 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/ii/paper-2/15f/i/solution

Codex  0 Created 2026-09-24 Updated 2026-10-06

S^{n} V

is the quotient of

V^{\otimes n}

identifying tensors which differ by permutation of the factors. The exterior power

Λ^{n} V

is the alternating quotient, with a tensor having repeated equal factors set to zero. Both quotients inherit the given group action. For a basis

e_{1}, \dots, e_{d}

, bases consist respectively of monomials

e_{1}^{a_{1}} \dots e_{d}^{a_{d}}

with

\sum a_{i} = n

, and wedge products

e_{i_{1}} \land \dots \land e_{i_{n}}

with

i_{1} < \dots < i_{n}

. Thus

dim S^{n} V = (n d + n - 1), dim Λ^{n} V = (n d),

with the second dimension zero for

n > d

A finite-order group element is diagonalizable over

C

, because its minimal polynomial divides

X^{m} - 1

, which has no repeated roots. Choose an eigenbasis with eigenvalues

λ_{1}, \dots, λ_{d}

. The induced eigenvalues, including multiplicities, are

λ_{1}^{a_{1}} \dots λ_{d}^{a_{d}} (a_{i} \geq 0, \sum a_{i} = n) on S^{n} V; λ_{i_{1}} \dots λ_{i_{n}} (i_{1} < \dots < i_{n}) on Λ^{n} V .

The zeroth powers are the one-dimensional trivial representation, with the empty product equal to

1

 New to topics? Read the docs here!