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Determinant-line curvature is the trace curvature

Statement

Let $M$ be a finite-dimensional smooth manifold, let $r \geq 1$ , and let $P : M \to M_{r} (C)$ be a $C^{2}$ orthogonal projector of constant rank $m$ , with $1 \leq m \leq r$ . Write $E_{P} = {(x, v) : P (x) v = v}$ and give it the projected connection $\nabla s = P d s$ . For $x \in M$ and $X, Y \in T_{x} M$ , define the curvature two-form with values in $M_{r} (C)$ by

$F_{x} (X, Y) : = P (x) (d P_{x} (X) d P_{x} (Y) - d P_{x} (Y) d P_{x} (X)) P (x) .$

Let $U \subseteq M$ be open and let $V : U \to C^{r \times m}$ be a $C^{2}$ orthonormal frame, meaning $V (x)^{*} V (x) = I_{m}$ and $V (x) V (x)^{*} = P (x)$ for all $x \in U$ . Put $A = V^{*} d V$ and $w = v_{1} \land \dots \land v_{m}$ , where $v_{1}, \dots, v_{m}$ are the columns of $V$ . The induced projected connection on $\det E_{P}$ has local connection one-form $a = tr A$ in the unit frame $w$ , and its scalar curvature is

$d a = tr (V^{*} F V) = tr F .$

Thus the curvature of the determinant line is the trace of the original bundle curvature, with the same orientation and sign convention.

Statement references

Determinant line and its projector

Projector curvature and frame covariance

Facts & Assumptions

Given: The smooth projector and the supplied local orthonormal frame in the Statement.

[F1]

Determinant line and its projector identifies $\det E_{P}$ with the range of $Q = ⋀^{m} P$ .

[F2]

The determinant-line projector is smooth and rank one proves that $Q$ is a smooth rank-one orthogonal projector.

[F3]

Projector curvature and frame covariance gives the frame connection $A = V^{*} d V$ and curvature $F_{f r} = d A + A \land A = V^{*} F V$ .

Proof

Given: [F1]–[F3].

1.1F1F2givenalgebra

Because the columns of $V$ are orthonormal, $w = v_{1} \land \dots \land v_{m}$ has unit norm and spans the determinant line. Differentiating by the product rule gives $d w = \sum_{j} v_{1} \land \dots \land d v_{j} \land \dots \land v_{m}$ . The projected connection applies $P$ to each differentiated factor. Since $P d v_{j} = \sum_{a} v_{a} A_{a j}$ , each term with $a \neq j$ vanishes from a repeated wedge factor, while the $a = j$ term is $A_{j j} w$ . Summing gives $Q d w = (tr A) w$ , so the determinant connection form is $a = tr A$ .

2.1F3step 1.1algebra∎

A line connection with scalar one-form $a$ has curvature $d a + a \land a = d a$ , since a scalar one-form wedges with itself to zero. By [F3], $tr F = tr (d A) + tr (A \land A)$ . In coordinates, $tr (A \land A) = \sum_{i, j} A_{i j} \land A_{j i} = 0$ : diagonal terms vanish and each off-diagonal pair cancels because one-forms anticommute. Therefore $d a = d tr A = tr F_{f r} = tr (V^{*} F V)$ . Finally, since $P F P = F$ and $V V^{*} = P$ , cyclicity of finite-dimensional trace gives $tr (V^{*} F V) = tr (F V V^{*}) = tr (F P) = tr F$ .

Depends on

Dependency tree · 2 levels

Sources

Di Xiao, Ming-Che Chang, and Qian Niu, Berry Phase Effects on Electronic Properties