Mike's lecture on vector spaces and dual spaces
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=LECTURE 1: Vector Spaces and Dual Spaces=
<span style="display: block; text-align: center;"><span class="MathJax"><span class="math"><span style="clip: rect(1.72em 1000em 2.742em -0.558em); display: inline-block; font-size: 120%; height: 0px; left: 0em; position: absolute; top: -2.538em; width: 1.731em;"><span class="mrow"><span class="mi" style="font-family: MathJax_Math;">//test//</span></span></span></span></span></span>
If you haven't seen monzos or vals before and are totally confused, please read the pages on [[xenharmonic/Monzos|Monzos]] and [[xenharmonic/Vals|Vals]] first!
If you have seen it, then to review, a **monzo** is a way to represent a JI interval that shows how it decomposes into a combination of simpler, "prime" intervals. It does so by directly representing an interval's prime factorization. A 5-limit monzo looks like \(\ket{a \s b \s c}\), where \(a\), \(b\), and \(c\) are the exponents for primes 2, 3, and 5, respectively. A 7-limit JI monzo looks like \(\ket{a \s b \s c \s d}\), where \(d\) represents the additional exponent for 7. The 11-limit gets you another coefficient and so on.
On the other hand, a **val** is a way to represent how JI intervals map to tempered steps along a chain of generator. A val does this by specifying the mapping for the primes, and in so doing ends up specifying the mapping for every JI interval as well: since every interval is a combination of primes, then we can find the mapping for any interval in some val by simply adding and subtracting the mapping for the primes in such a way that the original interval is recreated. A 5-limit val looks like \(\bra{x \s y \s z}\), where \(x\), \(y\), and \(z\) are the number of steps along the chain that primes 2, 3, and 5 map to, respectively. A 7-limit val looks like \(\bra{a \s b \s c \s d}\), where \(d\) represents the additional mapping for 7. Like with monzos, going to the 11-limit gets you another coefficient and so on.
Again, if this is confusing, please go back to the pages on [[xenharmonic/Monzos|Monzos]] and [[xenharmonic/Vals|Vals]] and read those first!
Assuming you understand that, then we've reached our first new idea, which will help us gain a geometric intuition into what some of these abstract entities mean. That idea, which will enable us to rediscover vals and monzos in a much stronger mathematical and geometric context, is this:
==1.1: A monzo can be viewed as a **VECTOR** in a **VECTOR SPACE**.==
For instance, the syntonic comma is \(\ket{\-4 \s 4 \s \-1}\). A geometric interpretation of this interval might be 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 z-axis. 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. Here's a widget that lets you plot vectors:
[[media type="custom" key="15537326"]]
Keep in mind that Wolfram Alpha is very fragile, so if you try to do anything fancy, it's going to break. But, 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/Erlich-MiddlePath.pdf]]
Now, 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." I can't get the exact colors I mentioned here, but I've cheated a bit to get Wolfram to change the colors, so you can plot covectors here:
[[media type="custom" key="15537360"]]
You can also plot more than one vector or covector by putting in a list of vectors separated by commas, something like \((12,19,28),\s(7,11,16)\). However, this will break the nice color properties I set up above. Also, if you put in too many entries, Wolfram Alpha has been known to break.
**So then, what's the point?**
This is all well and good by itself, but it doesn't mean anything unless you understand how covectors interact with vectors. Covectors are mathematical objects that are thought to //act on// vectors. When a covector "acts on" a vector, the interaction occurs by you taking the **dot product** of the two vectors.
======For example: say 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<span style="font-size: 80%; vertical-align: super;">[[Mike's Lecture on Vector Spaces and Dual Spaces#ref1|{1}]]</span> of the two is \(12 \cdot \-4 + 19 \cdot 4 + 28 \cdot \-1 = 0\). Thus, the result of \((12,19,28)^*\) acting on \((\-4,4,\-1)\) is \(0\).======
The action of a covector on a vector must, of course, be pictured as the different colored arrows lining up and exploding and spitting out a single number, or something. Wolfram unfortunately doesn't let me do nice explosion effects, so you'll have to imagine it.
Alternatively, if you're used to dot products, then you'll also be able to understand this operation in terms of the projection of one vector on another. However, if you choose to visualize it this way, you must understand that these two things you're projecting onto one another lie in __different__ spaces - one is a vector lying in a vector space, and the other is a covector lying in this new "dual space" you're learning about.
Thus, covectors are little mathematical machines - they take in vectors, do some simple dot product-ish stuff, and output a scalar. Simple! We'll see that they turn up again and again and again in tuning theory, too.
OK, so how do we use these things?
==1.2: Covectors mean stuff. (OR: YOU DON'T KNOW MONZO)==
One interesting way to think of covectors, since they're these dual vectors that "act on" normal vectors, is thus as functions - they take in a vector as input, multiply each coefficient of the vector by the corresponding coefficient of the covector, sum them up, and spit out a number. In other words, you know that the action of the covector \((12,19,28)^*\) on any arbitrary vector \((a,b,c)\) is going to be \(12a + 19b + 28c\). So, you can think of \((12,19,28)^*\) itself as a function looking something like \(f(\v{v}) = 12a + 19b + 28c\) for some vector of the form (a, b, c). Note that the formatting on \(\v{v}\) specifies that \(\v{v}\) is a vector that's being taken in as input.
Before we go on, however, let's clean up the notation a bit. In physics, the notation commonly used is to notate covectors \(\bratext{like this}\) and to notate vectors \(\kettext{like this}\). Physicists call this "bra-ket" notation, or sometimes "Dirac" notation. So instead of writing covectors as \((x,y,z)^*\), I'll just write \(\bra{x \s y \s z}\) from now on. And instead of writing vectors as \((a,b,c)), I'll just write \(\ket{|a \s b \s c>}\) 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 12-EDO?" 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 12-EDO 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 12-EDO do I arrive at?"
So, in 12-EDO, 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 12-EDO, 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 28|-4 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 28|-4 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...
[[#ref1]][1] - Note that some have raised technical concerns about this operation being called the "dot product," insisting that the dot product is something that's only done between two vectors, or two covectors, but never between one covector and one vector. Another term that's sometimes been used for this product in the "**bracket product**", for reasons we don't need to get into here. However, confusingly, the term bracket product has also been used for the ordinary dot product, and it's also very common to hear people call the thing I'm calling the dot product above. It's best at this point to just know that the two terms are out there. I'm going to continue calling it the dot product since its' something more people are familiar with.
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<!-- ws:start:WikiTextHeadingRule:6:<h1> --><h1 id="toc1"><a name="LECTURE 1: Vector Spaces and Dual Spaces"></a><!-- ws:end:WikiTextHeadingRule:6 -->LECTURE 1: Vector Spaces and Dual Spaces</h1>
<span style="display: block; text-align: center;"><span class="MathJax"><span class="math"><span style="clip: rect(1.72em 1000em 2.742em -0.558em); display: inline-block; font-size: 120%; height: 0px; left: 0em; position: absolute; top: -2.538em; width: 1.731em;"><span class="mrow"><span style="font-family: MathJax_Math;" class="mi"><em>test</em></span></span></span></span></span></span><br />
If you haven't seen monzos or vals before and are totally confused, please read the pages on <a class="wiki_link" href="http://xenharmonic.wikispaces.com/Monzos">Monzos</a> and <a class="wiki_link" href="http://xenharmonic.wikispaces.com/Vals">Vals</a> first!<br />
<br />
If you have seen it, then to review, a <strong>monzo</strong> is a way to represent a JI interval that shows how it decomposes into a combination of simpler, "prime" intervals. It does so by directly representing an interval's prime factorization. A 5-limit monzo looks like \(\ket{a \s b \s c}\), where \(a\), \(b\), and \(c\) are the exponents for primes 2, 3, and 5, respectively. A 7-limit JI monzo looks like \(\ket{a \s b \s c \s d}\), where \(d\) represents the additional exponent for 7. The 11-limit gets you another coefficient and so on.<br />
<br />
On the other hand, a <strong>val</strong> is a way to represent how JI intervals map to tempered steps along a chain of generator. A val does this by specifying the mapping for the primes, and in so doing ends up specifying the mapping for every JI interval as well: since every interval is a combination of primes, then we can find the mapping for any interval in some val by simply adding and subtracting the mapping for the primes in such a way that the original interval is recreated. A 5-limit val looks like \(\bra{x \s y \s z}\), where \(x\), \(y\), and \(z\) are the number of steps along the chain that primes 2, 3, and 5 map to, respectively. A 7-limit val looks like \(\bra{a \s b \s c \s d}\), where \(d\) represents the additional mapping for 7. Like with monzos, going to the 11-limit gets you another coefficient and so on.<br />
<br />
Again, if this is confusing, please go back to the pages on <a class="wiki_link" href="http://xenharmonic.wikispaces.com/Monzos">Monzos</a> and <a class="wiki_link" href="http://xenharmonic.wikispaces.com/Vals">Vals</a> and read those first!<br />
<br />
Assuming you understand that, then we've reached our first new idea, which will help us gain a geometric intuition into what some of these abstract entities mean. That idea, which will enable us to rediscover vals and monzos in a much stronger mathematical and geometric context, is this:<br />
<br />
<!-- ws:start:WikiTextHeadingRule:8:<h2> --><h2 id="toc2"><a name="LECTURE 1: Vector Spaces and Dual Spaces-1.1: A monzo can be viewed as a VECTOR** in a **VECTOR SPACE."></a><!-- ws:end:WikiTextHeadingRule:8 -->1.1: A monzo can be viewed as a <strong>VECTOR</strong> in a <strong>VECTOR SPACE</strong>.</h2>
<br />
For instance, the syntonic comma is \(\ket{\-4 \s 4 \s \-1}\). A geometric interpretation of this interval might be 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 z-axis. 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. Here's a widget that lets you plot vectors:<br />
<br />
<!-- ws:start:WikiTextMediaRule:1:<img src="http://www.wikispaces.com/site/embedthumbnail/custom/15537326?h=0&w=0" class="WikiMedia WikiMediaCustom" id="wikitext@@media@@type=&quot;custom&quot; key=&quot;15537326&quot;" title="Custom Media"/> --><script type="text/javascript" id="WolframAlphaScriptf5af8de6802460753a75a4692d255641" src="http://www.wolframalpha.com/widget/widget.jsp?id=f5af8de6802460753a75a4692d255641&output=lightbox">
</script><!-- ws:end:WikiTextMediaRule:1 --><br />
<br />
Keep in mind that Wolfram Alpha is very fragile, so if you try to do anything fancy, it's going to break. But, 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: <a class="wiki_link_ext" href="http://sethares.engr.wisc.edu/paperspdf/Erlich-MiddlePath.pdf" rel="nofollow">http://sethares.engr.wisc.edu/paperspdf/Erlich-MiddlePath.pdf</a><br />
<br />
Now, 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." I can't get the exact colors I mentioned here, but I've cheated a bit to get Wolfram to change the colors, so you can plot covectors here:<br />
<br />
<!-- ws:start:WikiTextMediaRule:2:<img src="http://www.wikispaces.com/site/embedthumbnail/custom/15537360?h=0&w=0" class="WikiMedia WikiMediaCustom" id="wikitext@@media@@type=&quot;custom&quot; key=&quot;15537360&quot;" title="Custom Media"/> --><script type="text/javascript" id="WolframAlphaScriptca79995ff5942e9f187c05cd2fce394b" src="http://www.wolframalpha.com/widget/widget.jsp?id=ca79995ff5942e9f187c05cd2fce394b">
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<br />
You can also plot more than one vector or covector by putting in a list of vectors separated by commas, something like \((12,19,28),\s(7,11,16)\). However, this will break the nice color properties I set up above. Also, if you put in too many entries, Wolfram Alpha has been known to break.<br />
<br />
<strong>So then, what's the point?</strong><br />
<br />
This is all well and good by itself, but it doesn't mean anything unless you understand how covectors interact with vectors. Covectors are mathematical objects that are thought to <em>act on</em> vectors. When a covector "acts on" a vector, the interaction occurs by you taking the <strong>dot product</strong> of the two vectors.<br />
<br />
<!-- ws:start:WikiTextHeadingRule:10:<h6> --><h6 id="toc3"><a name="LECTURE 1: Vector Spaces and Dual Spaces-1.1: A monzo can be viewed as a VECTOR** in a **VECTOR SPACE.----For example: say 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{1} of the two is \(12 \cdot \-4 + 19 \cdot 4 + 28 \cdot \-1 = 0\). Thus, the result of \((12,19,28)^*\) acting on \((\-4,4,\-1)\) is \(0\)."></a><!-- ws:end:WikiTextHeadingRule:10 -->For example: say 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<span style="font-size: 80%; vertical-align: super;"><a class="wiki_link" href="/Mike%27s%20Lecture%20on%20Vector%20Spaces%20and%20Dual%20Spaces#ref1">{1}</a></span> of the two is \(12 \cdot \-4 + 19 \cdot 4 + 28 \cdot \-1 = 0\). Thus, the result of \((12,19,28)^*\) acting on \((\-4,4,\-1)\) is \(0\).</h6>
<br />
The action of a covector on a vector must, of course, be pictured as the different colored arrows lining up and exploding and spitting out a single number, or something. Wolfram unfortunately doesn't let me do nice explosion effects, so you'll have to imagine it.<br />
<br />
Alternatively, if you're used to dot products, then you'll also be able to understand this operation in terms of the projection of one vector on another. However, if you choose to visualize it this way, you must understand that these two things you're projecting onto one another lie in <u>different</u> spaces - one is a vector lying in a vector space, and the other is a covector lying in this new "dual space" you're learning about.<br />
<br />
Thus, covectors are little mathematical machines - they take in vectors, do some simple dot product-ish stuff, and output a scalar. Simple! We'll see that they turn up again and again and again in tuning theory, too.<br />
<br />
OK, so how do we use these things?<br />
<br />
<!-- ws:start:WikiTextHeadingRule:12:<h2> --><h2 id="toc4"><a name="LECTURE 1: Vector Spaces and Dual Spaces-1.2: Covectors mean stuff. (OR: YOU DON'T KNOW MONZO)"></a><!-- ws:end:WikiTextHeadingRule:12 -->1.2: Covectors mean stuff. (OR: YOU DON'T KNOW MONZO)</h2>
<br />
One interesting way to think of covectors, since they're these dual vectors that "act on" normal vectors, is thus as functions - they take in a vector as input, multiply each coefficient of the vector by the corresponding coefficient of the covector, sum them up, and spit out a number. In other words, you know that the action of the covector \((12,19,28)^*\) on any arbitrary vector \((a,b,c)\) is going to be \(12a + 19b + 28c\). So, you can think of \((12,19,28)^*\) itself as a function looking something like \(f(\v{v}) = 12a + 19b + 28c\) for some vector of the form (a, b, c). Note that the formatting on \(\v{v}\) specifies that \(\v{v}\) is a vector that's being taken in as input.<br />
<br />
Before we go on, however, let's clean up the notation a bit. In physics, the notation commonly used is to notate covectors \(\bratext{like this}\) and to notate vectors \(\kettext{like this}\). Physicists call this "bra-ket" notation, or sometimes "Dirac" notation. So instead of writing covectors as \((x,y,z)^*\), I'll just write \(\bra{x \s y \s z}\) from now on. And instead of writing vectors as \((a,b,c)), I'll just write \(\ket{|a \s b \s c>}\) from now on.<br />
<br />
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).<br />
Now then, let's say you're going to ask a question like "does 81/80 vanish in 12-EDO?" But when you ask that question, what are you really asking?<br />
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 12-EDO do I arrive at?"<br />
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 12-EDO do I arrive at?"<br />
So, in 12-EDO, 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.<br />
1) first you go down 4 octaves, at a rate of 12 steps per octave, putting you underwater at -48 steps.<br />
<br />
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.<br />
<br />
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.<br />
So, if you mechanically work out the way that you'd compute how many steps 81/80 is in 12-EDO, you get 0 steps, meaning it's "equated with 1/1" and hence tempered out. No surprise there.<br />
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 28|-4 4 -1>.<br />
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 28|-4 4 -1> - same basic thing.<br />
So that's what the interpretation of the dot product is.<br />
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...<br />
<br />
<!-- ws:start:WikiTextAnchorRule:14:<img src="/i/anchor.gif" class="WikiAnchor" alt="Anchor" id="wikitext@@anchor@@ref1" title="Anchor: ref1"/> --><a name="ref1"></a><!-- ws:end:WikiTextAnchorRule:14 -->[1] - Note that some have raised technical concerns about this operation being called the "dot product," insisting that the dot product is something that's only done between two vectors, or two covectors, but never between one covector and one vector. Another term that's sometimes been used for this product in the "<strong>bracket product</strong>", for reasons we don't need to get into here. However, confusingly, the term bracket product has also been used for the ordinary dot product, and it's also very common to hear people call the thing I'm calling the dot product above. It's best at this point to just know that the two terms are out there. I'm going to continue calling it the dot product since its' something more people are familiar with.<br />
<br />
<br />
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