Character determines representation in characteristic zero
Suppose is a finite group and is a field of characteristic zero. Then, the character of any finite-dimensional representation of over completely determines the representation, i.e., no two inequivalent finite-dimensional representations can have the same character.
Note that does not need to be a splitting field.
- 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 copies of an irreducible representation to itself.
- Equivalent linear representations of finite group over field are equivalent over subfield in characteristic zero
Given: A group , two linear representations of with the same character over a field of characteristic zero.
To prove: and are equivalent as linear representations.
Proof: By Fact (2), both and are completely reducible, and are expressible as sums of irreducible representations. Suppose is a collection of distinct irreducible representations obtained as the union of all the representations occurring in a decomposition of into irreducible representations and a decomposition of into irreducible representations. In other words, there are nonnegative integers such that:
Let denote the character of and denote by the value (note: this would be 1 if 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||and for each||Fact (3)||Direct application of fact|
|2||and for each||has characteristic zero, so the manipulation makes sense||Step (1)|
|3||for each||has characteristic zero||Step (2)||[SHOW MORE]|
|4||and are equivalent||Step (3)|