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 |
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1 | ![]() ![]() ![]() |
Fact (3) | Direct application of fact | ||
2 | ![]() ![]() ![]() |
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Step (1) | ||
3 | ![]() ![]() |
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Step (2) | [SHOW MORE] | |
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Step (3) |