# Combining and decomposing representations

If $${G}$$ and $${H}$$ are groups or Lie groups with representations on vector spaces $${V}$$ and $${W}$$, we can define the direct sum of the representations as the representation of $${G\times H}$$ on $${V\oplus W}$$ defined by $${\left(g,h\right)\left(v,w\right)\equiv\left(g\left(v\right),h\left(w\right)\right)}$$. The Lie algebra of $${G\times H}$$ is $${\mathfrak{g}\oplus\mathfrak{h}}$$, and it then has a representation on $${V\oplus W}$$ similarly given by $${\left(A,B\right)\left(v,w\right)\equiv\left(A\left(v\right),B\left(w\right)\right)}$$.

Since every linear transformation leaves the origin invariant, no linear representation is transitive. However, one can ask that at least no vector subspace be invariant. An irreducible linear representation (AKA irrep) on $${V}$$ is defined as a group or algebra representation that has no non-trivial invariant subspace (AKA subrepresentation, or submodule if an algebra rep) $${0\subset W\subset V}$$ such that $${g\left(W\right)\subset W\:\forall g\in G}$$. A representation is completely reducible (AKA decomposable) if the orthogonal complement of every invariant subspace is also invariant; any finite-dimensional completely reducible representation can then be written as a direct sum of irreps. Referring back to the figure on group actions, the action of $${SO(2)}$$ on $${\mathbb{R}^{3}}$$ is completely reducible, and can be written as the direct sum of the identity irrep on the axis of rotation $${\mathbb{R}^{1}}$$ and the rotation irrep on the plane $${\mathbb{R}^{2}}$$ orthogonal to it.

Note that a representation can be reducible but not completely reducible, i.e. can have an invariant subspace and yet not be a direct sum of irreps. However, most representations of interest are either irreducible or completely reducible:

• Every representation of a finite group is completely reducible
• Every representation of a compact Lie group is completely reducible
• Every unitary representation is completely reducible
• Weyl’s theorem: every representation of a Lie algebra is completely reducible iff the Lie algebra is semisimple (semisimple will be defined when we cover compact Lie groups)
• Every representation of a connected semisimple Lie group is completely reducible

Again considering groups or Lie groups $${G}$$ and $${H}$$ with representations on $${V}$$ and $${W}$$, we can define the tensor product of the representations as the representation of $${G\times H}$$ on $${V\otimes W}$$ defined by $${(g,h)(v\otimes w)\equiv g(v)\otimes h(w)}$$. In this case the representation of the Lie algebra $${\mathfrak{g}\oplus\mathfrak{h}}$$ is given by $${\left(A,B\right)\left(v\otimes w\right)=A\left(v\right)\otimes I+I\otimes B\left(w\right)}$$, in order to make it linear on $${V\otimes W}$$.

The tensor product of two representations of the same group $${G}$$ can be viewed as a new representation of $${G}$$ on the vector space $${V\otimes W}$$ given by $${g\left(v\otimes w\right)\equiv g\left(v\right)\otimes g\left(w\right)}$$. Even if the two original representations are irreducible, this new tensor product representation may not be; decomposing it into a direct sum of irreps is called Clebsch-Gordan theory.

By noting that the kernel and image of an intertwiner are invariant subspaces, one arrives at Schur’s lemma, which states that any intertwiner between irreps is either zero or an isomorphism. This has several immediate consequences, which are sometimes referred to themselves as Schur’s lemma:

• Any self-intertwiner of a finite-dimensional complex irrep is a multiple of the identity map
• Any two intertwiners between finite-dimensional complex irreps differ by only a complex constant multiple
• Any matrix in the center of the image of a complex irrep is a multiple of the identity matrix
• A complex irrep maps any element in the center of a Lie group to a multiple of the identity transformation
• Any complex irrep of an abelian Lie group is one-dimensional (as a complex manifold)

For a real Lie algebra $${\mathfrak{g}}$$, we can consider its complexification $${\mathfrak{g}^{\mathbb{C}}}$$ as a complex Lie algebra. We can then ask, if $${\mathfrak{g}^{\mathbb{C}}}$$ has an irrep on $${\mathbb{C}^{n}}$$ (i.e. as an algebra with complex matrices in $${\mathbb{C}(n)}$$ as vectors and scalars in $${\mathbb{C}}$$), does this correspond to an irrep of $${\mathfrak{g}}$$ on $${\mathbb{C}^{n}}$$ (i.e. as an algebra of complex matrices in $${\mathbb{C}(n)}$$ as vectors and scalars in $${\mathbb{R}}$$)? The answer is yes; the irreps on $${\mathbb{C}^{n}}$$ of $${\mathfrak{g}}$$ are one to one with those of $${\mathfrak{g}^{\mathbb{C}}}$$ as a complex Lie algebra.