Lemma from Schur

The Schur's Lemma , named after Issai Schur , describes the homomorphisms between simple modules . It says that any such homomorphism except the null homomorphism is an isomorphism .

Formulations of the lemma

Schur's lemma in the modular theoretical version reads ( be a ring with 1): ${\displaystyle R}$

Let it be , simple link modules. Then: ${\displaystyle M}$ ${\displaystyle N}$ ${\displaystyle R}$

1. ${\displaystyle M\not \cong N\Rightarrow \operatorname {Hom} _{R}(M,N)=0}$
2. ${\displaystyle \operatorname {End} _{R}(M)}$is an oblique body .

In the representation theory version, Schur's lemma reads ( be a finite group, a body): ${\displaystyle G}$ ${\displaystyle K}$

Let them be irreducible representations of . Then: ${\displaystyle \rho :\ G\rightarrow \operatorname {GL} _{K}(V),\tau :\ G\rightarrow \operatorname {GL} _{K}(W)}$ ${\displaystyle G}$

1. It be with . Then: or is bijective (and in this case and are equivalent). ${\displaystyle f\in \operatorname {Hom} _{K}(V,W)}$ ${\displaystyle f\circ \rho (g)=\tau (g)\circ f\ ,\ \forall g\in G}$ ${\displaystyle f=0}$ ${\displaystyle f}$ ${\displaystyle \rho }$ ${\displaystyle \tau }$
2. ${\displaystyle Z(\rho )}$ is an oblique body.

The second statement also applies in reverse, so that an inclined body is exactly when the representation is irreducible. ${\displaystyle Z(\rho )}$ ${\displaystyle \rho }$

Due to the connection between the representations of over and KG modules, both versions mean the same thing. ${\displaystyle G}$ ${\displaystyle K}$

proof

The proof (of the representation theory version) only needs elementary linear algebra. Let it be invertible matrices, invertible matrices, and let it be a matrix. The following applies to the die products ${\displaystyle \rho (g)}$ ${\displaystyle n\times n}$ ${\displaystyle \tau (g)}$ ${\displaystyle m\times m}$ ${\displaystyle f}$ ${\displaystyle m\times n}$

${\displaystyle f\,\rho (g)=\tau (g)\,f\qquad \forall g\in G}$

Then the core of is an invariant subspace for the representation , because from follows . Because of the irreducibility of, there can only be the zero vector space or the entire vector space. In the first case it is invertible and provides a similarity transformation between the representation matrices and . In the second case is the zero matrix . ${\displaystyle f}$ ${\displaystyle \rho (g)}$ ${\displaystyle fv=0}$ ${\displaystyle f\rho (g)v=0}$ ${\displaystyle \rho }$ ${\displaystyle \operatorname {Kern} \,f}$ ${\displaystyle f}$ ${\displaystyle \rho }$ ${\displaystyle \tau }$ ${\displaystyle f}$

Applications

For practical purposes (tabulation) the matrices of an irreducible representation are occasionally standardized. For example, the common eigenvectors of rotations around a selected axis serve as a standard basis for the rotation group . In such cases the matrices are of irreducible representations and are either inequivalent or identical. This makes the following addition to Schur's Lemma relevant: ${\displaystyle \rho (g)}$ ${\displaystyle \tau (g)}$

Out for all follows , i.e. H. is a complex multiple of the identity matrix . ${\displaystyle f\rho (g)=\,\!\rho (g)f}$ ${\displaystyle g\in G}$ ${\displaystyle f=\lambda \,{\boldsymbol {1}}}$ ${\displaystyle f}$

Proof: Let it be a (complex) eigenvalue of and be an associated eigenvector. With the assumed equation also holds ${\displaystyle \lambda \,\!}$ ${\displaystyle f}$ ${\displaystyle e\,\!}$

${\displaystyle (f-\lambda {\boldsymbol {1}})\,\rho (g)=\rho (g)\,(f-\lambda {\boldsymbol {1}})\qquad \forall g\in G}$

Therefore the kernel of is an invariant subspace of the representation and can only be the null space or the whole space due to irreducibility. Since the eigenvector belongs to the kernel, only the second possibility remains. So it applies . ${\displaystyle f-\lambda {\boldsymbol {1}}}$ ${\displaystyle \rho (g)}$ ${\displaystyle e\neq 0}$ ${\displaystyle f-\lambda {\boldsymbol {1}}=0}$

A simple corollary of Schur's lemma is that any complex irreducible representation of an Abelian group must be one-dimensional.

See also

• Schur disassembly

Individual evidence

1. ↑ M. Chaichian, R. Hagedorn, Symmetries in quantum mechanics: from angular momentum to supersymmetry , Institute of Physics Publishing, Bristol 1998