Complexification

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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.

1 Formal definition
2 Basic properties
3 Examples
4 Complex conjugation
5 Linear transformations
6 Dual spaces and tensor products
7 See also
8 References

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

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

[edit] 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:

$\dim_{\mathbb C}V^{\mathbb C} = \dim_{\mathbb R}V.$

[edit] 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.

[edit] 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.

[edit] Linear transformations

Given a real linear transformation f : V → W be 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.

[edit] 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

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