# Unipotent inertia

My prediction at the end of the last post was complete nonsense! Here are some examples of actions where the stabilizer jumps in codimension 1:

1. The action G_a^n x P^n —> P^n given by (a_1, …, a_n), (x_0: …: x_n) maps to (x_0: x_1 + a_1x_0: … : x_n + a_nx_0). The generic stabilizer is trivial and over the divisor x_0 = 0 the stabilizer is G_a^n. So the dimension of the stabilizer can jump up arbitrarily high in codimension 1.
2. A special case of the example above is the case n = 1 which Jarod Alper pointed out. If y = x_0/x_1 then the action looks like y maps to y/(1 + ty) where t is the coordinate on G_a.
3. Note that there are many formal actions \hat{G_a} x \hat{A^1} —> \hat{A^1}, because if theta is the derivation ty^k(d/dy) acting on C[[y]] then if k > 1 we can exponentiate and get automorphisms phi_t = e^theta : C[[y]] —> C[[y]] which satisfy phi_t \circ phi_s = phi_{s + t}.
4. Another example due to Jarod is the action G_a x A^2 —> A^2 given by t, (x, y) maps to (x + ty, y). The locus of points where the stabilizer is G_a is y = 0. This action seems very different from the action in case 2, allthough it may not be so easy to prove.
5. Take the product P^1 x P^1 with the action of G_a which is trivial on the first component and as in example 2 on the second. Then we may blow up (several times) in invariant points. If you do this in a suitable manner you will find an exceptional curve E consisting of fixed points where the local ring of the blow up at the generic point of E looks like C[x, y]_{(y)} and where the action is given by (x, y) maps to (x/(1 + txy^n), y). This gives infinitely many actions which cannot be etale locally isomorphic since the action is trivial modulo y^n and not y^{n + 1}. Note that y is the uniformizer of the local ring in question.

The conclusion is that if you allow the jump of the inertia group to be non-reductive, then many examples exist (there may even be moduli in the examples).