Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 126 / 3 / ii

For

x : S \to G

T_{x} : G \times_{k} S \to G, (g, s) \mapsto x (s) g .

Then

T_{x / S} = (T_{x}, pr_{2})

is an

S

x^{- 1}

is its inverse.

Because an isomorphism preserves relative differentials and

T_{x / S}

lies over

S

T_{x / S}^{*} Ω_{G \times S / S} ≅ Ω_{G \times S / S} .

Using base change,

Ω_{G \times S / S} ≅ pr_{1}^{*} Ω_{G / k}

, while the left side is

T_{x}^{*} Ω_{G / k}

. Pulling this isomorphism back along the identity section

e \times 1_{S}

gives

x^{*} Ω_{G / k} ≅ O_{S} \otimes_{k} Ω_{G / k} (e) .

Finally take

S = G

and

x = 1_{G}

Ω_{G / k} ≅ O_{G} \otimes_{k} Ω_{G / k} (e),

Ω_{G / k}

O_{G}

-module.

Solved by gpt-5.6-sol high.

 Ancestors (10)

There are no discussions about this article yet.

There are currently no matching articles.