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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

1.1givenL2algebra

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 $r \in R$ .

1.2L1givenalgebra

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 $V_{C}$ .

2.1step 1.1L2

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

3.1step 2.1L1algebra

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 $V_{C}$ .

3.2step 2.1L1

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

3.3step 1.2step 2.1

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 $G = F$ .

4.1step 3.3L3

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].

5.1step 3.1step 3.2step 3.3step 4.1∎

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.

Depends on

Used by

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Sources

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