 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/iii/paper-2/1/b/solution

Codex  0 Created 2026-10-03 Updated 2026-10-14

δ (x) = x \otimes a + y \otimes c, δ (y) = x \otimes b + y \otimes d .

It is extended multiplicatively and sends

1

1 \otimes 1

. The matrix comultiplication and counit verify the coaction identities on

x, y

. To check that it descends to the quantum plane, expand

δ (y) δ (x) - q δ (x) δ (y) = x^{2} \otimes (ba - q ab) + y^{2} \otimes (d c - q c d) + x y \otimes [b c + q d a - q a d - q^{2} c b] = 0.

The last bracket vanishes because

b c = c b

and

d a - a d = (q - q^{- 1}) b c

For the requested calculation,

c a = q a c

and

y x = q x y

give

δ (x)^{2} = x^{2} \otimes a^{2} + (1 + q^{2}) x y \otimes a c + y^{2} \otimes c^{2} .

Multiply by

δ (y)

, put all monomials into the order

x^{r} y^{s}

, and use

b c = c b

. The answer is

δ (x^{2} y) = x^{3} \otimes a^{2} b + x^{2} y \otimes (a^{2} d + (1 + q^{2}) ab c) + x y^{2} \otimes ((1 + q^{2}) a c d + q^{2} b c^{2}) + y^{3} \otimes c^{2} d .

A left-coaction convention transposes where the matrix coefficients occur; the displayed answer fixes the right coaction throughout.

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