mathematical physics - If all conserved quantities of a system are known, can they be explained by symmetries?

If a system has $N$ degrees of freedom (DOF) and therefore $N$ independent¹ conserved quantities integrals of motion, can continuous symmetries with a total of $N$ parameters be found that deliver these conserved quantities by means of Noether's theorem? I think this is not exactly the opposite of Noether's theorem since I don't ask whether for each conserved quantity a symmetry can be retrieved, I ask about a connection between the whole set of conserved quantities and symmetries.

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¹⁾ or $2N-1$, or $N$, depending on definition and details that are irrelevant here. But let me expand on it anyway... I consider the number of DOFs equal to the number of initial conditions required to fully describe a system in Classical Mechanics. That means, velocities (or momenta) are considered individual DOFs, and not that each pair of coordinate + velocity make up only one DOF. Time is no DOF however, it is a parameter. Please discuss this in this question if you disagree.

Answer

There are already several good answers. However, the off-shell aspect related to Noether Theorem has not been addressed so far. (The words on-shell and off-shell refer to whether the equations of motion (e.o.m.) are satisfied or not.) Let me rephrase the problem as follows.

Consider a (not necessarily isolated) Hamiltonian system with $N$ degrees of freedom (d.o.f.). The phase space has $2N$ coordinates, which we denote $(z^1, \ldots, z^{2N})$.

(We shall have nothing to say about the corresponding Lagrangian problem.)

1) Symplectic structure. Usually, we work in Darboux coordinates $(q^1, \ldots, q^N; p_1, \ldots, p_N)$, with the canonical symplectic potential one-form

$$\vartheta=\sum_{i=1}^N p_i dq^i.$$

However, it turns out to be more efficient in later calculations, if we instead from the beginning consider general coordinates $(z^1, \ldots, z^{2N})$ and a general (globally defined) symplectic potential one-form

$$\vartheta=\sum_{I=1}^{2N} \vartheta_I(z;t) dz^I,$$

with non-degenerate (=invertible) symplectic two-form

$$\omega = \frac{1}{2}\sum_{I,J=1}^{2N} \omega_{IJ} \ dz^I \wedge dz^J = d\vartheta,\qquad\omega_{IJ} =\partial_{[I}\vartheta_{J]}=\partial_{I}\vartheta_{J}-\partial_{J}\vartheta_{I}. $$

The corresponding Poisson bracket is

$$\{f,g\} = \sum_{I,J=1}^{2N} (\partial_I f) \omega^{IJ} (\partial_J g), \qquad \sum_{J=1}^{2N} \omega_{IJ}\omega^{JK}= \delta_I^K. $$

2) Action. The Hamiltonian action $S$ reads

$$ S[z]= \int dt\ L_H(z^1, \ldots, z^{2N};\dot{z}^1, \ldots, \dot{z}^{2N};t),$$

where

$$ L_H(z;\dot{z};t)= \sum_{I=1}^{2N} \vartheta_I(z;t) \dot{z}^I- H(z;t) $$

is the Hamiltonian Lagrangian. By infinitesimal variation $$\delta S = \int dt\sum_{I=1}^{2N}\delta z^I \left( \sum_{J=1}^{2N}\omega_{IJ} \dot{z}^J-\partial_I H - \partial_0\vartheta_I\right)+ \int dt \frac{d}{dt}\sum_{I=1}^{2N}\vartheta_I \delta z^I, \qquad \partial_0 \equiv\frac{\partial }{\partial t},$$

of the action $S$, we find the Hamilton e.o.m.

$$ \dot{z}^I \approx \sum_{J=1}^{2N}\omega^{IJ}\left(\partial_J H + \partial_0\vartheta_J\right) = \{z^I,H\} + \sum_{J=1}^{2N}\omega^{IJ}\partial_0\vartheta_J. $$

(We will use the $\approx$ sign to stress that an equation is an on-shell equation.)

3) Constants of motion. The solution

$$z^I = Z^I(a^1, \ldots, a^{2N};t)$$

to the first-order Hamilton e.o.m. depends on $2N$ constants of integration $(a^1, \ldots, a^{2N})$. Assuming appropriate regularity conditions, it is in principle possible to invert locally this relation such that the constants of integration

$$a^I=A^I(z^1, \ldots, z^{2N};t)$$

are expressed in terms of the $(z^1, \ldots, z^{2N})$ variables and time $t$. These functions $A^I$ are $2N$ constants of motion (c.o.m.), i.e., constant in time $\frac{dA^I}{dt}\approx0$. Any function $B(A^1, \ldots, A^{2N})$ of the $A$'s, but without explicit time dependence, will again be a c.o.m. In particular, we may express the initial values $(z^1_0, \ldots, z^{2N}_0)$ at time $t=0$ as functions

$$Z^J_0(z;t)=Z^J(A^1(z;t), \ldots, A^{2N}(z;t); t=0)$$

of the $A$'s, so that $Z^J_0$ become c.o.m.

Now, let

$$b^I=B^I(z^1, \ldots, z^{2N};t)$$

be $2N$ independent c.o.m., which we have argued above must exist. The question is if there exist $2N$ off-shell symmetries of the action $S$, such that the corresponding Noether currents are on-shell c.o.m.?

Remark. It should be stressed that an on-shell symmetry is a vacuous notion, because if we vary the action $\delta S$ and apply e.o.m., then $\delta S\approx 0$ vanishes by definition (modulo boundary terms), independent of what the variation $\delta$ consists of. For this reason we often just shorten off-shell symmetry into symmetry. On the other hand, when speaking of c.o.m., we always assume e.o.m.

4) Change of coordinates. Since the action $S$ is invariant under change of coordinates, we may simply change coordinates $z\to b = B(z;t)$ to the $2N$ c.o.m., and use the $b$'s as coordinates (which we will just call $z$ from now on). Then the e.o.m. in these coordinates are just

$$\frac{dz^I}{dt}\approx0,$$

so we conclude that in these coordinates, we have

$$ \partial_J H + \partial_0 \vartheta_J=0$$

as an off-shell equation. [An aside: This implies that the symplectic matrix $\omega_{IJ}$ does not depend explicitly on time,

$$\partial_0\omega_{IJ} =\partial_0\partial_{[I}\vartheta_{J]}=\partial_{[I} \partial_0\vartheta_{J]}=-\partial_{[I}\partial_{J]} H=0.$$

Hence the Poisson matrix $\{z^I,z^J\}=\omega^{IJ}$ does not depend explicitly on time. By Darboux Theorem, we may locally find Darboux coordinates $(q^1, \ldots, q^N; p_1, \ldots, p_N)$, which are also c.o.m.]

5) Variation. We now perform an infinitesimal variation $\delta= \varepsilon\{z^{I_0}, \cdot \}$,

$$\delta z^J = \varepsilon\{z^{I_0}, z^J\}=\varepsilon \omega^{I_0 J},$$

with Hamiltonian generator $z^{I_0}$, where $I_0\in\{1, \ldots, 2N\}$. It is straightforward to check that the infinitesimal variation $\delta= \varepsilon\{z^{I_0}, \cdot \}$ is an off-shell symmetry of the action (modulo boundary terms)

$$\delta S = \varepsilon\int dt \frac{d f^0}{dt}, $$

where

$$f^0 = z^{I_0}+ \sum_{J=1}^{2N}\omega^{I_0J}\vartheta_J.$$ The bare Noether current is

$$j^0 = \sum_{J=1}^{2N}\frac{\partial L_H}{\partial \dot{z}^J} \omega^{I_0 J}=\sum_{J=1}^{2N}\omega^{I_0J}\vartheta_J,$$

so that the full Noether current

$$ J^0=j^0-f^0=-z^{I_0} $$

becomes just (minus) the Hamiltonian generator $z^{I_0}$, which is conserved on-shell $\frac{dJ^0}{dt}\approx 0$ by definition.

So the answer is yes in the Hamiltonian case.