For which hypersurfaces in projective space does the complement admit an algebraic group structure?

For example, if $H$ is a hyperplane, then $\mathbb{P}^n - H = \mathbb{A}^n$, which is a vector space.

If $n = m^2 - 1$, then we can regard $\mathbb{A}^{n+1}$ as the space of $m \times m$ matrices and take the hypersurface $H$ in $\mathbb{P}^n$ corresponding to the singular matrices. The complement $\mathbb{P}^n - H$ is $\mathbf{PGL}_n$.

If we restrict ourselves to irreducible $H$, are there any more examples besides the two above?

If we allow reducible hypersurfaces, then we can get a few more. We can realize the multiplicative group $\mathbb{G}_m$ as $\mathbb{P}^1$ minus two points, and removing the union of two distinct lines from $\mathbb{P}^2$ will give us $\mathbb{G}_m \times \mathbb{A}^1$. What can we say about the situation here?

The complement of a hypersurface is affine, so only linear algebraic groups will arise.

I haven't put much thought into the base field, so we can just start with $\mathbb{C}$.

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I definitely don't have a solution (would make this a comment if I had the reputation), but perhaps it would be helpful to consider which affine algebraic groups can be embedded in this way. So, for instance, SO(n) would have to go into n(n-1)/2 dimensional projective space minus a hypersurface. So, for instance, SO(2) would have to be P^1 minus a hypersurface, but this cannot be for irreducible H, as P^1 minus a point is simply-connected, but SO(2) is not. In fact, it can't be an open subset of P^1 at all. Haven't got too much more that I can add, except that we know how it works for A^1 and G_m, PGL(n), so we can start figuring it out piecemeal by looking at classes of groups and determining when they can be embedded in this way. Maybe there's more known about embeddings of linear algebraic groups as quasi-projective varieties?

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SO(2) (over the complex numbers) is isomorphic to G_m. – ulrich Oct 22 '09 at 15:25
Ahh, I was premature on the "not at all" part, but it still isn't P^1 minus an irreducible hypersurface. – Charles Siegel Oct 22 '09 at 17:30

Not an answer, but here's a Hodge-theoretic restriction on subvarieties $Z \subset \mathbf{P}^n$ such that $G = \mathbf{P}^n - Z$ admits the structure of a linear algebraic group.

Under the above conditions, the natural mixed Hodge structure (MHS) on $R\Gamma(G,\mathbf{Z})$ is of mixed Tate type (by the Bruhat decomposition, say); this condition on a Hodge structure means, roughly, that only (n,n) classes show up. As $G$ is smooth as a variety, it follows by duality that the same is true for compactly supported cohomology $R\Gamma_c(G,\mathbf{Z})$. On the other hand, there is an exact triangle of MHSs

$R\Gamma_c(G,\mathbf{Z}) \to R\Gamma(\mathbf{P}^n,\mathbf{Z}) \to R\Gamma(Z,\mathbf{Z})$

This means that the MHS on $R\Gamma(Z,\mathbf{Z})$ also has to be of mixed Tate type. This restriction rules out any Z with "interesting" cohomology.

Now if one further assumes that Z is smooth hypersurface, then the mixed Tate condition forces $h^{p,q}(Z) = 0$ unless p=q. Standard calculatons with Hodge numbers of hypersurfaces (see, eg, page 126 of "A Survey of the Hodge Conjecture" by Lewis) then show that Z has degree 1 or 2, at least when the ambient projective space is at least 6 dimensional.

(Edited to include degree restrictions in last paragraph.)

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In your examples, the action of G on itself extends to P^n. We should be able to classify such cases. Sato and Kimura classified group representations with a dense group orbit. It should not be hard too extend their work to classify projective representations with a dense orbit.

If you don't require that the action of G on itself extends to the projective space, then I suspect there are a lot more examples that we are not thinking of.

EDITED for clarity.

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One can get some restrictions on G by calculating H^i(P^n - H, Z) for i = 1,2.

If H is reducible, it follows then H^1 is infinite so G cannot be semisimple. In fact its solvable radical must contain a non-trivial torus. Another example with H reducible (besides tori) is GL_n itself; it is an open subset of A^{n^2} so also of P^{n^2).

If H is irreducible then H_1(P^n - H, Z) is Z/dZ where d is the degree of H. So the solvable radical of G must be nilpotent. The quotient Q of G by its radical is semisimple and has \pi_1 = H_1 = Z/dZ. If d > 4 it follows that this can only happen if Q has a non-trivial quotient of type A_l for some l in which case one can also bound d in terms of n (since the centre of SL_n is isomorphic to Z/nZ).

I would guess that the only semisimple groups that occur are PGL_n but I'm not really sure,

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