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Complexification is initial for real-linear maps into complex vector spaces, and is unique up to unique isomorphism

Statement

Let $V$ be a real vector space and let $ι : V \to V_{C}$ be the canonical real-linear embedding of Complexification as $C \otimes_{R} V$ with its canonical real-linear embedding. For every complex vector space $W$ and every real-linear map $f : V \to W$ there is a unique complex-linear map $F : V_{C} \to W$ with $F \circ ι = f$ , namely $F (z \otimes v) = z f (v)$ .

Moreover, if $U$ is a complex vector space and $ι^{'} : V \to U$ is a real-linear map with the same property, then there is a unique complex-linear isomorphism $u : V_{C} \to U$ with $u \circ ι = ι^{'}$ .

Facts & Assumptions

Given: A real vector space $V$ with canonical embedding $ι : V \to V_{C}$ , and a real-linear map $f : V \to W$ into a complex vector space $W$ .

[L1]

The complexification carries the scalar action $z^{'} \cdot (z \otimes v) = (z^{'} z) \otimes v$ and $ι v = 1 \otimes v$ (Complexification as $C \otimes_{R} V$ with its canonical real-linear embedding).

[L2]

Every $R$ -balanced map $b : C \times V \to A$ extends uniquely to an $R$ -linear map out of $C \otimes_{R} V$ (Universal property of the tensor product for balanced maps into abelian groups).

[L3]

Two pairs representing the same class of balanced maps are related by a unique isomorphism commuting with the structure maps, obtained by applying each universal property to the other pair (Tensor products are unique up to a unique isomorphism carrying elementary tensors to elementary tensors).

Proof

technique · direct

1.1

The map $b : C \times V \to W$ given by $b (z, v) = z f (v)$ is $R$ -bilinear and $R$ -balanced: additivity follows from the linearity of $f$ in $v$ and the distributivity of complex scalar multiplication, and $b (z r, v) = (z r) f (v) = z (r f (v)) = z f (r v) = b (z, r v)$ for $r \in R$ .

givenL2algebra

1.2

Uniqueness of any extension: if $G : V_{C} \to W$ is complex-linear with $G \circ ι = f$ , then $G (z \otimes v) = G (z \cdot (1 \otimes v)) = z G (1 \otimes v) = z f (v)$ on every elementary tensor by [L1], so its values are already determined, and elementary tensors generate $V_{C}$ .

L1givenalgebra

2.1

By [L2] there is a unique $R$ -linear map $F : V_{C} \to W$ with $F (z \otimes v) = z f (v)$ .

step 1.1L2

3.1

The map $F$ is complex-linear: by the scalar action of [L1], $F (z^{'} \cdot (z \otimes v)) = F ((z^{'} z) \otimes v) = (z^{'} z) f (v) = z^{'} (z f (v)) = z^{'} F (z \otimes v)$ , and additivity extends the identity to all of $V_{C}$ .

step 2.1L1algebra

3.2

The map $F$ satisfies $F \circ ι = f$ because $F (ι v) = F (1 \otimes v) = f (v)$ by [L1].

step 2.1L1

3.3

Uniqueness: steps 1.2 and 2.1 show that any complex-linear extension $G$ with $G \circ ι = f$ agrees with $F$ on every elementary tensor, hence everywhere, so $G = F$ .

step 1.2step 2.1

4.1

For the uniqueness up to unique isomorphism, apply the universal property of $(V_{C}, ι)$ to $ι^{'}$ to get a unique complex-linear $u : V_{C} \to U$ with $u \circ ι = ι^{'}$ , and the universal property of $(U, ι^{'})$ to $ι$ to get $v : U \to V_{C}$ with $v \circ ι^{'} = ι$ . Both $v \circ u$ and $id$ carry $ι$ to itself, and both $u \circ v$ and $id$ carry $ι^{'}$ to itself, so step 3.3 forces both composites to be identities; this is the two-application argument recorded in [L3].

step 3.3L3

5.1

Steps 3.1 and 3.2 prove the universal property, step 3.3 its uniqueness clause, and step 4.1 the uniqueness of the representing pair up to unique isomorphism.

step 3.1step 3.2step 3.3step 4.1∎

Depends on

Used by

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Sources

Keith Conrad, Complexification (notes) (standard reference, not scraped)