# Chernology for physicists: How to compute the Chern classes of $Sym^{n}$ of a rank 2 bundle?

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How to  compute the Chern classes of  $Sym^{n}$ of a rank 2 bundle $E$ in terms of the Chern classes of the  bundle $E$ ?.

We have that $rk(E) = 2$ and $c(E) =1 +c_{1}(E)+c_{2}(E)$.

edited Oct 24, 2016

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Let me describe the standard way to do this kind of computation. The key claim is that to obtain this kind of formulas, one can pretend that the vector bundle is a sum of line bundles (even if it is not true in general). This claim is called the splitting principle and can be justified but I will not do it here (unless if explicitely asked). Let me just show how it can be used in practice.

We have our rank two vector bundle $E$. Let us pretend that $E=L_1 \oplus L_2$ where $L_1$ and $L_2$ are line bundles. Then $c_1(E)=c_1(L_1)+c_1(L_2)$ and $c_2(E)=c_1(L_1)c_1(L_2)$ are the elementary symmetric functions in $c_1(L_1)$ and $c_1(L_2)$.

We have $Sym^n(E)=L_1^n \oplus (L_1^{n-1} \otimes L_2) \oplus \dots \oplus L_2^n$ hence

$c(Sym^n(E))=\prod_{i=0}^n c(L_1^{n-i} \otimes L_2^i)=\prod_{i=0}^n (1+(n-i)c_1(L_1)+ic_1(L_2))$

This expression is a symmetric polynomial in the variables $c_1(L_1)$ and $c_1(L_2)$ and so can be written as a polynomial in the elementary symmetric polynomials in these variables, i.e. as a polynomial in $c_1(E)$ and $c_2(E)$.

answered Oct 24, 2016 by (4,940 points)

Very nice explanation.  Many thanks.  The question remaining is how to expand the product and to rewrite the resulting expression in terms of $c_{1}(E)$ and $c_{2}(E)$.  Such algebraic work was done efficiently using Schubert.

Is this not true only for vector bundles over $\mathbb{P}_{\mathbb{C}}^1$?

@juancho: indeed, the last step uses the fact that any symmetric polynomial is a polynomial in the elementary symmetric functions. It is a classical fact, which can be proved by induction (for instance, introducing lexicographic order on monomials) in an constructive way. In particular, it is possible to do it by hand or to write a computer program to do it, and it is likely that it is done that way in the package you mentioned.

@conformal_gk It is true that to have all vector bundles as sum of line bundles is an exceptional case, which is rarely realized (it is the case on $\mathbb{P}^1$ but not on $\mathbb{P}^2$). The point of the "splitting principle" is that if you want to derive universal formulas about Chern classes, then you can pretend that you work with a sum of line bundles, even if it is not true.

@40227 I do not understand this :)

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One possible solution using the Maple package "schubert" is as follows:

restart:with(schubert):DIM:=4: b:=bundle(2,c);
> chern(b);
> for i from 1 to 4 do print(c[i](Sym^n*[E])=subs(c1=c[1](E),c2=c[2](E),factor(coeff(chern(Symm(n,b)),t,i)))) end do;

With such code we obtain

In the particular case with $n = 3$  we obtain

answered Oct 22, 2016 by (990 points)

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