User soulphysics - MathOverflow

Answer by soulphysics for Counterexamples in PDE

2013-01-08T01:42:57Z

A particularly simple example is Norton's dome, with height given as a function of radial distance on the surface of the dome by

$h = \frac{2}{3g}r^{3/2}$

where $g$ is the gravitational constant near the surface of the earth. The dome has a curvature singularity at the apex. And, if we model a mass at the ($r=0$) apex of this dome with zero velocity, we find that Newton's equation does not have a unique solution; the mass can "fall" at any arbitrary time $t$ for no reason at all.

Norton's paper about the dome: http://philsci-archive.pitt.edu/2943/

A helpful reply: http://philsci-archive.pitt.edu/3195/

"Complete" collections of Hamiltonian flows on symplectic manifolds

2012-11-13T00:31:03Z

Question for Math Overflow

Let $M$ be a smooth $2n$-dimensional manifold and $\Omega$ a symplectic form, and let $(q,p)=(q_1,...,q_n,p_1,...,p_n)$ be a local (Darboux) coordinate system on $M$. Let $Q_i:M\rightarrow\mathbb{R}$ be the projection function $Q_i(q,p)=q_i$, and similarly let $P_i(q,p)=p_i$. These projection functions have the following property: if a function $f:M\rightarrow\mathbb{R}$ Poisson commutes with all of them, in that $\{Q_i,f \}=\{P_i,f\}=0$ for each $i=1,...,n$, then it must be a constant function, $f(q,p)=k\in\mathbb{R}$.

I am interested in when the generators of a collection of Hamiltonian flows will have this property. Let $\varphi^{s_i}_\sigma$ and $\varphi^{r_i}_\rho$ be the local Hamiltonian flows generated by a set of functions $s_i$ and $r_i$, for $i=1,...,n$. (That is, $\varphi^{s_i}_\sigma$ is a group of symplectic transformations parametrized by $\sigma$, which threads the vector field on $M$ generated by $s_i$, and similarly for $\varphi^{r_i}_\rho$.) If the set of $2n$ functions $s_i$, $r_i$ satisfy the property above (that $\{s_i,f \}=\{r_i,f\}=0$ for all $i$ only if $f$ is constant), then I'll call this collection of flows complete. (This is different, of course, from the notion of a "complete Hamiltonian flow.")

When is a collection of Hamiltonian flows complete in this way? Has anyone heard of or worked on an interesting criterion?

For example, I thought it might be interesting if one of the groups were known to represent 'translations' of some interesting quantity. One way to have this would be to have a maximal set of Poisson commuting functions $q_1, ..., q_n$ such that for all $i,j\in\mathbb{R}$ and for all $\sigma\in\mathbb{R}$,

$q_i\circ\varphi^{s_j}_\sigma = q_i + \delta_{ij}\sigma$

$q_i\circ\varphi^{r_j}_\rho = q_i$

(where $\delta_{ij}=1$ if $i=j$ and $0$ otherwise). Is there something beyond this that might make $\varphi^{s_i}_\sigma$, $\varphi^{r_j}_\rho$ a complete collection in the sense above? I'd be curious to hear your thoughts.

When do commuting Hamiltonian flows have commuting generators?

2012-06-01T02:45:59Z

Let $(P,\Omega)$ be a symplectic manifold, and let $[\cdot,\cdot]$ be the natural Poisson bracket. Let $\varphi^h(a)$ be the Hamiltonian flow generated by the smooth function $h:P\rightarrow\mathbb{R}$, and let $\varphi^g(b)$ be the Hamiltonian flow generated by the smooth function $g:P\rightarrow\mathbb{R}$.

Suppose the two flows commute, $\varphi^g(b)\varphi^h(a) = \varphi^h(a)\varphi^g(b)$. Are there interesting circumstances under which it follows that the generators commute as well, $[g,h]=0$?

I know that the flows commuting implies that $[g,h]=k$ for some constant $k$; a reference for the proof of this can be found, for example, in Arnold's (1989) Mathematical Methods of classical mechanics, pg. 218 Cor. 9. It is also easy to show that if $[g,h]=0$, then $\varphi^g(b)\varphi^h(a) = \varphi^h(a)\varphi^g(b)$. But under what circumstances does the converse hold?

It's interesting that in quantum mechanics, the analogous relations hold in both directions. That is, if $G$ and $H$ are linear self-adjoint operators on a Hilbert space, and if $e^{ibG}$ and $e^{iaH}$ are the continuous one-parameter unitary groups they generate, then $e^{ibG}e^{iaH}=e^{iaH}e^{ibG}$ if and only if $[H,G]=0$ (where now $[\cdot,\cdot]$ is the commutator bracket).

Classical analogue of the Stone-von Neumann Theorem?

2011-10-19T12:40:36Z

Let $U_s$, $V_t$ be a pair of continuous $n$-parameter groups ($n < \infty$) of unitary operators on a complex Hilbert space $\mathcal{H}$. The Stone-von Neumann Theorem establishes that any such pair forming an irreducible representation of the Weyl relations,

$U_sV_t = e^{is\cdot t}V_tU_s$

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 $U$.)

Question: Is there a known analogue of this result in the context of classical Hamiltonian mechanics?

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!).

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,

$\{Q,P\}=1$

where $\{\cdot,\cdot\}$ is the Poisson bracket associated with $(\mathcal{M}, \omega)$, are related by a single canonical transformation.

Does this seem like a reasonable way to formulate the classical analogue? Is the status of this conjecture obvious? Your thoughts are appreciated!

Do unitary bijections act invariantly on irreducible representations?

2011-02-09T20:49:03Z

Let $\mathcal{A}$ be a $C^*$ algebra. Let $(\pi, \mathcal{H})$ be a faithful, irreducible, unitary, Hilbert space representation of $\mathcal{A}$; i.e., $\pi:\mathcal{A}\rightarrow\mathcal{B}(\mathcal{H})$ is an injective *-homomorphism from $\mathcal{A}$ into a subset of the bounded operators $\mathcal{B}(\mathcal{H})$ on a Hilbert space $\mathcal{H}$, such that the resulting representation is irreducible. (A representation is irreducible if whenever $\pi(\mathcal{A})$ acts invariantly on a subspace $\mathcal{H}^\prime \subseteq \mathcal{H}$, then $\mathcal{H}^\prime = \mathbf{0}$ or $\mathcal{H}^\prime = \mathcal{H}$.)

Let $U \in \mathcal{B}(\mathcal{H})$ be any unitary ($U^* = U^{-1}$) bijection on $\mathcal{H}$, not necessarily in $\pi(\mathcal{A})$.

Does it follow that $U\pi(\mathcal{A})U^{*} = \pi(\mathcal{A})$?

Is exp(rA) = (exp(A))^r for real r and A in a Banach space?

2010-12-24T15:29:40Z

Is $e^{(rA)} = (e^{A})^r$ when $r \in \mathbb{R}$ and $A$ is an element of a Banach algebra?

Clearly if $n$ is an integer, then

$e^{(nA)} = e^{A+A \cdots +A} = e^{A}e^{A}\cdots e^{A} = (e^{A})^n$,

where the second equality follows from the Baker-Hausdorff lemma and the fact that [A,A]=0. On the other hand, I think the equality is not generally true when $r \in \mathbb{C}$. But what about the reals?

Many thanks for your thoughts!

What is the outer automorphism group of SU(n)?

2010-09-30T19:43:49Z

All the automorphisms of $SU(2)$ seem to be inner, which would mean that $\mathrm{Out}$ $SU(2)$ is trivial. Is that correct? Is this true in general $SU(n)$? I can't quite see -- any thoughts would be helpful.

Answer by soulphysics for mathematics in nature

2010-07-19T11:50:39Z

A few quick examples:

Breeding bunnies $\rightarrow$ the Fibonacci sequence
Projectile/planetary motion $\rightarrow$ conic sections
Natural springs and sinks $\rightarrow$ Gauss' law
Galilean relativity + constant speed of light $\rightarrow$ non-Euclidean geometry
Relativistic gravity $\rightarrow$ intrinsic curvature
Invariances of Maxwell equations $\rightarrow$ Conformal transformations
Observables in quantum mechanics $\rightarrow$ Lie groups (a la Wigner)

EDIT:

Symmetries of the electron $\rightarrow$ Quaternions

Answer by soulphysics for Why are they called isothermal coordinates?

2010-07-16T13:57:42Z

According to Gray, Abbena and Salamon, that's the name given to such coordinate systems by Gabriel Lamé in his 1833 study of heat transfer. The reason is, if you've got a thermally isolated surface of constant heat conduction, the constant coordinate lines are isotherms iff the coordinates are isothermal.

How big is the center of an orthogonal group?

2010-07-16T12:56:07Z

How big is the center of an arbitrary orthogonal group $O(m,n)$?

In the special case of the "circle group" $O(2)$, it's clear that $|\zeta O(2)|$ = 1. In the case of $O(3)$, it seems clear that the center has two elements $\zeta O(3) = \lbrace 1, -1 \rbrace$. I can see this by visualizing a sphere in an arbitrary $(i, j, k)$ basis, and observing that both the identity and the "complete" reversal $(i, j, k) \mapsto (-i, -j, -k)$ commute with everthing.

But I'd like a simple way to see how the situation changes for more general orthogonal groups like the (inhomogeneous) Lorentz group $O(3,1)$.

Comment by soulphysics

2012-06-03T20:22:59Z

> There they commute iff... $[G,H]$ is a constant multiple of the identity. I understand $\mathbb{P}\mathcal{H}$ to contain the equivalence classes of vectors related by a multiplicative constant. But then, isn't calling $[G,H]$ a "constant multiple of the identity" the same as saying $[G,H]=0$? It seems to me that on the projective space $\mathbb{P}\mathcal{H}$, it is also the case that $[e^{ibG}, e^{iaH}]=0$ iff $[G,H]=0$. What am I missing? [P.S.: Thanks for the Roels and Weinstein reference, that is very helpful.]

Comment by soulphysics

2012-06-02T22:25:39Z

As it happens I am exceedingly lowbrow, touché! ;) Just in case, would you happen to have a reference for the Souriau approach?

Comment by soulphysics

2010-12-26T17:21:30Z

Oh dear -- you're of course correct, the claim is not generally true. Thanks!

Comment by soulphysics

2010-12-24T16:44:29Z

@fedja -- One way is the Taylor expansion: $A^r = \Sigma^\infty_{n=0}\frac{r^n \log^n(A)}{n!},$ where the logarithms are again defined by their Taylor expansions. @arsmath -- That's a way to define the operator $e^X$ (where $X$ is in the Banach algebra), but what's needed is $X^r$.

Comment by soulphysics

2010-09-30T20:06:52Z

Excellent, I can see that now. Thanks very much! And now it seems that complex conjugation sends the fundamental representation of $SU(2)$ to itself. So, am I correct to think that $\mathrm{Out}$ $SU(2)$ is trivial?

Comment by soulphysics

2010-07-21T12:12:34Z

@Victor -- obviously. But the question asks for natural phenomena that "perfectly illustrate" a mathematical concept, not "historically inspired". Sheesh.

Comment by soulphysics

2010-07-19T12:00:27Z

Hey, it's an idealization, ok? And anyway, if it was good enough for Fibonacci himself... ;)

Comment by soulphysics

2010-07-16T15:29:56Z

Thanks Robin -- I've convinced myself now. Although the Lorentz group has four interesting discrete transformations, only I and -I are central.

Comment by soulphysics

2010-07-16T14:06:01Z

Robin -- could you elaborate? I seem to recall from somewhere that the Lorentz group has a 4-element center...

Comment by soulphysics

2010-07-16T13:20:00Z

Thanks -- I was thinking about $O$ over the real number field, for which what I said is correct. But as you suggest, the question gets more interesting for $O$ over arbitrary fields!