A prehomogeneous vector space is a pair $(G,V)$ where $V$ is a finite dimensional $\mathbb{C}$-vector space of dimension $n$ and $G$ is a reductive group of complex dimension $n$, such that $G$ admits a group action on $V$. If we embed $G$ into $\text{GL}(V)$, then the subgroup $G^\circ$ whose image in $\text{GL}(V)$ has absolute determinant one acts on $V$ with a single open orbit. In this case the ring of polynomial invariants is isomorphic to $\mathbb{C}[x]$, and is generated by a polynomial which can be called the discriminant. The open orbit is given by the non-vanishing of the discriminant.

An example of a prehomogeneous vector space is given by $G = \text{GL}_2(\mathbb{C})$ and $V = \{a_3 x^3 + a_2 x^2 y + a_1 xy^2 + a_0 y^3 : a_i \in \mathbb{C}\}$ the space of binary cubic forms, with the group action given by substitution.

I am interested in the following generalization: let $V$ instead be a projective variety, and $G$ a reductive group acting on $V$, and such that $\dim_{\mathbb{C}} G = \dim_{\mathbb{C}} V$. If $V$ is geometrically irreducible then there will only be one open orbit, defined by the non-vanishing of some polynomial which then generates the ring of polynomial invariants.

Have such pairs $(G,V)$ been studied in the literature (I think of them as ``prehomogeneous varieties" as in the title of the question, but perhaps they go by a different name)?

Examples include the spaces of Klein forms of degrees $3,4,6,12,20$. Klein forms are binary forms which admit an exceptionally large finite subgroup of $\text{PGL}_2(\mathbb{C})$ as an automorphism group. All binary cubic forms are Klein forms, giving the example of a prehomogeneous vector space above, while quartic Klein forms are defined by the vanishing of the $I$-invariant. That is, if we put

$$\displaystyle V_4 = \{a_4 x^4 + \cdots + a_0 y^4 : a_i \in \mathbb{C}\}$$

then the set of quartic Klein forms is given by the variety $\mathcal{V}_4 = \{a_4 x^4 + \cdots + a_0 y^4 : 12 a_4 a_0 - 3 a_3 a_1 + a_2^2 = 0\}$, which is a quadric 3-fold in $V_4$. Similarly, Klein forms of degrees 6, 12, and 20 are given by intersections of quadric hypersurfaces (see this paper by Bennett and Dahmen). In each case there is only one open orbit under $\text{GL}_2(\mathbb{C})$. For example, all non-degenerate (i.e., with non-zero discriminant) quartic Klein forms are $\text{GL}_2(\mathbb{C})$-equivalent to the form $x(x^3 + y^3)$.

  • 7
    $\begingroup$ The assumption that $G$ should have dimension equal to $\dim(V)$ sounds unusual to me, and highly restrictive (for instance it excludes $\mathrm{SL}(V)$ acting on $V$ if $V\neq 0$). Also I'm confused by the assertion on invariants: if there is an open orbit, isn't the $\mathbf{C}$-algebra of invariants reduced to constants? $\endgroup$
    – YCor
    Nov 14 '20 at 14:00
  • 2
    $\begingroup$ A class of examples are (projective) toric varieties $\endgroup$ Nov 14 '20 at 16:30
  • $\begingroup$ Look up "equivariant compactifications of homogeneous spaces". $\endgroup$ Nov 14 '20 at 19:28
  • 3
    $\begingroup$ You're now writing that there is an open $G$-orbit as a consequence, but it seems to be an assumption. Also, purely for the terminology, it seems to be that requiring that $G$ has dimension equal to $\dim(V)$ is not usual in "prehomogeneous vector space". (Still it's a reasonable assumption.) $\endgroup$
    – YCor
    Nov 17 '20 at 14:30
  • 1
    $\begingroup$ AS @YCor says, you start off with a mere group action (which I think means linear action of a group—it seems like a strange way of saying it), and claim that it has an open orbit. This seems impossible to deduce merely as a consequence of the group action. For example, $G$ might act trivially. $\endgroup$
    – LSpice
    Nov 17 '20 at 15:22

Along the lines of toric varieties, there's also a very active field of spherical varieties $V$ on which a reductive group $G$ acts with an open dense orbit under a Borel subgroup $B$. They bear a resemblance to the theory of prehomegeneous vector spaces, but the lack of a vector space structure makes them much more difficult to study.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.