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 $E$.

My thoughts about this question are the following:

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?


2 Answers 2


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

                    currentvariety_ is Gc, DIM is 4


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

  • $\begingroup$ 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? $\endgroup$
    – Victory
    Dec 21, 2022 at 10:37

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