# Complexification

For the complexification of a real Lie group, see Complexification (Lie group).

In mathematics, the complexification of a vector space V over the field of real numbers (a "real vector space") yields a vector space VC over the complex number field, obtained by formally extending the scaling of vectors by real numbers to include their scaling ("multiplication") by complex numbers. Any basis for V (a space over the real numbers) may also serve as a basis for VC over the complex numbers.

## Formal definition

Let V be a real vector space. The complexification of V is defined by taking the tensor product of V with the complex numbers (thought of as a two-dimensional vector space over the reals):

${\displaystyle V^{\mathbb {C} }=V\otimes _{\mathbb {R} }\mathbb {C} .}$

The subscript R on the tensor product indicates that the tensor product is taken over the real numbers (since V is a real vector space this is the only sensible option anyway, so the subscript can safely be omitted). As it stands, VC is only a real vector space. However, we can make VC into a complex vector space by defining complex multiplication as follows:

${\displaystyle \alpha (v\otimes \beta )=v\otimes (\alpha \beta )\qquad {\mbox{for all }}v\in V{\mbox{ and }}\alpha ,\beta \in \mathbb {C} .}$

More generally, complexification is an example of extension of scalars – here extending scalars from the real numbers to the complex numbers – which can be done for any field extension, or indeed for any morphism of rings.

Formally, complexification is a functor VectR → VectC, from the category of real vector spaces to the category of complex vector spaces. This is the adjoint functor – specifically the left adjoint – to the forgetful functor VectC → VectR from forgetting the complex structure.

## Basic properties

By the nature of the tensor product, every vector v in VC can be written uniquely in the form

${\displaystyle v=v_{1}\otimes 1+v_{2}\otimes i}$

where v1 and v2 are vectors in V. It is a common practice to drop the tensor product symbol and just write

${\displaystyle v=v_{1}+iv_{2}.\,}$

Multiplication by the complex number a + ib is then given by the usual rule

${\displaystyle (a+ib)(v_{1}+iv_{2})=(av_{1}-bv_{2})+i(bv_{1}+av_{2}).\,}$

We can then regard VC as the direct sum of two copies of V:

${\displaystyle V^{\mathbb {C} }\cong V\oplus iV}$

with the above rule for multiplication by complex numbers.

There is a natural embedding of V into VC given by

${\displaystyle v\mapsto v\otimes 1.}$

The vector space V may then be regarded as a real subspace of VC. If V has a basis {ei} (over the field R) then a corresponding basis for VC is given by {ei ⊗ 1} over the field C. The complex dimension of VC is therefore equal to the real dimension of V:

${\displaystyle \dim _{\mathbb {C} }V^{\mathbb {C} }=\dim _{\mathbb {R} }V.}$

Alternatively, rather than using tensor products, one can use this direct sum as the definition of the complexification:

${\displaystyle V^{\mathbb {C} }:=V\oplus V,}$

where ${\displaystyle V^{\mathbb {C} }}$ is given a linear complex structure by the operator J defined as ${\displaystyle J(v,w):=(-w,v),}$ where J encodes the data of "multiplication by i". In matrix form, J is given by:

${\displaystyle J={\begin{bmatrix}0&-I_{V}\\I_{V}&0\end{bmatrix}}.}$

This yields the identical space – a real vector space with linear complex structure is identical data to a complex vector space – though it constructs the space differently. Accordingly, ${\displaystyle V^{\mathbb {C} }}$ can be written as ${\displaystyle V\oplus JV}$ or ${\displaystyle V\oplus iV,}$ identifying V with the first direct summand. This approach is more concrete, and has the advantage of avoiding the use of the technically involved tensor product, but is ad hoc.

## Complex conjugation

The complexified vector space VC has more structure than an ordinary complex vector space.[examples needed] It comes with a canonical complex conjugation map:

${\displaystyle \chi :V^{\mathbb {C} }\to {\overline {V^{\mathbb {C} }}}}$

defined by

${\displaystyle \chi (v\otimes z)=v\otimes {\bar {z}}.}$

The map χ may either be regarded as a conjugate-linear map from VC to itself or as a complex linear isomorphism from VC to its complex conjugate ${\displaystyle {\overline {V^{\mathbb {C} }}}}$.

Conversely, given a complex vector space W with a complex conjugation χ, W is isomorphic as a complex vector space to the complexification VC of the real subspace

${\displaystyle V=\{w\in W:\chi (w)=w\}.}$

In other words, all complex vector spaces with complex conjugation are the complexification of a real vector space.

For example, when W = Cn with the standard complex conjugation

${\displaystyle \chi (z_{1},\ldots ,z_{n})=({\bar {z}}_{1},\ldots ,{\bar {z}}_{n})}$

the invariant subspace V is just the real subspace Rn.

## Linear transformations

Given a real linear transformation f : VW between two real vector spaces there is a natural complex linear transformation

${\displaystyle f^{\mathbb {C} }:V^{\mathbb {C} }\to W^{\mathbb {C} }}$

given by

${\displaystyle f^{\mathbb {C} }(v\otimes z)=f(v)\otimes z.}$

The map fC is naturally called the complexification of f. The complexification of linear transformations satisfies the following properties

• ${\displaystyle (\mathrm {id} _{V})^{\mathbb {C} }=\mathrm {id} _{V^{\mathbb {C} }}}$
• ${\displaystyle (f\circ g)^{\mathbb {C} }=f^{\mathbb {C} }\circ g^{\mathbb {C} }}$
• ${\displaystyle (f+g)^{\mathbb {C} }=f^{\mathbb {C} }+g^{\mathbb {C} }}$
• ${\displaystyle (af)^{\mathbb {C} }=af^{\mathbb {C} }\quad \forall a\in \mathbb {R} }$

In the language of category theory one says that complexification defines an (additive) functor from the category of real vector spaces to the category of complex vector spaces.

The map fC commutes with conjugation and so maps the real subspace of VC to the real subspace of WC (via the map f). Moreover, a complex linear map g : VCWC is the complexification of a real linear map if and only if it commutes with conjugation.

As an example consider a linear transformation from Rn to Rm thought of as an m × n matrix. The complexification of that transformation is exactly the same matrix, but now thought of as a linear map from Cn to Cm.

## Dual spaces and tensor products

The dual of a real vector space V is the space V* of all real linear maps from V to R. The complexification of V* can naturally be thought of as the space of all real linear maps from V to C (denoted HomR(V,C)). That is,

${\displaystyle (V^{*})^{\mathbb {C} }=V^{*}\otimes \mathbb {C} \cong \mathrm {Hom} _{\mathbb {R} }(V,\mathbb {C} ).}$

The isomorphism is given by

${\displaystyle (\varphi _{1}\otimes 1+\varphi _{2}\otimes i)\leftrightarrow \varphi _{1}+i\varphi _{2}}$

where φ1 and φ2 are elements of V*. Complex conjugation is then given by the usual operation

${\displaystyle {\overline {\varphi _{1}+i\varphi _{2}}}=\varphi _{1}-i\varphi _{2}}$

Given a real linear map φ : VC we may extend by linearity to obtain a complex linear map φ : VCC. That is,

${\displaystyle \varphi (v\otimes z)=z\varphi (v).}$

This extension gives an isomorphism from HomR(V,C)) to HomC(VC,C). The latter is just the complex dual space to VC, so we have a natural isomorphism:

${\displaystyle (V^{*})^{\mathbb {C} }\cong (V^{\mathbb {C} })^{*}.}$

More generally, given real vector spaces V and W there is a natural isomorphism

${\displaystyle \mathrm {Hom} _{\mathbb {R} }(V,W)^{\mathbb {C} }\cong \mathrm {Hom} _{\mathbb {C} }(V^{\mathbb {C} },W^{\mathbb {C} }).}$

Complexification also commutes with the operations of taking tensor products, exterior powers and symmetric powers. For example, if V and W are real vector spaces there is a natural isomorphism

${\displaystyle (V\otimes _{\mathbb {R} }W)^{\mathbb {C} }\cong V^{\mathbb {C} }\otimes _{\mathbb {C} }W^{\mathbb {C} }.}$

Note the left-hand tensor product is taken over the reals while the right-hand one is taken over the complexes. The same pattern is true in general. For instance, one has

${\displaystyle (\Lambda _{\mathbb {R} }^{k}V)^{\mathbb {C} }\cong \Lambda _{\mathbb {C} }^{k}(V^{\mathbb {C} }).}$

In all cases, the isomorphisms are the “obvious” ones.