Complex structures and compatible metrics

Try

A quarter turn usually sends (1,0) to (0,1). Now try the real matrix J with rows (0,−2),(1/2,0). It sends e₁=(1,0) to (0,1/2), and applying it again sends e₁ to −e₁. In fact J²=−I on every vector. That algebra suggests multiplication by i, but the first step has halved ordinary length. Which part of the usual complex-plane picture survives?

Use the weighted measurement g(x,y)=x₁y₁+4x₂y₂. Its Gram matrix is G=diag(1,4); “Gram matrix” here means exactly that g(x,y)=xᵀGy. Under this measurement e₁ and Je₁ both have squared length 1. Write the two products g(Je₁,e₂) and −g(e₁,Je₂) before checking: both equal 2. Their ordinary-dot-product counterparts are 1/2 and 2, so they disagree.

The last pair has ordinary dot product −3/8 but weighted product 0. The metric determines the geometry; J²=−I alone determines an algebraic operation. In this lesson V is a finite-dimensional real vector space, I its identity operator, and all inner products called g are real, symmetric and positive definite. The zero-dimensional space is permitted.

Predict

Predict what happens if the 2 is changed to 3 and the reciprocal 1/2 to 1/3. Does the square of J change? Does the ordinary dot product suddenly become compatible? Which diagonal metric should replace diag(1,4)? Test your proposed metric on e₁ and Je₁ rather than guessing from the picture.

There are two logically different questions. First, can a real vector be multiplied by every complex number consistently? Second, can its lengths and angles be represented by a positive Hermitian inner product? We will build the first from J²=−I, then build the second from a compatible metric. Compatibility means g(Jx,y)=−g(x,Jy) for every x,y, also called g-skew-adjointness. A single perpendicular pair does not prove this universal equation.

Check the changed prediction

For J with rows (0,−3),(1/3,0), the square is still −I. The metric diag(1,9) makes e₁ and Je₁ equal in length and satisfies JᵀG+GJ=0. The ordinary metric does not: Jᵀ+J has off-diagonal entries −8/3. The reciprocal entries make the operator square work; the squared weight repairs the geometry.

Experiment

The playground starts with coordinate stretch a=2 and angle 45°. Before changing anything, compare its two dot products with the exact last row above. Change “Coordinate stretch a” to 1, then 3. Use “Angle θ in degrees” to compare 0°, 90° and 45°. Record a case where ordinary perpendicularity fails and one where a particular pair is perpendicular despite the operator not being ordinary-skew.

A quarter turn depends on how lengths are measured

Start at e₁ = (1, 0) and apply exp(θJ), where J has rows (0, −a), (1/a, 0). Since J² = −I, the result is x = (cos θ, sin θ/a). Predict whether x and Jx are perpendicular in the ordinary dot product before changing a.

Coordinate stretch a123 Angle θ in degrees

45°

Reset metric experiment

J = (0, −2; 1/2, 0). The metric matrix G = diag(1, 4) defines g(u,v) = u₁v₁ + 4u₂v₂. Angles in the formula use radians: θ = 45π/180.

Original coordinates

Coordinates that display the metric

Ordinary squared length xᵀx

0.625

Metric squared length xᵀGx

1

Ordinary dot product xᵀJx

-0.375

Metric product xᵀGJx

0

J²x = (-0.707107, -0.353553) = −x up to rounding. The off-diagonal entries of Jᵀ + J are -1.5: this J is not skew-adjoint for the ordinary dot product, even if a particular pair of vectors happens to have zero dot product. In contrast, JᵀG + GJ = 0 for every allowed a.

The general metric repair g₀(u,v) + g₀(Ju,Jv), starting from the ordinary dot product g₀, gives diag(1.25, 5). Here it is a positive scalar multiple of G, so it gives the same orthogonality.

Try a = 2 and θ = 90°. Ordinary squared length falls to 1/4 while metric squared length remains 1. Return to 45° to see nonzero ordinary dot product. Explain why J² = −I alone cannot guarantee Euclidean length preservation. Values use floating-point arithmetic and are rounded to six decimal places; the lesson proves the general identities. Controls are temporary and do not record completion.

The left plot uses the original coordinates with equal horizontal and vertical scales. The right plot explicitly changes coordinates to z=(x₁,a x₂). Its circle is evidence about that declared metric representation, not an undeclared stretching of the axes. For a=2 and 90°, ordinary squared length is 1/4 while weighted squared length remains 1. At 45°, ordinary dot product is −3/8. These are exact worked values; the live outputs are rounded numerical evaluations.

A manual check explains the whole orbit: x(θ)=cos θ e₁+sin θ Je₁=(cos θ,sin θ/a). Since J²=−I, the power series for exp(θJ), if already known, reduces to cos θ I+sin θ J. No exponential theorem is needed here: the displayed formula and addition identities already define the rotations and prove their composition law. A finite set of slider positions cannot establish the universal identities below.

Understand

Define multiplication by a complex scalar directly: (a+ib)x=ax+bJx, for real a,b. Expanding two successive multiplications gives

(aI+bJ)(cI+dJ)=(ac−bd)I+(ad+bc)J.

This is the complex multiplication rule because J²=−I. Real linearity gives both distributive laws and 1x=x. Thus V is a complex vector space, without requiring any metric. This is a new scalar structure on the same set of vectors, not the complexification V⊗C, which is a different construction.

Choose complex-independent vectors until they span. This process terminates: every complex-independent list e₁,…,eₘ gives a real-independent list e₁,Je₁,…,eₘ,Jeₘ, because a vanishing real combination is exactly a vanishing complex combination. Conversely, complex spanning gives real spanning of that paired list. Hence a complex basis gives a real basis of twice its size, and dim_R V=2 dim_C V. In particular an odd-dimensional real space cannot carry J²=−I. The empty basis handles V=0.

Suppose now that J is g-skew. Substitute Jy for y to obtain g(Jx,Jy)=g(x,y), and use symmetry to obtain g(Jx,x)=0. Define

h(x,y)=g(x,y)−i g(Jx,y).

Our convention is linear in the first variable. The sign matters: h(Jx,y)=g(Jx,y)+i g(x,y)=i h(x,y), whereas h(x,Jy)=−g(Jx,y)−i g(x,y)=−i h(x,y). Together with real bilinearity these prove complex linearity in the first slot and conjugate linearity in the second. Moreover g(Jy,x)=−g(Jx,y), so h(y,x) is the complex conjugate of h(x,y). Finally h(x,x)=g(x,x), positive for nonzero x. These verify every inner-product obligation; the real part of h recovers g.

For the worked metric, h(e₁,e₂)=−2i. As a check, h(Je₁,e₂)=2=i(−2i). Choosing the opposite sign would correspond to the other slot convention, not the one used here.

Every complex structure admits a compatible positive metric

Start with any real inner product g₀ and set g₁(x,y)=g₀(x,y)+g₀(Jx,Jy). It is symmetric and bilinear, and its first summand alone makes g₁(x,x)>0 for x≠0. Applying J to both arguments swaps the two summands, since J²=−I; therefore g₁(Jx,Jy)=g₁(x,y). Replacing y with −Jy proves skew-adjointness. This construction repairs a metric; it does not claim the original metric was compatible or that the repair is unique.

In the worked example, ordinary g₀ gives G₁=I+JᵀJ=diag(5/4,5), a positive multiple of G. For a general symmetric matrix with rows (r,s),(s,t), the equation JᵀG+GJ=0 forces s=0 and t=4r. Positivity then says r>0. Thus even this two-dimensional example has infinitely many compatible metrics.

Normalize a skew operator, keeping its kernel visible

The preceding lesson supplies an orthogonal decomposition of a real skew-adjoint S into a kernel and rotation planes. On a nonzero-frequency plane choose an orthonormal pair e,f with Se=λf and Sf=−λe, where λ>0. Define Je=f and Jf=−e: on that plane J=S/λ. Then J²=−I and J is skew-adjoint. Repeat on every nonzero plane. In invariant notation on W=(ker S)⊥, define T=√(−S²) by multiplying each frequency eigenspace by its positive λ; T is invertible there and J=ST⁻¹. This definition is independent of which orthonormal basis is chosen inside a repeated-frequency space, because T is scalar there. It commutes with S.

Do not divide by T on the kernel. W has even dimension. An extension to all V exists exactly when ker S has even dimension: necessity follows from the paired-basis dimension theorem; sufficiency follows by pairing an orthonormal basis of the kernel and placing a quarter turn on each pair. Such an extension remains skew and commutes with S, but the choices on the kernel are not determined by S. If S has frequencies 2 and 5, no one global scalar c can normalize it: S²/c² would have both values −4/c² and −25/c². Frequency-by-frequency normalization is essential. If S=0, there is no nonzero-frequency normalization; a compatible complex structure can still be chosen precisely in even dimension.

Apply

Change coordinates in the worked example to the basis whose column matrix is P with rows (1,1),(0,1). Old coordinates are P times new coordinates. Before computing, decide which transformation is a similarity and which is a congruence. Then calculate J′, G′ and h(e₁,e₂) in the new coordinates. Check the skew equation directly, not by the appearance of the entries.

Check the coordinate calculation and its meaning

J′=P⁻¹JP has rows (−1/2,−5/2),(1/2,1/2), while G′=PᵀGP has rows (1,1),(1,5). Multiplication gives G′J′ with rows (0,−2),(2,0), so J′ᵀG′+G′J′=0. The new h(e₁,e₂)=1−2i. In general the left side of the compatibility equation transforms to Pᵀ(JᵀG+GJ)P, proving the coordinate rule for any invertible P. Retaining G=I after an arbitrary basis change would instead change the geometric question.

Now prove a statement that covers every angle at once. For real a,b, expand g((aI+bJ)x,(aI+bJ)y). The cross terms cancel by skewness and the J–J term equals g(x,y), leaving (a²+b²)g(x,y). Thus on a nonzero space exactly the complex scalars of modulus 1 are isometries; other scalars multiply lengths by their modulus. This proof connects the complex-vector-space rule to the metric circle. It also explains why an ellipse in arbitrary coordinates need not indicate a failure of rotation geometry.

For a final changed case, take two nonzero rotation planes and a one-dimensional kernel. Explain why normalizing the planes gives a complex structure only on their four-dimensional sum. Add one more zero direction and construct an extension explicitly by pairing the two kernel basis vectors. The extension is extra structure, not new information recovered from S.

Continue

Return to real skew-adjoint rotation blocks to review the decomposition used in normalization. The six written tasks ask for proofs of the scalar laws and metric conditions, changed numerical metrics and coordinates, and kernel boundaries. A drawing of a circle or a few successful products is not a substitute for these universal claims.

Keep three conclusions separate in your notes: J²=−I supplies complex scalars; a compatible positive metric supplies a Hermitian product with the declared slot convention; a skew operator supplies a normalized J on its nonzero-frequency space, with an additional choice needed on an even-dimensional kernel. These distinctions survive coordinate changes when both J and the Gram matrix are transported correctly.