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

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

which, upon expanding and collecting like terms, gives

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

Answer 3

I would like to point out another way to compute the Chern class of the tangent bundle of $G(2,4)$. This space embeds in $P(\Lambda^2{\bf C}^4) \simeq P^5({\bf C})$ thanks to the map $$ \matrix{P(\Lambda^2{\bf C}^4) & \longrightarrow & P^5({\bf C}) \cr \hbox{vect}(v, w) & \mapsto & v \wedge w \cr} $$ A short computation shows that it is a quadric of degree 2 in $P^5$. As a result, the Chern class of its normal bundle $N$ is $1+2x$, where $x$ is the pull-back of the class of the tautological bundle of $P^5$. The Chern class of the tangent bundle of $P^5$ is given by $(1+x)^6$ and the isomorphism $$ i^*TP^5 \simeq TG(2,4) \oplus N $$ gives $$c(TG(2,4)) = {(1+x)^6 \over (1+2x)} = 1+4x+7x^2+6x^3+3x^4$$ Note that $x^4$ is twice the dual of the fundamental class of $G(2,4)$, giving an Euler class equal to 6, as expected.

The formulas displayed in the other answers to the question can be recovered by noting the relations in the cohomology ring of $G(2,4)$, $c_1^3 = 2c_1c_2$, $c_1^4 = 2c_2^2$ and the fact that $x= \sigma_1 = c_1(Q)=-c_1(T)$. Beware not to mix up the rank 2 tautological bundle $T$ over $G(2,4)$ which has $c_1(T)$ and $c_2(T)$ as its chern classes with the pull-back of the rank one tautological bundle over $P^5$ which has $x$ as its first chern class and 0 as its second chern class.