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Curvature two-form of a connection on a trivial plane bundle

Statement

Let $E = R^{2} \times R^{2} \to R^{2}$ be the trivial rank-two bundle, with base coordinates $(x, y)$ and its standard global frame. For fixed real $2 \times 2$ matrices $A, B$ , there is a connection with connection matrix

$ω = B y d x + A x d y,$

and its curvature matrix is

$Ω = (A - B + x y (B A - A B)) d x \land d y .$

Thus the quadratic term in the structure equation remembers the order of matrix multiplication.

Facts & Assumptions

Given: The displayed trivial bundle, fixed matrices $A, B$ , coordinates, and standard global frame.

[F1]

A connection is a real-linear operator on sections satisfying $\nabla (f s) = d f \otimes s + f \nabla s$ . Connection on a smooth vector bundle.

[F2]

In a frame with connection matrix $ω$ , the curvature matrix is $Ω = d ω + ω \land ω$ , with $(ω \land ω)^{i}_{j} = \sum_{k} ω^{i}_{k} \land ω^{k}_{j}$ in that order. Curvature two-form structure equation.

[F3]

If a form is written in coordinate wedges, its exterior derivative is obtained by differentiating the scalar coefficients. The local coordinate formula for the exterior derivative.

[F4]

The wedge product is the pointwise alternating product of forms. The wedge product of differential forms.

[F5]

Matrix multiplication uses the ordered entry formula $(C D)_{i j} = \sum_{k} C_{i k} D_{k j}$ . Rectangular matrix multiplication and the identity matrix $I_{n}$ , including zero-sized shapes.

Proof

technique · explicit construction and calculation

1.1

Write every section uniquely as $s = e u$ in the standard global frame $e$ and define $\nabla (e u) = e (d u + ω u)$ . This operator is real-linear. Moreover, $d (f u) = d f u + f d u$ and $ω (f u) = f ω u$ , so $\nabla (f e u) = d f \otimes (e u) + f \nabla (e u)$ . Hence [F1] makes it a connection. For a constant standard basis column $u_{j}$ , the derivative term vanishes and $\nabla (e u_{j}) = e ω u_{j}$ , so its connection matrix is the displayed $ω$ .

F1F5algebra

1.2

Apply [F3] entrywise. Since $d (B y) = B d y$ and $d (A x) = A d x$ , one gets $d ω = B d y \land d x + A d x \land d y = (A - B) d x \land d y$ .

F3F4algebra

1.3

Expand the ordered matrix-valued wedge product using [F4]–[F5]. The two self-products vanish because $d x \land d x = d y \land d y = 0$ , while the cross terms give $ω \land ω = x y (B A d x \land d y + A B d y \land d x) = x y (B A - A B) d x \land d y$ .

F4F5algebra

2.1

Substitution of steps 1.2–1.3 into [F2] proves $Ω = (A - B + x y (B A - A B)) d x \land d y$ .

F2step 1.2step 1.3algebra

2.2

The order-sensitive term can be genuinely nonzero. For $A = (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix})$ and $B = (\begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix})$ , direct multiplication gives $B A = (\begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix})$ and $A B = (\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix})$ , hence $B A - A B = diag (- 1, 1) \neq 0$ . Thus at every point with $x y \neq 0$ the quadratic summand is nonzero.

F5step 1.3algebra

3.1

The base and fibres are nonempty and have fixed dimension and rank two, so empty, zero-dimensional, rank-zero, and one-dimensional cases are inapplicable to this example. No inverse or division occurs: $x = 0$ , $y = 0$ , $A = 0$ , $B = 0$ , and commuting $A, B$ are all allowed and the same formula then specializes correctly. The base is all of $R^{2}$ , with no endpoint or manifold boundary. The frame, matrices, connection, and witness in step 2.2 are explicit, so no choice principle is used. No biconditional is asserted.

F1F2F3F4F5step 1.1step 1.2step 1.3step 2.1step 2.2∎

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Sources

Will J. Merry, Differential Geometry (2021) (standard reference, not scraped)
Ved Datar, Lectures on Riemannian Geometry (standard reference, not scraped)