Chern classes of tangent bundle over the Grassmannian G(2,4)

Question

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 $c_3$.

Answer 1

Let $E$ and $F$ be complex vector bundles over $B$ of ranks $r$ and $s$ respectively (or locally free sheaves if you prefer). One can compute the Chern classes of the tensor product $E\otimes F$ in terms of the Chern classes of $E$ and $F$. For example:

$$c_1(E\otimes F) = rc_1(F) + sc_1(E)$$

$$c_2(E\otimes F) = rc_2(F) + sc_2(E) + \binom{r}{2}c_1(F)^2 + \binom{s}{2}c_1(E)^2 + (rs-1)c_1(E)c_1(F)$$

In particular, the formula you wrote for $c_2(E\otimes F)$ is incorrect.

These formulae can be obtained as follows. Recall that the Chern character of $E\otimes F$ is given by

\begin{align*} \operatorname{ch}(E\otimes F) =&\ rs + c_1(E\otimes F) + \frac{1}{2}(c_1(E\otimes F)^2 - 2c_2(E\otimes F))\\ &\ + \frac{1}{6}(c_1(E\otimes F)^3 - 3c_1(E\otimes F)c_2(E\otimes F) + 3c_3(E\otimes F)) + \dots \end{align*}

On the other hand,

\begin{align*} &\ \operatorname{ch}(E\otimes F)\\ =&\ \operatorname{ch}(E)\operatorname{ch}(F)\\ =&\ \left[r + c_1(E) + \frac{1}{2}(c_1(E)^2 - 2c_2(E)) + \frac{1}{6}(c_1(E)^3 - 3c_1(E)c_2(E) + 3c_3(E)) + \dots\right]\\ &\ \left[s + c_1(F) + \frac{1}{2}(c_1(F)^2 - 2c_2(F)) + \frac{1}{6}(c_1(F)^3 - 3c_1(F)c_2(F) + 3c_3(F)) + \dots\right] \end{align*}

Comparing the degree one and two parts of these two expressions gives rise to the formulae for $c_1(E\otimes F)$ and $c_2(E\otimes F)$. To find the formula for $c_3(E\otimes F)$, we compare the degree three parts. For notational convenience, let $\alpha_i = c_i(E)$ and $\beta_i = c_i(F)$. After multiplying both sides by $6$, we see that

\begin{align*} &\ c_1(E\otimes F)^3 - 3c_1(E\otimes F)c_2(E\otimes F) + 3c_3(E\otimes F)\\ =&\ r(\beta_1^3 - 3\beta_1\beta_2 + 3\beta_3) + 3\alpha_1(\beta_1^2 - 2\beta_2) + 3\beta_1(\alpha_1^2 - 2\alpha_2) + s(\alpha_1^3 - 3\alpha_1\alpha_2 + 3\alpha_3). \end{align*}

Using the formulae for $c_1(E\otimes F)$ and $c_2(E\otimes F)$ we find that

$$(r\beta_1 + s\alpha_1)^3 -3(r\beta_1 + s\alpha_1)\left(r\beta_2 + s\alpha_2 + \textstyle\binom{r}{2}\beta_1^2 + \binom{s}{2}\alpha_1^2 + (rs-1)\alpha_1\beta_1\right) + c_3(E\otimes F)\\ = r(\beta_1^3 - 3\beta_1\beta_2 + 3\beta_3) + 3\alpha_1(\beta_1^2 - 2\beta_2) + 3\beta_1(\alpha_1^2 - 2\alpha_2) + s(\alpha_1^3 - 3\alpha_1\alpha_2 + 3\alpha_3)$$

which, upon expanding and collecting like terms, gives

\begin{align*} c_3(E\otimes F) =&\ 3\textstyle\binom{s}{3}\alpha_1^3 + 3\binom{r}{3}\beta_1^3 + 6\binom{s}{2}\alpha_1\alpha_2 + 6\binom{r}{2}\beta_1\beta_2 + 3s\alpha_3 + 3r\beta_3 + 3(rs-2)\alpha_2\beta_1\\ & + 3(rs-2)\alpha_1\beta_2 + \left(\tfrac{3}{2}rs - 1\right)(s-1)\alpha_1^2\beta_1 + \left(\tfrac{3}{2}rs-1\right)(r-1)\alpha_1\beta_1^2. \end{align*}

If I haven't made any mistakes, plugging $\alpha_i = c_i(E)$ and $\beta_i = c_i(F)$ into the above will result in a formula for $c_3(E\otimes F)$ in terms of the Chern classes of $E$ and $F$.

You can repeat this process to find $c_4(E\otimes F)$, $c_5(E\otimes F)$, and so on. In your case, you don't need to compute $c_4(\mathcal{S}^*\otimes\mathcal{Q})$ using such a formula because $\operatorname{Gr}(2, 4)$ has real dimension $8$, and hence the fourth Chern class is equal to the Euler class which is just $\chi(\operatorname{Gr}(2, 4)) = \binom{4}{2} = 6$ times the generator of $H^8(\operatorname{Gr}(2, 4); \mathbb{Z})$.

Answer 2

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 $t$.