## January 23, 2011

Posted by Dung Nguyen in Uncategorized.

$\sqrt{-1}$

## formula1November 28, 2010

Posted by Dung Nguyen in Uncategorized.

${\mathcal C}$

## formulaNovember 28, 2010

Posted by Dung Nguyen in Uncategorized.

$\mu^{-1}(H)_{|M_2}$

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

Posted by Dung Nguyen in Curves, Vector Bundles.
Consider a curve $C$ of degree $d$ and genus $g$ inside the projective space $\mathbb{P}^n$. Take a point $P$ and consider the line bundle $F$ on $C$ defined as follows : at each point $Q$ of $C$, the fiber of $F$ is the tangent line to the projective line $PQ$ at $Q$. Now what is the degree of $F$?
The solution I come up with is this : First we notice that we can define this line bundle over any subvarieties not passing through $P$ 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 $\mathbb{P}^n$ at $P$. This way the lines going through $P$ get separated and hence we can define the fan bundle on the whole space. Any rational equivalence of algebraic cycles away from $P$ is preserved so now we can claim that the degree of $F$ on $C$ is $d$ times that of $F$ on any line $L$. To compute the degree of the fan bundle on $L$, we restrict to two-dimensional setting: consider the plane $K$ generated by $P$ and $L$. Then the fan bundle on $L$ is just the normal bundle to $L$ in $K$, so has degree $1$. Thus, the fan bundle in $C$ has degree $d$, 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 $w(C,P)$ for plane curve, but I never learned vector bundles in topological setting ( I wish I did ) so I won’t dare to think.