1

I am trying to do the following:

Problem: Reformulate the condition that a line in $\mathbb{CP}^4$ lies on a quadratic hypersurface in terms of the Chern classes of a vector bundle defined over the grassmannian $G_2(\mathbb{C}^5)$.

My try: A line in $\mathbb{CP}^4$ corresponds to a $2$-plane in $\mathbb{C}^5$, hence a point in $G_2(\mathbb{C}^5)$. I think we have to define a vector bundle over $G_2(\mathbb{C}^5)$ that give these Chern classes, but I do not see how to do this or which conditions these Chern classes should satisfy.

Math Student 020
  • 1,599
  • I know the question has been already answered but let me add that a very good reference for this kind of question is the book by Harris and Eisenbud, 3264 and all that. –  Apr 15 '17 at 22:02

1 Answers1

2

Points of $G_2(\mathbb{C}^5)$ corresponds to planes in $\mathbb{C}^5$. So on $G_2(\mathbb{C}^5)$ there is a tautological rank 2 vector bundle $E$, namely the bundle such that the vector space over a point $P$ is the plane $P$ itself.

Now, this vector bundle is not exactly the one we are interested in, this is because we want to study a quadratic hypersurface. This quadratic hypersurface is defined by a homogeneous polynomial $Q(X_1,...,X_5)$ which can be restricted to every plane of $\mathbb{C}^5$. This give a quadratic form associated to every point of $G_2(\mathbb{C}^5)$. So we are lead to consider the vector bundle such that the vector space over a point $P$ is the space of quadratic form over the plane $P$. This vector bundle is $\operatorname{Sym}^2(E^\vee)$. This is a vector bundle of rank 3.

The point is, the polynomial $Q$ is now a section $s$ of $\operatorname{Sym}^2(E^\vee)$ by the procedure I wrote (namely $P\mapsto Q_{|P}$ where $Q_{|P}$ is the restriction of $Q$ to the plane $P$).

What does it mean for a line $l$ in $\mathbb{CP}^4$ to lie on the hypersurface $V(Q)$ defined by $Q$ ? This line correspond to a plane $P$ in $\mathbb{C}^5$, and the line is contained in $V(Q)$ if and only if $Q_{|P}$ is identically $0$. In other words, if $s(P)=0$.

Thus the lines of $\mathbb{CP}^4$ contained in $V(Q)$ correspond to the zero set of a section of the vector bundle $\operatorname{Sym}^2(E^\vee)$ over $G_2(\mathbb{C}^5)$.

The section $s$ having a zero (and more precisely, the zero set of $s$) can now be expressed in term of Chern classes...

Roland
  • 12,487
  • Thanks! One more question: we now look for $s^{-1}(0)$. I read that the homology class $[s^{-1}(0)]$ (which lies in $H^3(G_2(\mathbb{C}^5), \mathbb{Z})$) is equal to the Poincare dual of the top Chern class (= the Euler class) $e(\operatorname{Sym}^2(E^{\vee}) \in H_{4-3}(G_2(\mathbb{C}^4), \mathbb{Z})$. Does this help to describe the Chern classes? – Math Student 020 Mar 26 '17 at 16:16
  • 1
    @MathStudent020 I am a bit confused by your notations. The homology class $[s^{-1}(0)]$ lies in $H_{6-3}(G_2(\mathbb{C}^5),\mathbb{Z})$ (the dimension of the Grassmanian being 6). This is indeed the Poincare dual of the top Chern class $c_3(\operatorname{Sym}^2(E^\vee))=e(\operatorname{Sym}^2(E^\vee))\in H^3(G_2(\mathbb{C}^4,\mathbb{C})$. Finally, I would say that the best way to compute it is to use the splitting principle to compute the Chern classes of a symmetric power (from the Chern classes of the tautological bundle). – Roland Mar 26 '17 at 17:59