Relation of $G$-invariants and $g$ -invariants. Let $G$ be a Lie group and $g$ its Lie algebra. Let $H$ a Hopf algebra and $M$ an $H$-module. By definition, $m \in M$ is called invariant if $x.m=\epsilon(x)m$, $\forall x \in H$, where $\epsilon: H \to \mathbb{C}$ is the counit of $H$. Now let $H=U(g)$. The counit $\epsilon$ satisfies the property: $x = \epsilon(x_{(1)} x_{(2)}$ for all $x \in H$, where $x_{(1)} \otimes x_{(2)} = \Delta(x)$. When $H = U(g)$, we have $\Delta(x) = 1 \otimes x + x \otimes 1$. Therefore $x = \epsilon(x) + \epsilon(1) x$. It follows that $\epsilon(x)=0$. Therefore an element $m \in M$ is $U(g)$-invariant if $x.m=0$ for all $x \in U(g)$. 
Suppose that $M$ is a $U(g)$-module, does $M$ also have a $G$-module structure which comes from the $U(g)$-module structure? If $M$ is a $G$-module, is the $G$-invariants defined in the following way: $m \in M$ is $G$-invariant if $x.m=m$ for all $x \in G$? What is the relation between $G$-invariants and $U(g)$-invariants? Any help will be greatly appreciated!
 A: This is too long for a comment.  First, I think the Hopf-algebraic discussion seems digressive.  It seems to me that you are simply asking about the relationship between representations of a Lie group $G$ and its Lie algebra $\mathfrak{g}$.
In general, a representation of $G$ induces one of $\mathfrak{g}$ simply by differentiating at the identity, but a representation of $\mathfrak{g}$ need not lift to a representation of $G$ but only to a representation of some covering group of the identity component of $G$.  Of course, if $G$ is connected and simply-connected then there is an equivalence of categories of representations.  
Concerning the second question, a $G$-invariant is indeed just as you have defined it.  Now let both $G$ and $\mathfrak{g}$ act on $M$.  Then $G$-invariants are also $\mathfrak{g}$-invariants (by differentiation), but the converse need not be the case.  It is the case if $G$ is connected, since a connected Lie group is generated multiplicatively by the image of the exponential map and something left invariant by $\mathfrak{g}$ is also invariant under the action of anything in the image of the exponential.
