TITLE: Chern classes of tangent bundle over the Grassmannian G(2,4)
QUESTION [5 upvotes]: What are the Chern classes of the tangent bundle $\tau_G$ of the Grassmannian $G=G(2,4)$ of lines in $\mathbb{P}^3$? This is Exercise 5.37 on page 191 of 3264 & All That by Eisenbud and Harris. 
By Theorem 3.5, the tangent bundle $\tau_G$ is isomorphic to $\mathcal{S}^{*} \otimes \mathcal{Q}$, where $\mathcal{S}$ and $\mathcal{Q}$ are the universal sub and quotient bundles of $G$. From Section 5.6.2, we have $c(\mathcal{Q})=1+\sigma_{1}+ \cdots+ \sigma_{n-k}$ and $c(\mathcal{S}^{*})=1+\sigma_1+\sigma_{1,1}+\cdots+\sigma_{1,1,\dots,1}$. 
I found the following formulas here (https://stacks.math.columbia.edu/tag/02UK) for the first two Chern classes of a tensor product of vector bundles $\mathcal{E}$ and $\mathcal{F}$ which are finite locally free of ranks $r,s$. 
$$ c_1(\mathcal{E} \otimes \mathcal{F})=rc_1(\mathcal{F})+sc_1(\mathcal{E})$$ 
$$ c_2(\mathcal{E} \otimes \mathcal{F})=r^2c_2(\mathcal{F})+rsc_1(\mathcal{F})c_1(\mathcal{E})+s^2 c_2(\mathcal{E}).$$ 
So I think the first two Chern classes are
$$ c_1(\tau_G)=2\sigma_1+2\sigma_1=4\sigma_1$$ 
$$c_2(\tau_G)=4\sigma_2+4\sigma_2\sigma_{1,1}+4\sigma_{1,1}=4(\sigma_2+\sigma_{1,1}),$$
but I haven't been able to find a formula for $c_3$.

REPLY [3 votes]: In schubert, the maple package for enumerative geometry and intersection theory,
with(SF);
with(schubert);
Gc = grass(2, 4, c, tan);
chern(tangentbundle(Gc));


 $1+4\,{\it c1}\,t+7\,{{\it c1}}^{2}{t}^{2}+ \left( 8\,{{\it c1}}^{3}-4\,{\it c1}\,{\it c2} \right) {t}^{3}+ \left( 8\,{{\it c1}}^{4}-16\,{\it c2}\,{{\it c1}}^{2}+6\,{{\it c2}}^{2} \right) {t}^{4}$

For how the schubert package does the calculation $(Sc)^*\otimes Qc$ or Hom(4-Qc,Qc) of chern characters and how they relate to chern classes look at this:
chern(Qc)

$1+{\it c1}\,t+{\it c2}\,{t}^{2}$
Or in terms of the chern character (elementary symmetric to power sums in the chern roots)
Qc

$2+{\it c1}\,t+1/2\, \left( -2\,{\it c2}+{{\it c1}}^{2} \right) {t}^{2}
-1/6\, \left( 3\,{\it c1}\,{\it c2}-{{\it c1}}^{3} \right) {t}^{3}+1/
24\, \left( 2\,{{\it c2}}^{2}-4\,{\it c2}\,{{\it c1}}^{2}+{{\it c1}}^{
4} \right) {t}^{4}$
subs(t=-t,4-Qc)*Qc;

$4+4\,{\it c1}\,t+{{\it c1}}^{2}{t}^{2}-2\,{\it c1}\,{\it c2}\,{t}^{3}+
2/3\,{{\it c1}}^{3}{t}^{3}+1/12\,{{\it c1}}^{4}{t}^{4}-{{\it c2}}^{2}{
t}^{4}$
i.e. multiply the chern character of the dual of $Sc=4-Qc$ with the chern character of $Qc$ truncated above the dimension $(4),$ then convert from power sums back to elementary symmetric:
$1+4\,{\it c1}\,t+7\,{{\it c1}}^{2}{t}^{2}+ \left( 8\,{{\it c1}}^{3}-4
\,{\it c1}\,{\it c2} \right) {t}^{3}+ \left( 8\,{{\it c1}}^{4}-16\,{
\it c2}\,{{\it c1}}^{2}+6\,{{\it c2}}^{2} \right) {t}^{4}$
and read off the $i$-th chern class of the tangent bundle as the coefficients of the $i$-th power of $t$.