Mike Battaglia FAQ
For those of you who don't know, Mike Battaglia is a total badass when it comes to explaining complicated mathy temperament stuff in ways that make good intuitive sense and answer the question of "How could this possibly be useful?". This page is a repository for some of the explanations he's offered in the Xenharmonic Alliance Facebook Group.
Q (Spectra Ce): My first point of confusion...the idea of a using a generator to create a temperament IE what makes a temperament generator unique and/or given the temperament, how can I find the generators and how many generators there are (IE 1 for EDO, 2 for 2D, 3 for 3D...)?  I "know", for example, quarter comma meantone has 2 generators...but one is so damn close to the other (IE the octave as the second generator) I wonder what the point is.
Another...say a temperament is accurate with primes 2,3,5, and 11. Does this mean it has good 5:9:11, 9:10:11...chords and how do you deduce which of the chords are available and from what roots from the "subgroup" information (or what else is needed to deduce this)?
A (Mike Battaglia): "My first point of confusion...the idea of a using a generator to create a temperament IE what makes a temperament generator unique and/or given the temperament, how can I find the generators and how many generators there are (IE 1 for EDO, 2 for 2D, 3 for 3D...)?"
Right, that's exactly right. That's what the dimensionality means, actually. An EDO is 1D, and 1D means by definition that it has 1 generator. If you already know what the dimensionality or the "rank" of a temperament is, that tells you how many generators it has.
Calculating the actual generators is a pain in the ass to do by hand. You shouldn't have to do it. You should just use the tools that we already have to do it for you.
In order for me to tell you the easiest way to find the generators, you have to tell me what you're starting with. Is it a comma, like 250/243? Is it a page on the wiki? What does a "temperament" mean to you?
"I "know", for example, quarter comma meantone has 2 generators...but one is so damn close to the other (IE the octave as the second generator) I wonder what the point is."
What do you mean it's close, exactly? One generator is like 696 cents, and the other is about 1200 cents, so they're decently far apart...
"Another...say a temperament is accurate with primes 2,3,5, and 11. Does this mean it has good 5:9:11, 9:10:11...chords"
That's a tricky question, since there are lots of ways to screw up prime error measurements. But, all things considered, it's a decently good rule of thumb that if a temperament tunes the primes accurately, then in general, simple chords that use those primes are tuned decently accurately as well.
First off, I should note that lots of these mathematical questions involve computations that ARE NOT EASY for you to do by hand. That's why there are tools like Graham's temperament finder to do those things for you. It wasn't until last year that I learned the actual math behind most of this stuff, and I still use Graham's finder for most of the stuff I want to do  learning to do it yourself is only necessary if you actually want to learn the math or if you find it interesting or whatever.
Things like "overall tuning damage" measures are in this category. They're a pain in the ass to do by hand, and lots of the naive ways of doing them lead to a few snags. For instance, if you just look at unweighted, pureoctave, average prime error, you might end up in a situation where 5 is tuned flat and 3 is tuned sharp (like porcupine) and so 6/5 is tuned WAY sharp, which makes 10:12:15 much worse than you'd expect.
But if you go to Graham's finder and look at porcupine, it automatically works out one of the right ways to do this calculation and tells you that chords in porcupine are a bit more accurate, on average, than diminished temperament and a bit less accurate on average than augmented temperament.
But it's always true that the best way to figure out what chords sound best is in some tuning is to try some different things and see. For instance, 22EDO doesn't have perfect 9/8's, nor does it have perfect 7/4's, but its 9/7's are pretty damn close to absolutely perfect. So if harmonic accuracy is what you care about, you might really chords with lots 9/7's in them in 22EDO. And on the other hand, you might find that your ears don't give a damn about superfine harmonic accuracy after all, like some people, so then who cares about any of this, right?
Q (Spectra Ce): "What do you mean it's close, exactly? One generator is like 696 cents, and the other is about 1200 cents, so they're decently far apart..." I mean...that they intersect at the period assuming octave equivalence...that you can stack 696 cent intervals and end up extremely close to a multiple at 1200 (if I have it right).
"That's why there are tools like Graham's temperament finder" Which I still wonder how to apply. IE for simply looking up 81/80 in 7limit I get http://x31eq.com/cgibin/rt.cgi?ets=12_14c&limit=7, including a list of prime errors. I see what you mean about prime errors IE the 5 is flat and 7 is sharp, so 7/5 is especially sharp and that accuracy of chords using 7/5 probably wouldn't be the best.
However, say I want to find out where the 5:6:7 chords (or 6:7:10....) in that temperament are IE on which roots/basenotes and how many of them there are. How would I do that using Graham's temperament finder...or is it even possible?  The thing that gets me is, say, if I just go by primes and think "how many temperaments for a scale that hit the primes of 3, 5, and 7 accurately within 9 tones or less?"....I get more options than I can deal with as a composer and end up getting impatient with the theory and falling back on my ears.
The other thing is the concept of five Blackwood fifths hitting the period/"root tone" dead on IE that you can chain them and end up back at the "relaxed" root tone for a chord progression. When someone points it out it seems obvious...but actually taking a temperament and finding such relations without, say, just trying to take powers of every single interval in the scale and seeing which ones eventually "intersect the root", seems very confusing. Is there a simplified way to approach the process?
A: "What do you mean it's close, exactly? One generator is like 696 cents, and the other is about 1200 cents, so they're decently far apart..." I mean...that they intersect at the period assuming octave equivalence...that you can stack 696 cent intervals and end up extremely close to a multiple at 1200 (if I have it right)."
Right, but they end up "almost intersecting" at multiple places. They almost intersect at 12 generators, for instance. Then they intersect even more closely at 19 generators. And then they intersect even more closely at 31 generators. And then they intersect closely at 43 generators, and closely at 50 generators, and so on.
At any of these "close intersection points" you could, as you suggest above, just say "screw it, let's just make these things equal"  they're close enough. If you do so, then you'll end up with an equal temperament. But for an unequal temperament like meantone, there will be infinitely many close intersection points, depending on how close you care about.
""That's why there are tools like Graham's temperament finder" Which I still wonder how to apply. IE for simply looking up 81/80 in 7limit I get http://x31eq.com/cgibin/rt.cgi?ets=12_14c&limit=7 , including a list of prime errors. I see what you mean about prime errors IE the 5 is flat and 7 is sharp, so 7/5 is especially sharp and that accuracy of chords using 7/5 probably wouldn't be the best."
How did you get to this page? When you clicked on 81/80, did you see a list of temperaments and then click "Injera" for some reason?
The key thing here to note is something like the "fundamental theorem of tempering": if you start with any temperament, and you eliminate one comma, the temperament you get with is one dimension LESS than the one you started with. For instance, if you start with 7limit JI, that's a 4D system, because the four generating intervals are 2/1, 3/1, 5/1, and 7/1 (or 2/1, 3/2, 5/4, 7/4, etc). If you temper out one comma, like 81/80, you get to a 3D, or "rank 3" temperament.
Here's the search for 81/80 in the 7limit: http://x31eq.com/cgibin/uv.cgi?uvs=81%2F80&limit=7
Note at the bottom, under rank3, there's only one. This is the only 7limit, rank3 temperament that exists that tempers out 81/80.
See all of the "rank 2 temperaments?" Those can be thought of as CHILDREN of the above rank3 temperament, each of which tempers out one additional comma. So if you start at rank3, and you temper out one more comma, you get to rank2.
Injera, which you linked to above, doesn't just temper out 81/80, but 50/49 as well. That's the extra comma you have to add to get to Injera.
Q (Spectra Ce): ^ Right, and so here we get what sounds like an explanation of why 31EDO often is used for quarter comma meantone and, in fact, also land on 12, 19, and several other good accuracy EDOs for meantone that seem to represent a fair deal of the kind of EDOs many firsttime Xenharmonic musicians start with and find very accurate for quartercommameantonelike composition (even 12 seems good for surprisingly many, as we know).
So it isn't a stretch to say "temperament is what happens when you simplify JI and have at least one set of two primes act as one (and the different combinations/sets represent the different "children")?"
A: there's a lot of questions going on at once here. I'm going to condense them into one post:
"However, say I want to find out where the 5:6:7 chords (or 6:7:10....) in that temperament are IE on which roots/basenotes and how many of them there are."
Common misconception: a temperament is an infinite set of pitches, so there are an infinite amount of any type of chord in it. I assume you're only concerned about how many chords there are in a certain scale in that temperament, right?

"The other thing is the concept of five Blackwood fifths hitting the period/"root tone" dead on IE that you can chain them and end up back at the "relaxed" root tone for a chord progression. When someone points it out it seems obvious...but actually taking a temperament and finding such relations without, say, just trying to take powers of every single interval in the scale and seeing which ones eventually "intersect the root", seems very confusing. Is there a simplified way to approach the process?"
Are you saying that, given the comma 256/243, is there an easy way to see what things are equated?

"So it isn't a stretch to say "temperament is what happens when you simplify JI and have at least one set of two primes act as one (and the different combinations/sets represent the different "children")?""
In a very roundabout sense, you could say it means that "two primes act as one." You could also take it to mean that two composite ratios act as one. For instance, in meantone, 9/8 and 10/9 are equal and hence "act as one."
Q (Spectra Ce): "Are you saying that, given the comma 256/243, is there an easy way to see what things are equated?" More like...to see which kind of chord progressions are possible given a comma.
"I assume you're only concerned about how many chords there are in a certain scale in that temperament, right?" Right. And I figure I've made that mistake many times IE if there's a single 7 or 9 tone scale under a temperament, I tend to use that as a general measuring stick for how "good" a temperament is far as my compositional use (including larger scales under the temperament, for example).
A: "Are you saying that, given the comma 256/243, is there an easy way to see what things are equated?" More like...to see which kind of chord progressions are possible given a comma."
Well, any chord progression that would move you around by the comma will end up getting you back to the unison. You may be interested in chord progressions that move around by simple things like 3/2, 5/4, and 6/5.
Monzos can make it easy to see what sorts of basic chord movements go into a comma.
For instance, 256/243 is
8 5 0>
And 3/2 is
1 1 0>
But let's say we don't care about octave displacement at all, right? Then it doesn't matter if you move by 3/1 or 3/2. So we can just throw the 2coefficient completely away, and replace it with a * to show that we don't care about it:
* 5 0> = 256/243
* 1 0> = 3/2
And now it's pretty easy to see that going down by 5 3/2's moves you 256/243 away from where you started, counting octave equivalence. So if 256/243 is being tempered out, then it becomes equal to 1/1, right? Therefore, you can see above that going down by 5 3/2's brings you back to the tonic in Blackwood temperament.
Also, since 256/243 is tempered out, you know that 243/256 is also tempered out. So this comma also vanishes:
* 5 0> = 243/256
This also means that going up by 5 3/2's also gets you back to the tonic in Blackwood.
Of course, some chord progressions sound better than others. Which combination of 3/2's and 5/4's and 6/5's and other stuff will sound the best? I have no idea  there's no formula for that that I know of. Just figure out what you like and use it. Feel free to ask questions about different temperaments too  many of us have accumulated a database of cool things to do in them.
"I assume you're only concerned about how many chords there are in a certain scale in that temperament, right?" Right. And I figure I've made that mistake many times IE if there's a single 7 or 9 tone scale under a temperament, I tend to use that as a general measuring stick for how "good" a temperament is far as my compositional use (including larger scales under the temperament, for example)."
So what's your question then? For an arbitrary scale, the best way is to just poke around and see. If it's something like a rank2 temperament and you're using an MOS, you can automate some of it. But it's typically always easiest to just manually poke around and start practicing the scale.
(at least for me, anyway.)
If you really do only care about MOS and rank2, then you can find the "Graham complexity" of the chord you care about, and subtract that from the number of generators you're using and add 1. But before dealing with that  is that what you're actually talking about here? The scales you tend to use are often not MOS.
Q: "You may be interested in chord progressions that move around by simple things like 3/2, 5/4, and 6/5."..."And now it's pretty easy to see that going down by 5 3/2's moves you 256/243 away" That's exactly what I was talking about...I get it now, thank you. :) Igs also brought this up to me ages ago...that having the root of the chord move by estimates of simple ratios is a very important compositional feature of a scale.  I believe one really good way to explain this would be (if one doesn't already exist) a program to break a comma into a Monzo and then identify all possible fractions that satisfy primes used in the Monzo within a given limit. This way, if I have it right, someone could throw in a comma and get a list of root tones for possible "basic" chord progressions.
"But before dealing with that  is that what you're actually talking about here? The scales you tend to use are often not MOS." I usually don't. However, for the sake of ease of teaching this to people new to the art...let's assume they are MOS or rank 2. I'm really just trying to tackle the problem of "given x scale...how many relatively resolved chords (IE of similar consonance to a minor chord) are there and where can they be found
I figure given A) A list of chords good enough to be used as resolutions in a scale (and hopefully also just a full list of chords) B) A list of possible root tones for chord progression many musicians will simply be able to jump in and start composing...even if such compositions seem rather formulaic and "poplike". It would at least pop their minds open to think "hey, this isn't random alien music after all...I can do this!".
And given the number of "resolved" chords in the scale and the number of roottone progressions based on commas possible...composers can hopefully get a quick idea how much flexibility a scale has far as composition to eliminate some of the usual "too many options" issues of xenharmonic scales.
A: "But before dealing with that  is that what you're actually talking about here? The scales you tend to use are often not MOS." I usually don't. However, for the sake of ease of teaching this to people new to the art...let's assume they are MOS or rank 2. I'm really just trying to tackle the problem of "given x scale...how many relatively resolved chords (IE of similar consonance to a minor chord) are there and where can they be found."
Work out how many stacked generators you need to express the chord, which is also known as the "Graham complexity" of the chord. Then subtract that from the number of stacked generators needed to express the whole scale, and then add 1. The remainder is the number of times the chord appears in the scale.
For instance, 4:5:6 in meantone has a Graham complexity of 4, because you need to stack four generators (fifths) to get to it:
1) CG
2) GD
3) DA
4) AE
Now, say you want to find out how many 4:5:6's are in meantone[7]. Meantone[7] is actually generated by 6 stacked fifths:
1) CG
2) GD
3) DA
4) AE
5) EB
6) BF#
To find out how many 4:5:6's are in meantone[7], just do 64+1 = 3.
The above is measuring things in units of "# of stacked generators." You can also use units of "# of notes", not the number of generators. For example, meantone[7] has 7 notes:
CGDAEBF#
And you need a 5note chain to express a major chord:
CGDAE
So you can see above that there are going to be three different instances of this 5note chain in meantone[7], which will end up being:
CGDAE
....GDAEB
........DAEBF#
consequently, there are 3 major chords, which are here:
CGxxxxE
....GDxxxxB
........DAxxxxF#
If you subtract notes from notes, and add 1, you get the same thing, which is 75+1 = 3 major chords in meantone[7].
The above is just so you can conceptually see a bit of how it works (and if you're confused, just ask me to clarify more). Assuming you get the gist, you can also just use a shortcut and subtract the Graham complexity (in stacked generators) from the MOS size (in notes). Just don't confuse the two units or you'll screw yourself over.
"I figure given A) A list of chords good enough to be used as resolutions in a scale (and hopefully also just a full list of chords)"
To answer question A: I'm not sure I've heard any adequate explanation of how something like a "resolution" works. It's pretty clear to me that, as you say here, we need to figure it out. Dustin's been throwing around a lot of interesting ideas about this, especially in BP, which seems like it's a step in the right direction as far as pinning down a theory of melodic resolution is concerned.
However, there is one useful thing I and others have noticed in terms of chords that do, for whatever reason, tend to sound "resolved." I note that an important class of chords which tend to sound resolved in some meaningful sense are ROOTED chords, which are chords of the form a:b:c:d:... that have the property that a is a power of 2.
OK, wtf does that mean? It's actually simple  here are some chords with that property:
4:5:6
8:9:10:12
4:7:9:11
2:3:5:7:9:11:13
OK, you get the idea. What's the point? Psychoacoustic theories that explain what's going on here are a dime a dozen, but all I can say is to compare these chords and decide for yourself:
7:9:11
3.5:7:9:11
1.75:3.5:7:9:11
4:7:9:11
and see which one seems most consonant and stable to you. To my ears, it's the latter, with the former sounding more dissonant and "augmented" or whatever.
One possible explanation is that chords like these are set up so that you can double the lowest note in the chord down some number of octaves in the bass, and your bass note will end up aligning with the virtual fundamental of the chord. Another is that chords like these allow you to take advantage of some kind of "chroma"matching VF integration strategy like that one paper that Keenan Pepper found. A third is that magic elves do it. Which one you believe is up to you.
Also, another important question is: what about minor chords like 6:7:9 and 10:12:15, which aren't rooted but still sound resolved? And what about 8:11:13, which is rooted and can sound unresolved? And what about 1/19/75/3, tempered by 245/243, which sounds mysteriously consonant for no reason?
Yep, you're right. I have no idea. This rooted thing is just a useful rule of thumb. All I know is that a chord like 7:9:11:13 can suddenly become way more resolved if you turn it into 4:7:9:11:13. And 14:18:21, which is supposedly the most awful dissonant thing ever, suddenly becomes a benign and harmless Earth, Wind and Fire type chord if you turn it into 8:14:18;21. So use it for what you can  otherwise take it with a grain of salt and use it for what it's worth.
Now, is this possible with powers of 3 if you hear tritave equivalence? I dunno, why not? You BP people should be telling me!

"B) A list of possible root tones for chord progression many musicians will simply be able to jump in and start composing...even if such compositions seem rather formulaic and "poplike". It would at least pop their minds open to think "hey, this isn't random alien music after all...I can do this!"."
When you say "possible root tones," do you mean a list of trippy characteristic chord progressions for any temperament that sound immediately awesome? (e.g. comma pumps?) I think Petr Pařízek is the guy you want, but yeah, we need to make an autocomma pump generator ASAP. I totally agree with everything you just said, but it's a matter of time and resources.
So until then, feel free to just ask on XH if anyone knows of cool stuff to do with __ temperament and I'm sure you'll get an answer. (And, TBH, any autocomma pump generator is going to sound like crap compared to an actual composer who knows how to bring out the emotion and nuance involved by thinking outside the box a bit.)

"And given the number of "resolved" chords in the scale and the number of roottone progressions based on commas possible...composers can hopefully get a quick idea how much flexibility a scale has far as composition to eliminate some of the usual "too many options" issues of xenharmonic scales."
That's not a terrible way to rate scales, just so long as you realize that it's a pretty crappy way to write music to just stick to the notes in a single MOS series and never modulate ever. If we did that in meantone, we'd never have augmented fifths, and minor/maj7 chords would cease to exist.
Once you allow yourself to modulate more and allow the scale to be less of a prison, you end up with "too many options" again. Except now it's like, woo hoo, so many options!
There's no real formula to answer some of these questions. I mean, to a point we can answer them, but the general question of "what awesome things are there to do in a tuning" is only going to be answered by playing them more and writing guides like you said and writing actual music.
What I have is a bunch of random pieces of information in my head about random cool things you can do in different tunings, none of which fits nicely as of 2012 into any sort of magic formula that you can calculate using matrix algebra. So if you have questions about specific interesting things to do in a specific temperament, or "how to use it," why not just ask here? Keenan Pepper and Ryan Avella and everyone else should feel free to field those questions too.
Q (Tutim Dennsuul Wafiil): Thanks Mike Battaglia for your wish of simplificate the overabundant information about xenharmonic things (: (: (:
My question is it: What name definitively coin to the scale comforms by a generator of aprox. 428 ~ 436 Cents (or its inverse 764 ~ 772 Cents) in which forms the scale with form LLLsLLLLsLLLLs (being 25EDD the middle size for this scale 22212222122221)?
Because 'Witnots', or Tetradecimal Triatonic are the names that are temporary in the Xenwiki. You can help for this small tangle of concepts. 'Supertriatonic' maybe, but ¬^'
A: so you mean around on the spectrum between 14EDO and 11EDO then?
Doesn't look like it has a name. Flatter of 14EDO you get squares, sharper of 11EDO you get sensi. Right between 14EDO and 11EDO is a middle ground that seems to have no name yet.
Wanna name it? What are the ratios that you see here? I assume you think the generator is about 9/7, right?
Q (Paul Erlich): I challenge anyone to try to use Mike's recipe above to correctly count the chords in Blackwood[10] (which is an MOS according to him). Or better yet, to improve the recipe so that it allows for correct counting in such cases. The familiar diminished (octatonic) scale would be another example of where many/most would have trouble applying Mike's recipe correctly . . .
A: OK, for counting chords for fractionaloctave periods, it's a bit more complex. The formula "subtract Graham complexity from number of notes in scale" still works, but the whole "Graham complexity" thing gets a bit easier to get tripped up on.
You still need to ask yourself  how many generators does it require to generate the chord in question? So consider augmented temperament, and say we want to look at 4:5:6. Here's 0 generators, leaving just a "shell" of periods
C E G# C E G# C E G# ...
now here's one generator per period, giving us augmented[6]
C Eb E G G# B C Eb E G G# B C Eb E G G# B ...
(excuse the 12EDO meantone notation, it's just to make things simple)
You can see that one generator gets us a major chord. However, since there's 3 periods per octave, it's really that it requires three additional notes to get us the major chord, so you count it as 3.
Then if you want to know how many major chords are in augmented[9], just subtract 9  3 = 6.
You can do everything I said where you subtract generators from generators or notes from notes as well, just instead of adding 1, add the number of periods instead.
I don't have an easy way to draw a diagram of this, so maybe Andrew Heathwaite can come up with something elegant :) But if you understand the fulloctave period case well enough, you shouldn't find it too unintuitive to jump tot his.
Q (Spectra Ce): "now here's one generator per period, giving us augmented[6]
C Eb E G G# B C Eb E G G# B C Eb E G G# B ..." What's the generator? It seems like it could be anything IE C to Eb or Eb to B or...
A: period  1/3 octave, mapped as 5/4. Generator = ~300 cents, mapped as 6/5. (Or you could say it's ~300 cents, which is just 400  300)
Q (Spectra Ce): For commas...assume we're trying to find the root tones of all possible chords using the comma 81/80. The prime "basis" from factorization of 81 is 3,3,3,3 and the primes for 80 are 2,2,2 and 5. Given this, can we assume circles of 3/2, 9/5, 9/8 (and others?)...could be applied to chord progressions and what would the roots look like?
Another question...how does SkeeLo's "I Wish" use a comma other than 81/80 in its chord pattern? Yet another..what are, supposedly, 10 of the most used commas for chord progressions in modern pop music? Maybe we should start there as a building block before messing around with scale in other temperaments which temper out those same commas...
A (Keenan Pepper):The chord root goes down by a major third, up by a minor third, down by a major third, and then up by a perfect fourth. If you assume these intervals are supposed to be 6/5, 5/4, and 4/3, then it goes 1/1 8/5 48/25 192/125 128/125 and therefore returns to the same tonic only if 128/125 is tempered out.
A (Mike Battaglia): "For commas...assume we're trying to find the root tones of all possible chords using the comma 81/80. The prime "basis" from factorization of 81 is 3,3,3,3 and the primes for 80 are 2,2,2 and 5. Given this, can we assume circles of 3/2, 9/5, 9/8 (and others?)...could be applied to chord progressions and what would the roots look like?"
Sure, why not? Just "factorize" 81/80 into a series of things that you multiply to get it  anything will do.
[skip ahead to lessons leading up to wedgies and multivals]
(See Mike's Lectures On Regular Temperament Theory)
LESSON 1: VECTOR SPACES AND DUAL SPACES
First off, you already know that a "monzo" is a way to represent a JI interval. A 5limit monzo looks like a b c>, where a b c are the exponents for primes 2, 3, and 5, respectively. A 7limit JI monzo looks like a b c d>, where d represents the additional exponent for 7. The 11limit gets you another coefficient and so on.
As you may also know, a monzo can also be viewed as a VECTOR in a VECTOR SPACE. For instance, the syntonic comma is 4 4 1>. You can also think of this as a point in a space, like the point (4, 4, 1). You'd plot this point by going 4 steps on the x axis, 4 steps on the y axis, and 1 steps on the zaxis. And if you really want to think of it like a vector in the sense that some high school or college algebra courses teach it, you can also draw an arrow with a big arrowhead from the origin that connects to this point.
Paul's "A Middle Path" paper has so many good plots of this that I might as well just point anyone interested to take a look at it over there: http://sethares.engr.wisc.edu/paperspdf/ErlichMiddlePath.pdf
Now, here's the interesting part: in linear algebra, every vector space has a "dual space," which of course must be thought of as a bizarro universe for the vector space in which the background is black and the arrows and points are white. The elements in this space are called "covectors."
Covectors can "interact" with vectors, or rather "act on" them, by taking the dot product of the covector and a vector. So for instance, if your covector is (12, 19, 28)* (the star means it's in the dual space), and your vector is (4, 4, 1), then the dot product of the two is 12*4 + 19*4 * 28*1 = 0. This must of course be pictured as the black and white arrows lining up and exploding and spitting out a single number, or something.
In a drier sense, a covector can also be thought of as a type of function that takes in a vector and spits out a number. So (12, 19, 28)* can also be thought of as f(v) = 12a + 19b + 28c for some vector of the form (a, b, c).
OK, what the hell does all of that mean? END LESSON 1
LESSON 2: YOU DON'T KNOW MONZO
OK, I just talked about vectors and covectors. WTF was the point of all that?
First off, let's clean up the notation a bit. In physics, the notation commonly used is to notate covectors <like this and to notate vectors like this>. Physicists call this "braket" notation, or "Dirac" notation. So instead of writing covectors as (x, y, z)*, I'll just write <x y z from now on. And instead of writing vectors as (a, b, c), I'll just write a b c> from now on.
Technically, the application of <x y z to a b c> isn't called the dot product, for obscure mathematical reasons. It's sometimes called the "bracket product." But I've seen even Gene call it the "dot product" before, so I'm just going to informally use that usage for now because it's something everyone's familiar with (and it's basically the same exact thing).
Now then, let's say you're going to ask a question like "does 81/80 vanish in 12EDO?" But when you ask that question, what are you really asking?
Well, you're asking the question  "if I go up four 3/2's, then go down a 5/1, thus putting me at the syntonic comma  how many steps in 12EDO do I arrive at?"
Or, to represent everything in terms of primes, that's the same as saying "if I go up four 3/1's, then down a 5/1, and then down four 2/1's to reduce within the octave  how many steps in 12EDO do I arrive at?"
So, in 12EDO, 2/1 is best mapped to 12 steps, 3/1 is best mapped to 19 steps, and 5/1 is best mapped to 28 steps. Let's see what happens if we do it mechanically out by stacking and removing intervals from one another like lego pieces or something.
1) first you go down 4 octaves, at a rate of 12 steps per octave, putting you underwater at 48 steps.
2) then, you go up 4 tritaves, times 19 steps per tritave, giving you 76 steps. This lands you at a net of 48 + 76 = 28 steps.
3) finally, you go down one 5/1, times 28 steps per 5/1, putting you down 28 more steps. This lands you at a net of 28  28 steps = 0.
So, if you mechanically work out the way that you'd compute how many steps 81/80 is in 12EDO, you get 0 steps, meaning it's "equated with 1/1" and hence tempered out. No surprise there.
But, if you really think about it, you'd also find that you've just seemingly applied <12 19 28 to 4 4 1>. What you've just done is evaluate the expression 12*4 + 19*4 + 28*1: down four octaves, up four tritaves, down a 5/1. This is the same as multiplying <12 19 284 4 1>.
And, to answer Mike S's question, note also that 12*4 + 19*4 + 28*1 is also the dot product of the vectors (12, 19, 28) and (4, 4, 1). Or it's the bracket product of <12 19 284 4 1>  same basic thing.
So that's what the interpretation of the dot product is.
For some of you, this may be review, but it's meant to give a basic foundation of the mathematical reasoning underpinning some of these objects. Stay tuned for more...
LESSON 3: PREPARE FOR WEDGIE
So then  assuming you've understood my exposition thus far, now you see where all things like monzos and vals come from.
 Monzos, straightforwardly, are elements in a vector space  they're vectors  like the spaces you learned in high school algebra.
 Vals, straightforwardly, are elements in the "dual space" to that space  they're covectors  which is the new thing you just learned about and which you'd end up learning in college linear algebra.
Also, hopefully, now you see that this whole <covectorvector> thing is just a mathematical representation of the mechanical process you'd be doing if you sat down in scala for a while and stacked and subtracted intervals and figured out what you end up arriving at in some EDO. <12 19 28 is just a little machine that takes in JI intervals and spits out steps, and the mathematical name for this sort of little machine is "covector."
NOW THEN 
The concept of taking JI intervals and treating them like vectors isn't new. Theorists have been doing that for years and years. Neither Paul nor Gene nor Graham nor Joe Monzo first came up with the concept of plotting JI intervals as vectors on a JI lattice.
The real quantum leap in thought, here, is to consider the musical interpretation of the elements in the dual space  the covectors. It's is one of those things that appears to be completely meaningless from a musical standpoint unless you really think about it. Someone at some point figured out all of the above and I think it's the one of the best insights ever made in music theory. Looking at the dual space is, in a sense, the logical completion of the idea of putting JI into a "lattice," as per Tenney/Fokker/Wilson/etc.
Now, where do we go from here?
Well, if there's one thing mathematicians know a lot of random crap about, it's vectors and covectors. They've been exploring manipulations of these sorts of objects for literally thousands of years under the moniker of "linear algebra." This may be new to music theory, but it's definitely not new to math.
Some of these linearalgebraic manipulations may seem to have no musical purpose at all. Others do. Others may seem to have no purpose at all, at first, and then end up having a purpose that strikes you in a flash after a little bit of thought.
One of the manipulations that would appear to be totally useless is a deceptively simple extension of all of this called "exterior algebra." It introduces a single product, called the "wedge product." You can multiply vectors together using this the same way you'd multiply polynomials on paper in high school algebra or something. It introduces a new type of object, called a "multivector"  defined as the product of vectors. This is one of those things that seems totally useless unless you're Gene Smith, at which point you realize that multivals are higherrank temperaments, and that this random field of mathematics has a quite musical interpretation.
Phew!
LESSON 3.5: WEDGIE CROSSING
Now that you're here, there are basically two different paths you can go down, both of which are basically the same exact thing:
1) The exterior algebra/"wedgie" route, a la Gene.
2) The matrix algebra/"mapping" route, a la Graham Breed.
I think that both of these have different strengths. Gene's approach I find to be remarkably elegant in a certain kind of way, in that multivectors are very intuitive objects to think about. Graham's approach, on the other hand, isn't quite as "elegant," but may be a bit simpler to work out and immediately glean information from  mapping matrices are a bit easier to understand than wedgies.
Which one would you like me to go into detail about? I take it that you still want to know about wedgies, but I'll give you a second here to decide if you have any other questions first.
And please, if there's something I've said that's difficult for you to understand, please ask!
Till next time...
LESSON 4: NOW LET'S NOT LOSE OUR TEMPERS
Ok, you're all masters of vectors and covectors now. You all knew vectors anyway, and now you know what these mysterious things called covectors are. Now let's move on to bigger and grander things.
First, before getting into exterior algebra itself, I'll introduce a simple mathematical theorem which is basically like the First Law of Tempering. For this, all you need to know is the dimensionality of the original, pretempered JI space you're working in. For example, the 3limit is 2D, the 5limit is 3D, the 7limit is 4D, the 11limit is 5D.
Then, this is the law. PREPARE THYSELF!

FIRST LAW OF TEMPERING
Given an initial JI space of dimension d, if you temper out n commas, there is exactly one unique temperament eliminating only those commas  and it will be of rank (dn).

...OK, that seems harmless enough. Let's go over some examples of what that means:
 Say you're starting in 5limit JI, which is a 3D space. Now let's say you temper out 81/80. You started with a 3D space and tempered out one comma. The first law tells us that there is exactly one temperament eliminating 81/80 and nothing else, and that it's of rank 31=2. You know this temperament as meantone.
 Say you're in the 5limit again, and this time you're tempering out 81/80 and 128/125. There is only one temperament eliminating these commas and nothing else, and in this case it's rank 32 = 1. It just so happens that if you work it out, the rank1 temperament in question is 12EDO, specifically <12 19 28, which is the patent val for 12EDO.*
 Say you're in the 7limit and you eliminate 81/80 and nothing else. There is only one temperament eliminating only 81/80 in the 7limit, and it's of rank 41=3. If you work it out, you get something that's effectively meantone but with an extra 7/1 generator that's totally outside the circle of fifths.
 Say in the 7limit, you temper out 81/80 and 126/125. There is only one temperament eliminating these commas and nothing else, and it's rank 42=2. You know it as septimal meantone, where augmented sixths are 7/4.
 Say in the 7limit, you temper out 81/80 and 36/35. The only temperament eliminating both of these and nothing else will be of rank 42=2, and it's called dominant temperament.

OK, hopefully you get it. It's a simple idea, but it leads to some very strong implications right away.
The first important implication is that rank1 temperaments are no different from rank2 temperaments, even though rank2 temperaments tend to have fancy names involving animals and rank1 temperaments don't. That they're no different means that if you start in the 5limit or in any 3D system (like 3.5.7limit JI, for instance), and temper out 2 commas, you get a single unique rank1 temperament, just like you get a single unique rank2 temperament if you temper out only 1 comma from 5limit JI.
The second important realization is that if you care about rank2, e.g. things producing MOS's, then: it requires 1 comma to be tempered out from the 5limit to get to rank2; it requires 2 commas to be tempered from the 7limit to get to rank2; it requires 3 commas to be tempered out from the 11limit to get to rank2, etc.
The third important realization is that if you start in something like the 7limit (which is 4D) and temper out one comma, then you get to a 41= rank3 temperament. And then, if you temper out ANOTHER comma on top of that, you get to a 42=rank2 temperament. And then, if you temper out ANOTHER comma, you get to a 43=rank1 temperament.
The fourth important realization is there's no conceptual difference at all between something like the dimensionality of a 4D JI lattice and the dimensionality of a rank4 temperament  whichever you start with, if you temper out an additional comma, you lose a dimension and go down to rank3. A rank4 temperament and some 4D JI system are both just different lattices that require four generators, and are hence both "rank4." It can be very useful, in fact, to think of something like 7limit JI as itself being a special type of "temperament" which tempers out 0 commas (and, consequently, has 0.0 cents of error), and to think of temperaments as "deriving" from it. (Carl has called JI the "identity temperament" before.)
Now then, you'll note I said a lot of stuff like "if you work it out" above. The question is, HOW do you work it out? Also, you'll note that I slapped <12 19 28 above as the answer to the thing that eliminates 81/80 and 128/125. What are the actual mathematical answers to the other things? Since you've all chosen exterior algebra as your weapon, we'll be answering those questions in short order.
But first, we'll need to deal with what I call the "second law of tempering." Stay tuned...
* (You may be asking "what if I double 12EDO and get <24 38 56? Isn't that a second temperament, thus invalidating the first law?" The answer is no, because <24 38 56 is considered to be a "contorted" thing which "contains" or "implements" <12 19 28. I call the first "a contorsion" of the latter. If this confuses you for now, don't worry about it just yet  believe me, you'll be a contorsion expert by the time we're through all this.)
LESSON 4.1: DEEPER INTO THE FIRST LAW
OK, so hopefully you found the last lesson to make sense  it's just a simple idea which describes the drop in dimensionality you experience in a tuning system which occurs as you eliminate commas. And if you didn't make sense, ask questions and I shall answer!
Before I get into the second law of tempering, I'm going to point to two important things which are meant by "tempering out" a comma  things which may be obvious, but which need mentioning anyway:
1) If you temper out a comma like 81/80, then things like (81/80)^2 and (81/80)^3 and so on are also tempered out. Additionally, 80/81 and (80/81)^2 and so on are also tempered out. In general, (81/80)^n vanishes for any n, so n*4 4 1> vanishes for any n. (Or whatever commas you're using)
2) If you temper out more than one comma, say 81/80 and 25/24, then things like (81/80)*(25/24)=135/128 are also tempered out. Also, (81/80)/(25/24)=243/250 is also tempered out. Also, (81/80)^2/(25/24)=2187/2048 is also tempered out. In general (81/80)^n * (25/24)^m for any n and m are also tempered out, so n*4 4 1> * m*3 1 2> are also tempered out. (Or whatever commas you're using)
Note that in the above example, n and m can be negative too, so that things like (81/80)^10 * (25/24)^100 are also tempered out. I have no idea what that comma is, but it's tempered out if you temper both 81/80 and 25/24 out, that's for sure. Also, in fact, if you eliminate 81/80, then you can consider monstrosities like sqrt(81/80) or 2 2 0.5>, whatever that might mean, to also be tempered out. The whole shebang gets eliminated; we're not screwing around here.

So, this is the important part: consider the set of all intervals that end up actually being tempered out if you eliminate 81/80. As I said above, you eliminate 81/80, and (81/80)^2, (81/80)^3, and things like 80/81, and even things like (81/80)^0.5 as well. This corresponds to monzos 4 4 1>, 8 8 2>, 12 12 3>, 4 4 1>, and 2 2 0.5>, respectively. In general, anything of the form n*4 4 1> will vanish. Now, think about what this looks like geometrically. If you plot the set of all vanishing intervals that end up being generated by the 81/80 comma, what do you get?
Turns out that the set of all vectors of the form n*4 4 1> form a straight line, going through the origin at 0 0 0>, and passing through 4 4 1> and going on forever. Every one of the intervals above will be a point on that line. Also, note that line doesn't terminate at 0 0 0>, but continues past it in the opposite direction through to 80/81 or 4 4 1>, and continues going on in that direction forever as well. So you can imagine three axes in your head, and imagine a random line going through the origin, to picture this.
Now think about what happens if you add another comma. What if you temper out 25/24 as well? What does that look like geometrically? Well, the first step in imagining what happens is to consider what happens if you temper out 25/24 by itself. You end up with (25/24)^2 and (25/24)^3 and so on vanishing, which you can think of as a different line through the origin, this time passing through 3 1 2> and continuing on forever. So if you throw this on top of the first one, now you have two lines, representing (81/80)^n and (25/24)^n vanishing.
However, you also need to note that if (81/80)^n and (25/24)^n are both being tempered out, then commas which are combinations of the two are ALSO tempered out. So things like [(81/80)^2*(25/24)^1] vanish, [(81/80)^1*(25/24)^2] vanishes, [(81/80)^1*(25/24)^2] vanishes, and huge compound commas like [(81/80)^100*(25/24)^99] vanish too.
Therefore, what you really care about is the set of all vectors of the form (n*4 4 1> + m*3 1 2>). In case it's not immediately clear what this is going to look like, what you actually want is to take those two lines representing (81/80)^n and (25/24)^n and imagine the PLANE going through the origin and containing both of them.
That's the basic rule:
 if you temper out a single comma, you're not just tempering out a single vector or a single point, but an entire line of intervals which represent the powers of that comma.
 if you temper out two commas, you're not just tempering out two vectors or two points, but an entire plane of intervals which represent all possible ways to combine those commas.
 if you temper out three commas, you're not just tempering out three vectors, but the entire 3D space of intervals which represent all possible ways to combine those commas.
In general, however many commas you temper out, you actually end up tempering out the SUBSPACE of the JI vector space that's generated by combining those commas in every shape or form. This subspace is called the NULL SPACE of the temperament, sometimes also referred to on the wiki as the KERNEL, and it's one of the most useful and important concepts in regular temperament theory! (Something you might have realized: tempering out three commas in 5limit JI is a bad idea, because the "subspace" tempered out fills all of JI and leaves nothing to NOT be tempered out! If you were in 7limit JI though, it'd make more musical sense  but is harder to visualize.)
Phew! Wrapping up the first law next.
LESSON 4.2: WRAPPING UP THE FIRST LAW
Now then, now that you know all of that, the first law can be restated more concisely:
REAL FIRST LAW
The dimensionality of the temperament itself (its "rank"), plus the dimensionality of the temperament's null space (its "codimension" or "nullity"), equals the dimensionality of the JI system that you started with.
OR
rank(temperament) + codimension(temperament) = dimensionality(original JI)

Again, remember that the dimensionality of the temperament's null space, or its "codimension" is just the number of commas you're tempering out.
Note that this also means that if you're starting with, say, 7limit JI, which is 4dimensional, and you temper out 2 commas, then the first law still holds true  you get the equation rank(temperament) + 2 = 4, so you know that rank(temperament) = 2.
Note also that you can't cheat and try to break my law by tempering out 1/1. 1/1 has codimension 0, so you're not subtracting anything from JI by tempering it out. In fact, since to "temper something out" means to equate it with 1/1, and 1/1 is always equated with itself no matter what, then 1/1 is always "tempered out" in every temperament, and even JI itself!
(This is another reason JI can thus be thought of as a mathematical object which is like a temperament  one which "tempers out" nothing but 1/1 and which has 0 cents of error.)
If you got all of that, then congratulations: you now understand the "ranknullity theorem" of linear algebra: http://en.wikipedia.org/wiki/Ranknullity_theorem, which notsorandomly has a musical interpretation. Who'd have thought?
Anyway, this is more than enough for now. I'll continue next time with the "second law," which then explains the other side of the coin, which is how all of this applies to vals  which in a way is far more interesting. Once you understand that, we'll finally be ready to move into how these concepts can be represented by linear algebra.
Also, I realize that this was much more mathematical than the last few things I've written. Please let me know if you have any questions! It's very very important that you be able to picture all of this and understand it conceptually, because that's REALLY how temperaments work  wedgies and all of that are just a convenient notation for these concepts I'm talking about here, which are what you really need to get to know wtf wedgies mean. Once you can imagine vectors forming lines and planes in your head, it's pretty simple  please feel free to keep asking questions if you don't get something I've written.
DIVERSION 1: WE'LL NEVER GET TO WEDGIES AT THIS RATE
Now then, to answer your question: the insight here is that when you give a temperament a tuning, what you're really saying is:
 every 2/1 maps to xxxx.xxx cents
 every 3/1 maps to yyyy.yyy cents
 every 5/1 maps to zzzz.zzz cents
 etc.
Then, when you want to work out nonprime "compound intervals," like 3/2, it's easy  you know that 3/2 maps to yyyy.yyy cents  xxxx.xxx cents. And 5/4 is zzzz.zzz cents  2*xxxx.xxx cents.
So in general, if your interval is a b c>, the "tuned" version of it is given by
xxxx.xxx*a + yyyy.yyy*b + zzzz.zzz*c (think about it)
OK, now you've thought about it. Well, THAT looks familiar! Yep, it's the dot product again. Hooray! What does it all mean? It means that we've somehow ended up back at covectors and vectors again. The above expression is the same as this:
<xxxx.xxx yyyy.yyy zzzz.zzza b c>
Looks like we're back to vals and monzos again, except this time the quote unquote "val" is some weird thing that has real numbers in it.
In fact, it's probably best to not think of this sort of covector as a "val" at all. For example, here's a covector that you can arrive at by starting with 12EDO and "optimizing" the octave stretch, so that the damage to 2/1, 3/1, and 5/1 is spread out as evenly as possible. If you do this, 2/1 takes a small hit as well:
<1198.440 1897.531 2796.361
Despite the superficial mathematical resemblance, you're really NOT supposed to interpret this thing as some sort of mapping of primes in an EDO or along some chain of generators. Or well, you can, if you're into masochism, but it probably makes more sense to just think of this as a separate sort of mathematical object which also just so happens to also be a type of covector.
This sort of mathematical object which represents tunings, and is hence a covector but not really a val in the usual sense, has a name. And that name is
TUNING MAP
(sound effect: http://www.youtube.com/watch?v=eg7NXRJwYXs)
So hopefully now you see what's going on: there's more than one sort of process which can lead you back to looking at covectors:
 There's the concept of considering VALS, e.g. prime mappings along a generator chain, which as we've already thoroughly addressed leads to the realization that vals are covectors.
 And now we can see there's the concept of considering TUNINGS, which we're now seeing leads to the realization that tuning maps are also covectors!
So you can see that we now have a nice happy family, like this:
COVECTORS

+VALS

+TUNING MAPS

+?????????????? (here be dragons)
Who knows what other sorts of thought processes might lead to you realizing you've just found another use for the covectors acting on the space of intervals? I have no idea at all, but this is neat, because while we've known about interval vectors (monzos) for a long time, we're now finally starting to really crank some mileage out of this whole covector dual space thing.
You may be curious why different ideas keep bringing us back to covectors again and again and again. We'll find out next
DIVERSION 1.1: SERIOUSLY, WHAT HAPPENED TO WEDGIES
OK, why do we keep coming back to covectors? It seems like every time someone so much as thinks about some random thing in tuning theory, we end up talking about covectors again. Why might that be?
This is because covectors are actually super general objects: they're literally just functions, like f(v) which take in a vector and spit out a number. Unlike the sorts of functions that you might be used to from high school algebra, they're not just taking in numbers on the xaxis and spitting out other numbers, but taking in VECTORS and spitting out numbers. So this might make it harder to visualize, but there's no reason such mathematical machines can't exist.
There's one other thing that makes covectors important: they also have to have the magical property that, for any number of vectors a, b, c, etc, f(a+b+c+etc) has to equal f(a) + f(b) + f(c) + f(etc). Anything else is just a function but not a covector (tm).
Holy crap, what the hell does THAT mean?
It means that covectors are functions which have the property that f(3/2) = f(3/1)  f(2/1). Think about that.
In other words, you're all very familiar with various tuning situations breaking down into the realization that "well, if I just work this problem out for the primes, I can then add and subtract the results for those primes to solve the problem for compound intervals too."
This realization comes up all the time. If you want to know how intervals are tuned, assign a tuning to the primes and then combine your tuning in various ways to get the tuning for compound intervals. If you want to know how intervals are mapped, assign a mapping to the primes and then combine your mapping to get the mapping for compound intervals.
Thus, ANY time you start considering a concept in tuning theory which, after a good thinking session, you realize reduces into this situation
1) the situation involves taking in an interval and spitting out a number representing some important thing
2) you realize that your situation can be worked out for the primes and then generalized by just adding and subtracting the results for the primes in various ways
you've found that you're now back to covectors. Hooray!
And who says that vals and tuning maps are the only way to arrive back at covectors? ANY time that you come up with some bright idea which ends up boiling down to the fact that you want to take in an interval and spit out a number  and also which ends up neatly reducing to just that you need to work it out for the primes in the way I wrote above  you're back to covectors. There are definitely a million thought processes that lead you back to the same idea.
(In fact, if you're the sort of person who wants to use mathematics to solve the world's problems, then it may be useful for you to think of "covectors" as actually just being a name for a specific type of thought process, one which you've had many times before. Abstract algebra is similarly awesome in this regard, which is also why I find the attitude that math and art are in opposition to one another to be repugnant. metacognition ftw.)
End abstract algebra rant!
DIVERSION 1.2: SOME FINAL NOTES ON TUNING MAPS AND COVECTORS
Hopefully you understood the last thing I wrote about abstract algebra. If you didn't, you might now be asking questions like "given a covector, how do I know if it's a val or if it's a tuning map?"
The answer is that you can interpret any covector as either a val OR a tuning map; they are both types of interpretations you can apply to a covector. A covector is just the name of a type of mathematical object that's at the terminal junction of a certain type of thought process.
For instance, <12 19 28 definitely makes sense as a val, but you could also think of it as a tuning map like <12.000 19.000 28.000 in which 2/1 is tuned, unfortunately, to 12 cents. Why you'd ever want to consider such tuning maps is beyond me.
And <1198.440 1897.531 2796.361 makes sense as a tuning map, but if you're REALLY selfloathing you can think of it as a really f'd up val in which 2/1 maps to a fractional number of generator steps. This is only recommended if you're really just miserable with life these days.
And <1200 1900 2800 makes sense as both a tuning map AND a val.
One thing I'm considering doing, in order to differentiate between the two interpretations, is to notate tuning maps like this (this is TE meantone):
{1201.397 1898.446 2788.196
as opposed to this:
<1201.397 1898.446 2788.196
Note that I've changed the < to a {.
Keep in mind that both of these are still covectors, but I'm using the { to indicate that one ought to think of the first in the sense of being a tuning map, whereas the second ought to be thought of in the sense of being a val, in which capacity it doesn't seem to make much sense.
Either way, you can still apply dot products in the usual sense, I'm just changing the notation. So this all still works:
2/1: {1201.397 1898.446 2788.1961 0 0> = 1201.397
3/1: {1201.397 1898.446 2788.1960 1 0> = 1898.446
5/1: {1201.397 1898.446 2788.1960 0 1> = 2788.196
3/2: {1201.397 1898.446 2788.1961 1 0> = 697.049
5/4: {1201.397 1898.446 2788.1962 0 1> = 385.402
81/80: {1201.397 1898.446 2788.1964 4 1> = 0.000
Of course, the question is, given any temperament, how the hell do we come up with a characteristic tuning map? Well, obviously there's an infinite amount of tuning maps, some of which are going to be of absurdly high error. But before we can even figure out which ones are of low error, how do we know which tuning maps are "valid" for any particular temperament to begin with?
The answer is contained above: you can see that under the above tuning map 81/80 maps to 0.000 cents. So, inasmuch as tuning maps are like vals, we can see that the above tuning map "tempers out 81/80." That's good. In general, if you have a temperament that eliminates some commas, all of the "valid tuning maps" for that temperament will be the ones that send those commas to 0 cents.
Now then, how do we come up with GOOD tuning maps, e.g. ones which are low in error? Well, that's a question for another day. (The short answer is you want to find ones which are close to {1200 1901.955 2786.314).