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The kernel of a finite direct sum is the intersection of the kernels

Statement

Let $(V_{i}, ρ_{i})_{i \in I}$ be a finite family of finite-dimensional representations of a group $G$ over a field $k$ . On $V = ⨁_{i \in I} V_{i}$ , the formula $ρ (g) ((v_{i})_{i}) = (ρ_{i} (g) v_{i})_{i}$ defines a finite-dimensional representation, and $\ker ρ = ⋂_{i \in I} \ker ρ_{i}$ . For $I = \emptyset$ , $V = 0$ and the empty intersection is understood inside $G$ , so both sides are $G$ .

Facts & Assumptions

Given: A finite index set $I$ and homomorphisms $ρ_{i} : G \to GL (V_{i})$ with each $V_{i}$ finite-dimensional over $k$ .

[F1]

A finite-dimensional representation is a homomorphism to the group of invertible linear maps of a finite-dimensional space (A finite-dimensional representation $ρ : G \to GL (V)$ over a field, and its degree).

[F2]

The kernel consists of elements mapped to the identity of the target group (The kernel and image of a group homomorphism).

[F3]

The direct sum consists of finitely supported tuples with coordinatewise operations and has coordinate inclusions; the empty sum is zero (The direct sum of an indexed family of modules).

Proof

technique · direct

1.1

Because $I$ is finite, every tuple has finite support, so $V = \prod_{i \in I} V_{i}$ with the operations in F3. Choose a finite basis of each $V_{i}$ . This uses only finitely many existential witnesses. Their coordinate inclusions form a finite basis of $V$ : they span because each coordinate has a basis expansion, and a zero combination has all coefficients zero by projecting to each coordinate and using that coordinate’s independence. Thus $V$ is finite-dimensional. Zero-dimensional coordinates contribute empty bases.

F3given

1.2

For $g \in G$ , define $ρ (g)$ by the displayed coordinate formula. The equality $ρ_{i} (g) (a v_{i} + b w_{i}) = a ρ_{i} (g) v_{i} + b ρ_{i} (g) w_{i}$ in every coordinate proves linearity. Also $ρ_{i} (g) ρ_{i} (g^{- 1}) = {id}_{V_{i}}$ and the reversed product is the identity, so $ρ (g^{- 1})$ is a two-sided inverse of $ρ (g)$ . Hence $ρ (g) \in GL (V)$ .

F1F3givenalgebra

2.1

For every tuple $v$ , the $i$ -coordinate of $ρ (g h) v$ is $ρ_{i} (g h) v_{i} = ρ_{i} (g) ρ_{i} (h) v_{i}$ , the $i$ -coordinate of $ρ (g) ρ (h) v$ . Similarly $ρ (e) v = v$ . Therefore $ρ$ is a homomorphism and, with step 1.1, a finite-dimensional representation as in F1.

F1step 1.1step 1.2

2.2

If $g \in \ker ρ$ , F2 gives $ρ (g) = {id}_{V}$ . For any $i$ and $v_{i} \in V_{i}$ , apply this equality to the coordinate inclusion $ȷ_{i} (v_{i})$ . Its $i$ -coordinate yields $ρ_{i} (g) v_{i} = v_{i}$ . Since $v_{i}$ was arbitrary, $ρ_{i} (g) = {id}_{V_{i}}$ , so $g \in \ker ρ_{i}$ for every $i$ .

F2F3step 1.2

3.1

Conversely, if $g \in \ker ρ_{i}$ for every $i$ , then for every tuple $v$ , $ρ (g) v = (ρ_{i} (g) v_{i})_{i} = (v_{i})_{i} = v$ . Hence $ρ (g) = {id}_{V}$ and $g \in \ker ρ$ . This proves the intersection formula.

F2step 1.2step 2.2

4.1

For $I = \emptyset$ , F3 gives $V = 0$ . Its unique endomorphism is its identity and is invertible, so every $g$ acts identically and $\ker ρ = G$ . The condition that $g$ lie in each kernel is vacuous, giving the same $G$ . For a one-element family steps 2.2–3.1 give the single coordinate kernel. For a zero coordinate its kernel is $G$ , so it places no further restriction on the intersection.

F2F3step 2.2step 3.1∎

Sources

Etingof et al., Chapter 4 opening, p. 61, supplies the representation convention. The coordinate action, its finite-dimensionality, and both kernel containments are derived locally from the direct-sum definition.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

Etingof et al., Introduction to Representation Theory (standard reference, not scraped)