Classical analogue of the Stone-von Neumann Theorem? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T11:28:21Z http://mathoverflow.net/feeds/question/78573 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/78573/classical-analogue-of-the-stone-von-neumann-theorem Classical analogue of the Stone-von Neumann Theorem? soulphysics 2011-10-19T12:40:36Z 2011-10-26T22:25:04Z <p>Let $U_s$, $V_t$ be a pair of continuous $n$-parameter groups ($n &lt; \infty$) of unitary operators on a complex Hilbert space $\mathcal{H}$. The <a href="http://en.wikipedia.org/wiki/Stone%E2%80%93von_Neumann_theorem" rel="nofollow">Stone-von Neumann Theorem</a> establishes that any such pair forming an irreducible representation of the Weyl relations,</p> <p>$U_sV_t = e^{is\cdot t}V_tU_s$</p> <p>is unitarily equivalent to the Schrödinger representation, and hence that all such representations are unitarily equivalent. (Note: the Weyl relations in this context are equivalent to the canonical commutation relations (CCRs) $[Q,P]\psi=i\psi$ for all $\psi$ in the common dense domain of $Q$ and $P$, where $Q$ and $P$ are the generators of $V$ and $U$.)</p> <p><strong>Question:</strong> Is there a known analogue of this result in the context of classical Hamiltonian mechanics?</p> <p>I don't know of a classical analogue of the Weyl relations. But there is a classical analogue of the CCRs, which is the Poisson bracket $\{q,p\}=1$. So, here's how I imagine a classical analogue of the Stone-von Neumann theorem might look (just a rough attempt, really!).</p> <p>Let $\mathcal{M}$ be a smooth $2n$-dimensional manifold and $\omega$ a symplectic form on $\mathcal{M}$. Let $\xi = (q,p)$ be any global coordinate system on $\mathcal{M}$, and let $Q:\mathcal{M}\rightarrow\mathbb{R}$ and $P:\mathcal{M}\rightarrow\mathbb{R}$ be the projections onto $q$ and $p$, respectively. Then (conjecture): all such pairs ($Q$, $P)$ satisfying,</p> <p>$\{Q,P\}=1$</p> <p>where $\{\cdot,\cdot\}$ is the Poisson bracket associated with $(\mathcal{M}, \omega)$, are related by a single canonical transformation.</p> <p>Does this seem like a reasonable way to formulate the classical analogue? Is the status of this conjecture obvious? Your thoughts are appreciated!</p> http://mathoverflow.net/questions/78573/classical-analogue-of-the-stone-von-neumann-theorem/79202#79202 Answer by Richard Montgomery for Classical analogue of the Stone-von Neumann Theorem? Richard Montgomery 2011-10-26T22:25:04Z 2011-10-26T22:25:04Z <p>Expanding on Chervov's comment: the Jacobian conjecture for two variables conjectures that if a polynomial map $(x,y) \to (X,Y)$ has for its Jacobian $\partial(X,Y)/\partial(x,y)$ a nonzero constant, then this polynomial map has a polynomial inverse. (The chain rule, plus the fact that over ${\mathbb C}$ polynomials have zeros, yields the truth of the converse: if the map has polynomial inverse, then its Jacobian is a nonzero constant.) If $(x,y) = (q,p), (X,Y) = (Q,P)$ then ${Q, P }$ is the Jacobian of this transformation. So: an affirmative answer for the Jacobian conjecture would precisely yield a 'yes' to your 'reasonable formulation' in the polynomial category when $n =1$.</p> <p>Going back to Stone-von-Neumann and the Weyl relations you wrote down suggests adding the hypothesis that the flows of the Hamiltonian vector fields for both $Q$ and $P$ are complete. With that addition, you have a chance of a 'yes' answer when $n=2$, perhaps in the smooth category. </p> <p>Note: your hypothesis imply that your manifold is ${\mathbb R}^{2n}$. </p>