For variables $x_1,\ldots,x_m$, the <elementary symmetric polynomial> and <power-sum symmetric polynomial> are
$$
e_r=\sum_{1\leq i_1<\cdots<i_r\leq m}x_{i_1}\cdots x_{i_r},
\qquad
p_r=\sum_{i=1}^m x_i^r.
$$
Put $E(t)=\prod_i(1+x_it)=\sum_{r\geq0}e_rt^r$. Its logarithmic derivative is
$$
\frac{E'(t)}{E(t)}
=\sum_i\frac{x_i}{1+x_it}
=\sum_{r\geq1}(-1)^{r-1}p_rt^{r-1}.
$$
Comparing the coefficient of $t^{n-1}$ in $E'(t)=E(t)E'(t)/E(t)$ gives the <Newton identities>
$$
\boxed{p_n-e_1p_{n-1}+e_2p_{n-2}-\cdots+(-1)^{n-1}e_{n-1}p_1+n(-1)^ne_n=0.}
$$

The given <Bockstein homomorphism> is the first Steenrod square. A real line bundle is pulled back from the universal line bundle over $\mathbb{RP}^{\infty}$, where the degree-one generator $x$ satisfies $\beta(x)=x^2$. By naturality,
$$
\boxed{\beta(w_1(L))=w_1(L)^2.}
$$

Apply the <splitting principle for real vector bundles>, writing the pulled-back bundle as $L_1\oplus\cdots\oplus L_r$ and $x_i=w_1(L_i)$. The pullback in cohomology is injective, $w_j(E)=e_j(x_1,\ldots,x_r)$, and the <Bockstein derivation rule> gives
$$
\beta(e_j)=\sum_i x_i^2e_{j-1}(x_1,\ldots,\widehat{x_i},\ldots,x_r).
$$
In $e_1e_j$, the terms for which the index from $e_1$ already lies in the $j$-element subset give this sum, while each square-free monomial of degree $j+1$ occurs $j+1$ times. Over $\mathbb F_2$ this yields
$$
\boxed{\beta(w_j(E))=w_1(E)w_j(E)+(j+1)w_{j+1}(E).}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 142 / 2 / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-142/2/solution