Complexification, Realification and Real Structures: Examples and Counterexamples

The list $(e_1,\dots ,e_n)$ is a real basis of ${\mathbb{R}}^n$ by definition of the standard basis.

By [L2], $\mathbb{C}{\otimes }_{\mathbb{R}}{\mathbb{R}}^n\cong {\mathbb{R}}^n\oplus i{\mathbb{R}}^n$ through $\mathrm{\Phi }$, and the assignment $(x,y)\mapsto x+iy$, taken coordinatewise, is a complex-linear isomorphism ${\mathbb{R}}^n\oplus i{\mathbb{R}}^n\to {\mathbb{C}}^n$: complex scalar multiplication $(a+bi)\cdot (x,y)=(ax-by,ay+bx)$ is sent to $(ax-by)+i(ay+bx)=(a+bi)(x+iy)$.

Composing, the element $z\otimes e_j$ maps to $z(e_j,0)$ and then to the $j$th standard complex vector scaled by $z$; additivity extends this to every tensor.

For $x=(x_1,\dots ,x_n)$ one has $\iota x=1\otimes x={\sum }_j{x}_j(1\otimes e_j)$, which step 2.1 sends to ${\sum }_j{x}_j{e}_j=x$; this is exactly the standard inclusion of ${\mathbb{R}}^n$ into ${\mathbb{C}}^n$, and by [L3] the images $\iota e_j=e_j$ form the complex basis of the complexification.

Steps 1.2 and 3.1 identify the complexification with ${\mathbb{C}}^n$ and the canonical embedding with the standard inclusion.

The monomials $1,x,\dots ,x^d$ form a real basis of $\mathbb{R}[x{]}_{\le d}$ and a complex basis of $\mathbb{C}[x{]}_{\le d}$.

The real-linear inclusion $f:\mathbb{R}[x{]}_{\le d}\hookrightarrow \mathbb{C}[x{]}_{\le d}$ extends by [L3] to a unique complex-linear map $F$ with $F(z\otimes p)=zp$; by the scalar action of [L1] this is the map $z\cdot (1\otimes p)\mapsto zf(p)$ on the tensor model.

By [L2], every element of $\mathbb{C}{\otimes }_{\mathbb{R}}\mathbb{R}[x{]}_{\le d}$ is $1\otimes p+i\otimes q$ with $p,q\in \mathbb{R}[x{]}_{\le d}$, and $F$ sends it to $p+iq\in \mathbb{C}[x{]}_{\le d}$; the monomial images $F(1\otimes x^j)=x^j$ are the complex basis of step 1.1, so $F$ is a complex-linear isomorphism.

Degree bound: $p+iq$ has degree at most $d$ because $p$ and $q$ do, so no degree bound is lost; the embedding $\iota p=1\otimes p$ maps to $p$ itself, the inclusion of the real polynomials.

Steps 1.2 through 3.1 identify the complexification with $\mathbb{C}[x{]}_{\le d}$ and the canonical embedding with the inclusion.

The standard basis has $n$ elements, so ${\dim }_{\mathbb{C}}{\mathbb{C}}^n=n$.

The displayed list spans $({\mathbb{C}}^n{)}_{\mathbb{R}}$: every $w={\sum }_j{z}_j{e}_j$ writes $z_j=a_j+i{b}_j$ with real $a_j,b_j$, and by [L1] scalar multiplication by the real parts is real scalar multiplication, giving $w={\sum }_j{a}_j{e}_j+{\sum }_j{b}_j(i{e}_j)$.

By [L2], ${\dim }_{\mathbb{R}}({\mathbb{C}}^n{)}_{\mathbb{R}}=2n$.

The list is real-linearly independent: ${\sum }_j{a}_j{e}_j+{\sum }_j{b}_j(i{e}_j)=0$ means ${\sum }_j(a_j+i{b}_j)e_j=0$ in ${\mathbb{C}}^n$, so complex independence of $(e_1,\dots ,e_n)$ forces every $a_j=b_j=0$.

Steps 1.2 and 2.2 exhibit the displayed list as a real basis with $2n$ entries, matching the dimension of step 2.1; the coordinate map identifies it with ${\mathbb{R}}^{2n}$.

The characteristic polynomial is ${\chi }_T(x)=\det (xI-A)=x^2+1$, and by [L1] the complexified operator has the same polynomial, which factors as $(x-i)(x+i)$ over $\mathbb{C}$.

For $\lambda =i$, the equation $(A-iI)w=0$ is $-ix-y=0$ and $x-iy=0$, so $y=-ix$ and $w_i=(1,-i)$ is an eigenvector; symmetrically $w_{-i}=(1,i)$ is an eigenvector for $\lambda =-i$.

The vectors $w_i$ and $w_{-i}$ are complex-linearly independent, so they form a complex basis of ${\mathbb{C}}^2$ in which $T_{\mathbb{C}}$ has the diagonal matrix $\mathrm{diag}(i,-i)$.

Conjugation satisfies $\sigma (w_i)=(1,i)=w_{-i}$, the conjugate-pair behaviour recorded in [L2] with exponent $1$.

A real eigenvector $v\mathrm{\ne }0$ would carry a real eigenvalue $\lambda$ with $Av=\lambda v$; taking a nonzero coordinate of $v$ shows $\lambda$ is real, and then step 1.1 gives ${\lambda }^2+1=0$, which has no real solution.

Steps 2.1 and 2.3 together prove the example: diagonalisation after complexification with no real eigenvector beforehand.

The characteristic polynomial is $\det (xI-A)=(x-1{)}^2+1=x^2-2x+2$, whose roots are $1\pm i$, both nonreal.

For $\lambda =1+i$, the equation $(A-\lambda I)w=0$ is $-ix-y=0$ and $x-iy=0$, so $y=-ix$; the choice $x=1$ gives $w=(1,-i)=u+iv$ with $u=(1,0)$ and $v=(0,-1)$.

Applying [L1] with $a=b=1$: $u$ and $v$ are $\mathbb{R}$-linearly independent, ${\mathrm{span}}_{\mathbb{R}}\{u,v\}={\mathbb{R}}^2$ is $T$-invariant, and in the ordered basis $(u,-v)=(e_1,e_2)$ the matrix of $T$ is $\left(\begin{array}{cc}a& -b\\ b& a\end{array}\right)=\left(\begin{array}{cc}1& -1\\ 1& 1\end{array}\right)=A$.

The conjugate vector $\sigma (w)=(1,i)$ is an eigenvector with eigenvalue $1-i$, as recorded in [L1].

Steps 2.1 and 2.2 reconstruct the invariant plane and the block from the single nonreal eigenvector.

The map ${\sigma }_0$ is a conjugation: it is additive, ${\sigma }_0(\lambda z)=(\overline{\lambda z_1},\overline{\lambda z_2})=\bar{\lambda }{\sigma }_0(z)$, and applying it twice is the identity.

The map ${\sigma }^{\mathrm{\prime }}$ is also a conjugation: additivity is clear, ${\sigma }^{\mathrm{\prime }}(\lambda z)=(\overline{\lambda z_2},\overline{\lambda z_1})=\bar{\lambda }{\sigma }^{\mathrm{\prime }}(z)$, and ${\sigma }^{\mathrm{\prime }}{\sigma }^{\mathrm{\prime }}(z)=(z_1,z_2)$.

The fixed set of ${\sigma }_0$ is $\{(z_1,z_2):{\bar{z}}_1=z_1,{\bar{z}}_2=z_2\}={\mathbb{R}}^2$, the real coordinate plane inside $({\mathbb{C}}^2{)}_{\mathbb{R}}$.

The fixed set of ${\sigma }^{\mathrm{\prime }}$ is $\{(z_1,z_2):z_1={\bar{z}}_2\}=\{(w,\bar{w}):w\in \mathbb{C}\}=\{(a+bi,a-bi):a,b\in \mathbb{R}\}$, a real two-plane.

The two fixed sets are different real subspaces: $(1,0)$ is fixed by ${\sigma }_0$ but ${\sigma }^{\mathrm{\prime }}(1,0)=(0,1)\mathrm{\ne }(1,0)$. By [L1] each is a real form whose complexification recovers ${\mathbb{C}}^2$, and by [L2] the distinct conjugations give distinct real forms.

Steps 1.1 through 3.1 exhibit two different conjugations and two different fixed real forms on one complex vector space.

The map ${\sigma }_0$ is a conjugation, and by [L1] its fixed real form is $\{(z_1,z_2):{\bar{z}}_1=z_1,{\bar{z}}_2=z_2\}={\mathbb{R}}^2$.

The map $T$ is complex-linear: $T(\lambda z_1,\lambda z_2)=(i\lambda z_1,\lambda z_2)=\lambda T(z_1,z_2)$.

The two maps do not commute: $T{\sigma }_0(1,0)=T(1,0)=(i,0)$, while ${\sigma }_{0}T(1,0)={\sigma }_0(i,0)=(-i,0)$.

By [L2], $T$ does not come from any real operator on ${\mathbb{R}}^2$; concretely $T(1,0)=(i,0)\mathrm{\notin }{\mathbb{R}}^2$, so $T$ does not even preserve the chosen real form as a set.

Steps 1.2, 1.3 and 2.1 refute the claimed universality: a complex-linear map can fail to preserve a chosen real form.

By [L2], ${\dim }_{\mathbb{C}}{V}_{\mathbb{C}}={\dim }_{\mathbb{R}}V$ for every finite-dimensional real $V$: the embedded image of a real basis is already a complex basis of the complexification.

The concrete witness $V=\mathbb{R}$ confirms the correct value: by [L1] with $n=1$, $V_{\mathbb{C}}\cong \mathbb{C}$, so ${\dim }_{\mathbb{C}}{V}_{\mathbb{C}}=1={\dim }_{\mathbb{R}}\mathbb{R}$, not $2$.

The doubling behaviour belongs to realification, the reverse construction, which replaces complex scalars by real ones; complexification keeps the numerical dimension unchanged.

Steps 1.1 and 1.2 contradict the claimed factor of $2$, so the displayed statement is false.

Multiplication by $i$ is a complex-linear automorphism of $W$. If it preserved $W_0$, then $iu\in W_0$ for every $u\in W_0$.

Choose $0\mathrm{\ne }u\in W_0$. Under the preservation assumption of step 1.1, $iu\in W_0$, so $iu=(iu)+i0=0+iu$ would be two decompositions of the same vector with real and imaginary parts in $W_0$, contradicting uniqueness in [L1].

Thus no real form of a nonzero complex vector space is invariant under all complex-linear automorphisms. The complex structure alone therefore singles out no preferred real form, and the claim is false.

By [L1], the operator $T$ is complex-linear but satisfies $T{\sigma }_0\mathrm{\ne }{\sigma }_{0}T$, with the concrete witness $T{\sigma }_0(1,0)=(i,0)\mathrm{\ne }(-i,0)={\sigma }_{0}T(1,0)$.

By [L2], the failure of commutation means $T$ is not of the form $\theta \circ S_{\mathbb{C}}\circ {\theta }^{-1}$ for any real operator $S$ on ${\mathbb{R}}^2$; hence $T$ does not descend to this chosen real form.

Steps 1.1 and 2.1 exhibit one complex-linear operator and one chosen real form for which descent fails, refuting the claimed universal statement.

By [L1], complexification creates a complex eigenvector: $(1,-i)$ is an eigenvector of $T_{\mathbb{C}}$ with eigenvalue $i$.

By [L1], $T$ has no real eigenvector: any real eigenvector $v\mathrm{\ne }0$ would carry a real eigenvalue $\lambda$ with ${\lambda }^2+1=0$, which is impossible in $\mathbb{R}$.

Steps 1.1 and 1.2 provide a case where a complex eigenvector is created with no accompanying real eigenvector, contradicting the claimed implication.