
This week I'll get back to explaining some serious math: the relation between associative, commutative, Lie and Poisson algebras, and how this relates to quantization. There's some beautiful algebra and combinatorics that shows up here: linear operads, their generating functions, and Stirling numbers of the first kind.
But first: the astronomy picture of the week! Lately we've been exploring the moons of Saturn  first Enceladus in "week272" and "week273", and then Phoebe and Iapetus in "week281". Someday we should talk about Rhea  a moon of Saturn with its own rings. But first let's take a big detour and sail in to Mercury.
In fact, the Messenger probe sailed in to Mercury starting on August 3, 2004. It's flown past this planet several times, and in March 2011 it's scheduled to orbit Mercury for a whole year. It's already taken some detailed photos:
1) Messenger, Image gallery, http://messenger.jhuapl.edu/gallery/sciencePhotos/
Superficially Mercury looks like the Moon, and thus not very exciting. But it's actually very different. First of all, parts of Mercury get really hot: about 430 Celsius near the equator during the day  considerably above the melting point of lead. Second, permanently shaded regions near the poles are not only cold, they actually have lots of ice! Third, Mercury had a violent past. For example, the Caloris basin on Mercury is one of the solar system's largest impact basins. Formed by a huge asteroid impact long ago, it's about 1,500 kilometers across:
2) NASA, New discoveries at Mercury, August 3, 2008. http://science.nasa.gov/headlines/y2008/03jul_mercuryupdate.htm
Fourth, Mercury is the densest planet, with the highest percentage of iron. Why is this? There are various theories. The most widely accepted is reminiscent of the "giant impact theory" for how our Moon formed (see "week273"). It goes like this. Once upon a time Mercury was over twice the size it is now, with a more ordinary chemical composition. Then it was hit by another body about 1/6 its own mass! This stripped off a lot of its crust and mantle, leaving a smaller Mercury, whose iron core now accounts for a greater percentage of its mass.
Fifth, and a direct consequence of the previous point, Mercury has a strong magnetic field  like Earth, and unlike Venus, Mars, or our Moon. And this brings me to the picture I really want you to stare at: a diagram of how Mercury's magnetic field interacts with the solar wind, which is very powerful so near the Sun. The Messenger probe learned a lot about this when it flew past Mercury on October 6th, 2008:
3) NASA, Magnetic tornadoes could liberate Mercury's tenuous atmosphere, http://www.nasa.gov/mission_pages/messenger/multimedia/magnetic_tornadoes.html
The pink stuff in this picture is the "magnetopause"  the zone where the solar wind crashes into Mercury's magnetic field. And see the spirals? These are "flux transfer events". Every so often, the solar magnetic field lines reconnect with those of Mercury and ions in the solar wind penetrate the magnetopause and rain down on Mercury's north and south poles. Similar flux transfer events happen here on Earth about every 8 minutes:
4) NASA, Magnetic portals connect Sun and Earth, October 30, 2008. http://science.nasa.gov/headlines/y2008/30oct_ftes.htm?list179029
The physics is complex and just starting to be understood: the basic equations governing the interaction of plasma (that is, ionized gas) and electromagnetism are devilishly nonlinear and hard to deal with. This is one reason fusion reactors that use magnetic confinement are so hard to develop. In particular, there's been a lot of recent work on "reconnection", where magnetic fields pointing in opposite directions crosslink and accelerate plasma in a "magnetic slingshot". Here's a great article on that subject:
5) James L. Burch and James F. Drake, Reconnecting magnetic fields, American Scientist 97 (2009), 392399. Also available at http://mms.space.swri.edu/AmSciReconnection.pdf
Finally: do you see the yellow "plasmoid" in the picture above? That's a coherent blob of plasma and magnetic field which forms in the the long "magnetotail" behind Mercury. Again, these also form near the Earth. And again, they're complex and mathematically interesting. So, while Mercury may look dead and boring, it's rich in activity if you know where to look!
Next, some math.
Today I'd like to talk about 4 of my favorite kinds of algebras: associative algebras, commutative algebras, Lie algebras and Poisson algebras. They're all important in quantum mechanics and quantization, and they fit together in a very nice way. There's a lot to say about this, but I just want to explain one thing: how the relation between these 4 kinds of algebras gives a pretty pattern involving Stirling numbers.
If you don't know what Stirling numbers are, don't worry! They'll show up on their own accord, and then we'll see why.
First: the four kinds of algebra. Let's review them.
An "associative algebra" is a vector space equipped with an identity element 1 and a binary operation called multiplication that's linear in each argument. We demand that these obey a few rules:
1x = x
x1 = x
(xy)z = x(yz)
In physics, associative algebras often show up as "algebras of observables"  their elements are things you can measure about a physical system.
A "commutative algebra" is an associative algebra that obeys one extra rule:
xy = yx
In classical mechanics, the algebras of observables are always commutative. The big deal about quantum mechanics is that we drop this rule and allow more general associative algebras. This wreaks havoc on our intuitions about physics, but in a very nice way.
A "Lie algebra" is a vector space with a "bracket" operation which is linear in each argument. We demand that this obeys two rules:
[x,y] = [y,x]
[x,[y,z]] = [[x,y],z] + [y,[x,z]]
These rules seem a lot scarier than the rules above  at least when you first meet them! The reason is that while ordinary numbers form an associative and even commutative algebra, they don't form a Lie algebra in any interesting way. Sure, you can define [x,y] = 0, and it works, but it's dull. The first nontrivial Lie algebra we meet in school may be the space of vectors in 3d space, where the bracket is the cross product. But most students don't even remember that the cross product satisfies the second rule above  the socalled "Jacobi identity". So to get comfortable with Lie algebras, most people need to start with an associative algebra that's not commutative, and then define the bracket by:
[x,y] = xy  yx
This is called the "commutator", and it's very important in quantum mechanics, in part because it tells you how far things are from being classical. In classical mechanics, the commutators are zero!
There's also a deeper and more important reason why commutators and Lie algebras are important in quantum theory: they show up when we study symmetries of physical systems. But that's another story, tangential to today's tale.
Anyway, it's fun  or at least good for your moral development  to check that the associative law for multiplication implies the Jacobi identity when we define the bracket by [x,y] = xy  yx.
So, we've got a recipe for turning an associative algebra into a Lie algebra. We've also seen a pathetically easy recipe for turning a commutative algebra into an associative one: just forget that it's commutative!
In the language of category theory, both these recipes are called "forgetful functors", because they lose information. So, we've got forgetful functors
CommAlg → AssocAlg → LieAlg
and this little diagram is the crux of our tale.
But to see why, I need to introduce the fourth character: Poisson algebras. The idea here is to realize that classical mechanics isn't really true: the world is quantum mechanical. So, even when we think our algebra of observables is commutative, it's probably not. This is probably just an approximation. It's not really true that the commutator [x,y] is zero. Instead, it's just tiny.
How do we formalize this? Well, in reality [x,y] is often proportional to a tiny constant called Planck's constant, h. When this happens, we can write
[x,y] = h{x,y}
where {x,y} is some other element of our associative algebra.
Mathematically, it's more convenient to treat h as a variable than as a fixed number. So, let's suppose we have an associative algebra A with a special element h that commutes with everything. And let's suppose that A is equipped with a new bracket operation {x,y} that satisfies the above equation.
Then let's consider the algebra A/hA, which we define by taking A and imposing the relation h = 0. This amounts to neglecting quantum effects, so A/hA is called the "classical limit" of our original algebra A.
What is this new algebra like?
Well, first of all, it's associative. Second, it's commutative, since [x,y] was proportional to h, but now we're setting h equal to zero. And third, it inherits from A a bracket operation {x,y}, called the "Poisson bracket".
What rules does the Poisson bracket satisfy? Well, since
[x,y] = [y,x]
we know that
h{x,y} = h{y,x}
in A. So it seems plausible that
{x,y} = {y,x}
in A/hA. Unfortunately I can't derive this from my meager assumptions thus far, since I'm not allowed to divide by h. So let me also assume that multiplication by h is onetoone in A. Then I know
{x,y} = {y,x}
in A and thus also in A/hA.
Similarly, from the Jacobi identity for the commutator
[x,[y,z]] = [[x,y],z] + [y,[x,z]]
we know that
{x,{y,z}} = {{x,y},z} + {y,{x,z}}
in A, and thus also in A/hA. The same sort of argument also shows that {x,y} is linear in each argument.
There's one more rule, too! Note that in A we have
[x,yz] = xyz  yzx
= xyz  yxz + yxz  yzx
= [x,y]z + y[x,z]
and thus
{x,yz} = {x,y}z + y{x,z}
So, this rule holds in A/hA too. This rule says that the operation "bracketing with x" obeys the product rule, just like a derivative.
And so, we've been led to the definition of a Poisson algebra! It's a commutative algebra with an extra operation, the Poisson bracket, which is linear in each argument and obeys these rules:
{x,y} = {y,x}
{x,{y,z}} = {{x,y},z} + {y,{x,z}}
{x,yz} = {x,y}z + y{x,z}
Physically, the idea here is that the Poisson bracket is the extra structure that we get from the fact that classical mechanics arises from quantum mechanics by neglecting quantities proportional to Planck's constant.
Mathematically, the idea is that a Poisson algebra is both a commutative algebra and a Lie algebra (with the Poisson bracket as its bracket), obeying the compatibility condition
{x,yz} = {x,y}z + y{x,z}
So, besides the forgetful functors I've already drawn, we have two more:
PoissonAlg → CommAlg
and
PoissonAlg → LieAlg
But you'll notice that in my above argument I got ahold of the Poisson algebra axioms starting from an associative algebra of a special sort: roughly, one that's "noncommutative, but only up to terms of order h". This suggests that Poisson algebras are a halfway house between associative and commutative algebras. And I'd like to make this more precise!
Technically, these special associative algebras are called "deformations" of commutative algebras. And there's a whole branch of mathematical physics called "deformation quantization" that studies them. So, some experts reading the previous paragraph may think I'm about to explain deformation quantization. But much as I'd love to talk about that, I won't now! That theme will have to remain lurking in the background.
Instead, I just want to show how the concept of Poisson algebra emerges from the forgetful functor
CommAlg → AssocAlg
And to do this, I'll need operads. I explained these back in "week191", so if the word "operad" fills you with bewilderment or terror instead of delight, please reread that. But today I'll be using linear operads, so let me explain those.
The concepts of associative algebra and commutative algebra and Lie algebra and Poisson algebra have a lot in common. In every case we start with a vector space and equip it with a bunch of nary operations that are linear in each argument. Moreover, these operations are required to satisfy equations where each variable shows up exactly once in each term, like
{x,{y,z}} = {{x,y},z} + {y,{x,z}}
or
{x,yz} = {x,y}z + y{x,z}
And this is precisely what linear operads are designed to handle!
More precisely, a linear operad O consists of a vector space O_{n} for each natural number n. We call this the space of "nary operations". They're not operations on anything yet  they're just "abstract" operations, with names like "multiplication" or "Poisson bracket". We can draw an nary operation as a little black box with n wires coming in and one coming out:
\  / \  / \  /      We're allowed to compose these operations in a treelike fashion:
\ / \  /  \ / \  /              \  / \  / \  / \  / \  / \  / \  /      Here we are feeding the outputs of n operations g_{1},..,g_{n} into the inputs of an nary operation f, obtaining a new operation which we call
f o (g_{1},...,g_{n})
Since we're doing linear operads todqay, we demand that this composition operation be linear in each argument. Moreover we demand that there be a unary operation serving as the identity for composition, and we impose an "associative law" that makes a composite of composites like this welldefined:
\ /  \  / \ / \ /  \  / \ /                 \  / / \  / / \  / /             \  / \  / \  / \  / \  / \  / \  /      (This picture has a 0ary operation in it, just to emphasize that this is allowed.) Furthermore, we can permute the inputs of an nary operation and get a new operation:
\ / / / / / \ / / / / / \ \  /      We demand that this give an action of the permutation group on the O_{n}. And finally, we demand that these permutation group actions be compatible with composition in two ways.
The first way is easy to draw:
\  /  \ / \\\ / / / \  /  \ / \\/ / /    /\\ / /  a   b   c  / \\/ /    / / / \ / / / / /\\ \ / / /   \\\ \ / / /   \\\ / /    / \ / =  b   c   a  / /    / / \ \  / \  / \  /    d   d       We can permute the wires leading into d and then compose it with the operations a,b,c, or compose them in a different order and then permute the wires.
The second way is harder to draw, because both sides of the equation look exactly the same! For example:
\ /   \ / \ /   / /   / \ / \   / \ \  /  \ / \  /  \ /     a   b   c     \  / \  / \  /   d    Here we can either compose the operations a,b,c with d and then permute the wires leading into the result, or apply permutations to the wires leading into a,b, and c and then compose the resulting operations with d. We get the same answer either way, and indeed the pictures look exactly the same.
We use operads to describe algebras. An "algebra" for a linear operad O is a vector space V together with maps that turn elements of O_{n} into nary operations on V that are linear in each argument. If you like representations of groups you might prefer to call this a "representation" of O on V, since the idea is that elements of O_{n} are getting represented as actual operations on the vector space V. Of course we demand that composing operations in O and permuting their arguments get along with this process.
Let's look at 4 examples.
First, there's an operad Assoc, whose algebras are associative algebras. This operad is generated by one binary operation, called multiplication, and one nullary operation, called 1. We'll write these as if they were actual functions, though it's is not really true until we choose an algebra for this operad. So, we'll write them as
(x,y) → xy
and
() → 1
The second operation looks funny: it's a "nullary operation", one that takes no inputs. A nullary operation is also known as a "constant", because its output doesn't depend on anything.
Starting from these two operations we can generate lots more by composition and taking linear combinations. Then we impose some relations. First we impose one saying that these two ternary operations are equal:
(x,y,z) → (xy)z
and
(x,y,z) → x(yz)
I can say this faster, as follows:
(xy)z = x(yz)
But remember: now I'm not talking about the associative law in any particular algebra  I'm talking about an equation that holds in the operad Assoc, and thus in every algebra of this operad. We also also impose these laws:
1x = x
x1 = x
This completes our "generators and relations" description of the linear operad Assoc. We could also describe it by saying what all the nary operations are, and how to compose them. Either way, it's clear that the dimension of the space of nary operations is n factorial:
dim(Assoc_{n}) = n!
For example, here's a basis of the space of 3ary operations:
(x,y,z) → xyz
(x,y,z) → xzy
(x,y,z) → yxz
(x,y,z) → yzx
(x,y,z) → zxy
(x,y,z) → zyx
Second, there's an operad Comm, whose algebras are commutative algebras. This is just like Assoc except we impose one extra relation:
xy = yx
As a result, all the ways of multiplying n things in different orders become equal, and we get
dim(Comm_{n}) = 1
Third, there's an operad Lie, whose algebras are Lie algebras. This is generated by one binary operation
(x,y) → [x,y]
satisfying the relations
[x,y] = [y,x]
[x,[y,z]] = [[x,y],z] + [y,[x,z]]
It's harder to work out the dimension of the space of nary operations in the Lie operad, but the answer is beautiful:
dim(Lie_{n}) = (n1)!
Why is this true? I'll give a proof later on!
Fourth, there's an operad Poisson, whose algebras are Poisson algebras. This is generated by two binary operations and one nullary operation:
(x,y) → xy
(x,y) → {x,y}
() → 1
which obey the relations we've already seen: the commutative algebra relations for xy, the Lie algebra relations for [x,y], and the product rule
{x,yz} = {x,y}z + y{x,z}
What's the dimension of the space of nary operations now? I'll leave this as puzzle. It will be very easy if you pay close attention to what I'm saying.
Okay. Now, you'll notice that we got the operad Comm from the operad Assoc by adding an extra relation. So, every operation in Assoc maps to one in Comm. This map is linear, and it preserves composition. So, we say there's a homomorphism of linear operads
Assoc → Comm
Quite generally, whenever we have an operad homomorphism
O → O'
we get a way to turn O'algebras into Oalgebras, since every operation in O can be reinterpreted as one in O'. So, we get a functor
O'Alg → OAlg
In particular, the homomorphism
Assoc → Comm
gives the forgetful functor we've already seen:
CommAlg → AssocAlg
It's really just another way of talking about this functor!
With the main characters introduced, now our tale begins in earnest. Let's use the homomorphism
Assoc → Comm
to construct the Poisson operad.
To do this, first note that linear operads are a lot like rings. In particular, we can talk about the "kernel" of an operad homomorphism, and this is always an "ideal". The "kernel" consists of operations that go to zero under the homomorphism. Saying it's an "ideal" means that if you compose any operation with one in the ideal, you get one in the ideal. For example, in a composite like this:
\  /  \ / \  /  \ /     a   b   c     \  / \  / \  /   d    if any one of the operations a,b,c,d is in the ideal, the whole composite is in the ideal. So if you think of operations as apples and the operations in the ideal as rotten apples, the rule for ideals is "one rotten apple spoils the whole tree".
Let's take the operad homomorphism
Assoc → Comm
and call its kernel I. I_{n} is the space of nary operations for associative algebras that go to zero when we think of them as operations for commutative algebras. Let's see what it's like! The first interesting case is I_{2}. Assoc_{2} is 2dimensional, with this basis:
(x,y) → xy
(x,y) → yx
and I_{2} is 1dimensional, with this basis:
(x,y) → xy  yx
since this is what's zero for commutative algebras. The quotient Assoc_{2}/I_{2} is the same as Comm_{2}: it's a 1dimensional space, and in this space we have identified the operations
(x,y) → xy
and
(x,y) → yx
Indeed, if you're used to rings, you shouldn't be surprised that the quotient of a linear operad by an ideal is always another operad, and since the homomorphism Assoc → Comm is onto, we have
Comm = Assoc/I
where I is the kernel of this homomorphism.
Let me quickly say how we use this to get the Poisson operad, and then work through the details a bit more slowly.
As with rings, we can take products of operad ideals. Given ideals J and K, their product JK consists of all linear combinations of composites f o (g_{1}, ..., g_{n}) where f is in J and at least one of the g_{i}'s in in K. So, given our ideal I, we get a sequence of ideals
I^{0}, I^{1}, I^{2}, I^{3}, ....
each containing the next. Here we set I^{0} = Assoc and I^{1} = I to get things going. We say the operad Assoc is "filtered" by this sequence of operad ideals. In highbrow terms, this means it's an operad in the category of filtered vector spaces. In lowbrow terms: each vector space in the list above contains the next, and
I^{m} I^{n} ⊆ I^{m+n}
As with rings, this lets us form the "associated graded" operad gr(Assoc), which is this direct sum:
gr(Assoc) = I^{0}/I^{1} + I^{1}/I^{2} + I^{2}/I^{3} + …
And this is the Poisson operad!
I won't prove this; I'll just sketch the idea, and I'm afraid what I say will only make sense if you have a good intuition for the difference between "filtered" and "graded" things, and how the "associated graded" construction converts the former to the latter.
Operations in I are those that contain at least one appearance of the bracket [x,y] = xy  yx: these are precisely the operations that vanish in a commutative algebra. For example:
(x,y,z) → [xy,z]
or
(x,y,z) → z[x,y]
Operations in I^{2} contain at least two appearances of the bracket like this:
(x,y,z) → [x,[y,z]]
And so on. But the operad Assoc is just "filtered", not "graded", because there's no way to say exactly how many appearances of the bracket a given operation contains  at least, no way that's compatible with composition and taking linear combinations. For example, you might say these operations contain 0 appearances of the bracket:
(x,y) → xy
and
(x,y) → yx
But their difference is the bracket!
The "associated graded" construction is designed precisely to cure this sort of problem: operations in I^{k}/I^{k+1} contain exactly k appearances of the bracket. And if we look at our example again, we'll see what this achieves. In gr(Assoc), the operations
(x,y) → xy
and
(x,y) → yx
live in I^{0}/I^{1}, but now they're equal, because they differ by the commutator, which lives in I^{1}. So, multiplication becomes commutative! Meanwhile, the operation
(x,y) → xy  yx
lives in I^{1}/I^{2}... but now we can call it the Poisson bracket:
(x,y) → {x,y}
And it's easy to check that these rules hold in gr(Assoc):
{x,y} = {y,x}
{x,{y,z}} = {{x,y},z} + {y,{x,z}}
{x,yz} = {x,y}z + y{x,z}
So  waving my hands rapidly here  we see that
gr(Assoc) = Poisson
But the fun isn't done! All this abstract nonsense is just the warmup to a very nice concrete calculation of how the nary operations in gr(Assoc) break up into grades I^{k}/I^{k+1}. And here is where the Stirling numbers show up.
Let's look at n = 3. The space of 3ary operations in Assoc has dimension 6. There's a 2d subspace of operations that live in I^{2}  that is, where the bracket shows up at least twice:
(x,y,z) → [x,[y,z]]
(x,y,z) → [y,[x,z]]
You might think it was a 3d subspace, but don't forget the Jacobi identity! There's a 5d subspace of operations that live in I  that is, where the bracket shows up at least once. For example, we can take the above two together with these three:
(x,y,z) → [x,y]z
(x,y,z) → [y,z]x
(x,y,z) → [x,z]y
And that leaves one more, for a total of 6:
(x,y,z) → xyz
A lot of nice patterns show up if you work out more examples. Here's the dimension of the space of nary operations in the Poisson operad that lie in I^{k}/I^{k+1}:
k = 5 k = 4 k = 3 k = 2 k = 1 k = 0 n = 1 1 n = 2 1 1 n = 3 2 3 1 n = 4 6 11 6 1 n = 5 24 50 35 10 1 n = 6 120 274 225 85 15 1If you're a true expert on combinatories, you'll instantly recognize these as "Stirling numbers of the first kind":
6) Wikipedia, Stirling numbers of the first kind, http://en.wikipedia.org/wiki/Stirling_numbers_of_the_first_kind
But even if you're like me, you'll still see some nice patterns!
First of all, when k = 0 we just get 1. This is the dimension of space of nary operations in the Poisson operad that don't use the bracket at all. Or in other words, operations in Comm:
I^{0}/I^{1} = Assoc/I = Comm
And we know this space is 1dimensional. For example, for n = 4 it has this basis vector:
(w,x,y,z) → wxyz
Second, when k = 1 we get the triangle numbers 1,3,6,10,.... This is the dimension of the space of nary operations in the Poisson operad that use the bracket exactly once. This makes sense if you think about it: for n = 4 here's a basis:
(w,x,y,z) → {w,x}yz
(w,x,y,z) → {w,y}xz
(w,x,y,z) → {w,z}xy
(w,x,y,z) → {x,y}wz
(w,x,y,z) → {x,z}wy
(w,x,y,z) → {y,z}wx
We're getting 4 choose 2 different operations.
Third, the numbers in the nth row add to n!. That's because the dimension of a filtered vector space equals that of the associated graded vector space. So, the total dimension of Poisson_{n} equals the dimension of Assoc_{n}, which is n factorial.
Fourth, the nth number along the diagonal is (n1)!. This is the dimension of the space of nary operations that use the bracket the maximum number of times: namely, n1 times. For example, when n = 3 this is a 2d space with basis
(x,y,z) → {x,{y,z}}
(x,y,z) → {y,{x,z}}
These are precisely the operations in the Lie operad! So now we're seeing the operad inclusion
Lie → Assoc
which gives the forgetful functor
AssocAlg → LieAlg
Indeed, quite generally, you can check that any operad O with an ideal I has a suboperad whose nary operations are those lying in I^{n1}.
Finally, when you learn about Stirling numbers, you see the general pattern. Stirling numbers count the number of permutations of n elements that have a fixed number of disjoint cycles. For example, these permutations of 4 elements have 3 disjoint cycles:
(w x) (y) (z)
(w y) (x) (z)
(w z) (x) (y)
(x y) (w) (z)
(x z) (w) (y)
(y z) (w) (x)
These correspond to the following 4ary operations in the Poisson operad:
(w,x,y,z) → {w,x}yz
(w,x,y,z) → {w,y}xz
(w,x,y,z) → {w,z}xy
(w,x,y,z) → {x,y}wz
(w,x,y,z) → {x,z}wy
(w,x,y,z) → {y,z}wx
As you can see, there's a lot of fun and mysterious stuff going on here. Todd Trimble wrote a legendary paper "Notes on the Lie operad" which would probably shed a lot of light on this stuff. But unfortunately, the reason I call it "legendary" is that it's almost impossible to find! If I ever get a copy I'll let you know.
For now, I'll wrap the story by proving that the Stirling numbers are really related to the Poisson operad as claimed.
The first step is to show that
dim(Lie_{n}) = (n1)!
For this we can use a famous argument, which is probably in Trimble's paper. First consider the forgetful functors:
AssocAlg → LieAlg → Vect
where Vect is the category of vector spaces. These forgetful functors have left adjoints. The first forms the free Lie algebra on a vector space V. Let's call this Lie(V):
Lie: Vect → LieAlg
The second forms the free associative algebra on a Lie algebra L. This is called its "universal enveloping algebra", U(L):
U: LieAlg → AssocAlg
If we compose these two functors, we get a functor that forms the free associative algebra on a vector space V. This is usually called its "tensor algebra", but let's write it as Assoc(V), for reasons soon to become clear:
Assoc: Vect → AssocAlg
So, we have an canonical isomorphism
Assoc(V) ≅ U(Lie(V))
But the PoincaréBirkhoffWitt theorem gives a canonical isomorphism of vector spaces between the universal enveloping algebra U(L) of a Lie algebra L and its "symmetric algebra"  that is, the free commutative algebra on its underlying vector space. Let's write this symmetric algebra as Comm(L). So, we get a vector space isomorphism
Assoc(V) ≅ Comm(Lie(V))
(Admittedly, the standard ugly proof of the PBW theorem does not give a canonical isomorphism. But the good proof does  see "week212".)
Next, let's use some wellknown black magic to describe the above functors using operads. The free Lie algebra on a vector space V is given by
Lie(V) ≅ ⊕_{n} Lie_{n} ⊗ V^{⊗n}
where we tensor over the action of the symmetric group. Similarly, the free associative algebra on a vector space V is given by
Assoc(V) ≅ ⊕_{n} Assoc_{n} ⊗ V^{⊗n}
Likewise, the free commutative algebra on V is given by
Comm(V) ≅ ⊕_{n} Comm_{n} ⊗ V^{⊗n}
These are categorified versions of formal power series. That's because linear operads are a special case of linear "species", or "structure types". So, we can decategorify them and get formal power series called their generating functions. I explained this in "week185", "week190", and "week102", but not in the linear case. It's no big deal: where we used cardinalities before, now we use dimensions! We get these generating functions:
Lie(x) = ∑_{n} dim(Lie_{n}) x^{n}/n!
Assoc(x) = ∑_{n} dim(Assoc_{n}) x^{n}/n!
= ∑_{n} x^{n}
= 1/(1x)
Comm(x) = ∑_{n} dim(Comm_{n}) x^{n}/n!
= ∑_{n} x^{n}/n!
= exp(x)
Now, by general abstract nonsense our isomorphism
Assoc(V) ≅ Comm(Lie(V))
gives an equation
Assoc(x) = Comm(Lie(x))
or
1/1x = exp(Lie(x))
so
Lie(x) = ln(1/1x)
= ∑_{n} x^{n}/n
but we saw
Lie(x) = ∑_{n} dim(Lie_{n}) x^{n}/n!
so
dim(Lie_{n}) = (n1)!
This is a beautiful way of counting the number of nary operations in the Lie operad.
Note also that (n1)! is also the number of permutations of n things with a single cycle. So, the Stirling numbers are already showing up.
Next let's use the fact that for any Lie algebra L, the symmetric algebra Comm(L) is not just a commutative algebra: it's a Poisson algebra! It has a Poisson bracket, called the KostantKirillov Poisson structure. Indeed, it's the free Poisson algebra on the Lie algebra L.
This implies that Comm(Lie(V)) is the free Poisson algebra on the vector space V:
Comm(Lie(V)) ≅ &oplus_{n} Poisson_{n} ⊗ V^{⊗n}
To get a basis of Poisson_{n}, it's therefore enough to consider commuting products of terms built using Poisson brackets, like this:
(a,b,c,d,e,f,g,h,i,j) → {{a,b},c} {d,e} {f,g} h i j
Any expression like this can be reinterpreted as a permutation:
(a b c) (d e) (f g) (h) (i) (j)
So, by what we've already seen, the dimension of the space of nary operations that involve a product of j terms is the same as the number of permutations of n things with j cycles. That's a Stirling number! And this dimension is also the dimension of the space of nary operations that live in I^{k}/I^{k+1}, where j + k = n.
That last fact was not supposed to be instantly obvious. But if you look at the example above, you'll see it works:
n = 10, since there are 10 letters
j = 6, since we've got a product of 6 terms built using Poisson brackets
k = 4, since we're using Poisson brackets 4 times
If you think a while, you'll see it always works like this.
To summarize: the dimension of the space of nary operations in I^{k}/I^{k+1} is the same as the number of permutations of n things with nk disjoint cycles.
Even if you didn't follow this argument, I hope you see that associative, commutative, Lie and Poisson algebras are involved in a beautiful web of relationships.
I didn't get to say much about what all this means for quantization. Indeed, I haven't really figured it all out yet! For example, it must be important that the universal enveloping algebra of a Lie algebra is a deformation quantization of its symmetric algebra. This should be a central part of the story I'm telling, especially because it's crucial to the proof of the PoincaréBirkoffWitt theorem mentioned in "week212". But I didn't fully integrate this stuff into the story.
I also didn't talk about the relation between the Lie operad and the homology of the poset of partitions of a finite set, described at the beginning of this paper:
7) Benoit Fresse, Koszul duality of operads and homology of partition posets, in Homotopy Theory: Relations with Algebraic Geometry, Group Cohomology, and Algebraic Ktheory, eds. Paul Gregory Goerss and Stewart Priddy, Contemp. Math 346, 2004, AMS, Providence, Rhode Island, pp. 115215. Also available at http://math.univlille1.fr/~fresse/PartitionHomology.html
and first discovered by Joyal:
8) André Joyal, Foncteurs analytiques et especes de structures, in Combinatoire Enumerative, Springer Lecture Notes in Mathematics 1234, Springer, Berlin (1986), 126159.
Nor did I bring the homology of the little kcubes operad into the game  the relation of this to the Poisson operad was described in "week220", but you'll notice that this only talks about k > 1. The story I'm discussing now concerns the case k = 1, because the algebra Assoc is the homology of the little 1cubes operad. For higher k, I especially recommend this paper:
9) Dev Sinha, The homology of the little disks operad, available as arXiv:math/0610236.
Finally, here are three side remarks that would have been too distracting earlier:
When I said "linear operads are a lot like rings", I could have been more precise. Linear operads are a lot like associative algebras  and indeed, an associative algebra is the same as a linear operad with only unary operations! But since we were talking about the linear operad for associative algebras, I didn't want to blow your mind by pointing out that an associative algebra also is a linear operad. We could also consider operads whose spaces of operations are abelian groups, with composition being a group homomorphism in each argument. An operad like this with only unary operations is the same as a ring. And this is the precise sense in which operad theory generalizes ring theory.
When I said "one rotten apple spoils the whole tree", I felt like saying "cherry" instead of "apple", since Boardman and Vogt talked about "cherry trees" in their work on operads. Unfortunately, the proverb "one rotten apple spoils the whole barrel" requires apples! For more, see this nice historical survey:
10) James Stasheff, Grafting Boardman's cherry trees to quantum field theory, in Homotopy Invariant Algebraic Structures: A Conference in Honor of J. Michael Boardman, eds. JeanPierre Meyer, Jack Morava, and W. Stephen Wilson, AMS, Providence, Rhode Island, 1999. Also available as http://www.math.unc.edu/Faculty/jds/boardman.ps
When I said "Indeed, quite generally, you can check that any operad O with an ideal I has a suboperad whose nary operations are those lying in I^{n1}", you might have been puzzled by the "1". Here's the point. All ways of composing operations can be built up from ways like this:
 \ /   \ /         \  / \  / \  / \  /      where we compose an mary operation and an nary operation (together with some identity operations). The result is an (m+n1)ary operation! For example, above I'm composing a 3ary operation and a 2ary operation and getting a 4ary operation.
So, if we take an mary operation in I^{m1} and compose it with an nary operation in I^{n1}, we get an (n+m)ary operation which lies in I^{n+m2} ⊆ I^{n+m1}. So we get a suboperad whose nary operations are those lying in I^{n1}.
You might enjoy working out what other ways there are to get suboperads from an operad with an ideal. Take all nary operations lying in I^{f(n)}. For what functions f do these form a suboperad?
Also, you might enjoy answering these questions, most of which I haven't tried:
Addenda: I thank James Dolan and Urs Schreiber for catching some mistakes. Allen Knutson adds to my list of questions:
Here's another: if gr(Assoc) = Poisson, what is the meaning of the Rees and blowup algebras associated to this filtration?Of course he means "operad" where he writes "algebra" or "ring"  while the constructions he describes are most familiar for algebras or rings, they work for operads too!(Given a filtration R = R_{0} ⊇ R_{1} ⊇..., e.g. by powers of I, you can look at the subring of R[t] that has t^{n} R_{n} in the nth degree piece; that's the blowup algebra. If you include t^{n} R in the negative powers, that's the Rees algebra. If you mod out Rees by (t  c), you get R for any nonzero c, and gr(R) for c=0.)
David Corfield points out:
Wikipedia wants to tag your Stirling numbers as 'unsigned', and yet notes that "that nearly all the relations and identities given on this page are valid only for unsigned Stirling numbers". Also the link with exponential generating functions goes through the unsigned version.So why deal with the signed version? Is it because
The Stirling numbers of the first and second kind can be understood to be inverses of oneanother, when taken as triangular matrices.
For more discussion visit the nCategory Café. In particular, Toby Bartels raised an important question: what's the physical meaning of treating Planck's constant as a variable instead of a number in deformation quantization?
The worthwhile problems are the ones you can really solve or help solve, the ones you can really contribute something to.  Richard Feynman
© 2009 John Baez
baez@math.removethis.ucr.andthis.edu
