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) |
