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I'm reading the paper Rational Analogs in projective planes by Zhixu Su. I am trying to work out the example to calculate the form $e_2$ for dimension 16 in the paper. I am however not sure how to proceed from the step before the last one to the last one. I tried grouping things with $p_j=\sum_{i_1<...<i_j}^{k/2}x_{i_1}^2...x_{i_j}^2$, the jth Pontryagin class, without success. In the expression, $k$ represents the dimension of the manifold. Here are the steps that are featured in the paper:

$$ e_2 = \sigma_2 (e^{x_1} + e^{-x_1}-2,x^{x_2} + e^{-x_2}-2,...) \\ =\sum_{j,k}(e^{x_j}+e^{-x_j}-2)(e^{x_k}+e^{-x_k}-2) \\ =\sum_{j,k}(x_j^2+\frac{x_j^4}{12}+\frac{x_j^6}{360}+\frac{x_j^8}{20160} + O(x_j^9))(x_k^2+\frac{x_k^4}{12}+\frac{x_k^6}{360}+\frac{x_k^8}{20160} + O(x_k^9)) \\ =\sum_{j,k}(x_j^2x_k^2 + \frac{x_j^4x_k^4}{144}+\frac{x_j^2x_k^6}{360}+\frac{x_j^6x_k^2}{360})+\text{terms of degree other than 8 and 16} \\ =p_2+\frac{p_2^2}{720}+\frac{p_4}{360} $$

Where we set beforehand $p_1=p_3=0$.

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  • $\begingroup$ Do you mean $j$s instead of $k$s in the RHS of your formula for $p_j$? $\endgroup$ – user10354138 Jun 12 at 14:37
  • $\begingroup$ @user10354138, sorry, what $ks$? $\endgroup$ – Alonso Perez Lona Jun 12 at 16:25
  • $\begingroup$ All occurrence of $k$ in $p_j=\sum_{i_1<\dots<i_k}^{k/2}x_{i_1}^2\dots x_{i_k}^2$, RHS doesn't depend on $j$ but on this mysterious $k$? $\endgroup$ – user10354138 Jun 12 at 16:27
  • $\begingroup$ Oh, true, thanks for pointing out. $\endgroup$ – Alonso Perez Lona Jun 12 at 16:30
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Hmm.. you are using $k$ for two different things, the dummy in $\sum_{j,k}$, and the dimension of manifold? Anyway, let's forget about the dimension of manifold and let every letter appearing in $\sum_{\dots}$ be dummy.

We have, by Newton-Girard formulae (and imposing $p_1=p_3=0$), \begin{align*} \sum_j x_j^4 &= p_1^2 - 2p_2=-2p_2\\ \sum_j x_j^6 &= p_1(p_1^2-2p_2)-p_2p_1+3p_3\\ &=p_1^3-3p_1p_2+3p_3=0\\ \sum_j x_j^8 &= p_1(p_1^3-3p_1p_2+3p_3)-(p_1^2 - 2p_2)p_2+p_3p_1-4p_4\\ &=p_1^4-4p_1^2p_2+4p_1p_3 + 2p_2^2-4p_4\\ &=2p_2^2-4p_4 \end{align*} So $$ \sum_{j<k} x_j^4x_k^4=\frac{\left(\sum_j x_j^4\right)^2-\sum_j x_j^8}2 =p_2^2+2p_4 $$ and $$ \sum_{j<k} (x_j^2x_k^6+x_j^6x_k^2)=\left(\sum_j x_j^2\right)\left(\sum_k x_k^6\right)-\sum_j x_j^8= 4p_4-2p_2^2 $$ Hence \begin{align*} &\sum_{j<k}\left(x_j^2x_k^2+\frac{x_j^4x_k^4}{144}+\frac{x_j^2x_k^6+x_j^6x_k^2}{360}\right)\\ &=p_2+\frac{p_2^2+2p_4}{144}+\frac{4p_4-2p_2^2}{360} \\ &=p_2+\frac{p_2^2}{720}+\frac{p_4}{40}. \end{align*}

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