Linear complex structure

A complex structure on a real vector space ${\displaystyle V}$ is a real linear transformation $${\displaystyle J:V\to V}$$ such that $${\displaystyle J^{2}=-{\text{id}}_{V}.}$$ Here ${\displaystyle J^{2}}$ means ${\displaystyle J}$ composed with itself and ${\displaystyle {\text{id}}_{V}}$ is the identity map on ${\displaystyle V}$. That is, the effect of applying ${\displaystyle J}$ twice is the same as multiplication by ${\displaystyle -1}$. This is reminiscent of multiplication by the imaginary unit, ${\displaystyle i}$. A complex structure allows one to endow ${\displaystyle V}$ with the structure of a complex vector space. Complex scalar multiplication can be defined by $${\displaystyle (x+iy){\vec {v}}=x{\vec {v}}+yJ({\vec {v}})}$$ for all real numbers ${\displaystyle x,y}$ and all vectors ${\displaystyle {\vec {v}}}$ in V. One can check that this does, in fact, give ${\displaystyle V}$ the structure of a complex vector space which we denote ${\displaystyle V_{J}}$.

Going in the other direction, if one starts with a complex vector space ${\displaystyle W}$ then one can define a complex structure on the underlying real space by defining ${\displaystyle Jw=iw~~\forall w\in W}$.

More formally, a linear complex structure on a real vector space is an algebra representation of the complex numbers ${\displaystyle \mathbb {C} }$, thought of as an associative algebra over the real numbers. This algebra is realized concretely as $${\displaystyle \mathbb {C} =\mathbb {R} [x]/(x^{2}+1),}$$ which corresponds to ${\displaystyle i^{2}=-1}$. Then a representation of ${\displaystyle \mathbb {C} }$ is a real vector space ${\displaystyle V}$, together with an action of ${\displaystyle \mathbb {C} }$ on ${\displaystyle V}$ (a map ${\displaystyle \mathbb {C} \rightarrow {\text{End}}(V)}$). Concretely, this is just an action of ${\displaystyle i}$, as this generates the algebra, and the operator representing ${\displaystyle i}$ (the image of ${\displaystyle i}$ in ${\displaystyle {\text{End}}(V)}$) is exactly ${\displaystyle J}$.

If ${\displaystyle V_{J}}$ has complex dimension ${\displaystyle n}$, then ${\displaystyle V}$ must have real dimension ${\displaystyle 2n}$. That is, a finite-dimensional space ${\displaystyle V}$ admits a complex structure only if it is even-dimensional. It is not hard to see that every even-dimensional vector space admits a complex structure. One can define ${\displaystyle J}$ on pairs ${\displaystyle e,f}$ of basis vectors by ${\displaystyle Je=f}$ and ${\displaystyle Jf=-e}$ and then extend by linearity to all of ${\displaystyle V}$. If ${\displaystyle (v_{1},\dots ,v_{n})}$ is a basis for the complex vector space ${\displaystyle V_{J}}$ then ${\displaystyle (v_{1},Jv_{1},\dots ,v_{n},Jv_{n})}$ is a basis for the underlying real space ${\displaystyle V}$.

A real linear transformation ${\displaystyle A:V\rightarrow V}$ is a complex linear transformation of the corresponding complex space ${\displaystyle V_{J}}$ if and only if ${\displaystyle A}$ commutes with ${\displaystyle J}$, i.e. if and only if $${\displaystyle AJ=JA.}$$ Likewise, a real subspace ${\displaystyle U}$ of ${\displaystyle V}$ is a complex subspace of ${\displaystyle V_{J}}$ if and only if ${\displaystyle J}$ preserves ${\displaystyle U}$, i.e. if and only if $${\displaystyle JU=U.}$$

The collection of ${\displaystyle 2\times 2}$ real matrices ${\displaystyle \mathbb {M} (2,\mathbb {R} )}$ over the real field is 4-dimensional. Any matrix

${\displaystyle J={\begin{pmatrix}a&c\\b&-a\end{pmatrix}},~~a^{2}+bc=-1}$

has square equal to the negative of the identity matrix. A complex structure may be formed in ${\displaystyle \mathbb {M} (2,\mathbb {R} )}$: with identity matrix ${\displaystyle I}$, elements ${\displaystyle xI+yJ}$, with matrix multiplication form complex numbers.

The fundamental example of a linear complex structure is the structure on R²ⁿ coming from the complex structure on Cⁿ. That is, the complex n-dimensional space Cⁿ is also a real 2n-dimensional space – using the same vector addition and real scalar multiplication – while multiplication by the complex number i is not only a complex linear transform of the space, thought of as a complex vector space, but also a real linear transform of the space, thought of as a real vector space. Concretely, this is because scalar multiplication by i commutes with scalar multiplication by real numbers ${\displaystyle i(\lambda v)=(i\lambda )v=(\lambda i)v=\lambda (iv)}$ – and distributes across vector addition. As a complex n×n matrix, this is simply the scalar matrix with i on the diagonal. The corresponding real 2n×2n matrix is denoted J.

Given a basis ${\displaystyle \left\{e_{1},e_{2},\dots ,e_{n}\right\}}$ for the complex space, this set, together with these vectors multiplied by i, namely ${\displaystyle \left\{ie_{1},ie_{2},\dots ,ie_{n}\right\},}$ form a basis for the real space. There are two natural ways to order this basis, corresponding abstractly to whether one writes the tensor product as ${\displaystyle \mathbb {C} ^{n}=\mathbb {R} ^{n}\otimes _{\mathbb {R} }\mathbb {C} }$ or instead as ${\displaystyle \mathbb {C} ^{n}=\mathbb {C} \otimes _{\mathbb {R} }\mathbb {R} ^{n}.}$

If one orders the basis as ${\displaystyle \left\{e_{1},ie_{1},e_{2},ie_{2},\dots ,e_{n},ie_{n}\right\},}$ then the matrix for J takes the block diagonal form (subscripts added to indicate dimension): $${\displaystyle J_{2n}={\begin{bmatrix}0&-1\\1&0\\&&0&-1\\&&1&0\\&&&&\ddots \\&&&&&\ddots \\&&&&&&0&-1\\&&&&&&1&0\end{bmatrix}}={\begin{bmatrix}J_{2}\\&J_{2}\\&&\ddots \\&&&J_{2}\end{bmatrix}}.}$$ This ordering has the advantage that it respects direct sums of complex vector spaces, meaning here that the basis for ${\displaystyle \mathbb {C} ^{m}\oplus \mathbb {C} ^{n}}$ is the same as that for ${\displaystyle \mathbb {C} ^{m+n}.}$

On the other hand, if one orders the basis as ${\displaystyle \left\{e_{1},e_{2},\dots ,e_{n},ie_{1},ie_{2},\dots ,ie_{n}\right\}}$, then the matrix for J is block-antidiagonal: $${\displaystyle J_{2n}={\begin{bmatrix}0&-I_{n}\\I_{n}&0\end{bmatrix}}.}$$ This ordering is more natural if one thinks of the complex space as a direct sum of real spaces, as discussed below.

The data of the real vector space and the J matrix is exactly the same as the data of the complex vector space, as the J matrix allows one to define complex multiplication. At the level of Lie algebras and Lie groups, this corresponds to the inclusion of gl(n,C) in gl(2n,R) (Lie algebras – matrices, not necessarily invertible) and GL(n,C) in GL(2n,R):

The inclusion corresponds to forgetting the complex structure (and keeping only the real), while the subgroup GL(n,C) can be characterized (given in equations) as the matrices that commute with J: $${\displaystyle \mathrm {GL} (n,\mathbb {C} )=\left\{A\in \mathrm {GL} (2n,\mathbb {R} )\mid AJ=JA\right\}.}$$ The corresponding statement about Lie algebras is that the subalgebra gl(n,C) of complex matrices are those whose Lie bracket with J vanishes, meaning ${\displaystyle [J,A]=0;}$ in other words, as the kernel of the map of bracketing with J, ${\displaystyle [J,-].}$

Note that the defining equations for these statements are the same, as ${\displaystyle AJ=JA}$ is the same as ${\displaystyle AJ-JA=0,}$ which is the same as ${\displaystyle [A,J]=0,}$ though the meaning of the Lie bracket vanishing is less immediate geometrically than the meaning of commuting.

If V is any real vector space there is a canonical complex structure on the direct sum V ⊕ V given by $${\displaystyle J(v,w)=(-w,v).}$$ The block matrix form of J is $${\displaystyle J={\begin{bmatrix}0&-I_{V}\\I_{V}&0\end{bmatrix}}}$$ where ${\displaystyle I_{V}}$ is the identity map on V. This corresponds to the complex structure on the tensor product ${\displaystyle \mathbb {C} \otimes _{\mathbb {R} }V.}$

If B is a bilinear form on V then we say that J preserves B if $${\displaystyle B(Ju,Jv)=B(u,v)}$$ for all u, v ∈ V. An equivalent characterization is that J is skew-adjoint with respect to B: $${\displaystyle B(Ju,v)=-B(u,Jv).}$$

If g is an inner product on V then J preserves g if and only if J is an orthogonal transformation. Likewise, J preserves a nondegenerate, skew-symmetric form ω if and only if J is a symplectic transformation (that is, if ${\textstyle \omega (Ju,Jv)=\omega (u,v)}$). For symplectic forms ω an interesting compatibility condition between J and ω is that $${\displaystyle \omega (u,Ju)>0}$$ holds for all non-zero u in V. If this condition is satisfied, then we say that J tames ω (synonymously: that ω is tame with respect to J; that J is tame with respect to ω; or that the pair ${\textstyle (\omega ,J)}$ is tame).

Given a symplectic form ω and a linear complex structure J on V, one may define an associated bilinear form g_J on V by $${\displaystyle g_{J}(u,v)=\omega (u,Jv).}$$ Because a symplectic form is nondegenerate, so is the associated bilinear form. The associated form is preserved by J if and only if the symplectic form is. Moreover, if the symplectic form is preserved by J, then the associated form is symmetric. If in addition ω is tamed by J, then the associated form is positive definite. Thus in this case V is an inner product space with respect to g_J.

If the symplectic form ω is preserved (but not necessarily tamed) by J, then g_J is the real part of the Hermitian form (by convention antilinear in the first argument) ${\textstyle h_{J}\colon V_{J}\times V_{J}\to \mathbb {C} }$ defined by $${\displaystyle h_{J}(u,v)=g_{J}(u,v)+ig_{J}(Ju,v)=\omega (u,Jv)+i\omega (u,v).}$$

Given any real vector space V we may define its complexification by extension of scalars:

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

This is a complex vector space whose complex dimension is equal to the real dimension of V. It has a canonical complex conjugation defined by

${\displaystyle {\overline {v\otimes z}}=v\otimes {\bar {z}}}$

If J is a complex structure on V, we may extend J by linearity to V^C:

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

Since C is algebraically closed, J is guaranteed to have eigenvalues which satisfy λ² = −1, namely λ = ±i. Thus we may write

${\displaystyle V^{\mathbb {C} }=V^{+}\oplus V^{-}}$

where V⁺ and V⁻ are the eigenspaces of +i and −i, respectively. Complex conjugation interchanges V⁺ and V⁻. The projection maps onto the V^± eigenspaces are given by

${\displaystyle {\mathcal {P}}^{\pm }={1 \over 2}(1\mp iJ).}$

So that

${\displaystyle V^{\pm }=\{v\otimes 1\mp Jv\otimes i:v\in V\}.}$

There is a natural complex linear isomorphism between V_J and V⁺, so these vector spaces can be considered the same, while V⁻ may be regarded as the complex conjugate of V_J.

Note that if V_J has complex dimension n then both V⁺ and V⁻ have complex dimension n while V^C has complex dimension 2n.

Abstractly, if one starts with a complex vector space W and takes the complexification of the underlying real space, one obtains a space isomorphic to the direct sum of W and its conjugate:

${\displaystyle W^{\mathbb {C} }\cong W\oplus {\overline {W}}.}$

Let V be a real vector space with a complex structure J. The dual space V* has a natural complex structure J* given by the dual (or transpose) of J. The complexification of the dual space (V*)^C therefore has a natural decomposition

${\displaystyle (V^{*})^{\mathbb {C} }=(V^{*})^{+}\oplus (V^{*})^{-}}$

into the ±i eigenspaces of J*. Under the natural identification of (V*)^C with (V^C)* one can characterize (V*)⁺ as those complex linear functionals which vanish on V⁻. Likewise (V*)⁻ consists of those complex linear functionals which vanish on V⁺.

The (complex) tensor, symmetric, and exterior algebras over V^C also admit decompositions. The exterior algebra is perhaps the most important application of this decomposition. In general, if a vector space U admits a decomposition U = S ⊕ T, then the exterior powers of U can be decomposed as follows:

${\displaystyle \Lambda ^{r}U=\bigoplus _{p+q=r}(\Lambda ^{p}S)\otimes (\Lambda ^{q}T).}$

A complex structure J on V therefore induces a decomposition

${\displaystyle \Lambda ^{r}\,V^{\mathbb {C} }=\bigoplus _{p+q=r}\Lambda ^{p,q}\,V_{J}}$

where

${\displaystyle \Lambda ^{p,q}\,V_{J}\;{\stackrel {\mathrm {def} }{=}}\,(\Lambda ^{p}\,V^{+})\otimes (\Lambda ^{q}\,V^{-}).}$

All exterior powers are taken over the complex numbers. So if V_J has complex dimension n (real dimension 2n) then

${\displaystyle \dim _{\mathbb {C} }\Lambda ^{r}\,V^{\mathbb {C} }={2n \choose r}\qquad \dim _{\mathbb {C} }\Lambda ^{p,q}\,V_{J}={n \choose p}{n \choose q}.}$

The dimensions add up correctly as a consequence of Vandermonde's identity.

The space of (p,q)-forms Λ^p,q V_J* is the space of (complex) multilinear forms on V^C which vanish on homogeneous elements unless p are from V⁺ and q are from V⁻. It is also possible to regard Λ^p,q V_J* as the space of real multilinear maps from V_J to C which are complex linear in p terms and conjugate-linear in q terms.

See complex differential form and almost complex manifold for applications of these ideas.

• Almost complex manifold
• Complex manifold
• Complex differential form
• Complex conjugate vector space
• Hermitian structure
• Real structure