# Solution space of a differential equation with 3D rotational symmetry

We know that the space of solutions will be invariant under 3D rotations, but why can we say that the space of solutions will constitute a representation of the rotation group $SO(3)$? We know that a Lie group representation is just a Lie group homomorphism from this Lie group to the space of linear transformations on some vector space: $$\Pi:G\longrightarrow GL(V)$$ Then if we are going to think of the solution space as a representation of $SO(3)$, what exactly is the solution space acting on or what exactly is the map for the homomorphism?

• Is the equation linear? – Valter Moretti Nov 10 '14 at 13:04
• is the linearity relevant? actually I don't really know much about solving DEs using Lie theory. what I described above is a statement taken from the Lie group book by Brian Hall, which intends to illustrate some examples for the applications of group representations. – M. Zeng Nov 10 '14 at 13:16
• Related: physics.stackexchange.com/q/19751/2451 , physics.stackexchange.com/q/53462/2451 and links therein. – Qmechanic Nov 10 '14 at 13:21
• is it very related? i don't quite see it. could you be more specific? – M. Zeng Nov 10 '14 at 13:28
• @Timo I asked about linearity just to understand if you refer to linear representations of $SO(3)$ or more general representations. – Valter Moretti Nov 10 '14 at 13:31

I assume to deal with an autonomous, first order (at least $C^1$ or smooth) system of ordinary differential equations and that the hypotheses sufficient for existence and uniqueness of maximal solutions are satisfied.

You may always reduce to the case of a first order system by adding auxiliary variables, $\dot{x}$, to the initial system of differential equations and adding trivial equations like $\frac{dx}{dt} = \dot{x}$.

The system of differential equations is assigned in some manifold $M$, for instance $\mathbb R^n$.

As far as I understand there is a natural action of $SO(3)$ on $M$ made of diffeomorphisms. In other words there is a map $$SO(3) \ni R \mapsto \phi_R \in Diff(M)$$ such that $\phi_R \phi_{R'}= \phi_{RR'}$ and $\phi_I = id$. I also expect that $M\times SO(3) \ni (p,R) \mapsto \phi_R(p) \in M$ is jointly smooth.

If $q\in M$ there is exactly one maximal solution through $q$ for $t=0$: $$\gamma_q: I_q \ni t \mapsto \gamma_q(t) \in M$$ with $I_q$ an open interval of $\mathbb R$ including $0$.

The space of solutions $S$ can be defined as $\{\gamma_q\}_{q\in M}$. (Actually we should also take the quotient with respect to the equivalence relation $\gamma_q \sim \gamma_{q'}$ iff $\gamma_q = \gamma_{q'}$. The quotient space is the true space of solutions if we want that $q\in M$ faithfully label the solutions.)

Next, saying the system of differential equations is $SO(3)$ invariant, means that $\phi_R \circ \gamma_q \in S$ if $\gamma_q \in S$. Obviously, due to the uniqueness theorem $$\phi_R \circ \gamma_q = \gamma_{\phi(q)}\quad \mbox{and}\quad I_q = I_{\phi_R(q)}$$ This way gives rise to a representation of $SO(3)$ on $S$ defined by $SO(3) \ni R \mapsto g_R$ with $$(g_R \gamma_q) := \gamma_{\phi_R(q)} = \phi_R \circ \gamma_q\:.$$ It is easy to prove that $g_I =id$ and that $g_Rg_{R'}= g_{RR'}$. Finally the action of $g$ is smooth as well. I mean that that the map $$I_q \times SO(3) \ni (t, R) \mapsto g_R(\gamma_q)(t) \in M$$ is smooth (at least $C^1$) in view of the smooth (or at least $C^1$) dependence theorem of the maximal solutions of a ODE from the initial conditions. Actually $A := U_{q\in M} \{q\} \times I_q \subset M\times \mathbb R$ is an open set (as a general result of ODE) and thus one can focus on the map $$A \times SO(3) \ni (q, t, R) \mapsto g_{R}(\gamma_q)(t) \in M$$ which turns out to be smooth (at least $C^1$) as well.