I recently got interested in the Galilean group and its central extension and found a paper "Quantization on a Lie group: Higher-order Polarizations" by Aldaya, Guerrero and Marmo.

Before asking my question, I will give a brief refresher on the subject. Consider a non relativistic particle moving in one dimension. The Lagrangian is

$$ L = \frac{1}{2} m \dot x^2. $$

A Galilean boost sends

$$ x \mapsto x + vt $$ or infinitesimally, $$ \delta x = \varepsilon t. $$ If $\varepsilon$ is a constant, then $$ \delta L = m \dot x \varepsilon = \frac{d}{dt} ( m x \varepsilon) $$ i.e., $L$ changes by a total derivative. This means we can use this symmetry variation to find a conserved quantity by Noether's theorem. (In the paper I referenced, the fact that $L$ changes by a total derivative called the "semi-invariance" of $L$.)

If we now say $\varepsilon = \varepsilon(t)$ is time dependent, we can carry out the Noether procedure to find the associated conserved quantity. So let's do that. Now that $\varepsilon$ is time dependent we have

$$ \delta L = m \dot x \frac{d}{dt} (\varepsilon t) = m \dot x \varepsilon + m \dot x \dot \varepsilon t. $$

On solutions to the equations of motion, we have \begin{align*} 0 &= \delta S \\ &= \int dt \delta L \\ &= \int dt \big( m \dot x \varepsilon + m \dot x \dot \varepsilon t \big) \\ &= \int dt \ \varepsilon \frac{d}{dt} \big( mx - m\dot x t\big) \end{align*}

Because $\delta S = 0$ no matter what $\varepsilon(t)$ is, we know that the "boost charge" $$ m \dot x t - mx $$ is conserved. Often this is rewritten as $$ p t - mx. $$ This conserved quantity comes off as weird and its physical interpretation is unclear. I think the best way to think of it is that, when your system has Galilean boost invariance, your particle is guaranteed to travel with a constant velocity. (Literally speaking, $mx - pt$ is $m$ times the initial position of your particle at $t = 0$. The fact that the initial position can be calculated in this simple way is due to the fact that the velocity is a constant $p/m$.)

Now, Noether's theorem is a two way street. In the Lagrangian formalism, you can use symmetries to calculated conserved quantities. In the Hamiltonian formalism, those conserved quantities will "generate" the original symmetry. The conserved quantity $Q$ generates the symmetry given by $$ \delta q_i = \frac{\partial Q}{\partial p_i} {\hspace 1 cm} \delta p_i = -\frac{\partial Q}{\partial q_i}. $$

Define the "boost charge" $K$ on phase space as

$$ K = p t - m q. $$ This generates $$ \delta q = \frac{\partial K}{\partial p} = t{\hspace 1 cm} \delta p = -\frac{\partial K}{\partial q} = m $$ which are indeed the infinitesimal changes to $q$ and $p$ by a boost.

The boost charge $K$ has an interesting relationship with the translation generator $p$ (the momentum). Note that the Poisson bracket of the two is the constant $m$. $$ \{p, K\} = m $$ The Galilean group is the group of space translations, rotations, time translations, and boosts (all the symmetries of non relativistic mechanics.) Its Lie algebra has the central extension, as written above. For the quantum particle, performing a translation, a boost, the inverse translation, and the inverse boost, leaves behind a constant phase proportional to $m$. (This can also be seen as a limiting case of the Poincare alegbra, $[K_i, P_i] = P_0$, where $P_0$ is the energy. In the $v \ll c$ limit, $P_0 = m$.)

Okay, the preamble is now over, here is my question. In the paper I linked to above, they seem to imply that the the "semi invariance" of boosts (that $L$ changes by a total derivative and not by $0$) is somehow related to the central charge, but I can't figure out if a concrete reason is given.

Does anyone know is there is a relationship between symmetries which change $L$ by a total derivative and central charges?


The quasisymmetry of Galilean boosts and the central extension of the Galilean algebra into the Bargmann algebra is discussed in e.g. Refs. 1-2 and this Phys.SE post.

FWIW, it is not true that any quasisymmetry is associated with a central extension. Counterexample: Consider the quasisymmetry $\delta x^k=\varepsilon \dot{x}^k$ of the Lagrangian $L(x,\dot{x})=\frac{m}{2}\dot{x}^2-V(x)$. The conserved quantity is just the energy, i.e. $H$ in the Hamiltonian formulation.


  1. V. Aldaya, J. Guerrero & G. Marmo, Quantization on a Lie group: Higher-order Polarizations, arXiv:physics/9710002; p. 6-8.

  2. R. Andringa, E. Bergshoeff, S. Panda & M. de Roo, Newtonian Gravity and the Bargmann Algebra, arXiv:1011.1145; p. 11.

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  • $\begingroup$ In the question you link to, David Bar Moshe points out that boosts do not leave $\theta = p dq$ invariant. I was thinking that maybe this had something to do with the prequantization algebra $Q(f) = - i \hbar X_f - \theta(X_f) + f$ which is a central extension of the Lie algebra of the vector fields. $\endgroup$ – user1379857 Sep 29 '19 at 0:47

Aha. I have figured out the answer, and it is quite pleasing. (The answer is yes, total derivatives have everything to do with central extensions.) Funnily enough it goes back to a question I asked a while ago.

In this question, I show that under the tiny transformation generated by a conserved quantity $Q$ (which satisfies $\{Q, H\} = 0$), the Lagrangian $L$ will change by

$$ \delta L = \frac{d}{dt} \Big( p_i \frac{\partial Q}{\partial p_i} - Q\Big). $$ (The repeated index $i$ is implicitly summed over from $1$ to $N$.) Therefore, if $Q$ satisfies $$ Q = p_i \frac{\partial Q}{\partial p_i} $$ then $\delta L =0 $. For lack of a better word, we will call any function $Q$ satisfying the above equality an "invariant" function, because it leaves $L$ invariant. Meanwhile, if $$ Q \neq p_i \frac{\partial Q}{\partial p_i} $$ then $\delta L$ is a total derivative, and not invariant.

(I have defined this term "invariant" in order to solve the question at hand. "Invariant" functions seem to have quite a few nice properties. Does anyone know if there is a proper name for these functions, and if more is known about them?)

Invariant functions are always of the form $$ Q = p_i f_i(q) $$ for arbitrary functions of position $f_i(q)$. Momentum $(Q = p)$ and angular momentum $(Q = p_1 q_2 - p_2 q_1)$ are examples of invariant functions. A random Hamiltonian $(Q = H)$ is not invariant, and in that case $\delta L = \tfrac{d}{dt} L$, as expected.

Let me now give a quick review of symplectic geometry. For any quantity $Q$, we can make it's Hamiltonian vector field $X_Q$, defined to be $$ X_Q = -\frac{\partial Q }{\partial p_i} \partial_{q_i} + \frac{\partial Q }{\partial q_i} \partial_{p_i}. $$ It satisfies $$ X_Q = \{ Q, \cdot \}. $$ From the Jacobi identity, $$ \{ \{g, h\}, f \} + \{ \{h, f\}, g \} + \{ \{f, g\}, h \} = 0 $$ which can be rearranged to be $$ \{ \{g, h\}, f \} = \{g, \{h, f\} \} - \{ h, \{g, f\} \} $$ $$ X_{ \{g, h\} } (f) = X_g (X_h (f) ) - X_h (X_g (f)) $$ $$ X_{ \{g, h\} } = [X_g, X_h]. $$ So $Q \mapsto X_Q$ is a Lie algebra homomorphism where the Poisson bracket is sent to the Lie Bracket. Note that a constant function is sent to the $0$ vector field, which is the origin of how the Lie algebra of vector fields becomes centrally extended.

We define the symplectic form to be $$ \omega = d q_i \wedge d p_i. $$ Note that $$ \omega(X_f, \cdot) = df $$ and therefore $$ \omega(X_f, X_g) = \{f, g\}. $$

Okay, the review is over. What does this have to do with our "invariant" functions? Well, note that the 1-form

$$ \theta = p_i d q_i $$ satisfies $$ \omega = - d \theta. $$ It is therefore called the "symplectic potential." Its relevance to our story is that \begin{align*} \theta (X_Q) &= p_i dq_i \big( -\frac{\partial Q }{\partial p_j} \partial_{q_j} + \frac{\partial Q }{\partial q_j} \partial_{p_j} \big) \\ &= - p_i \frac{\partial Q }{\partial p_i} \end{align*} Therefore, we can see that invariant functions satisfy $$ Q = - \theta (X_Q). $$

Now let's talk about central extensions. As discussed in this other question, $\omega(X_f, X_g)$, evaluated at an arbitrary point $m_0$, gives us the central extension from vector fields back to functions on phase space. The ambiguity in choosing $m_0$ reflects the ambiguity of adding "the boundary of a 1-cocycle to our central extension," i.e. a linear function $b([X_f, X_g])$. This is because

\begin{align*} \omega( [X_f, X_g], \cdot ) &= \omega( X_{\{f, g\} }, \cdot ) \\ &= d (\{f, g\} ) \end{align*} and so \begin{align*} b([X_f, X_g]) &= \int_{m_0}^{m_1} \omega( [X_f, X_g], \cdot ) \\ &= \int_{m_0}^{m_1}d (\{f, g\} ) \\ &= \{f, g\}\biggr\rvert_{m_1} - \{f, g\}\biggr\rvert_{m_0} \\ &= \omega(X_f, X_g)\biggr\rvert_{m_1} - \omega(X_f, X_g)\biggr\rvert_{m_0} \end{align*} as desired.

Claim: If $f$ and $g$ are invariant functions, i.e. they satisfy $$ f = - \theta(X_f) \hspace{1 cm} g = - \theta(X_g), $$ then $$\{f, g\}$$ is also an invariant function.

Proof 1 (for babies): We know we can write $$ f = p_i f_i(q) \hspace {1 cm} g = p_i g_i(q) $$ for some functions $f_i(q)$ and $g_i(q)$ which are only functions of $q$. A simple computation yields $$ \{f, g\} = p_i (f_i' g_i - f_i g_i'). $$ Because $(f_i' g_i - f_i g_i')$ is only a function of $q$, we can see that $\{f, g\} $ is also invariant.

Proof 2 (for grown-ups): Using the coordinate free expression for the exterior derivative, \begin{align*} \{f, g\} &= \omega(X_f, X_g) \\ &= - d \theta (X_f, X_g) \\ &= - \Big( X_f (\theta(X_g) ) - X_g (\theta(X_f)) - \theta([X_f, X_g]) \Big) \\ &= X_f(g) - X_g(f) + \theta([X_f, X_g]) \\ &= 2 \{f, g\} + \theta([X_f, X_g]) \end{align*} which implies $$ \{f, g\} = - \theta([X_f, X_g]) = - \theta(X_{\{f, g\} }) $$ showing $\{f, g\}$ is invariant, as desired.

Claim: If $f$ and $g$ are invariant functions then $\omega(X_f, X_g)$ can be written as some linear function of $[X_f, X_g]$, and is therefore the "boundary of a 1-cocycle," and hence is a trivial central extension. (THEREFORE, the central extension of two invariant functions (functions that do not change $L$ by a total derivative) is always trivial, as I suspected!)

Proof: Because $\{f, g\}$ is also invariant, \begin{align*} \omega(X_f, X_g) &= \{f, g\} \\ &= - \theta ( X_{ \{f, g\} } ) \\ &= - \theta ( [X_f, X_g ] ) \end{align*} thus concluding the proof.

By the way, if you choose the basepoint $m_0$ to be the origin, where all $p_i = 0$, then the symplectic form evaluated there is $0$ if both $f$ and $g$ are invariant, giving us an explicit realization that the central charge really is zero: $$ \omega(X_f, X_g)\biggr\rvert_{m_0} = \{f, g\}\biggr\rvert_{m_0} = p_i (f_i' g_i - f_i g_i')\biggr\rvert_{m_0} = 0. $$

We have now developed a nice little theory of these "invariant" functions. We can see that the central extension arising from the Poisson bracket of any two invariant functions is always trivial. Therefore, if any two functions DO gain a central extension, then one of the two functions can't be invariant. For instance, $\{q, p\} = 1$ implies that one of $q$ or $p$ can't be invariant. And of course, it is obviously $q$ that is the non invariant one. Likewise, in the case of $\{K, p\} = m$, we can see that $K$ is also not an invariant function because it changes the Lagrangian by a total derivative.

(Doesn't it seem like this story is relevant to the story of the Brown Henneaux central charge? Small diffeomorphisms do not develop a central charge, but large ones do.)

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