Covariant derivative

Revision as of 11:37, 15 March 2013

1 Definition

Let ${{Authors|Jost Eschenburg}}{{Stub}} == Definition == <wikitex>; Let $E \to M$ be a vector bundle over a smooth manifold $M$ and $\Gamma E$ the space of smooth sections. A {\it covariant derivative} on $E$ is a bilinear map $\nabla : \Gamma TM \times \Gamma E \to \Gamma E$, $(X,s)\mapsto\nabla_Xs$, which is a [[tensor]] (linear over $C^\infty(M)$) in the first argument and a {\it derivation} in the second argument: {{equation|$\begin{matrix}\nabla_{(fX)}s &=& f\nabla_Xs\, \ \nabla_{X}(fs) &=& (Xf)s + f\nabla_Xs \,,\end{matrix}$|1}} where $f$ is a smooth function and $X$ a vector field on $M$ and $s$ a section of $E$, and where $Xf = \partial_Xf = df.X$ is the ordinary derivative of the function $f$ in the direction of $X$. By these properties, $\nabla$ is defined locally and even pointwise regarding the first argument: For any $v\in T_pM$ we may define $\nabla_xs := (\nabla_Xs)_p$ where $X$ is any (local) vector field with $X_p = x$. A covariant derivative has all properties of the ordinary directional derivative for functions with exception of the commutativity: For functions $f$ we have $\partial_i\partial_jf = \partial_j\partial_if$ with $\partial_if = \frac{\partial(f \circ \phi)}{\partial u_i} = \partial_{\phi_i}f$, where $\phi : \R^n\to M$ is a local parametrization of $M$ and $\phi_i := \partial_i\phi$ its $i$-th partial derivative. Instead, for covariant derivatives, the commutator $[\nabla_i,\nabla_j]$ (with $\nabla_i = \nabla_{\phi_i}$) is nonzero in general, but it is only a ''tensor'' (rather than a differential operator), the [[Curvature tensor and second derivative|curvature tensor]]. Covariant derivatives are not ''tensors'' since they are derivations in the second argument, but two covariant derivative $\nabla,\tilde\nabla$ on $E$ differ by a tensor: $A = \tilde\nabla-\nabla$ defines a tensor $A \in \textup{Hom}(TM\otimes E,E)$. </wikitex> == References == {{#RefList:}} [[Category:Definitions]]E \to M$ be a vector bundle over a smooth manifold $M$ and $\Gamma E$ the space of smooth sections. A {\it covariant derivative} on $E$ is a bilinear map $\nabla : \Gamma TM \times \Gamma E \to \Gamma E$, $(X,s)\mapsto\nabla_Xs$, which is a tensor (linear over $C^\infty(M)$) in the first argument and a {\it derivation} in the second argument:

$\begin{matrix}\nabla_{(fX)}s &=& f\nabla_Xs\, \\ \nabla_{X}(fs) &=& (Xf)s + f\nabla_Xs \,,\end{matrix}$(1)

where $f$ is a smooth function and $X$ a vector field on $M$ and $s$ a section of $E$, and where $Xf = \partial_Xf = df.X$ is the ordinary derivative of the function $f$ in the direction of $X$. By these properties, $\nabla$ is defined locally and even pointwise regarding the first argument: For any $v\in T_pM$ we may define $\nabla_xs := (\nabla_Xs)_p$ where $X$ is any (local) vector field with $X_p = x$.

A covariant derivative has all properties of the ordinary directional derivative for functions with exception of the commutativity: For functions $f$ we have $\partial_i\partial_jf = \partial_j\partial_if$ with $\partial_if = \frac{\partial(f \circ \phi)}{\partial u_i} = \partial_{\phi_i}f$, where $\phi : \R^n\to M$ is a local parametrization of $M$ and $\phi_i := \partial_i\phi$ its $i$-th partial derivative. Instead, for covariant derivatives, the commutator $[\nabla_i,\nabla_j]$ (with $\nabla_i = \nabla_{\phi_i}$) is nonzero in general, but it is only a tensor (rather than a differential operator), the curvature tensor.

Covariant derivatives are not tensors since they are derivations in the second argument, but two covariant derivative $\nabla,\tilde\nabla$ on $E$ differ by a tensor: $A = \tilde\nabla-\nabla$ defines a tensor $A \in \textup{Hom}(TM\otimes E,E)$.

2 References