What's the superposition principle for group action?

[Jianbing_Shao]

A very simple question: if given a vector $v(t_0)$ and two group functions $G(t)$ and $G'(t)$, here $t$ is the parameter of time, the two group functions act on $v(t_0)$ simultaneously, then we can get a vector field $v(t)$, then how to get $v(t)$?

 

[Orodruin]

It is unclear to me what you mean. Can you be more specific?

 

[Jianbing_Shao]

It is unclear to me what you mean. Can you be more specific?

ok. for example,if $G(t)=exp(\theta^1(t)X_1)$ is a rotation around $x^1$ axis, and $G'(t)=exp(\theta^2(t)X_2)$ is a rotation around $x^2$ axis,
Now if we only act $G(t)$ on a vector $v(t_0)$, we can get a vector rotate around $x^1$ axis, we act $G'(t)$ on $v(t_0)$, we can get a vector rotate around $x^2$ axis,
If we act $G(t)$ and $G'(t)$ on $v(t_0)$ at the same time, then what vector field shall we get?

 

[Orodruin]

ok. for example,if $G(t)=exp(\theta^1(t)X_1)$ is a rotation around $x^1$ axis, and $G'(t)=exp(\theta^2(t)X_2)$ is a rotation around $x^2$ axis,
Now if we only act $G(t)$ on a vector $v(t_0)$, we can get a vector rotate around $x^1$ axis, we act $G'(t)$ on $v(t_0)$, we can get a vector rotate around $x^2$ axis,
If we act $G(t)$ and $G'(t)$ on $v(t_0)$ at the same time, then what vector field shall we get?

Again, it is unclear what you mean by ”acting on v at the same time”. I suspect that this is where your confusion lies. There are several things I could imagine this meaning. Which one you are after would depend on what exactly you are trying to do.

 

[Jianbing_Shao]

Again, it is unclear what you mean by ”acting on v at the same time”. I suspect that this is where your confusion lies. There are several things I could imagine this meaning. Which one you are after would depend on what exactly you are trying to do.

What I mean is that between time $t$ and $t+\delta t$, the change of vector only contains two parts: $v(t)exp((\partial_t(\theta^1(t))X_1)\delta t)$ and $v(t)exp((\partial_t(\theta^2(t))X_2)\delta t)$.
Do you think ”acting on v at the same time”can have other meaning?

 

[DrGreg]

What I mean is that between time $t$ and $t+\delta t$, the change of vector only contains two parts: $v(t)exp((\partial_t(\theta^1(t))X_1)\delta t)$ and $v(t)exp((\partial_t(\theta^2(t))X_2)\delta t)$.
Do you think ”acting on v at the same time”can have other meaning?

But how do combine the two parts into one?

In general if I have two operators A and B, combining them could mean AB or BA or A+B or something else.

Are you, perhaps, referring to the Lie product formula or the related Trotter product formula?

 

[Jianbing_Shao]

But how do combine the two parts into one?

In general if I have two operators A and B, combining them could mean AB or BA or A+B or something else.

Are you, perhaps, referring to the Lie product formula or the related Trotter product formula?

In my opinion. perhaps Lie product formula is also not valid here, I think we can use differential equations here,
to the example mentioned above, we can describe $v(t)$ in such a way:
$$\partial_t v(t)=v(t)(\partial_t\theta^1(t)X_1+\partial_t\theta^2(t)X_2)$$
Then we can get $v(t)$ in the formula of time-ordered product:
$$v(t)=v(t_0)T\{exp\int(\partial_t\theta^1(t)X_1+\partial_t\theta^2(t)X_2)dt\}$$
This formula can guarantee that the local change of vector field only contains two Infinitesimal rotations。