Decomposing a Representation from Its Character
Build the character table of a small finite group over the complex numbers and use weighted character inner products to decompose finite-dimensional representations.
What a learner can do afterwards
- Construct the character table of a small non-abelian group
- Decompose finite-dimensional complex representations, including permutation representations, using inner products with irreducible characters, and check dimensions from the multiplicities
- Read a normal subgroup off the table from the kernel of a character
1 · Read
Work with finite-dimensional complex representations of a finite group G. As in the previous topic, characters are traces and are constant on conjugacy classes. A character table has one row per irreducible character and one column per conjugacy class. The identity entry chi(e) is the representation's dimension.
For classes C_j with representatives c_j, define <chi,psi>=(1/|G|) sum_j |C_j| chi(c_j) conjugate(psi(c_j)). Conjugation changes a+bi to a-bi and leaves real values unchanged. If psi is irreducible, <chi,psi> is its multiplicity in the representation with character chi. Align classes, conjugate psi, multiply by class size, sum and divide by |G|. Multiplicities are nonnegative integers.
S3 permutes three letters. Its classes are identity, transpositions and 3-cycles, of sizes 1,3,2. Use that order throughout. A transposition swaps two letters; a 3-cycle moves all three cyclically. The permutation character counts fixed letters, giving P=(3,1,0). The trivial row is (1,1,1). The sign row is (1,-1,1), since transpositions are odd and 3-cycles are even.
The permutation space splits into constant vectors and the plane x1+x2+x3=0. Subtract trivial (1,1,1) from P to get S=(2,0,-1). For a known representation character, squared norm equals the sum of squared irreducible multiplicities, so norm 1 means irreducible. Here <S,S>=(4+0+2)/6=1. Trivial and sign are distinct one-dimensional irreducibles. S supplies a third, and three conjugacy classes mean the table is complete.
For S3, class sizes are 1,3,2 and the group order is 6. Take chi=(5,1,2). Pairing with trivial (1,1,1) gives (5+3+4)/6=2 copies. Pairing with sign (1,-1,1) gives (5-3+4)/6=1 copy. Pairing with standard (2,0,-1) gives (10+0-4)/6=1 copy. Thus chi contains two trivial copies, one sign copy and one standard copy.
Multiplicity counts copies; chi(e) gives total dimension. If irreducible chi_i has dimension d_i and multiplicity m_i, the number of constituents, counting copies, is sum_i m_i. Dimension is sum_i m_i d_i. The trivial character has value 1 everywhere, so its multiplicity is (1/|G|) sum_j |C_j| chi(c_j), not generally chi(e).
The regular representation permutes basis vectors indexed by G through left multiplication. Its character is |G| at the identity and 0 elsewhere. Pairing it with an irreducible of dimension d gives |G|*d/|G|=d copies. Since the regular representation has dimension |G|, the squares of all irreducible dimensions sum to |G|. This checks the dimension data of a completed table.
For a character chi of dimension d, ker(chi)={g: chi(g)=d} equals the representation kernel and is a normal subgroup. To see the equality, each representing matrix has finite order, so it is diagonalizable with eigenvalues of modulus 1. Their sum can equal d only when every eigenvalue is 1. Diagonalizability then makes the matrix the identity. Conversely, the identity matrix has trace d.
The character kernel equals the representation kernel. If rho(g)=I, then rho(hgh^-1)=rho(h)I rho(h)^-1=I, so conjugation preserves the kernel. As a homomorphism kernel, it is a subgroup and therefore normal. An element lies in a direct-sum kernel exactly when it acts as the identity on every summand, so intersect the summand kernels. For a kernel K, the quotient group has |G|/|K| elements.
In S3, the trivial kernel is all of S3. The sign kernel is A3={e,(123),(132)}, because those are exactly its elements with value 1. The standard kernel is {e}, since only its identity value equals its dimension 2. A single irreducible row need not reveal every normal subgroup; intersections of row kernels may be needed.
The norm-1 irreducibility test assumes a known representation character. A class function can have norm 1 without being a character. Its expansion in irreducible rows must have nonnegative integer coefficients to be a representation character. If valid character data give a fractional multiplicity, check class alignment, conjugation, class-size weights, division by |G| and arithmetic.
Weighted character inner products count irreducible copies. Dimensions weight those copies, and character kernels are normal subgroups.
This lesson, its practice questions and flashcards were written for this dataset.
The readings, practice questions and flashcards were written for this dataset. The two linked eigenvalue videos are identified by Khan Academy records; their reuse licenses have not been verified.
More to read
Constructing the S3 character table
Put the identity, transpositions and 3-cycles in that column order. These classes contain 1, 3 and 2 elements. A permutation matrix has a diagonal 1 for each fixed letter, so the permutation character is P=(3,1,0).
The constant vectors form a trivial subrepresentation with character T=(1,1,1). The plane of vectors whose coordinates sum to zero is also invariant. These spaces form a direct sum, so the plane has character P-T=(2,0,-1).
Its weighted squared norm is (4+0+2)/6=1. Since it is already a representation character, this proves irreducibility. The sign representation gives another row, (1,-1,1). Trivial, sign and standard are three distinct irreducibles. S3 has three conjugacy classes, so no further irreducible row is missing.
The standard representation is two-dimensional. Decomposition produces invariant subspaces, which need not be individual axes.
Why traces identify the representation kernel
Let rho be a d-dimensional complex representation of a finite group, with character chi. If g has order n, then rho(g)^n=I. The polynomial z^n-1 has distinct complex roots, so rho(g) is diagonalizable and its eigenvalues have modulus 1.
Suppose chi(g)=d. This trace is the sum of d eigenvalues. Each eigenvalue has real part at most 1. Their real parts sum to d, so each must be 1. On the unit circle, that forces every eigenvalue to equal 1. Diagonalizability then gives rho(g)=I.
The converse follows by taking the trace of I. Thus chi(g)=chi(e) exactly on the representation kernel. If g lies there, rho(hgh^-1)=rho(h)I rho(h)^-1=I for every h, proving invariance under conjugation.
For a direct sum, each group matrix has one block for each summand. It is the identity exactly when every block is the identity. This explains why direct-sum kernels are intersections.
Checking a character decomposition
For finite-dimensional complex representations of a finite group, irreducible multiplicities are character inner products. Class sizes matter because a column stands for every element in that class. Conjugate the irreducible values, multiply matching values and class sizes, sum and divide by the group order.
On S3, use class sizes 1,3,2 and class order identity, transpositions, 3-cycles. For chi=(8,0,2), the trivial coefficient is (8+0+4)/6=2. The sign coefficient is also 2. The standard coefficient is (16+0-4)/6=2.
There are six constituents counting copies. The dimension is 2*1+2*1+2*2=8, matching chi(e). These are different checks: one counts summands, while the other counts vector-space dimensions.
A norm of 1 proves irreducibility only after the function is known to be a representation character. For an arbitrary class function, inspect its irreducible coefficients. A negative or non-integer coefficient rules out a representation character, even if its squared norm happens to be 1.
Explore more
2 · Watch
Take it off screen
Where it sits
This opens up
Nothing builds on it yet.
18 questions wait behind this lesson, each with its answer explained. Every answer feeds the sky: stars light as they are learned, and dim when it is time to come back.