Symplectic Mechanics and Symmetry « The Everything Seminar

Symplectic Mechanics and Symmetry

In my last post, I outlined how solving Newton’s $F=ma$ for a single object moving in a static force field could be restated as flowing along a vector field in the tangent bundle (=cotangent bundle). However, if I had written out the details of what flowing along the vector field meant, it would have been clear that it was just a fancy way of saying the particle accelerates in the direction of the force. Is this just elaborate computational legerdemain, where I have used big words and new concepts to misdirect you from the fact that I am just moving the difficulty around?

The aim of this post is to demonstrate that the symplectic perspective has its own merits, by describing the technique of ’symplectic reduction’. The principle here is a simple and beautiful one: if the mechanical system under consideration has a (good) symmetry, then there is some quantity one can assign to the possible states of the system which is conserved for all time. This meta-theorem is called Noether’s Theorem, though we will not be using it in full generality. Once a ‘conserved quantity’ is found (think of something like angular momentum), together with the symmetry it came from, the mechanical system can be reduced to one on a lower dimensional space.

The starting point is the following observation:

Observation. A function $H$ on a symplectic manifold $M$ is preserved by the flowing along its symplectic gradient $X_H$ . That is, if $\phi_t:M\rightarrow M$ is the the flow after time $t$ , then $\phi_t^*H=H$ .

Proof. A point $x$ flows to $x+X_H(x)\epsilon$ in time $\epsilon$ . The difference in $H$ at these two points is $dH(X_H)$ , the differential of $H$ along $X_H$ . However, by the definition of symplectic gradient, this is $\{X_H,X_H\}$ , which is zero by skew-symmetry. Hence, $H$ is constant on the images of $x$ after all time. End Proof.

This implies what I mentioned last time, that the total energy function $E$ is conserved by time evolution. It is also worth noting that the symplectic form itself is preserved under all symplectic gradient flows.

Ok, so lets see how to use this to our advantage in mechanics. We have a symplectic manifold $N$ (which is the tangent/cotangent bundle to space $M$ ), and we have a total energy function $E$ on $N$ , and we want to flow along $X_E$ .

Suppose that we have a 1-dimensional symmetry of this setup; that is, an action of $\mathbb{R}^1$ on $N$ that fixes $E$ and the symplectic form. If we act by a very small number $\epsilon$ , points in $N$ move a very small amount away from themselves, and so in the limit as $\epsilon \rightarrow 0$ , we get a vector field $v$ on $N$ . This vector field generates the action of $\mathbb{R}^1$ , in the sense that the action of a number $t$ is the same as flowing along $v$ for time $t$ .

In general, it is hard to say anything, but suppose further that the vector field $v$ is the symplectic gradient $X_{H}$ of some function $H$ on $N$ . Then the action of $\mathbb{R}^1$ is just the flow along this vector field. Since the action of $\mathbb{R}^1$ fixes $E$ , the infinitesmal action of $X_H$ also fixes $E$ , so

$0=dE(X_H)=\{X_E,X_H\}=-\{X_H,X_E\}=-dH(X_E)$ .

Hence, the flow of $X_E$ fixes $H$ . Thus, the value of $H$ for a particle is conserved as it moves through time; we call such a function a conserved quantity.

Since its value is conserved, we no longer have to think about the whole space $N$ when trying to see where a particle starting at point $x$ will go; we can restrict our attention to $H^{-1}(\lambda)$ for $\lambda=H(x)$ . We have a problem, though, in that the space $H^{-1}(\lambda)$ is not a symplectic manifold, since it is an odd dimensional manifold (for basic reasons, symplectic manifolds must be even dimensional).

However, we still have one piece of information we haven’t used yet: the symmetry itself. Since the action of $\mathbb{R}^1$ preserves $H$ by the Observation, $\mathbb{R}^1$ acts on $H^{-1}(\lambda)$ . We can quotient $H^{-1}(\lambda)$ by the action of $\mathbb{R}^1$ , and the resulting manifold will inherit a well-defined symplectic form. This is is denoted $N//_\lambda\mathbb{R}^1$ and called the symplectic reduction at $\lambda$ . The special case of $\lambda=0$ is called the symplectic quotient of $N$ by $\mathbb{R}^1$ .

Because the action of $\mathbb{R}^1$ preserved $E$ , $E$ and $X_E$ descend to the quotient to give a function $E'$ and its symplectic gradient $X_{E'}$ . Now, if we can figure out where a point in $N//_\lambda\mathbb{R}^1$ goes when we flow along $X_{E'}$ , we can lift this flow to a solution in $N$ . Hence, we’ve reduced the mechanical system to a (hopefully) easier problem.

(Note: the following example didn’t come out very well; you might want to skip it)

Lets see how this works for the example I mentioned last time. We had a planet $P$ moving in a two dimensional space $M$ around a fixed sun $S$ . If we write the coordinates of $TM$ as $(x,y,x',y')$ , the total energy function looks like:

$E(x,y,x',y')=-\frac{Gm_Sm_P}{\sqrt{x^2+y^2}}+m_P\frac{x'^2+y'^2}{2}$

We need a symmetry to work with; the obvious one is rotational symmetry:

$\theta_r (x,y,x',y')=(cos(r)x-sin( r)y, sin(r)x+cos(r)y,$
$cos( r)x'-sin( r)y', sin( r)x'+cos(r)y')$

To find the infinitesmal action, we find the derivative with respect to $r$ at $r=0$ :

$\frac{d\theta_r}{dr}_{r=0}(x,y,x',y')=(-y,x,-y',x')$ .

Finding a function $L$ whose symplectic gradient is this vector field is most easily done with guess and check, so I will leave the computation to the reader. The resulting $L$ is:

$L(x,y,x',y')=xy'-x'y$

The reader might recognize this as the angular momentum of the planet $P$ around the sun (without the mass term). Therefore, we have discovered that (without some unaccounted for force acting on it), a planet’s angular momentum around its sun will be constant. Woo!

Ok, now lets assume we know our planet’s starting value for $L$ is $\lambda$ . The first step in symplectic reduction is to restrict our attention to the subset $L^{-1}(\lambda)$ of $TM$ given by $xy'-x'y=\lambda$ .

Next, we need to quotient out by the action of rotation. This is best done by changing coordinates. The two variables $x$ and $y$ get replaced by the radius $r$ , which is the only quantity that can be discerned after the quotient. Similarly, $x'$ and $y'$ are replaced by $r'$ and $\theta'$ . In these new coordinates, the restriction that $L=\lambda$ becomes:

$L(r,r',\theta')=r^2\theta'=\lambda$

Thus, $\theta'=\lambda r^{-2}$ , and so the only free variables are $r$ and $r'$ . Thus, the symplectic reduction is isomorphic to $\mathbb{R}^+\times \mathbb{R}$ .

Now, if we were doing things by the book, we would see how the vector field $X_E$ descends to the reduction, and solve the flow there. However, I have a lazier idea. We know that the value of $E$ will be preserved by the flow, and the reduction is a two dimensional space. Thus, we know the flow will stay in the one dimensional level sets of $E$ , but then there’s no room for anything interesting to happen! The particle will move around these level sets.

The function $E$ with the change of coordinates is:

$E(r,r',\theta')=-\frac{Gm_Pm_S}{r}+m_P\frac{r'^2+r'^2\theta'^2}{2}$

We then add the fact that $\theta'=\lambda r'^{2}$ ,

$E(r,r')=-\frac{Gm_Pm_S}{r}+m_P\frac{r'^2+\lambda^{2}r'^{-2}}{2}$

Declaring that the total energy of the planet is $\gamma$ , we get:

$2r\gamma+2Gm_Pm_S=m_Prr'^2+\lambda^{2}m_Prr'^{-2}$

This complicated looking formula nevertheless carves out a path in the heavens that the planet must take.

Notice that the symplectic reduction is a new symplectic manifold that isn’t necessarily the cotangent space of any regular manifold. This means we should care about symplectic gradient flows on arbitrary symplectic manifolds (since they might arise as symplectic reductions). We are starting to see that symplectic manifolds are inherently interesting guys, at least as far as classical mechanics is concerned.

Another advantage is that we can potentially find another symplectic reduction. This would amount to finding another symmetry of the system; however, in order for this new symmetry to descend to the symplectic reduction by an old symmetry, the two symmetries must commute. This is equivalent to having a $\mathbb{R}^2$ symmetry.

In fact, for $\mathbb{R}^n$ symmetries of our system, we don’t have to reduce successive times. We can upgrade our techniques so that we can just do one big reduction. Say, for instance, that we have $n$ functions $H_i$ such that the flows along $X_{H_i}$ :

1) preserves the energy function $E$ , and

2) commute with each other.

The functions are said to be ‘in involution with each other and $E$ ‘. What we want to do is come up with a way to specify the fixed values of all of the $H_i$ at once.

If we have such functions, we can cook up the following function $\mu:N\rightarrow Hom(\mathbb{R}^n,\mathbb{R})$ :

$\mu_x(a_1,...a_n)=a_1H_1(x)+...+a_nH_n(x)$

This map is called a moment map of the system (it is not unique since the $H_i$ ’s aren’t unique, they can each vary by a constant).

If we pick an element in $\lambda\in Hom(\mathbb{R}^n,\mathbb{R})$ , then the set $\mu^{-1}(\lambda)$ is exactly the intersection of the level sets of each of the $H_i$ . This is just a fancy way of picking all the fixed values. By the same logic as above, $\mathbb{R}^n$ acts on $\mu^{-1}(\lambda)$ , and we can define the symplectic reduction at $\lambda$ to be the quotient of this action. The function $E$ and its gradient descend to this new space, which is likely a much simpler space.

Phew, that was a fair amount of talking, and we didn’t even say anything about the case of a non-abelian symmetry group. Suffice it to say, there is a version of moment maps, conserved quantities and symplectic reduction that works here too and is quite beautiful. I might say more about this stuff in a later post, or I might share with you some of the fun facts about torus actions.

Tags: math.MP, math.SG

This entry was posted on October 18, 2007 at 6:45 pm and is filed under Basic Grad Student, Greg. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

8 Responses to “Symplectic Mechanics and Symmetry”

John Armstrong Says:
October 18, 2007 at 7:14 pm
The special case of $\lambda=0$ is called the symplectic quotient of $N$ by $\mathbb{R}^1$ .

Is this really at all a special case? You came up with $H$ as a function whose symplectic gradient generates the symmetry. But we can add any constant to $H$ and still get the same gradient. That is, the condition that $\lambda$ be zero is not given by the action of $\mathbb{R}^1$ . The definition seems to involve more information than the terms given.
Greg Muller Says:
October 18, 2007 at 8:46 pm
Thats a good point. The reason I considered as a special case is because it is actually special for the action of a non-abelian group $G$ . In that case, the moment map $\mu:N\rightarrow \frak{g}^*$ can not be moved around freely, because there is a requirement that the map be $G$ -equivariant (with respect to the coadjoint action on the right side). Then, the symplectic reduction at $\lambda=0$ is special because you must reduce with respect to an entire coadjoint orbit, and those are frequently larger than a point.
John Armstrong Says:
October 18, 2007 at 8:56 pm
Here’s another view: the fact that you can add an arbitrary constant signals that we’re not looking at functions with values in $\mathbb{R}$ , but rather in an $\mathbb{R}$ -torsor. Will there in general be a $G$ -torsor floating around in the nonabelian case?
Greg Muller Says:
October 18, 2007 at 11:49 pm
I think that in general, there won’t be a $G$ -torsor sitting around, but in some nice cases, there will be a subgroup of $G$ that moves between equivalent moment maps.

A good example is the Heisenberg group $\mathcal{H}$ , the strictly upper triangular 3×3 matrices. Its isomorphic as a vector space to $\mathbb{R}P+\mathbb{R}Q+\mathbb{R}C$ , with the multiplication:
$(Q,P,C)(Q',P',C')=(Q+Q',P+P',QP'+C+C')$
Its Lie algebra $\frak{h}$ is generated by $q$ , $p$ , and $c$ , with the relations that $\langle q , p \rangle = c$ and $c$ is central (I am using angle brackets for Lie brackets, since hard brackets keep giving me parsing errors).

Consider $T\mathbb{R}^1=\mathbb{R}^2$ with coordinates $x$ and $x'$ with symplectic form $\{x,x'\}=1$ . Let $\mathcal{H}$ act on $T\mathbb{R}$ by
$tQ\cdot(x,x')=(x+t,x')$
$tP\cdot(x,x')=(x,x'+t)$
$tC\cdot(x,x')=(x,x')$

A Hamiltonian for $q$ (that is, a function whose symplectic gradient generates the action of $Q$ ) looks like $H_q=-x'+\alpha$ , for some constant $\alpha$ . Similarly, $H_p=x+\beta$ and $H_c=\gamma$ .

However, in the non-abelian case, we aren’t free to choose each constant independantly. Specifically, the assignment $H_*:\frak{h}\rightarrow (T\mathbb{R})^*$ must be a map of Lie algebras, where $(T\mathbb{R})^*$ is a Lie algebra by:
$\langle f,g\rangle=X_f(g)-X_g(f)$
This is the canonical Poisson algebra structure on the dual to any symplectic space. Note that constant functions are central.

So the condition that $H_*$ is a map of Lie algebras implies that
$\langle H_q,H_p\rangle=H_{\langle q,p\rangle}$

$\langle H_q,H_p\rangle=\langle-x'+\alpha,x+\beta\rangle=\langle -x',x\rangle$
$=-X_{x'}(x)+X_x(x')=\partial_x(x)+\partial_{x'}(x')=2$

$H_{\langle q,p\rangle}=H_c=\gamma$

Therefore, $H_c=2$ , there’s no choice about it. It is interesting to note, however, that there is an action of $H$ on the space of all moment maps which is isomorphic to its action on the original space $T\mathbb{R}$ .

Oh, and as a footnote to this example, notice the importance of the trivial action of $C$ on $T\mathbb{R}$ . Certainly the quotient $\mathcal{H}/\mathbb{R}C$ also acts on $T\mathbb{R}$ , but as was just demonstrated, $H_*$ can never be a map of Lie algebras. This is a nice example of a group action such that every infinitesmal action is the symplectic gradient of some function, but no moment map exists. However, this can usually be fixed by a 1-dimensional central extension.
John Baez Says:
October 23, 2007 at 1:47 am
Your first “Observation” has some LaTeX that didn’t get LaTeXed: $M$ .

Feel free to delete this!
Greg Muller Says:
October 23, 2007 at 2:29 am
Yeah, I don’t know what the deal is with that… periodically I get latex that doesn’t get recognized as code, and sometimes the code which isn’t processed changes on subsequent edits. Also annoying is when code that parsed successfully suddenly starts having parsing errors, without having been edited at all.

Ah well, not much I can do about it, though. Heck, back in the dim antiquity of the internet, people had to talk about math without embedded latex compilers at all!
Hamiltonian Mechanics « Rigorous Trivialities Says:
October 24, 2007 at 5:42 pm
[...] you’ve been keeping up with The Everything Seminar, the good people over there have discussed this very topic recently (with some of the Lagrangian formalism thrown in), but I feel it never hurts to have [...]
Paul J. Udoh Says:
November 28, 2007 at 12:25 pm
I am interested in this research programme on simplectic algebra of symmetry. I do not have the grasp of the subject. I would want to know more about it so that I can apply it to my mechanics of thermoelasticity. The major problems being those of cylindrical symmetry of deformation.
Please, direct me accordingly.
Thank you

The Everything Seminar

Symplectic Mechanics and Symmetry

8 Responses to “Symplectic Mechanics and Symmetry”

Leave a Reply