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

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 $c_3$.

## 1 Answer

Let $E$ and $F$ be complex vector bundles over $B$ of ranks $r$ and $s$ respectively (or locally free sheaves if you prefer). Recall that the Chern character of $E\otimes F$ is given by

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

On the other hand,

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

Comparing the degree one and two parts of these two expressions gives rise to the formulae for $c_1(E\otimes F)$ and $c_2(E\otimes F)$ you mention.

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

$$c_1(E\otimes F)^3 - 3c_1(E\otimes F)c_2(E\otimes F) + 3c_3(E\otimes F) = r(\beta_1^3 - 3\beta_1\beta_2 + 3\beta_3) + 3\alpha_1(\beta_1^2 - 2\beta_2) + 3\beta_1(\alpha_1^2 - 2\alpha_2) + s(\alpha_1^3 - 3\alpha_1\alpha_2 + 3\alpha_3).$$

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

$$(r\beta_1 + s\alpha_1)^3 -3(r\beta_1 + s\alpha_1)(r^2\beta_2 + rs\alpha_1\beta_1 + s^2\alpha_2) + c_3(E\otimes F) = r(\beta_1^3 - 3\beta_1\beta_2 + 3\beta_3) + 3\alpha_1(\beta_1^2 - 2\beta_2) + 3\beta_1(\alpha_1^2 - 2\alpha_2) + s(\alpha_1^3 - 3\alpha_1\alpha_2 + 3\alpha_3).$$

Expanding, we find that.

$$r^3\beta_1^3 + s^3\alpha_1^3 - 3r^3\beta_1\beta_2 - 3rs^2\alpha_2\beta_1 - 3r^2s\alpha_1\beta_2 - 3s^2\alpha_1\alpha_2 + c_3(E\otimes F) = r\beta_1^3 - 3r\beta_1\beta_2 + 3r\beta_3 + 3\alpha_1\beta_1^2 - 6\alpha_1\beta_2 + 3\alpha_1^2\beta_1 - 6\alpha_2\beta_1 + s\alpha_1^3 - 3s\alpha_1\alpha_2 + 3s\alpha_3.$$

Rearranging, we see that

$$c_3(E\otimes F) = (r-r^3)\beta_1^3 + 3(r^3-r)\beta_1\beta_2 + 3r\beta_3 + 3\alpha_1\beta_1^2 + (3r^2s - 6)\alpha_1\beta_2 + 3\alpha_1^2\beta_1 + (3rs^2- 6)\alpha_2\beta_1 + (s-s^3)\alpha_1^3 + 3(s^3-s)\alpha_1\alpha_2 + 3s\alpha_3.$$

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 the decomposition of $c_3(E\otimes F)$ into 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 $H^8(\operatorname{Gr}(2, 4); \mathbb{Z})$.