abelian group structure. I've been waiting for
> someone to tell us why they wanted to know about the ring
> structure of Z^2, but apparently that isn't going to
> happen.

What happened is that I said we were looking at a group or
lattice, and not a ring. Then you said "It's not anything
unless you define it." and defined a ring.

Graham

Carl Lumma <carl@lumma.org> wrote:

> Wikipedia mentioned that the basis of a ring must have a
> unit determinant.

Exactly what does Wikipedia say? This may be true, but I'm
guessing you're paraphrasing wrongly. And I'm guessing
that, because you led me to misinterpret this quote:

"In general, a square matrix over a commutative ring is
invertible iff its determinant is a unit in that ring."

For our purposes, the ring here is Z, not Z^2. The
observation you started the thread with seems to be stated
here:

http://en.wikipedia.org/wiki/Lattice_%28group%29#Lattices_over_general_vector-spaces

The period/generator pair is a basis for the lattice. A
pair of steps is also a basis. If the matrix defining one
in terms of the other has a unit determinant, the lattices
are isomorphic, hence the bases are for the same lattice.

Graham

Graham wrote:

>> Wikipedia mentioned that the basis of a ring must have a
>> unit determinant.
>
>Exactly what does Wikipedia say?

I've provided links and posted a text excerpt twice.

>"In general, a square matrix over a commutative ring is
>invertible iff its determinant is a unit in that ring."

There it is again!

>For our purposes, the ring here is Z, not Z^2.

That's what Gene said.

>The observation you started the thread with seems to be stated
>here:
>
> http://bit.ly/hq75Bl
>
>The period/generator pair is a basis for the lattice. A
>pair of steps is also a basis. If the matrix defining one
>in terms of the other has a unit determinant, the lattices
>are isomorphic, hence the bases are for the same lattice.

I'm not sure what you mean.

-Carl

Carl Lumma <carl@lumma.org> wrote:
> Graham wrote:
>
> >> Wikipedia mentioned that the basis of a ring must have
> >> a unit determinant.
> >
> >Exactly what does Wikipedia say?
>
> I've provided links and posted a text excerpt twice.

I know. Please treat me with enough respect to assume I
tried to find the reference in good faith before I asked
you for it.

> >"In general, a square matrix over a commutative ring is
> >invertible iff its determinant is a unit in that ring."
>
> There it is again!
>
> >For our purposes, the ring here is Z, not Z^2.
>
> That's what Gene said.

Yes, I know. I read that as well. So I know that you know
that the quote isn't talking about the basis of a ring.
Will you answer my question and tell me where you see a
statement that does?

> >The observation you started the thread with seems to be
> >stated here:
> >
> > http://bit.ly/hq75Bl
> >
> >The period/generator pair is a basis for the lattice. A
> >pair of steps is also a basis. If the matrix defining
> >one in terms of the other has a unit determinant, the
> >lattices are isomorphic, hence the bases are for the
> >same lattice.
>
> I'm not sure what you mean.

At which point do you lose the thread? A vague, one
sentence response to a dense paragraph's explanation
doesn't give me much to go on.

Graham

Graham wrote:
>I know. Please treat me with enough respect to assume I
>tried to find the reference in good faith before I asked
>you for it.

I don't mean any disrespect but I need to communicate my
limits. I do far more than my share of providing (shortened)
links, making an effort at legible posts with a proper amount
of quoted context, and so on. This is a short thread,
everything's in the original message at the top. Always a
good place to start when in doubt.

>> >"In general, a square matrix over a commutative ring is
>> >invertible iff its determinant is a unit in that ring."
>>
>> There it is again!
>>
>> >For our purposes, the ring here is Z, not Z^2.
>>
>> That's what Gene said.
>
>Yes, I know. I read that as well. So I know that you know
>that the quote isn't talking about the basis of a ring.

Indeed. That's why it says "invertible matrix" with respect
to a ring. In a section on bases of a vector space over a
field. The key point, which you can see I'd already mostly
figured out in my original post, is that in a field the only
noninvertible element is zero while in a ring the only
invertible elements are +/- 1.

>Will you answer my question and tell me where you see a
>statement that does?
>
>> >The observation you started the thread with seems to be
>> >stated here:
>> >
>> > http://bit.ly/hq75Bl
>> >
>> >The period/generator pair is a basis for the lattice. A
>> >pair of steps is also a basis. If the matrix defining
>> >one in terms of the other has a unit determinant, the
>> >lattices are isomorphic, hence the bases are for the
>> >same lattice.
>>
>> I'm not sure what you mean.
>
>At which point do you lose the thread? A vague, one
>sentence response to a dense paragraph's explanation
>doesn't give me much to go on.

Which period/generator are you speaking of? The temperament's
or the keyboard's? Hopefully the latter because the former's
irrelevant. An example would have made it clear.

-Carl

--- In tuning-math@yahoogroups.com, Carl Lumma <carl@...> wrote:

> Indeed. That's why it says "invertible matrix" with respect
> to a ring. In a section on bases of a vector space over a
> field. The key point, which you can see I'd already mostly
> figured out in my original post, is that in a field the only
> noninvertible element is zero while in a ring the only
> invertible elements are +/- 1.

In a ring, all units are invertible, and there can be far more of these than +-1. But I don't see that any ring but Z is relevant.

Gene wrote:
>> Indeed. That's why it says "invertible matrix" with respect
>> to a ring. In a section on bases of a vector space over a
>> field. The key point, which you can see I'd already mostly
>> figured out in my original post, is that in a field the only
>> noninvertible element is zero while in a ring the only
>> invertible elements are +/- 1.
>
>In a ring, all units are invertible,

Are you referring to the additive inverses? I was referring
to the monoidic side.

>But I don't see that any ring but Z is relevant.

The relationship between a basis in Z^2, and something to do
with Z, is still not clear to me.

-Carl

--- In tuning-math@yahoogroups.com, Carl Lumma <carl@...> wrote:

> >In a ring, all units are invertible,
>
> Are you referring to the additive inverses? I was referring
> to the monoidic side.

No, "units" is in reference to multiplication.

> >But I don't see that any ring but Z is relevant.
>
> The relationship between a basis in Z^2, and something to do
> with Z, is still not clear to me.

I don't see that the ring structure is relevant; it's Z^2 as an abelian group which is significant.

Gene wrote:
>> >In a ring, all units are invertible,
>>
>> Are you referring to the additive inverses? I was referring
>> to the monoidic side.
>
>No, "units" is in reference to multiplication.

So other than +/- 1, what are some multiplicative inverses
in Z (a ring)?

>I don't see that the ring structure is relevant; it's Z^2 as an
>abelian group which is significant.

Recall that we're trying to determine whether a given pair
of vectors in the vector space associated with Z^2 is a basis.
Apparently their determinant (or the determinant of their
transpose, which for a 2x2 matrix just gives the additive
inverse of that) must be 1. How does the abelian group structure
of Z^2 help us understand this?

-Carl

On 12 December 2010 11:40, Carl Lumma <carl@lumma.org> wrote:
> Gene wrote:

>>But I don't see that any ring but Z is relevant.
>
> The relationship between a basis in Z^2, and something to do
> with Z, is still not clear to me.

What's not clear? I hope it's clear that there's a Z in Z^2. The
determinant of the matrix that transforms one basis to another must be
+/- 1 so that you don't introduce contorsion. Contorsion is a problem
because you can't add or subtract fractions of a generator. Fractions
mean division. Division of integers isn't closed because integers
generally don't have a multiplicative inverse. Hence integers are a
ring where floats are a field. If you were dealing with R^2, you
wouldn't need to worry about contorsion, and so any finite determinant
would do.

Graham

--- In tuning-math@yahoogroups.com, "genewardsmith" <genewardsmith@...> wrote:

> I don't see that the ring structure is relevant; it's Z^2 as an abelian group which is significant.

Actually, a ring structure which might be useful if we are working with hexagonal lattices is

[a1 b1] * [a2 b2] = [a1a2 - b1b2 a1b2 + b1a2 - b1b2]

On 12 December 2010 11:54, Carl Lumma <carl@lumma.org> wrote:
> Gene wrote:
>>> >In a ring, all units are invertible,
>>>
>>> Are you referring to the additive inverses?  I was referring
>>> to the monoidic side.
>>
>>No, "units" is in reference to multiplication.
>
> So other than +/- 1, what are some multiplicative inverses
> in Z (a ring)?

Gene said all units are invertible, at the top there, and he's right.
He meant multiplicative inverses -- but they still have additive
inverses, along with every element in a ring. So we understand that
units have multiplicative inverses. I don't think any other elements
in Z have multiplicative inverses in Z and I didn't notice anybody
suggest that they did.

>>I don't see that the ring structure is relevant; it's Z^2 as an
>>abelian group which is significant.
>
> Recall that we're trying to determine whether a given pair
> of vectors in the vector space associated with Z^2 is a basis.
> Apparently their determinant (or the determinant of their
> transpose, which for a 2x2 matrix just gives the additive
> inverse of that) must be 1.  How does the abelian group structure
> of Z^2 help us understand this?

You may be trying to determine something. The rest of us are happy
that it's been determined.

It isn't a vector space. It's a group. It may be a lattice or even a
ring, but all you need to know in this case is that it's a group.
That helps because you don't need to start a discussion about rings.
But group theory can get unnecessarily complicated so we could talk
about lattices, and I gave you a reference about lattices.

We're not determining if it "is a basis". We want to know if it's a
basis of the same lattice we started with. That is, if it's an
isomophism. Group theory will tell you what an isomorphism is. The
reference I gave tells you when two lattice bases are isomorphic.

Graham

combining messages here; Graham wrote:

>I don't think any other elements
>in Z have multiplicative inverses in Z and I didn't notice anybody
>suggest that they did.

Gene said "there can be far more of these than +-1". I guess
I screwed up and clipped that bit. Anyway, that's what I was
asking for an example of: units in Z besides +/- 1.

>That helps because you don't need to start a discussion about rings.

For the last time, Wikipedia said ring structure is important,
not me.

>We're not determining if it "is a basis". We want to know if it's a
>basis of the same lattice we started with. That is, if it's an
>isomophism. Group theory will tell you what an isomorphism is. The
>reference I gave tells you when two lattice bases are isomorphic.

I don't follow you all. It's like you're writing Greek or
something. What lattice did we start with? There's only one
lattice here: the keyboard, Z^2.

>What's not clear? I hope it's clear that there's a Z in Z^2.

Not really, no. None of the elements of Z are in Z^2.

>Contorsion is a problem
>because you can't add or subtract fractions of a generator. Fractions
>mean division. Division of integers isn't closed because integers
>generally don't have a multiplicative inverse. Hence integers are a
>ring where floats are a field. If you were dealing with R^2, you
>wouldn't need to worry about contorsion, and so any finite determinant
>would do.

This bit sounds like what I mentioned in my first post with
the parallelograms.

-Carl

On 12 December 2010 11:20, Carl Lumma <carl@lumma.org> wrote:

> I don't mean any disrespect but I need to communicate my
> limits.  I do far more than my share of providing (shortened)
> links, making an effort at legible posts with a proper amount
> of quoted context, and so on.  This is a short thread,
> everything's in the original message at the top.  Always a
> good place to start when in doubt.

You said you have a reference for the basis of a ring having a unit
determinant. You said it was important. Yes, I checked that first
message, but I didn't find the reference. If you want the discussion
to progress, you'll have to provide a quote with a citation.
Otherwise, you're wasting our time.

>>> >"In general, a square matrix over a commutative ring is
>>> >invertible iff its determinant is a unit in that ring."
<snip>
>>Yes, I know.  I read that as well.  So I know that you know
>>that the quote isn't talking about the basis of a ring.
>
> Indeed.  That's why it says "invertible matrix" with respect
> to a ring.  In a section on bases of a vector space over a
> field.  The key point, which you can see I'd already mostly
> figured out in my original post, is that in a field the only
> noninvertible element is zero while in a ring the only
> invertible elements are +/- 1.

Right, it's not about the basis of a ring. It's not the quote we're
looking for.

>>> >The period/generator pair is a basis for the lattice.  A
>>> >pair of steps is also a basis.  If the matrix defining
>>> >one in terms of the other has a unit determinant, the
>>> >lattices are isomorphic, hence the bases are for the
>>> >same lattice.
>>>
>>> I'm not sure what you mean.
>>
>>At which point do you lose the thread?  A vague, one
>>sentence response to a dense paragraph's explanation
>>doesn't give me much to go on.
>
> Which period/generator are you speaking of?  The temperament's
> or the keyboard's?  Hopefully the latter because the former's
> irrelevant.  An example would have made it clear.

What temperament are you proposing that has a different
period/generator to the keyboard?

You provided an example here:

"Example: For pajara, suppose we use [2,-1] for the fourth
(generator) and [2, 1] for the half-octave (period). The
determinant is -4 (or 4) and it seems we can only map 1 of
every 4 buttons on the keyboard. The reasoning is, the
vectors define a parallelogram of volume 4 and can't address
points inside it. That seems to make sense."

Has it stopped making sense?

Graham

On 12 December 2010 12:26, Carl Lumma <carl@lumma.org> wrote:
> combining messages here; Graham wrote:
>
>>I don't think any other elements
>>in Z have multiplicative inverses in Z and I didn't notice anybody
>>suggest that they did.
>
> Gene said "there can be far more of these than +-1".  I guess
> I screwed up and clipped that bit.  Anyway, that's what I was
> asking for an example of: units in Z besides +/- 1.

I think he was talking about other rings. Not Z.

>>That helps because you don't need to start a discussion about rings.
>
> For the last time, Wikipedia said ring structure is important,
> not me.

Then once again, do you have a citation?

>>We're not determining if it "is a basis".  We want to know if it's a
>>basis of the same lattice we started with.  That is, if it's an
>>isomophism.  Group theory will tell you what an isomorphism is.  The
>>reference I gave tells you when two lattice bases are isomorphic.
>
> I don't follow you all.  It's like you're writing Greek or
> something.  What lattice did we start with?  There's only one
> lattice here: the keyboard, Z^2.

One lattice with different bases. If the transformation's isomorphic,
the two lattices are the same. One basis is the period/generator
pair.

>>What's not clear?  I hope it's clear that there's a Z in Z^2.
>
> Not really, no.  None of the elements of Z are in Z^2.

No, but each element of Z^2 is a pair of elements in Z. That should
be clear. It's obvious. So whatever's tripping you up must be
something else.

>>Contorsion is a problem
>>because you can't add or subtract fractions of a generator.  Fractions
>>mean division.  Division of integers isn't closed because integers
>>generally don't have a multiplicative inverse.  Hence integers are a
>>ring where floats are a field.  If you were dealing with R^2, you
>>wouldn't need to worry about contorsion, and so any finite determinant
>>would do.
>
> This bit sounds like what I mentioned in my first post with
> the parallelograms.

Of course it is. That's what the thread's about.

Graham

Graham wrote:

>You provided an example here:
>"Example: For pajara, suppose we use [2,-1] for the fourth
>(generator) and [2, 1] for the half-octave (period). The
>determinant is -4 (or 4) and it seems we can only map 1 of
>every 4 buttons on the keyboard. The reasoning is, the
>vectors define a parallelogram of volume 4 and can't address
>points inside it. That seems to make sense."
>Has it stopped making sense?

My question was about the text on wikipedia, which I cited.
I'm not sure what else I can do.

>I think he was talking about other rings.

You're probably right, since he said "can be".

>> Which period/generator are you speaking of? The temperament's
>> or the keyboard's? Hopefully the latter because the former's
>> irrelevant. An example would have made it clear.
>
>What temperament are you proposing that has a different
>period/generator to the keyboard?

I'm not proposing anything, just attempting to understand
what you wrote. I can use any temperament with any keyboard
basis.

>One basis is the period/generator pair.

Again, are you talking about the period/generator pair over
just intonation? It almost sounds like you are.

>No, but each element of Z^2 is a pair of elements in Z. That should
>be clear. It's obvious. So whatever's tripping you up must be
>something else.

It "tripped up" Mike too. They're saying the determinant has
to be a unit but it isn't, it's a unit in a different ring.

-Carl

On 12 December 2010 12:58, Carl Lumma <carl@lumma.org> wrote:
> Graham wrote:
>
>>You provided an example here:
>>"Example: For pajara, suppose we use [2,-1] for the fourth
>>(generator) and [2, 1] for the half-octave (period).  The
>>determinant is -4 (or 4) and it seems we can only map 1 of
>>every 4 buttons on the keyboard.  The reasoning is, the
>>vectors define a parallelogram of volume 4 and can't address
>>points inside it.  That seems to make sense."
>>Has it stopped making sense?
>
> My question was about the text on wikipedia, which I cited.
> I'm not sure what else I can do.

No, your question was about the text that I cited in response to your
original question.

>>> Which period/generator are you speaking of?  The temperament's
>>> or the keyboard's?  Hopefully the latter because the former's
>>> irrelevant.  An example would have made it clear.
>>
>>What temperament are you proposing that has a different
>>period/generator to the keyboard?
>
> I'm not proposing anything, just attempting to understand
> what you wrote.  I can use any temperament with any keyboard
> basis.

I'm talking about the obvious period and generator pair that's
relevant to the question. I don't know how to explain that I'm not
talking about something else unless you tell me what that other thing
is.

Your second sentence as it stands is too vague to be meaningful. What
is "any keyboard basis"? You can't, on the face of it, use any
temperament because it has to be regular and the ranks have to agree.

>>One basis is the period/generator pair.
>
> Again, are you talking about the period/generator pair over
> just intonation?  It almost sounds like you are.

There's no "again" about just intonation. You haven't mentioned it
before in this thread. How can a generator pair be over it? I was
talking about the period/generator pair.

>>No, but each element of Z^2 is a pair of elements in Z.  That should
>>be clear.  It's obvious.  So whatever's tripping you up must be
>>something else.
>
> It "tripped up" Mike too.  They're saying the determinant has
> to be a unit but it isn't, it's a unit in a different ring.

The determinant belongs to the ring the matrix is over. As far as I'm
aware, that's a correct usage of the word "over". It's a technical
term that might trip you up. But this was explained way back. It
shouldn't be a problem now.

Graham

Graham wrote:
>> Again, are you talking about the period/generator pair over
>> just intonation? It almost sounds like you are.
>
>There's no "again" about just intonation. You haven't mentioned it
>before in this thread. How can a generator pair be over it? I was
>talking about the period/generator pair.

Nothing about the temperament itself is relevant that I can see
(other than, obviously, it's rank 2 like the thread says).
If you're claiming otherwise please demonstrate. -C.

--- In tuning-math@yahoogroups.com, "genewardsmith" <genewardsmith@...> wrote:

> Actually, a ring structure which might be useful if we are working with hexagonal lattices is
>
> [a1 b1] * [a2 b2] = [a1a2 - b1b2 a1b2 + b1a2 - b1b2]

Expanding on this, we find

[1 0] * [a b] = [a b]

so [1 0] is the identity. If we set w = [0 -1], then

w = [0 -1]
w^2 = [-1 -1]
w^3 = [-1 0]
w^4 = [0 1]
w^5 = [1 1]
w^6 = [1 0]

Hence the powers of w are invertible, and we get six units. Multiplying by one of these units rotates the lattice. In general, multiplication both rotates and expands. Rotations are a special sort of lattice isomorphism: if we multiply [a b] by w, we get [-b -a-b], which means the matrix of the transformation is [[0 -1], [-1 -1]] which has determinant -1; in general the transformation we get by a unimodular (+-1 determinant) matrix will not preserve angles and distances.