I would like to know whether there is an example of a reductive algebraic group $G$ (say, over the complex numbers $\mathbb{C}$) and a finite dimensional representation $V$ of $G$ such that dim$(V//G)$ is different from dim$(V^\*//G)$. Here $V//G=Spec\ \mathbb{C}[V]^G$ is the categorical quotient and $V^\*$ is the dual representation of $G$. 


More generally, I am interested in the relations between the structure of $G$-orbits of $V$ and $V^\*$, so if everyone knows a good reference about this topic, please let me know. 


Thank you in advance.



Is the dimension of $V//G$ always the same as the dimension of $V^*//G$?
If $G$ is connected and $T\subseteq G$ is a maximal torus then there is an involution $\theta:G\to G$ with $\theta(t)=t^{-1}$ for all $t\in T$. It has the property that as a representation $V^\*$ is isomorphic to the $\theta$-twisted representation $V$, i.e., to $G\overset\theta\to G\to GL(V)$. To see this just look at the highest weights. Thus there is an isomorphism $\phi:V\overset\sim\to V^\*$ with $\phi(gv)=\theta(g)\phi(v)$. This induces $V^\*/\!/G=V/\!/\theta(G)=V/\!/G$. Moreover, the $G$-orbit structures of $V^\*$ and $V$ are the same.


If $G$ is not connected, I am not so sure what to do. If the ground field is $\mathbb C$ then one can argue as follows: Let $K\subseteq G$ be a maximal compact subgroup and choose a $K$-invariant Hermitian scalar product on $V$. Since $K$ is Zariski dense in $G$ one has $\tau(G)=G$ where $\tau(g):=(g^{-1})^\dagger$ and $g^\dagger$ is the adjoint with respect to the scalar product. This induces an isomorphism $\phi:\overline V\to V^\*$ with $\phi(gv)=\tau(g)\phi(v)$. So, topologically, $V^\*$ and $V$ are equivariantly the same. In particular $V^\*/\!/G=\overline{V}/\!/G=V/\!/G$ where the last equality is an antiholomorphic isomorphism.


**Edit:** Here is a general


**Theorem**. Let $k$ be any field, $G$ a reductive $k$-group and $V$ a finite dimensional representation of $G$. Then
$\dim V/\!/G=\dim V^\*/\!/G$.


First reduction: One may assume that $k$ is algebraically closed field since field extension commutes with taking invariants.


Second reduction: One may assume assume that $G$ is connected (obvious).


Now observe that the involution $\theta$ above exists over $k$. The same argument shows that the assertion holds if $V$ is irreducible or, more generally, if $V$ is completely reducible. Now the theorem follows from


**Theorem.** Same assumptions as above. Let $U\_1,\ldots,U\_r$ be the composition factors of $V$ and $\overline V:=\bigoplus\_{i=1}^rU\_i$. Then $\dim V/\!/G=\dim\overline V/\!/G$.


This is seen by a deformation argument. Let $0=V\_0\subset V\_1\subset\ldots\subset V\_r=V$ be a composition series. Put $$\tilde V:=\oplus\_{i=1}^\infty V\_i\otimes\_kt^i\subseteq V\otimes\_kk[t]$$ where $V\_i=V$ for $i\ge r$. Then $\tilde V$ is a $G$-vector bundle over $\mathbf A^1$ whose general fiber is $V$ and whose zero-fiber is $\overline V$. Let $\mathbf V\to\mathbf A^1$ be its geometrical realization such that $\mathbf V\_1=V$ and $\mathbf V\_0=\overline V$. The quotient $\mathbf V/\!/G$ is faithfully flat over $\mathbf A^1$. In particular, the fiber dimensions are constant: $\dim(\mathbf V/\!/G)\_1=\dim(\mathbf V/\!/G)\_0$. Geometric reductivity implies that the morphisms $V/\!/G\to(\mathbf V/\!/G)\_1$ and $\overline V/\!/G\to(\mathbf V/\!/G)\_0$ are completely inseparable. In particular the dimensions are the same. This shows the claim.


