Complexification

In mathematics, the complexification of a real vector space V is a vector space V^C over the complex number field obtained by formally extending scalar multiplication to include multiplication by complex numbers. Any basis for V over the real numbers serves as a basis for V^C 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):

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

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 V^C is only a real vector space. However, we can make V^C into a complex vector space by defining complex multiplication as follows:

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

Basic properties

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

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

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

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

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

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

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

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

with the above rule for multiplication by complex numbers.

There is a natural embedding of V into V^C given by

$v\mapsto v\otimes 1.$

The vector space V may then be regarded as a real subspace of V^C. If V has a basis {e_i} then a corresponding basis for V^C is given by {e_i⊗1}. The complex dimension of V^C is therefore equal to the real dimension of V:

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

Examples

• The complexification of real coordinate space Rⁿ is complex coordinate space Cⁿ.
• Likewise, if V consists of the m×n matrices with real entries, V^C would consist of m×n matrices with complex entries.

Complex conjugation

The complexified vector space V^C has more structure than an ordinary complex vector space. It comes with a canonical complex conjugation map:

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

defined by

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

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

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

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

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

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

the invariant subspace V is just the real subspace Rⁿ.

Linear transformations

Given a real linear transformation f : V → W between two real vector spaces there is a natural complex linear transformation

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

given by

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

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

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

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 f^C commutes with conjugation and so maps the real subspace of V^C to the real subspace of W^C (via the map f). Moreover, a complex linear map g : V^C → W^C 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 R^m thought of as an m × n matrix. The complexification of that transformation is the exact same matrix, but now thought of as a linear map from Cⁿ to C^m.

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 Hom_R(V,C)). That is,

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

The isomorphism is given by

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

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

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

Given a real linear map φ : V → C we may extend by linearity to obtain a complex linear map φ : V^C → C. That is,

$\varphi (v\otimes z)=z\varphi (v).$

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

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

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

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

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

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

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

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

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

See also

• Linear complex structure

References

• Roman, Steven (2005). Advanced Linear Algebra. Graduate Texts in Mathematics 135 ((2nd ed.) ed.). New York: Springer. ISBN 0-387-24766-1.