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\author{Gabriel C. Drummond-Cole}
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\title{Math Physics\\October 19, 2006}
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\maketitle
So this is another one in the series of what William said.

He's been talking about the representation theory of the Poincar\'e group.

We're working with $M^n,$ which is affine Minkowski space over a Lorentzian inner product space $V.$  Let $x^0,\ldots,x^{n-1}$ be an affine, orthonormal set of coordinates.  So $g$ is $dx^0\otimes dx^0 -dx^1\otimes dx^1-\ldots-dx^{n-1}\otimes dx^{n-1}.$

No we have that $\frac{\partial}{\partial x^0}$ is timelike, meaning $||\frac{\partial}{\partial x^0}||^2>0,$ and the other coordinates $\frac{\partial}{\partial x^i}$ is spacelike, meaning $||\frac{\partial}{\partial x^i}||<0.$  

So on $V$ you have the light cone $C$ which we use to define certain subgroups.  Let $SO(V)$ denote the identity component of $O(V)$ with respect to the Lorentzian metric.  So these are $\{A\in O(V)|\det A=1$ and $A$ preserves the components of $C\}.$

So then there's the Poincar\'e group $P_n$, which is the identity component of $Isom(M^n),$ which is in the short exact sequence
$$
1\to V\to P_n\to SO(V)\to 1$$
Physicists prefer to work with representations of the lie algebra.  Let $\la{p}_n=Lie(P_n).$  Then we have the short exact sequence
$$
0\to V\to \la{p}_n\to \la{so}(V)\to 0$$

[How is $V$ a Lie algebra?]

It's the Lie algebra of the Lie group $V.$  It's Abelian.  It's not like Bill Gates, who's Abelianaire

Anyway, William wants to talk about representations of $\la{p}_n.$  Before that he talked about the structure of this group.  He did this in a very physics way.  All right.  So $P_n$ acts on $M^n.$  It also acts, in a natural way, on $T^*M^n.$  I'm doing this because this is how William was presenting this.  He was using Poisson brackets, so I need to do this in the symplectic case.  If $A\in P_n,$ then $A(x,p)=(Ax,(dA^{-1})^*p).$  With respect to the coordinates that we're using, we have coordinates on $T^*M^n$ which are $\{x^0,\ldots, x^{n-1}\}$ and then the $\{p_0,\ldots,p_{n-1}\}$ where $p_j:T^*M^n\to\rR$ via $p_j(dx^i)=\delta^i_j,$ with the form $\omega=dx^i\wedge dp_i.$  We have this symplectic structure so we have the Poisson bracket $\{f,g\}=\frac{\partial f}{\partial x_i}\frac{\partial g}{\partial p_i}-\frac{\partial f}{\partial p_j}\frac{\partial g}{\partial x_j}.$  We have this explicitly because we've chosen coordinates.

Recall that if $G$ is a Lie group that acts as symplectomorphisms of $(M,\omega),$ then there is a map from $(\la{g},[\ ,\ ])$ into $(C^\infty(M),\{\ ,\ \}).$  That map is the Noether charge, where you take $j$ to the Noether charge of $j,$ which is a smooth function.  

William was essentually using this paradigm when he was talking about the Poincar\'e group.  Any questions so far?  Well, let's see.  the Noether charge for travel along the $e_i$ direction is $p_i.$  The basis element $e_i$ gets sent to the function $p_i\in C^\infty(T^*M^n).$  Here $e_i$ corresponds to $x^i.$

So William was using the fact that translations are represented by momentum functions.  So in particular, if you want to track how translatinons act on $C^\infty(T^*M^n),$ if $f$ is such an object, then $Lie(\frac{\partial}{\partial x^i})f=-\{p_i,f\}.$  Okay?

So those are the translations.  What about $\la{so}(V).$  What does this Lie algebra look like, more or less?  Say $a\in \la{so}(V).$  Then I can infinitessimally exponentiate this to $\exp(ta).$  Then $g(\exp(ta)v,\exp t_a w)=g(v,w).$  So differentiating you get $g(av,w)+g(v,aw)=0.$  So this algebra should be naturally isomorphic with $\wedge^2 V,$ where $v\wedge w\mapsto g(v)\otimes w-g(w)\otimes v,$ where here $g$ is a map $V\to V^*.$

This brings us to $\{e_i\}\leadsto \{e_i\wedge e_j\},$ a basis for $\wedge^2 V,$ which leads to $\{e^i\wedge e_j-e^j\wedge e_i\}.$  Now define $M_{ij}$ to be the Noether charge for $e_i\wedge e_j.$  Did we compute this?  The answer was $x_ip^j-x_jp^i.$ 

Let $\xi_{ij}$ be the vector field of $T^*M^n$ induced by $e_i\wedge e_j,$ then $Lie(\xi_{ij})f=-\{M_{ij},f\}.$  Here $$\xi_{ij}=x^i\frac{\partial}{\partial x^j}-x^j\frac{\partial}{\partial x^i}$$

The Poincar\'e group acts, I claim, on the tangent space, this is homework.  The homomorphism we have from $(\la{p}_n,[\ ,\ ])\to (C^\infty(T^*M^n),\{\ ,\ \})$ is injective.  Therefore, to look at the structure in $\la{p}_n$ it suffices to look at its image in $C^\infty(T^*M^n).$  Physicists think of symmetries as functions, associate them with charges.  This is a very physics-y thing to do.  Should it be physicsie?

[Like Dixie?]

[Dixie is spelled with a y.]

No it's not, you should know, you're from Texas.  Texas, with its football, and its\ldots, presidential candidates, and its death penalty for retards.

We've picked a basis.  For translations we have $e^i$ and for rotations we have $e^i\wedge e^j.$  So let me say that again.  A basis of $V,$ which are the $e^i,$ determine a basis for the Lie algebra $\la{p}_n,$ where translations have as basis $\{e_i\}$ and as rotations $\{e_i\wedge e_j\}.$  This works because there is a natural splitting of the short exact sequence on the level of Lie algebras (into a semidirect product $\la{so}(V)\rtimes V$), where $[a,v]=av.$  

[What is William's curly $\delta?$]

It's the Lie derivative.  Physicists usually are thinking about one particular vector field all the time when they say that.

Okay, so the whole point was that William starts with a basis on $V.$  This gives me a basis on the Lie algebra, and then we have $\la{g},[\ ,\ ],$ and a basis $\{T_i\}.$  Then $[T_i,T_j]=\Gamma_{ij}^kT_k,$ and these are called the structure constants of $\la{g}$ with respect to the basis $\{T_i\}.$

[How do those transform with change of basis?]

How it transforms depends on where it lives, so it's in $\wedge^2 \la{g}^*\otimes \la{g},$ which explains how it transforms with respect to automorphisms of $\la{g}.$

Because William works with a basis, he expresses everything in terms of structure constants, and does it on the function side.

[What about representations?]

William hasn't talked about those yet much.  The map into $C^\infty(T^*M^n),$ that's a representation.  If you want it to act on vectors, you choose some matrices in $\la{gl}(n)$ with the same structure constants.  You have $V$ and its tensor algebra, you get those actions of $\la{so}(V)$ for free.  If you want the Lie algebra action to be unitary, you get into quantum field theory.

There's a theorem that for noncompact groups there are might not be a finite dimensional unitary representations.  Those functions are fields in physics language.

[Are the functions Lie algebras of an obvious Lie group?]

Hamiltonian diffeomorphisms.

Okay, I was going to say, now we can write $\delta f(x).$  Of course, with respect to some set of constants $\omega^{ab}$ which are describing something about the Lorentz transformation.  So $\omega^{01}$ would be some kind of boost, and $\omega^{12}$ would be a rotation around the $z$ axis.

So $$\delta f(x)=\{\frac{1}{2}\omega^{ab}M_{ab},f\},$$ where $M_{ab}=x_ap_b-x_bp_a=x_{[a}p_{b]}.$  So this whole thing is $$\frac{1}{2}\omega^{ab}\{x_{[a}p_{b]},f(x),$$ of which only one term can survive, which is $$\frac{1}{2}\omega^{ab}x_{[a}(-\frac{\partial p_{b]}}{\partial p_c}\frac{\partial f}{\partial x^c})=
-\frac{1}{2}\omega^{ab}(x_a\delta^c_b\frac{\partial f}{\partial x^c}-x_b\delta^c_a\frac{\partial f}{\partial x^c}$$$$
=\frac{-1}{2}\omega^{ab}\underbrace{(x_a\frac{\partial}{\partial x^b}-x_b\frac{\partial}{\partial x^a})}_{\xi_{ab}}f.$$

So you can replace $M_{ab}$ with $\xi_{ab}$ and $\{\ ,\ \}_{\text{P.B.}}$ with $[\ ,\ ].$

I think we went over again.

[But only by a minute.]

I should say that if you compute $\{M,M\}=fM,$ you'll get the same $f$ with $[\xi,\xi]=f\xi.$  That's the point.  They're isomorphic as Lie algebras.


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