Schur's lemma: Difference between revisions

Latest revision as of 14:34, 5 November 2023

This fact is related to: linear representation theory
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Statement

Here, $G$ is a group and $k$ is a field. In the abstract formulation, $V_{1},V_{2}$ are vector spaces over $k$ .

Statement no.	Condition	Conclusion in abstract formulation for vector spaces: $\rho _{1}:G\to GL(V_{1}),\rho _{2}:G\to GL(V_{2})$ are linear representations of $G$ over a field $k$ . $f:V_{1}\to V_{2}$ is a homomorphism from $\rho _{1}$ to $\rho _{2}$ .	Conclusion in concrete matrix formulation: $\rho _{1}:G\to GL(m,k)$ , $\rho _{2}:G\to GL(n,k)$ are representations. $F\in \operatorname {Mat} _{n\times m}$ is a homomorphism of representations
1	$\rho _{1}$ is irreducible	$f$ is either the zero map or is injective.	$F$ is either the zero matrix or has full column rank, and $n\geq m$ .
2	$\rho _{2}$ is irreducible	$f$ is either the zero map or is surjective	$F$ is either the zero matrix or has full row rank, and $m\geq n$ .
3	$\rho _{1},\rho _{2}$ are both irreducible.	$f$ is either the zero map or is bijective and hence an isomorphism of representations.	$F$ is either the zero matrix or is invertible, and $m=n$ .
4	$\rho _{1}=\rho _{2}$ , it is irreducible, and $k$ is an algebraically closed field	$f$ is a scalar multiplication map.	$F$ is a scalar matrix and $m=n$ .

Applications

Central implies image under every irreducible representation is scalar
Irreducible representation over splitting field surjects to matrix ring
Character orthogonality theorem: Schur's lemma is used crucially to show that certain matrix averages are zero and certain others are scalars.
Grand orthogonality theorem: Schur's lemma is used crucially to show that certain matrix averages are zero and certain others are scalars.
Irreducible complex representation of abelian group is one dimensional
Irreducible real representation of abelian group is one or two dimensional
Sum of irreducible representation on conjugacy class is scalar multiple of identity matrix

Proof

Verbal proof

For this, we use the fact that the kernel of any homomorphism of representations is an invariant subspace. If $\rho _{1}$ is irreducible, the kernel is either the whole of $V_{1}$ (in which case we have the zero map) or the zero subspace (in which case we have an injective map).
For this, we use the fact that the image of any homomorphism of representations is an invariant subspace. If $\rho _{2}$ is irreducible, the image is either the whole of $V_{2}$ (in which case we have a surjective map) or the zero subspace (in which case we have the zero map).
This follows from the previous two parts.
Any homomorphism from a representation on that field to itself can be represented by a linear operator from the vector space to itself. This linear operator must have an eigenvalue, because the field is algebraically closed. Subtracting that eigenvalue times the identity matrix, we now get a linear operator that has zero as an eigenvalue, which is also a homomorphism of representations.

But part (3) tells us that any homomorphism of representations is either the zero map or an isomorphism. An isomorphism cannot have zero as an eigenvalue (since it is an invertible linear operator) and hence the new linear operator we get must be the zero map.

Thus the original linear operator must have been a scalar times the identity.

@@ Line 3: / Line 3: @@
 ==Statement==
-Let <math>G</math> be a [[group]] and <math>\rho_1: G \to GL(V_1), \rho_2: G \to GL(V_2)</math> be [[linear representation]]s of <math>G</math> over a [[field]] <math>k</math>. Then the following hold:
+Here, <math>G</math> is a [[group]] and <math>k</math> is a [[field]]. In the abstract formulation, <math>V_1,V_2</math> are vector spaces over <math>k</math>.
+{| class="sortable" border="1"
+! Statement no. !! Condition !! Conclusion in abstract formulation for vector spaces: <math>\rho_1: G \to GL(V_1), \rho_2: G \to GL(V_2)</math> are [[linear representation]]s of <math>G</math> over a [[field]] <math>k</math>. <math>f:V_1 \to V_2</math> is a [[fact about::homomorphism of representations|homomorphism]] from <math>\rho_1</math> to <matH>\rho_2</math>. !! Conclusion in concrete matrix formulation: <math>\rho_1:G \to GL(m,k)</math>, <math>\rho_2:G \to GL(n,k)</math> are representations. <math>F \in \operatorname{Mat}_{n \times m}</math> is a [[homomorphism of representations]]
+|-
+| 1 || <math>\rho_1</math> is [[fact about::irreducible linear representation|irreducible]] ||<math>f</math> is either the zero map or is injective. || <math>F</math> is either the zero matrix or has full column rank, and <math>n \ge m</math>.
+|-
+| 2 || <math>\rho_2</math> is irreducible || <math>f</math> is either the zero map or is surjective || <math>F</math> is either the zero matrix or has full row rank, and <math>m \ge n</math>.
+|-
+| 3 || <math>\rho_1, \rho_2</math> are both irreducible. || <math>f</math> is either the zero map or is bijective and hence an isomorphism of representations. || <math>F</math> is either the zero matrix or is invertible, and <math>m = n</math>.
+|-
+| 4 ||<math>\rho_1 = \rho_2</math>, it is irreducible, and <math>k</math> is an [[algebraically closed field]] || <math>f</math> is a scalar multiplication map. || <math>F</math> is a scalar matrix and <math>m = n</math>.
+|}
-# If <math>\rho_1</math> is [[irreducible linear representation|irreducible]], then any [[homomorphism of representations]] from <math>\rho_1</math> to <math>\rho_2</math> is either injective, or the zero map
+==Applications==
-# If <math>\rho_2</math> is irreducible, then any [[homomorphism of representations]] from <math>\rho_1</math> to <math>\rho_2</math> is either surjective, or the zero map
-# If both are irreducible, then any homomorphism of representations from <math>\rho_1</math> to <math>\rho_2</math> is either an isomorphism of representations, or the zero map
+* [[Central implies image under every irreducible representation is scalar]]
-# If <math>\rho_1 = \rho_2</math> is irreducible, and the field is an [[algebraically closed field]], then any homomorphism from <math>\rho_1</math> to <math>\rho_2</math> is a scalar multiplication map
+* [[Irreducible representation over splitting field surjects to matrix ring]]
+* [[Character orthogonality theorem]]: Schur's lemma is used crucially to show that certain matrix averages are zero and certain others are scalars.
+* [[Grand orthogonality theorem]]: Schur's lemma is used crucially to show that certain matrix averages are zero and certain others are scalars.
+* [[Irreducible complex representation of abelian group is one dimensional]]
+* [[Irreducible real representation of abelian group is one or two dimensional]]
+* [[Sum of irreducible representation on conjugacy class is scalar multiple of identity matrix]]
 ==Proof==
@@ Line 15: / Line 31: @@
 # For this, we use the fact that the kernel of any homomorphism of representations is an [[invariant subspace for a linear representation|invariant subspace]]. If <math>\rho_1</math> is irreducible, the kernel is either the whole of <math>V_1</math> (in which case we have the zero map) or the zero subspace (in which case we have an injective map).
 # For this, we use the fact that the image of any homomorphism of representations is an invariant subspace. If <math>\rho_2</math> is irreducible, the image is either the whole of <math>V_2</math> (in which case we have a surjective map) or the zero subspace (in which case we have the zero map).
 # This follows from the previous two parts.
 # Any homomorphism from a representation on that field to itself can be represented by a linear operator from the vector space to itself. This linear operator must have an eigenvalue, because the field is algebraically closed. Subtracting that eigenvalue times the identity matrix, we now get a linear operator that has zero as an eigenvalue, which is also a homomorphism of representations.