6.2 Complex numbers and complex structures

中文

The complex numbers $\mathbb{C}$ form a two-dimensional commutative and associative algebra over the real numbers, with a basis $\{1,i\}$ having the defining relations

$$1^2=11=1,i1=1i=i,i^2=ii=-1.$$

Setting $E_1=1,E_2=i$ the structure constants are

$$\begin{array}{lll}{C}_{11}^1=1& C_{12}^1=C_{21}^1=0& C_{22}^1=-1\\ C_{11}^2=0& C_{12}^2=C_{21}^2=1& C_{22}^2=0.\end{array}$$

It is common to write the typical element $x{E}_1+y{E}_2=x1+xi$ simply as $x+iy$ and Eq. (6.1) gives the standard rule for complex multiplication,

$$(u+iv)(x+iy)=ux-vy+i(uy+vx).$$

Exercise

Verify that this algebra is commutative and associative.

Every non-zero complex number $\alpha =x+iy$ has an inverse ${\alpha }^{-1}$ with the property $\alpha {\alpha }^{-1}={\alpha }^{-1}\alpha =1$ . Explicitly,

$${\alpha }^{-1}=\frac{\overline{\alpha }}{|\alpha {|}^2},$$

where

$$\overline{\alpha }=x-iy$$

and

$$|\alpha |=\sqrt{\alpha \overline{\alpha }}=\sqrt{x^2+y^2}$$

are the complex conjugate and modulus of $\alpha$, respectively.

Any algebra in which all non-zero vectors have an inverse is called a division algebra, since for any pair of elements $A$, $B\left(B\ne O\right)$ it is possible to define $A/B=A{B}^{-1}$ . The complex numbers are the only associative, commutative division algebra of dimension $>1$ over the real numbers $\mathbb{R}$.

Exercise

Show that an associative, commutative division algebra is a field.

Example 6.4

There is a different, but occasionally useful representation of the complex numbers as matrices. Let $I$ and $J$ be the matrices

$$\mathsf{I}=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right),\mathsf{J}=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right).$$

It is a trivial matter to verify that

$$\mathsf{JI}=\mathsf{IJ}=\mathsf{J},{\mathsf{I}}^2=\mathsf{I},{\mathsf{J}}^2=-\mathsf{I},\text{\hspace{1em}(6.3)}$$

and the subalgebra of ${\mathcal{M}}_2$ generated by these two matrices is isomorphic to the algebra of complex numbers. The isomorphism can be displayed as

$$x+iy⟷x\mathsf{I}+y\mathsf{J}=\left(\begin{array}{cc}x& y\\ -y& x\end{array}\right).$$

Exercise

Check that the above map is an isomorphism by verifying that

$$(u+iv)(x+iy)⟷(u\mathsf{I}+v\mathsf{J})(x\mathsf{I}+v\mathsf{J}).$$

Complexification of a real vector space

Define the complexification $V^C$ of a real vector space $V$ as the set of all ordered pairs $w=(u,v)\in V\times V$ with vector addition and scalar product by complex numbers defined as

$$\begin{array}{c}(u,v)+(u^{\prime },v^{\prime })=(u+u^{\prime },v+v^{\prime }),\\ (a+ib)(u,v)=(au-bv,bu+av),\end{array}$$

for all $u,u^{\prime },v,v^{\prime }\in V$ and $a$, $b\in \mathbb{R}$ . This process of transforming any real vector space into a complex space is totally natural, independent of choice of basis.

Exercise

Verify that the axioms (VS1)–(VS6) in Section 3.2 are satisfied for $V^C$ with the complex numbers $\mathbb{C}$ as the field of scalars. Most axioms are trivial, but (VS4) requires proof:

$$(c+id)((a+ib)(u,v))=((c+id)(a+ib))(u,v).$$

Essentially what we have done here is to ‘expand’ the original vector space by permitting multiplication with complex scalars. There is no ambiguity in adopting the notation $w=u+iv$ for $w=(u,v)$ , since

$$(a+ib)(u+iv)\equiv (a+ib)(u,v)=(au-bv,bu+av)\equiv au-bv+i(bu+av).$$

If $V$ is finite dimensional and $n=\dim V$ then $V^C$ is also finite dimensional and has the same dimension as $V$. For, let $\{e_i|i=1,\dots ,n\}$ be any basis of $V$. These vectors clearly span $V^C$ , for if $\mathbf{u}={\mathbf{u}}^j{\mathbf{e}}_j$ and $v=v^j{e}_j$ are any pair of vectors in $V$ then

$$u+iv=(u^j+i{v}^j)e_j.$$

Furthermore, the vectors $\{e_j\}$ are linearly independent over the field of complex numbers, for if $(u^j+i{v}^j)e_j=0$ then $(u^j{e}_j,v^j{e}_j)=(0,0)$ . Hence $u^j{e}_j=0$ and $v^j{e}_j=0$ , so that $u^j=v^j=0$ for all $j=1,\dots ,n$ . Thus $\{e_1,e_2,\dots ,e_n\}$ also forms a basis for $V^C$

In the complexification $V^C$ of a real space $V$ we can define complex conjugation by

$$\overline{w}=\overline{u+iv}=u-iv.$$

In an arbitrary complex vector space, however, there is no natural, basis-independent, way of defining complex conjugation of vectors. For example, if we set the complex conjugate of a vector $u=u^j{e}_j$ to be $\overline{u}=\overline{u^j}{e}_j$ , this definition will give a different answer in the basis $\{i{e}_j\}$ since

$$u=(-i{u}^j)(i{e}_j)⟹\overline{u}=(i\overline{u^j})(i{e}_j)=-\overline{u^j}{e}_j.$$

Thus the concept of complex conjugation of vectors requires prior knowledge of the ‘real part’ of a complex vector space. The complexification of a real space has precisely the required extra structure needed to define complex conjugation of vectors, but there is no natural way of reversing the complexification process to produce a real vector space of the same dimension from any given complex vector space.

Complex structure on a vector space

One way of creating a real vector space $V^R$ from a complex vector space $V$ is to forget altogether about the possibility of multiplying vectors by complex numbers and only allow scalar multiplication with real numbers. In this process a pair of vectors $u$ and $iu$ must be regarded as linearly independent vectors in $V^R$ for any non-zero vector $u\in V$ . Thus if $V$ is finite dimensional and $\dim V=n$ , then $V^R$ is $2n$-dimensional, for if $\{e_1,\dots ,e_n\}$ is a basis of $V$ then

$$e_1,e_2,\dots ,e_n,i{e}_1,i{e}_2,\dots ,i{e}_n$$

is readily shown to be a l.i. set of vectors spanning $V^R$

To reverse this ‘realification’ of a complex vector space, observe firstly that the operator $J:V^R\to V^R$ defined by $Jv=iv$ satisfies the relation $J^2=-\mathrm{i}{\mathbf{d}}_{V^R}$ . We now show that given any operator on a real vector space having this property, it is possible to define a passage to a complex vector space. This process is not to be confused with the complexification of a vector space, but there is a connection with it (see Problem 6.2).

If $V$ is a real vector space, any operator $J:V\to V$ such that $J^2=-{\mathrm{id}}_V$ is called a complex structure on $V$. A complex structure $J$ can be used to convert $V$ into a complex vector space $V_J$ by defining addition of vectors $u+v$ just as in the real space $V$ , and scalar multiplication of vectors by complex numbers through

$$(a+ib)v=av+bJv.$$

It remains to prove that $V_J$ is a complex vector space; for example, to show axiom (VS4) of Section 3.2

$$\begin{array}{rl}(a+ib)((c+id)v)& =a(cv+dJv)+bJ(cv+dJv)\\ & =(ac-bd)v+(ad+bc)Jv\\ & =(ac-bd+i(ad+bc))v\\ & =((a+ib)(c+id))v.\end{array}$$

Most other axioms are trivial.

A complex structure is always an invertible operator since

$$J{J}^3=J^4=(-{\mathrm{id}}_V{)}^2={\mathrm{id}}_V⟹J^{-1}=J^3.$$

Furthermore if $\dim V=n$ and $\{e_1,\dots ,e_n\}$ is any basis of $V$ then the matrix $\mathsf{J}=[J_i^j]$ defined by $J{e}_i=J_i^j{e}_j$ satisfies

$${\mathsf{J}}^2=-\mathsf{I}.$$

Taking determinants gives

$$(\det \mathrm{J}{)}^2=\det (-\mathsf{I})=(-1{)}^n,$$

which is only possible for a real matrix $J$ if $n$ is an even number, $n=2m$ . Thus a real vector space can only have a complex structure if it is even dimensional.

As a set, the original real vector space $V$ is identical to the complex space $V_J$ , but scalar multiplication is restricted to the reals. It is in fact the real space constructed from $V_J$ by

the above realification process,

$$V=(V_J{)}^R.$$

Hence the dimension of the complex vector space $V_J$ is half that of the real space from which it comes, $\dim V_J=m=\frac{1}{2}\dim V$.

Problems

Problem 6.1

The following is an alternative method of defining the algebra of complex numbers. Let $\mathcal{P}$ be the associative algebra consisting of real polynomials on the variable $x$, defined in Example 6.3. Set $\mathcal{C}$ to be the ideal of $\mathcal{P}$ generated by $x^2+1;$ i.e., the set of all polynomials of the form $f(x)(x^2+1)g(x)$ . Show that the linear map $\varphi :\mathbb{C}\to \mathcal{P}/\mathcal{C}$ defined by

$$\varphi (i)=[x]=x+\mathcal{C},\varphi (1)=[1]=1+\mathcal{C}$$

is an algebra isomorphism.

Which complex number is identified with the polynomial class $[1+x+3x^2+5x^3]\in \mathcal{P}/\mathcal{C}?$

Problem 6.2

Let $J$ be a complex structure on a real vector space $V$, and set

$$V(J)=\{v=u-iJu\mid u\in V\}\subseteq V^C,\bar{V}(J)=\{v=u+iJu\mid u\in V\}.$$

(a) Show that $V(J)$ and $\bar{V}(J)$ are complex vector subspaces of $V^C$

(b) Show that $v\in V(J)\Rightarrow Jv=iv\mathrm{and}v\in \bar{V}(J)\Rightarrow Jv=-iv$.

(c) Prove that the complexification of $V$ is the direct sum of $V(J)$ and $\bar{V}(J)$ ,

$$V^C=V(J)\oplus \bar{V}(J).$$

Problem 6.3

If $V$ is a real vector space and $U$ and $\bar{U}$ are complex conjugate subspaces of $V^C$ such that $V^C=U\oplus \bar{U}$ , show that there exists a complex structure $J$ for $V$ such that $U=V(J)$ and $\bar{U}=\bar{V}(J)$ , where $V(J)$ and $\bar{V}(J)$ are defined in the previous problem.

Problem 6.4

Let $J$ be a complex structure on a real vector space $V$ of dimension $n=2m$ . Let $u_1,u_2,\dots ,u_m$ be a basis of the subspace $V(J)$ defined in Problem 6.2, and set

$$u_a=e_a-i{e}_{m+a}\text{ where }{e}_a,e_{m+a}\in V(a=1,\dots ,m).$$

Show that the matrix ${\mathsf{J}}_0=[J_i^j]$ of the complex structure, defined by $J{e}_i=J_i^j{e}_j$ where $i=1,2,\dots ,n=2m$ , has the form

$$J_0=\left(\begin{array}{cc}O& I\\ -I& O\end{array}\right).$$

Show that the matrix of any complex structure with respect to an arbitrary basis has the form

$$\mathsf{J}={\mathsf{AJ}}_0{\mathsf{A}}^{-1}.$$