# Character determines representation in characteristic zero

## Statement

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.

## 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 copies of an irreducible representation to itself.

### Applications

## Facts used

## Proof

**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:

and

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) |