About the map $S(\mathfrak{g}^ * )^G\rightarrow S(\mathfrak{h}^ * )^H$ for $H < G$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T05:16:10Z http://mathoverflow.net/feeds/question/107389 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/107389/about-the-map-s-mathfrakg-g-rightarrow-s-mathfrakh-h-for-h About the map $S(\mathfrak{g}^ * )^G\rightarrow S(\mathfrak{h}^ * )^H$ for $H < G$ Zhaoting Wei 2012-09-17T15:13:04Z 2012-09-18T09:39:38Z <p>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 $H&lt; G$ we have the same construction and get $S(\mathfrak{h}^ * )^H$.</p> <p>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 $$\phi: S(\mathfrak{g}^ * )^G \rightarrow S(\mathfrak{h}^ * )^H.$$</p> <p>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).</p> <p>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?</p> http://mathoverflow.net/questions/107389/about-the-map-s-mathfrakg-g-rightarrow-s-mathfrakh-h-for-h/107418#107418 Answer by Will Sawin for About the map $S(\mathfrak{g}^ * )^G\rightarrow S(\mathfrak{h}^ * )^H$ for $H < G$ Will Sawin 2012-09-17T19:27:06Z 2012-09-17T19:27:06Z <p>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.</p> http://mathoverflow.net/questions/107389/about-the-map-s-mathfrakg-g-rightarrow-s-mathfrakh-h-for-h/107450#107450 Answer by Alexander Premet for About the map $S(\mathfrak{g}^ * )^G\rightarrow S(\mathfrak{h}^ * )^H$ for $H < G$ Alexander Premet 2012-09-18T09:39:38Z 2012-09-18T09:39:38Z <p>Here is a complete answer to this question: the map $\varphi$ is an embedding if and only if the group $H$ contains a maximal torus of $G$. I'm assuming (as in the question) that all groups are complexified but originate from a compact group. </p> <p>The map $\varphi$ is injective if and only if no nonzero $G$-invariant function on $\mathfrak g$ vanishes on $X:=G\cdot \mathfrak{h}$. This happens if and only if $X$ is Zariski dense in $\mathfrak g$ (if $X$ is not dense then $f(X)=0$ for some nonzero invariant $f\in S(\mathfrak{g}^*)^G$ by an old result of Nagata). Since the original group was compact the group $H$ is reductive (as $H(\mathbb{R})$ is anisotropic). Let $T_0$ be a maximal torus of $H$ and $\mathfrak{t}_0={\rm Lie}(T_0)$, a Cartan subalgebra of $\mathfrak{h}$. As all maximal tori in $H$ are conjugate and the semisimple elements of $\mathfrak{h}$ are dense in $\mathfrak{h}$, the set $H \cdot \mathfrak{t}_0$ is dense in $\mathfrak{h}$. So $\varphi$ is injective if and only if $G\cdot\mathfrak{t}_0=G\cdot (H\cdot\mathfrak{t}_0)={(G\cdot\mathfrak{h})}_s=X_s$ is dense in $\mathfrak{g}$. Now $T_0$ is contained in a maximal torus of of $G$, say $T$, and it follows from the theorem of the dimension of fibres of a morphism that $\dim X = \dim G\cdot \mathfrak{t}_0\le \dim G+\dim \mathfrak{t}_0-\dim T$. So $X$ is dense in $\mathfrak g$ if and only if $T_0=T$ as claimed. </p>