Enumeratively Relevant

The degree of the “fan bundle” June 7, 2010

I came across this cute little problem while I was trying to solve an enumerative problem. The problem looks like it should be trivially easy, but for some reason it took me quite a while to figure it out ( without using tools from enumerative geometry ). This shows that my intuition about vector bundles is extremely bad.

The problem is as follows :

Consider a curve of degree and genus inside the projective space . Take a point and consider the line bundle on defined as follows : at each point of , the fiber of is the tangent line to the projective line at . Now what is the degree of ?

The solution I come up with is this : First we notice that we can define this line bundle over any subvarieties not passing through and we call this the fan bundle. ( I did a google search to make sure that this name hasn’t been used elsewhere ). We would expect that the degree of the fan bundle is proportional to the degree of the variety it is defined on. But the problem is we don’t have a line bundle on the whole projective space.

One way to get around this is to blow up at . This way the lines going through get separated and hence we can define the fan bundle on the whole space. Any rational equivalence of algebraic cycles away from is preserved so now we can claim that the degree of on is times that of on any line . To compute the degree of the fan bundle on , we restrict to two-dimensional setting: consider the plane generated by and . Then the fan bundle on is just the normal bundle to in , so has degree . Thus, the fan bundle in has degree , the genus is irrelevant.

I’m also curious to what the answer would be like in the real setting. My guess is that the answer is trivial for space curve, and is the winding number for plane curve, but I never learned vector bundles in topological setting ( I wish I did ) so I won’t dare to think.