Complex representation

A complex representation of a group $G$ is a homomorphism $\rho :G\to \mathrm{GL}(V)$ for a complex vector space $V$.

A finite-dimensional continuous representation of a topological group is a continuous homomorphism $\rho :G\to \mathrm{GL}(V)$, where $V$ is a finite-dimensional complex vector space.

The degree of a finite-dimensional representation $\rho :G\to \mathrm{GL}(V)$ is $\dim V$.

Representations $\rho :G\to \mathrm{GL}(V)$ and ${\rho }^{\prime }:G\to \mathrm{GL}(V^{\prime })$ are isomorphic when an invertible linear map $T:V\to V^{\prime }$ intertwines the actions: $T\rho (g)={\rho }^{\prime }(g)T$ for every $g\in G$.

On polynomials of degree at most one, let the generator of $\mathbb{Z}$ act by $p(x)\mapsto p(x+1)$. In the basis $(1,x)$ its matrix isThis nontrivial Jordan block is not diagonalizable, so the representation is not a direct sum of one-dimensional subrepresentations.

A representation is faithful when its kernel is trivial.

The left regular representation acts on the basis $(e_h{)}_{h\in G}$ by $\rho (g)e_h=e_{gh}$. It is faithful.

If ${\rho }_1,\dots ,{\rho }_k$ are the complex irreducible representations of a finite group $G$ and $n_j=\dim {\rho }_j$, decomposition of the regular representation gives

By Maschke's theorem, the complex regular representation of a finite group is a direct sum of irreducibles. Its kernel is the intersection of their kernels. If the group is simple, each kernel is either trivial or the whole group; faithfulness of the regular representation therefore forces at least one irreducible constituent to be faithful.

For a $p$-cycle $g\in S_p$, each $g^k$ with $1\le k<p$ is conjugate to $g$. In a faithful complex representation, $\rho (g)$ has a nontrivial $p$th-root eigenvalue $\lambda$, and similarity of $\rho (g)$ and $\rho (g{)}^k$ puts all $p-1$ values ${\lambda }^k$ in its spectrum. Hence the representation has dimension at least $p-1$.

Commuting complex involutions are simultaneously diagonalizable, so $(C_2{)}^n$ can act faithfully in dimension $d$ only if $n\le d$. A group containing $(C_2{)}^n$ for arbitrarily large $n$, such as the full permutation group of $\mathbb{N}$, has no faithful finite-dimensional complex representation.

Let $G$ be the real upper unitriangular three-by-three group, let $Z$ be its centre, and let $Z_0$ be the integer subgroup of $Z$. Every finite-dimensional continuous complex representation of $G/Z_0$ kills the central circle $Z/Z_0$, and hence is not faithful.

The image of a one-dimensional representation lies in the abelian group ${\mathbb{C}}^{\times }$, so its kernel contains the commutator subgroup of the represented group.

Given a quotient group map $\pi :G\to G/N$ and a representation $\rho$ of $G/N$, its inflation to $G$ is the representation $\rho \circ \pi$. Conversely, a representation of $G$ factors through $G/N$ exactly when its kernel contains $N$.

Every irreducible complex representation of a finite abelian group is one-dimensional. After making the representation unitary, the commuting representing matrices are simultaneously diagonalizable; any common eigenspace is invariant, so irreducibility leaves only one dimension.