From: baez@math.removethis.ucr.andthis.edu (John Baez)
Subject: This Week's Finds in Mathematical Physics (Week 282)
Organization: University of California, Riverside
Sender: baez@math.removethis.ucr.andthis.edu (John Baez)
Newsgroups: sci.physics.research,sci.physics,sci.math
Also available at http://math.ucr.edu/home/baez/week282.html
October 29, 2009
This Week's Finds in Mathematical Physics (Week 282)
John Baez
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, 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 the product of I^m and
I^n is contained in 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 1
If 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 the
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 Todd'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) = Comm(Lie(V))
But the PoincareBirkhoffWitt 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) = sum_n Lie_n tensor V^{tensor 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) = sum_n Assoc_n tensor V^{tensor n}
Likewise, the free commutative algebra on V is given by
Comm(V) = sum_n Comm_n tensor V^{tensor 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 idea 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) = sum_n dim(Lie_n) x^n / n!
Assoc(x) = sum_n dim(Assoc_n) x^n / n!
= sum_n x^n
= 1/(1x)
Comm(x) = sum_n dim(Comm_n) x^n / n!
= sum_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)
= sum_n x^n / n
but we saw
Lie(x) = sum_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 S(Lie(V)) is the free Poisson algebra on
the vector space V:
S(Lie(V)) = sum_n Poisson_n tensor V^{tensor 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 PoincareBirkoffWitt 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) Andre 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 5ary operation.
So, if we take an mary operation in I^{m1} and compose it with an
mary operation in I^{n1}, we get an (n+m)ary operation in
I^{n+m2}, which in turn lies in 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:
If I is the ideal in Assoc for which Assoc/I = Comm, what sort of
algebras are described by the operad I^k?
What about the operad I^k/I^{k+1}? What about the operad I^k/I^{k+2},
and so on?
And what about the suboperads of Assoc concocted as in the previous
paragraph?

Quote of the Week:
The worthwhile problems are the ones you can really solve or help solve,
the ones you can really contribute something to.  Richard Feynman

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?
(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.)
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!
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 Cafe, here:
http://golem.ph.utexas.edu/category/2009/10/this_weeks_finds_in_mathematic_43.html
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?

Previous issues of "This Week's Finds" and other expository articles on
mathematics and physics, as well as some of my research papers, can be
obtained at
http://math.ucr.edu/home/baez/
For a table of contents of all the issues of This Week's Finds, try
http://math.ucr.edu/home/baez/twfcontents.html
A simple jumpingoff point to the old issues is available at
http://math.ucr.edu/home/baez/twfshort.html
If you just want the latest issue, go to
http://math.ucr.edu/home/baez/this.week.html