Let $G$ be a compact connected semisimple Lie group, $\mathfrak{g}$ be its complexified Lie algebra and $\mathfrak{g}^\*$ its complex dual space. We can form the symmetric algebra $S(\mathfrak{g}^ \* ) $ and its $G$-invariant subalgebra $S(\mathfrak{g}^ \* ) ^G$. For a closed connected subgroup $H< G$ we have the same construction and get $S(\mathfrak{h}^ \* )^H$.


Since $\mathfrak{h} \hookrightarrow \mathfrak{g}$ we get $\mathfrak{g}^\*\twoheadrightarrow \mathfrak{h}^ \* $ hence $S(\mathfrak{g}^ \* ) \twoheadrightarrow S(\mathfrak{h}^ \* )$. Take invariant subalgebra we get the map 
$$
\phi: S(\mathfrak{g}^ \* )^G \rightarrow S(\mathfrak{h}^ \* )^H.
$$


Notice that $\phi$ is not always a projection map. For example when $H=T$ is a Cartan subalgebra, then it is well-known that $S(\mathfrak{g}^ \* )^G \cong S(\mathfrak{t}^ \* )^W$ is the invariant subalgebra under the Weyl group action and $S(\mathfrak{t}^ \* )^T=S(\mathfrak{t}^ \* )$ since $T$ is abelian. Therefore $\phi: S(\mathfrak{g}^ \* )^G \rightarrow S(\mathfrak{t}^ \* )^T$ is the embedding (injective but not surjective).


My question is: for which $H$, the map $\phi$ is injective? We know that $H=G$ itself or $H$ is a Cartan subgroup makes $\phi$ an injection. Are these the only cases?



About the map $S(\mathfrak{g}^ * )^G\rightarrow S(\mathfrak{h}^ * )^H$ for $H <  G$
The symmetric algebra can be viewed as polynomial functions on $\mathfrak g$. The kernel of the map consists of $G$-invariant polynomial functions that vanish on $\mathfrak h$. Thus the kernel is trivial when the $G$-orbit of $\mathfrak h$ is Zariski-dense in $\mathfrak g$, for instance, when the $G$-orbit of $H$ is Zariski-dense in $G$. This accounts for both the examples you gave.


