Character determines representation in characteristic zero: Difference between revisions

Latest revision as of 14:07, 21 July 2011

Statement

Suppose $G$ is a finite group and $K$ is a field of characteristic zero. Then, the character of any finite-dimensional representation of $G$ over $K$ completely determines the representation, i.e., no two inequivalent finite-dimensional representations can have the same character.

Note that $K$ does not need to be a splitting field.

Related facts

Opposite facts

Character does not determine representation in any prime characteristic: The problem is that we can construct representations whose character is identically zero simply by adding $p$ copies of an irreducible representation to itself.

Applications

Equivalent linear representations of finite group over field are equivalent over subfield in characteristic zero

Facts used

Proof

Given: A group $G$ , two linear representations $ρ_{1}, ρ_{2}$ of $G$ with the same character $χ$ over a field $K$ of characteristic zero.

To prove: $ρ_{1}$ and $ρ_{2}$ are equivalent as linear representations.

Proof: By Fact (2), both $ρ_{1}$ and $ρ_{2}$ are completely reducible, and are expressible as sums of irreducible representations. Suppose $φ_{1}, φ_{2}, \dots, φ_{s}$ is a collection of distinct irreducible representations obtained as the union of all the representations occurring in a decomposition of $ρ_{1}$ into irreducible representations and a decomposition of $ρ_{2}$ into irreducible representations. In other words, there are nonnegative integers $a_{11}, a_{12}, \dots, a_{1 s}, a_{21}, a_{22}, \dots, a_{2 s}$ such that:

$ρ_{1} ≅ a_{11} φ_{1} \oplus a_{12} φ_{2} \oplus \dots \oplus a_{1 s} φ_{s}$

and

$ρ_{2} ≅ a_{21} φ_{1} \oplus a_{22} φ_{2} \oplus \dots \oplus a_{2 s} φ_{s}$

Let $χ_{i}$ denote the character of $φ_{i}$ and denote by $m_{i}$ the value $⟨ χ_{i}, χ_{i} ⟩_{G}$ (note: this would be 1 if $K$ were a splitting field, and in general it is the sum of squares of multiplicities of irreducible constituents over a splitting field).

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	$m_{i} a_{1 i} = ⟨ χ, χ_{i} ⟩_{G}$ and $m_{1} a_{2 i} = ⟨ χ, χ_{i} ⟩_{G}$ for each $1 \leq i \leq s$	Fact (3)			Direct application of fact
2	$a_{1 i} = \frac{1}{m} ⟨ χ, χ_{i} ⟩_{G}$ and $a_{2 i} = \frac{1}{m} ⟨ χ, χ_{i} ⟩_{G}$ for each $1 \leq i \leq s$		$K$ has characteristic zero, so the manipulation makes sense	Step (1)
3	$a_{1 i} = a_{2 i}$ for each $1 \leq i \leq s$		$K$ has characteristic zero	Step (2)	[SHOW MORE] Note that if $K$ had characteristic $p$ , we could only conclude equality modulo $p$ , and not equality as nonnegative integers.
4	$ρ_{1}$ and $ρ_{2}$ are equivalent			Step (3)

@@ Line 44: / Line 44: @@
 | 2 ||<math>a_{1i} = \frac{1}{m}\langle \chi,\chi_i \rangle_G</math> and <math>a_{2i} = \frac{1}{m} \langle \chi, \chi_i \rangle_G</math> for each <math>1 \le i \le s</math> || || <math>K</math> has characteristic zero, so the manipulation makes sense || Step (1) ||
 |-
-| 3 || <math>a_{1i} = a_{2i}</math> for each <math>1 \le i \le s</math> || || <math>K</math> has characteristic zero || || <toggledisplay>Note that if <math>K</math> had characteristic <math>p</math>, we could only conclude equality modulo <math>p</math>, and not equality as nonnegative integers.</toggledisplay>
+| 3 || <math>a_{1i} = a_{2i}</math> for each <math>1 \le i \le s</math> || || <math>K</math> has characteristic zero || Step (2) || <toggledisplay>Note that if <math>K</math> had characteristic <math>p</math>, we could only conclude equality modulo <math>p</math>, and not equality as nonnegative integers.</toggledisplay>
 |-
 | 4 || <math>\rho_1</math> and <math>\rho_2</math> are equivalent || || || Step (3) ||
 |}