nLab exterior covariant derivative

Idea

The combination of an exterior derivative with a covariant derivative.

Definition

Definition on vector bundles

Let $p: E \to M$ be a vector bundle with a linear connection given by the covariant derivative $\nabla: \Gamma(T M) \times \Gamma(E) \to \Gamma(E)$. We let $\Omega(M, E)$ be the space of all differential forms with values in? $E$. To define the exterior covariant derivative, we take the explicit formula of the exterior derivative, and replace the usual derivative with the convariant derivative.

Definition

The exterior covariant derivative $d_\nabla: \Omega^p (M, E) \to \Omega^{p + 1}(M, E)$ is defined by the following formula: given a $p$-form $\Phi \in \Omega^p (M, E)$, its exterior covariant derivative is given by

$$\begin{aligned} (\mathrm{d}_\nabla \Phi)(X_0, \dots, X_p) =& \displaystyle\sum_{i = 0}^p (-1)^i \nabla_{X_i}\Phi (X_0, \dots, \hat{X_i}, \dots, X_p) \\ &+ \displaystyle\sum_{i \lt j} (-1)^{i + j} \Phi([X_i, X_j], X_0, \dots, \hat{X_i}, \dots, \hat{X_j}, \dots, X_p), \end{aligned}$$

where each $X_i$ is a vector field on $M$, and $\hat{X}$ means omission of $X$.

In the case of the trivial bundle with the trivial connection, this gives the usual exterior derivative of a vector-valued differential form.

Definition on principal $G$-bundles

Let $p: P \to M$ be a principal $G$-bundle. Let $\omega \in \Omega^1(P,\mathfrak{g})$ be a connection on $P$. Let $H: T P \to T P$ be the horizontal projection given by $\omega$.

We let $V$ be a vector space, and $\rho: G \to \GL(V)$ be a representation of $G$ on $V$.

Definition

A differential form $\psi \in \Omega^k(P,V)$ is called:

• horizontal, if $\psi_q(v_1,\dots,v_k)=0$ whenever one of the vectors $v_i$ is vertical.

• equivariant, if $r_g^{*}\psi = \rho(g^{-1},\psi)$ for all $g\in G$, where $r_g: P \to P$ denotes the right action of $G$ on $P$.

We denote by $\Omega^k_{\rho}(P,V)$ the space of horizontal and equivariant forms. Note that $\Omega_{\rho}(P,V)$ is in general not closed under the ordinary exterior derivative. There is a canonical isomorphism

$$\Omega^k_{\rho}(P,V) \cong \Omega^k(M,P \times_{\rho}V),$$

where $P \times_{\rho}V$ is the vector bundle associated to $P$ via the representation $\rho$.

Definition

The exterior covariant derivative for forms on $P$

$$\mathrm{d}_{\omega}: \Omega^k(P,V) \to \Omega^{k+1}(P,V)$$

is defined by

$$(\mathrm{d}_{\omega}\psi)_q(v_1,\dots,v_k) := (\mathrm{d}\varphi)_q(H v_1,\dots,H v_k).$$

Proposition

Every form in the image of $\mathrm{d}_{\omega}$ is horizontal. If a form $\psi$ is equivariant, $\mathrm{d}_{\omega} \psi$ is also equivariant.

Proposition

The restriction of $\mathrm{d}_{\omega}$ to $\Omega^k_{\rho}(P,V)$ can be described in terms of the connection 1-form $\omega\in \Omega^1(P,\mathfrak{g})$ and the derivative $\mathrm{d}\rho: \mathfrak{g} \times V \to V$ of the representation $\rho$:

$$\mathrm{d}_{\omega}(\psi) = \mathrm{d}\psi + \omega \wedge_{\mathrm{d}\rho} \psi\text{.}$$

Here we have used the following general notation: if $U,V,W$ are vector spaces, $\varphi \in \Omega^p(M,V)$, $\psi\in\Omega^q(M,W)$ and $f: V \times W \to U$ is a linear map, we have $\varphi \wedge_{f} \psi \in \Omega^{p+q}(M,U)$.

Proposition

Unlike the usual exterior derivative, the exterior covariant derivative need not be nilpotent in general. Instead, we have

$$\mathrm{d}_{\omega}(\mathrm{d}_{\omega}(\psi)) = \Omega \wedge_{\mathrm{d}\rho} \psi.$$

In particular, $\mathrm{d}_{\omega} \circ \mathrm{d}_{\omega} = 0$ if $\omega$ is flat.

Definition

The exterior covariant derivative for forms on $M$

$$\mathrm{D}_{\omega}: \Omega^k(M,P \times_{\rho} V) \to \Omega^{k+1}(M,P \times_{\rho}V)$$

is the map induced from $\mathrm{d}_\omega$ under the isomorphism $\Omega^k_{\rho}(P,V) \cong \Omega^k(M,P \times_{\rho}V)$.

Note that a connection on the principal bundle $p: P \to M$ induces? a connection on the associated vector bundle $P \times_\rho V$. Then the exterior covariant derivative in the sense of Definition coincides with this exterior covariant derivative.

Example

• The connection $\omega$ itself is not in $\Omega_{\mathrm{Ad}}^1(P,\mathfrak{g})$: it is not horizontal.

• The curvature? of $\omega$ is $\Omega := \mathrm{d}_\omega(\omega)\in \Omega^2_{\mathrm{Ad}}(P,\mathfrak{g})$. Since $\mathrm{d}(\mathrm{Ad})=[-,-]$, we have

  $$\Omega = \mathrm{d}\omega + [\omega \wedge \omega]\text{.}$$

  The Bianchi identity is $\mathrm{d}_{\omega}\Omega=0$.

Under the isomorphism $\Omega^k_{\rho}(P,V) \cong \Omega^k(M,P \times_{\rho}V)$, the curvature $\Omega$ corresponds to a 2-form $F_{\omega} \in \Omega^2(M,\mathrm{Ad}(P))$, and the Bianchi identity corresponds to $\mathrm{D}_{\omega}F_{\omega} = 0$.

Applications

• In Yang-Mills theory…

Related Concepts

• differential form

• connection on a bundle

• covariant derivative

• Yang-Mills theory