Computing $\exp(tT)$ where $T(b)=\{\frac{1}{2m}p^2,b\}$

Question

I've recently been reading Baez's paper on Noether's theorem and I was trying to test the simplest case I could think of: a mass moving in one dimension in a space with potential $0$. The Hamiltonian in this case would just be the kinetic energy $H(q,p)=\frac{1}{2m}p^2$. The manifold is $M=\mathbb R$, the cotangent bundle is $T^\ast M =\mathbb R\times \mathbb R$.

I'm doing this to test out and reconcile the usual statement of Noether's theorem for Hamiltonians with Baez's exposition using observables (elements of $C^\infty(T^\ast M)$) which form a Poisson algebra under the bracket $\{a,b\}:=\frac{\partial a}{\partial p}\frac{\partial b}{\partial q}-\frac{\partial a}{\partial q}\frac{\partial b}{\partial p}$.

Now translation by $t$ is a one-parameter subgroup $\Phi_t(q)=q+t$ of diffeomorphisms of $M$. According to the classical statement of Noether's theorem, the conserved quantity is $I(q,p)=\left. p\cdot \frac{d\Phi_t(q)}{d t}\right|_{t=0}=p$, which makes complete sense: momentum is conserved.

In terms of Baez's exposition, $H$ and $I$ are two observables such that $\{H,I\}=\{0\}$, and according to his formulation of Noether's theorem, $H$ should generate symmetries of $I$ and $I$ should generate symmetries of $H$. In this context, given two observables $a,b$, the observable $a$ is said to generate symmetries of $b$ if there's a flow $F^a_t:C^\infty(T^\ast M)\to C^\infty(T^\ast M)$ such that $\frac{d}{dt}F^a_t(b)=\{a,F^a_t(b)\}$ and $F_t^a(b)=b$. In this case, $F^a_t$ can be computed as $\exp(tT)$ where $T(c):=\{a,c\}$.

Now one quickly discovers that for $a=I$ above, $T=\frac{\partial}{\partial q}$. After a brief search I seemed to find something which I think says that $\exp(tT)(b)(q,p)=b(q+t, p)$. This makes sense because $b(q+t, p)$ is the result of lifting the original one-parameter group to $T^\ast M$ results in $\Phi_t^\ast(q,p)=(q+t,p)$ and then lifting it to be on $C^\infty(T^\ast M)$ just uses precomposition. So in short, $\exp(tT)$ corresponds exactly with my original flow $\Phi_t$.

Finally, to test the symmetry of the relationship between $H$ and $I$ fully, I wanted to compute the symmetries of $I$ that are generated by $H$. To do so, we'd need to look at $T(c):=\{H, c\}$ which I believe works out to $T(c)=\frac{p}{m}\frac{\partial c}{\partial q}$, so $T=g\frac{\partial}{\partial q}$ where $g$ is the function $g(q,p)=\frac{p}{m}$.

The problem is, I have no idea how to get a handle on what $\exp(tT)$ is in that case. At least I'm interested in knowing what the flow on $T^\ast M$ looks like.

Is there some way to sort it out to something recognizable? Or is it unexpectedly complicated?

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Update

I started to wonder if it isn't just $\exp(tT)(h)=h(q+t\frac{p}{m}, p)$. It feels heuristically right, since $t\frac{p}{m}$ is as constant with respect to $q$ as $t$ is alone. But more importantly, the flow $\Phi_t(q)=q+t\frac{p}{m}=q+t\dot q$ makes a ton of sense. It's just the displacement after time $t$.

Also importantly, its cotangent lift is $(q,p)\mapsto (q+\frac{p}{m}, p)$ again, and that fixes the observable $I$ above, as hoped.

However, when I compute its conserved quantity, $I'(q,p)=\left. p\cdot \frac{d\Phi_t(q)}{d t}\right|_{t=0}=p\cdot \frac{p}{m}$ which is twice the original Hamiltonian. I suppose that modifying an observable by scaling does not change the level sets, and so it tells you the same information.

I'm still sort of looking for confirmation of whether or not my conclusions are on solid ground.

Answer 1

Your conclusions are essentially correct. Some useful operator identities to have around in this context are

$$\exp\left( a \frac{d}{dx} \right)(f)(x) = f(x+a), \qquad \exp\left( bx \frac{d}{dx} \right)(f)(x) = f(e^b x)$$

when $a, b$ do not depend on $x$, and their generalization to multiple variables

$$\exp\left( \mathbf{a} \cdot \nabla \right)(f)(\mathbf{x}) = f(\mathbf{x}+\mathbf{a}), \qquad \exp\left( \mathbf{x}^\top B^\top \nabla \right)(f)(\mathbf{x}) = f(e^B \mathbf{x}).$$

where $\mathbf{a}$ is a vector and $B$ is a matrix (and $e^B$ denotes a matrix exponential). The latter identity becomes particularly useful when dealing with rotational or dilational symmetries. In fact, an alternative way to arrive at the expression in your question is to note that

$$T = \frac{p}{m} \frac{\partial}{\partial q}=\mathbf{z}^\top \begin{pmatrix} 0 & 0\\ \frac{1}{m} & 0\end{pmatrix} \nabla_{\mathbf{z}},$$

where $\mathbf{z} = (q, p)^\top$, so that

$$\exp\left(tT \right)(h)(\mathbf{z}) = h\left(\exp\begin{pmatrix} 0 & \frac{1}{m}\\ 0 & 0\end{pmatrix}\: \mathbf{z} \right) = h\left(\begin{pmatrix} 1 & \frac{1}{m}\\ 0 & 1\end{pmatrix} \mathbf{z} \right) = h\left(q+\frac{p}{m},p\right)$$

using the identity above and the fact that $e^B =I+B$ for $B$ nilpotent. In physics this kind of transformation is called a Galilean transformation or Galilean boost, and can be thought of as the limiting case of a relativistic pseudo-rotation when we formally send the speed of light $c$ to $\infty$.