Intro

In this work, the symetries of a system of partial differential equations were constructed from their generators X∈Γ(TE)X\in\Gamma\left(T\mathcal{E}\right), where (E,π,C3+1)\left(\mathcal{E},\pi,C^{3+1}\right) is a fiber bundle with typical fiber composed by the electric and magnetic fields and base manifold C3+1C^{3+1}, the Carrollian manifold in four dimensions. This was accomplished in a two step process. First we need to find a curve γ:R⟶E\gamma:\mathbb{R}\longrightarrow\mathcal{E} whose tangent vector γ˙(λ)\dot{\gamma}(\lambda) is equal to the generator XX evaluated along the curve, denoted by Xγ(λ)X_{\gamma(\lambda)}, for any point γ(λ)∈E\gamma(\lambda)\in\mathcal{E}, this amounts to solving the system of ordinary differential equations given by

γ˙(λ)=Xγ(λ).\begin{align} \dot{\gamma}(\lambda)=X_{\gamma(\lambda)}. \end{align}

Solutions to this system are denoted by γX\gamma^{X}. Furthermore, solutions with initial conditions γX(0)=p\gamma^{X}(0)=p are denoted by γpX\gamma^{X}_{p}. Once these curves have been obtained, the flows hX:R×E⟶Eh^{X}:\mathbb{R}\times\mathcal{E}\longrightarrow\mathcal{E} were constructed as

hX:R×E⟶E(λ,e)⟼hX(λ,e):=γeX(λ),\begin{align} h^{X}:\mathbb{R}\times\mathcal{E}&\longrightarrow\mathcal{E}\\ (\lambda,e)&\longmapsto h^{X}\left(\lambda,e\right):=\gamma^{X}_{e}(\lambda), \end{align}

these maps can be rewritten in a more convenient manner as

hλX:E⟶Ee⟼hλX(e):=hX(λ,e),\begin{align} h^{X}_{\lambda}:\mathcal{E}&\longrightarrow\mathcal{E}\\ e&\longmapsto h^{X}_{\lambda}(e):=h^{X}(\lambda,e), \end{align}

where the maps hλXh^{X}_{\lambda} form a one-parameter group for each generator XX under composition, this is hλ1X∘hλ2X=hλ1+λ2Xh^{X}_{\lambda_{1}}\circ h^{X}_{\lambda_{2}}=h^{X}_{\lambda_{1}+\lambda_{2}}.

OK, I recognize this may be a little too abstract, so here's a working example:

Working example: rotations on a plane

Consider a vector field X∈Γ(TM)X\in\Gamma\left(TM\right), where M=R2M=\mathbb{R}^{2}, given by

X=y∂∂x−x∂∂y,\begin{align} X=y\dfrac{\partial}{\partial x}-x\dfrac{\partial}{\partial y}, \end{align}

and a curve γ:R⟶R2\gamma:\mathbb{R}\longrightarrow\mathbb{R}^{2} written as γ(λ)=(x(λ),y(λ))\gamma(\lambda)=(x(\lambda),y(\lambda)), then the vector field XX evaluated along γ\gamma is just

Xγ(λ)=y(λ)∂∂x−x(λ)∂∂y,\begin{align} X_{\gamma(\lambda)}=y(\lambda)\frac{\partial}{\partial x}-x(\lambda)\frac{\partial}{\partial y}, \end{align}

and the tangent of the curve is

γ˙(λ)=x˙(λ)∂∂x+y˙(λ)∂∂y.\begin{align} \dot{\gamma}(\lambda)=\dot{x}(\lambda)\frac{\partial}{\partial x}+\dot{y}(\lambda)\frac{\partial}{\partial y}. \end{align}

With this, (1) is written as

x˙(λ)∂∂x+y˙(λ)∂∂y=Xγ(λ)=y(λ)∂∂x−x(λ)∂∂y,\begin{align} \dot{x}(\lambda)\frac{\partial}{\partial x}+\dot{y}(\lambda)\frac{\partial}{\partial y}= X_{\gamma(\lambda)}=y(\lambda)\frac{\partial}{\partial x}-x(\lambda)\frac{\partial}{\partial y}, \end{align}

given that the vectors ∂∂x\dfrac{\partial}{\partial x} and ∂∂y\dfrac{\partial}{\partial y} are linearly independent, this system of ordinary differential equations is equivalent to the following pair of equations

x˙(λ)=y(λ)y˙(λ)=−x(λ).\begin{align} \dot{x}(\lambda)&=y(\lambda)\\ \dot{y}(\lambda)&=-x(\lambda). \end{align}

Solving these equations is a pretty known procedure and therefore omitted. We consider now the initial conditions that tells us where the curve is when the parameter λ\lambda is equal to zero, this is

γX(0)=(x0,y0),\begin{align} \gamma^{X}(0)=\left(x_{0},y_{0}\right), \end{align}

we conclude, then, that γ(x0,y0)X\gamma^{X}_{\left(x_{0},y_{0}\right)} must be

γ(x0,yo)X(λ)=(x0cos⁡λ+y0sin⁡λ,y0cos⁡λ−x0sin⁡λ).\begin{align} \gamma^{X}_{\left(x_{0},y_{o}\right)}(\lambda)=\left(x_{0}\cos\lambda+y_{0}\sin\lambda,y_{0}\cos\lambda-x_{0}\sin\lambda\right). \end{align}

From this curve, we construct the flow hXh^{X} associated to the vector field XX as

hX:R×R2⟶R2(λ,x,y)⟼hX(λ,x,y):=γ(x,y)X(λ)=(xcos⁡λ+ysin⁡λ,ycos⁡λ−xsin⁡λ),\begin{align} h^{X}:\mathbb{R}\times\mathbb{R}^{2}&\longrightarrow\mathbb{R}^{2}\\ \left(\lambda,x,y\right)&\longmapsto h^{X}\left(\lambda,x,y\right):=\gamma^{X}_{\left(x,y\right)}(\lambda)=\left(x\cos\lambda+y\sin\lambda,y\cos\lambda-x\sin\lambda\right), \end{align}

wich, of course, is just a rotation of angle λ\lambda of the point (x,y)(x,y). The group structure of the flow is evidenced by virtue of writing it as

hλX:R2⟶R2(x,y)⟼hλX(x,y):=hX(λ,x,y)=(xcos⁡λ+ysin⁡λ,ycos⁡λ−xsin⁡λ),\begin{align} h^{X}_{\lambda}:\mathbb{R}^{2}&\longrightarrow\mathbb{R}^{2}\\ (x,y)&\longmapsto h^{X}_{\lambda}(x,y):=h^{X}\left(\lambda,x,y\right)=\left(x\cos\lambda+y\sin\lambda,y\cos\lambda-x\sin\lambda\right), \end{align}

where it can be easily verified that (hλ1X∘hλ2X)(x,y)=hλ1+λ2X(x,y)\left(h^{X}_{\lambda_{1}}\circ h^{X}_{\lambda_{2}}\right)(x,y)=h^{X}_{\lambda_{1}+\lambda_{2}}(x,y) for any (x,y)∈R2(x,y)\in\mathbb{R}^{2}.