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In particular, in the OP’s question, the intersection of (C-p) with (W(g-1)-(g-1).p), the divisor (D-E) = (g-2).p is effective, so the intersection is improper, but the pulled back line bundle still equals O(K-(g-2).p). In the example cited by Francesco, (C-p).(W(g-1)-(K+(1-g).p)), the divisor D-E = K-g.p, in general is not effective (e.g if p is not a "Weierstrass" point), so the intersection divisor = g.p in that case, and in every case the pulled back line bundle is O(gp).

In particular, in the OP’s question, the intersection of (C-p) with (W(g-1)-(g-1).p), the divisor (D-E) = (g-2).p is effective, so the intersection is improper, but the pulled back line bundle still equals O(K-(g-2).p). In the example cited by Francesco, (C-p).(W(g-1)-(K+(1-g).p)), the divisor D-E = K-g.p, in general is not effective (e.g if p is not a "Weierstrass" point), so the intersection divisor = g.p in the general case, and in every case the pulled back line bundle is O(gp).

In particular, in the OP’s question, the intersection of (C-p) with (W(g-1)-(g-1).p), the divisor (D-E) = (g-2).p is effective, so the intersection is improper, but the pulled back line bundle still equals O(K-(g-2).p). In the example cited by Francesco, (C-p).(W(g-1)-(K+(1-g).p)), the divisor D-E = K-g.p, in general should not be effective, so the intersection divisor = g.p in that case, and in every case the pulled back line bundle is O(gp).

In particular, in the OP’s question, the intersection of (C-p) with (W(g-1)-(g-1).p), the divisor (D-E) = (g-2).p is effective, so the intersection is improper, but the pulled back line bundle still equals O(K-(g-2).p). In the example cited by Francesco, (C-p).(W(g-1)-(K+(1-g).p)), the divisor D-E = K-g.p, in general is not effective (e.g if p is not a "Weierstrass" point), so the intersection divisor = g.p in that case, and in every case the pulled back line bundle is O(gp).

In particular, in the OP’s question, the intersection of (C-p) with (W(g-1)-(g-1).p), the divisor (D-E) = (g-2).p is effective, so the intersection is improper, but the pulled back line bundle still equals O(K-(g-2).p). In the example cited by Francesco, (C-p).(W(g-1)-(K+(1-g).p)), the divisor D-E = K-g.p, in general should not be effective, so the intersection divisor = g.p in that case, and in very case the pulled back line bundle is O(gp).

In particular, in the OP’s question, the intersection of (C-p) with (W(g-1)-(g-1).p), the divisor (D-E) = (g-2).p is effective, so the intersection is improper, but the pulled back line bundle still equals O(K-(g-2).p). In the example cited by Francesco, (C-p).(W(g-1)-(K+(1-g).p)), the divisor D-E = K-g.p, in general should not be effective, so the intersection divisor = g.p in that case, and in every case the pulled back line bundle is O(gp).