This is a write-up of an exercise I did in my office with Alex Perry and Will Sawin. Namely we made an example where you can’t flip a Weil divisor. I couldn’t immediately find one by googling; I hope this helps those who google; all mistakes are mine. For the exact notion of flip, please see below (it may not be the same as your notion of flip).<\/p>\n\n\n\n
Let C be an elliptic curve. Let E = L_1 \u2295 L_2 be a direct sum of two invertible modules of degree 1 on C. Let L be a third invertible module of degree 1 on C. We will assume L, L_1, L_2 are Z-lineary independent in Pic(C). Let p : X = P(E) —-> C be the corresponding projective bundle which with my normalization means that p_*O_X(1) = E. Observe that O_X(1) is ample on the surface X because E is an ample vector bundle on C.<\/p>\n\n\n\n
Let A = \u2a01 H^0(X, O_X(n)). Then Z = Spec(A) is the projective cone on X wrt O_X(1). Thus X gives a nice threefold singularity. Denote U the complement of the vertex in Z. There is a morphism U —> X. The pullback of L via the composition U —> X —> C is of the form O_U(D) for some Weil divisor (class) D on Z. If we take the closure Y of the graph of U —> C in Z x C then we see that D pulls back to a Cartier divisor on Y which is moreover ample on Y (equivalently relatively ample with respect to Y —> Z). Finally, note that the fibre of Y —> Z over the vertex has dimension 1.<\/p>\n\n\n\n
Another way to construct Y is to consider the graded A-algebra\n\n\n\n
\nB+ =\n\u2a01 d \u2265 0 <\/sub> H^0(U, O(dD)) =\n\u2a01d, n \u2265 0 <\/sub> H^0(X, O_X(n) \u2297 p^*L^d)\n<\/p><\/blockquote>\n\n\n\n
(with grading given by d) and then Y = Proj(B+). Proof omitted.<\/p>\n\n\n\n
OK, so now we can ask: can we flip (Y —> Z, D)? What I take this to mean is that we want to find a proper morphism Y’ —> Z which is an isomorphism over U, whose fibre over the vertex has dimension < 2 and such that -D determines a Q-Cartier divisor on Y’ which is ample on Y’. Note the sign in front of D!<\/p>\n\n\n\n
It turns out that Y’ exist if and only if the algebra<\/p>\n\n\n\n
\nB- =\n\u2a01 d \u2265 0 <\/sub> H^0(U, O(-dD)) =\n\u2a01d, n \u2265 0 <\/sub> H^0(X, O_X(n) \u2297 p^*L^-d)\n<\/p><\/blockquote>\n\n\n\n
is finitely generated; you can find this in the literature when you google the question. Using p_*O_X(n) = Sym^n(E) this becomes<\/p>\n\n\n\n
\nB- = \u2a01a, b, d \u2265 0<\/sub> H^0(C, L_1^a \u2297 L_2^b \u2297 L^-d).\n<\/p><\/blockquote>\n\n\n\n
Thus we get a natural Z^3-grading for this algebra. By our choice of L_1, L_2, L above we see that we have nonzero elements in the graded piece with (a, b, d) = (0, 0, 0) and in the graded pieces corresponding to (a, b, d) with a + b – d > 0. Thus B- is not finitely generated, because the elements in degrees (a, b , a + b – 1) are all needed as generators for the algebra B-.<\/p>\n","protected":false},"excerpt":{"rendered":"
This is a write-up of an exercise I did in my office with Alex Perry and Will Sawin. Namely we made an example where you can’t flip a Weil divisor. I couldn’t immediately find one by googling; I hope this … Continue reading