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

Solution

For variables

x_{1}, \dots, x_{m}

e_{r} = \sum_{1 \leq i_{1} < \dots < i_{r} \leq m} x_{i_{1}} \dots x_{i_{r}}, p_{r} = \sum_{i = 1}^{m} x_{i}^{r} .

Put

E (t) = \prod_{i} (1 + x_{i} t) = \sum_{r \geq 0} e_{r} t^{r}

\frac{E ^{'} ( t )}{E ( t )} = \sum_{i} \frac{x _{i}}{1 + x _{i} t} = \sum_{r \geq 1} (- 1)^{r - 1} p_{r} t^{r - 1} .

Comparing the coefficient of

t^{n - 1}

E^{'} (t) = E (t) E^{'} (t) / E (t)

p_{n} - e_{1} p_{n - 1} + e_{2} p_{n - 2} - \dots + (- 1)^{n - 1} e_{n - 1} p_{1} + n (- 1)^{n} e_{n} = 0.

RP^{\infty}

, where the degree-one generator

x

satisfies

β (x) = x^{2}

. By naturality,

β (w_{1} (L)) = w_{1} (L)^{2} .

L_{1} \oplus \dots \oplus L_{r}

and

x_{i} = w_{1} (L_{i})

. The pullback in cohomology is injective,

w_{j} (E) = e_{j} (x_{1}, \dots, x_{r})

β (e_{j}) = \sum_{i} x_{i}^{2} e_{j - 1} (x_{1}, \dots, x_{i}, \dots, x_{r}) .

e_{1} e_{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

F_{2}

this yields

β (w_{j} (E)) = w_{1} (E) w_{j} (E) + (j + 1) w_{j + 1} (E) .

There are no discussions about this article yet.

There are currently no matching articles.