# Euler number of symmetric cube of the tautological bundle

Let $$E$$ -is a tautological two-dimensional bundle (rank $$n=2$$) over complex Grassmannian $$\operatorname{Gr}(2, 4)$$ ($$2$$-dimensional planes in $$C^4$$). I'm trying to compute the Euler number $$\oint_{\operatorname{Gr}(2, 4)}{e(S^3E)}$$.

Where $$S^3E$$ -is a symmetric cube of the bundle $$E$$.

I know that the Euler class of a complex vector bundle is always equal to the top Chern class.

For tautological vector bundle $$E$$ the top Chern class is $$c_4(\operatorname{Gr}(2, 4)) = 6c_2(Q)^2$$ where $$c_2(Q)^2$$ is the generator of $$H^8(\operatorname{Gr}(2, 4); \mathbb{Z})$$. Also we have the splitting principle $$c(S^{p} E)=\prod_{1 \leq i_1 \leq i_2 \leq \ldots \leq i_p \leq n} (1+x_{i_1}+\ldots+x_{i_p})$$ (possibly it may be useful in this situation). I don't know how to continue computations for symmetric cube. Please, can you explain these computations in more details?

By splitting principle $$c_4(S^3E) = 3x_1(2x_1+x_2)(x_1+2x_2)3x_2 = 9x_1x_2(2x_1^2 + 5x_1x_2 + 2x_2^2) = 9x_1x_2(2(x_1 + x_2)^2 + x_1x_2) = 18c_1(E)^2c_2(E) + 9c_2(E)^2.$$ It remains to note that $$c_1(E)^2c_2(E) = c_2(E)^2 = 1$$.

In schubert for maple

grass(2,4,c,tan);
currentvariety_ is Gc, DIM is 4

chi(symm(3,Qc));
20


The 27 lines on a cubic surface in $${\Bbb P}^3$$.

integral(chern(4,symm(3,Qc)));
27

• Thank you for your answer. In general I want to undertstand why this euler number is equal to 20. Please can you explain this in more details? Dec 21, 2022 at 10:37