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The generic definition of a symmetry in terms of an action is simply some transformation leaving the action invariant. If we have some configuration space $\mathrm{Conf}$, for instance for a point particle

\begin{equation} \mathrm{Conf} = C^\infty(\mathbb{R}, \mathbb{R}^n) \end{equation}

and the action is some functional

\begin{equation} S : \mathrm{Conf} \to \mathbb{R}. \end{equation}

Then a symmetry is just some automorphism on $\mathrm{Conf}$ leaving the action invariant :

\begin{equation} f \in \mathrm{Aut}(\mathrm{Conf}),\ S \circ f= S. \end{equation}

If we have our action $S$, the set of physical configurations is the critical locus of the $1$-form $\delta S$, ie the set on which this $1$-form vanishes :

\begin{equation} \mathrm{Crit}(S) = \{ \phi \in \mathrm{Conf} | \delta S_\phi = 0 \}. \end{equation}

A specific case of symmetries we can consider is the type generated by some vector on $T \mathrm{Conf}$. If we pick a configuration $\phi \in \mathrm{Conf}$, and a one-parameter group $\gamma : \mathbb{R} \to \mathrm{Aut}(\mathrm{Conf})$, such that every element is a symmetry of $S$, this generates a flow of $\phi$ along a curve, giving a family of configurations

\begin{equation} \gamma(0) = \phi,\ \gamma(t) = \phi_t \end{equation}

such that $S(\phi_t)$ is a constant for all values of $t$. This gives rise to a vector $v = \dot{\gamma}(0)$, for which we have \begin{equation} \delta S[v] = v[S] = 0 \end{equation}

ie the derivative in the $v$ direction is zero, since the action is constant along it.

By the Morse lemma, we have that any critical point $\phi \in \mathrm{Crit}(S)$ is isolated if the Hessian of $S$ is non-degenerate, $\det \delta^2 S \neq 0$ (those corresponding to actions without gauge symmetries), in the sense that there is a neighbourhood $U \subset \mathrm{Conf}$ for which $\phi \in U$ and there are no other critical points in $U$, meaning that for a non-gauge theory, the shell in the configuration space is totally disconnected.

Does that mean that any non-gauge symmetry, including "parametrized" ones like translation etc are discontinuous in the configuration space? For instance if we take some time translation symmetry

\begin{eqnarray} T : \mathrm{Conf} \times \mathbb{R} &\to& \mathrm{Conf}\\ (\phi, a) &\mapsto& \phi(- + a) \end{eqnarray}

which, on an object in the shell, will map to other objects in the shell, as the new derivative also vanishes

\begin{equation} \delta_{\phi(- + a)} S = \delta_{\phi} S = 0. \end{equation}

Is the action of this symmetry entirely discontinuous in $\mathrm{Conf}$?

Qmechanic
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Slereah
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1 Answers1

6

  1. Let us assume that the Lagrangian depends on time derivatives, so that we have a dynamical (rather than a static) physical system. Then we need to impose boundary conditions (BCs) to have a well-posed variational problem.

  2. Critical points are determined by the Euler-Lagrange (EL) equations plus the BCs. In point mechanical applications, there are typically only a discrete number of critical points/instanton sectors.

  3. Note that a symmetry transformation (ST) does typically not map a critical point into a critical point as a ST does typically not preserve BCs, cf. e.g. this Phys.SE post.

  4. Example: A free particle $L=\frac{m}{2}\dot{q}^2$ with Dirichlet BCs $$ q(t_i)~=~q_i\quad\text{and}\quad q(t_f)~=~q_f;$$ and translational symmetry $q\to q+a.$

Qmechanic
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