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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?

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    $\begingroup$ Is the equation linear? $\endgroup$ – Valter Moretti Nov 10 '14 at 13:04
  • $\begingroup$ 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. $\endgroup$ – M. Zeng Nov 10 '14 at 13:16
  • $\begingroup$ Related: physics.stackexchange.com/q/19751/2451 , physics.stackexchange.com/q/53462/2451 and links therein. $\endgroup$ – Qmechanic Nov 10 '14 at 13:21
  • $\begingroup$ is it very related? i don't quite see it. could you be more specific? $\endgroup$ – M. Zeng Nov 10 '14 at 13:28
  • $\begingroup$ @Timo I asked about linearity just to understand if you refer to linear representations of $SO(3)$ or more general representations. $\endgroup$ – Valter Moretti Nov 10 '14 at 13:31
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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.

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