Why the Lie group and its Lie algebra have the same action? Let $\phi$ be a $g$ equivariant vector valued function that is $\phi(p \triangleleft g)=g^{-1}\triangleright \phi(p)$ and $\phi(p) \in V$ where $V$ is some vector space.
Now let $g=\exp(At)$ be a matrix group where  $A$ is an element of its lie algebra. In this lecture
Covariante derivatives  Professor Schuller calculates the derivative
$$\phi(p \triangleleft \exp(At))'(0)=\exp(-At)'\triangleright \phi(p)(0)=-A\triangleright \phi(p)$$
My question why is that the action $\triangleright$ in  $\exp(-At)'\triangleright \phi(p)(0)$ is the same as action $\triangleright$ in $-A\triangleright \phi(p)$ ?
 A: It's not, but it's related, so one often uses the same notation for both. In general, an action of a group $G$ on a vector space $V$ is a homomorphism $\Psi:G\to \operatorname{GL}(V)$, but instead of writing $\Psi(g)(v)$, we write $g\triangleright v$. The corresponding action of the Lie algebra $\mathfrak{g} = T_eG$ on $V$ is the derivative of this map at the identity, i.e.
$$
T_e\Psi:T_e G\to T_I\operatorname{GL}(V)
$$
Now $T_I\operatorname{GL}(V)= \mathfrak{gl}(V)$ is isomorphic to $L(V)$, the set of linear (but not necessarily invertible) maps on $V$. So we can say
$$
T_e\Psi:\mathfrak{g}\to L(V).
$$
Instead of writing $T_e\Psi(A)(v)$ for $A\in\mathfrak{g}$ and $v\in V$, the lecturer is writing $A\triangleright v$ for simplicity.
A: Let $gl(n,C)$ be the algebra of complex matrix, $B$  a lie subalgebra of $gl(n,C)$  and $A$ the lie algebra of a matrix group $G$ where $G$ as set is contained in $gl(n,C)$.
Now suppose that $A$ and $B$ are isomorphic as a lie algebra and that $g(t)=\exp(bt)$ is a curve in $G \subset gl(n,C) $ with $b \in B$. So since $g(t)$ and $b$ are element of the same  algebra of matrices we can define the same action.
Note people often say that for matrix group  $\frac{d}{dt} \exp(At)(0)=A$, but we can not define this operation in a matrix group because in a matrix group there is no addition and so you have to go to the algebra of matrices  $gl(n,C)$ to define this operation
