coordinate basis

From Mathematics wiki

Also called a holonomic basis, a coordinate basis is a set of basis vectors that are derivatives along coordinate curves:

$\vec{e}_i \equiv \frac{\partial}{\partial x^i}\,$ .

In writing a general vector as

$\vec{V} = \frac{d}{d\lambda} = V^i\frac{\partial}{\partial x^i}\,$

we have implicitly made use of the coordinate basis vectors $\vec{e}_i\,$ , so that we can write

$\vec{V} = V^i \vec{e}_i\,$ .

Note that acting $\vec{V} = \frac{d}{d\lambda}\,$ on an arbitrary function $f\,$ , we obtain a new function $\vec{V}(f)\,$ . If $f\,$ is suitably differentiable, then we can act upon this function with another vector field $\vec{W} = \frac{d}{d\mu}\,$ :

$\vec{W}\left(\vec{V}(f)\right)\,$	$= \frac{d}{d\mu}\left( \frac{df}{d\lambda} \right)\,$ ,
	$= \frac{d}{d\mu}\left( V^i \frac{\partial f}{\partial x^i} \right)\,$ ,
	$= W^j \frac{\partial}{\partial x^j}\left( V^i \frac{\partial f}{\partial x^i} \right)\,$ ,
	$= W^j \frac{\partial V^i}{\partial x^j}\frac{\partial f}{\partial x^i} + W^j V^i \frac{\partial^2f}{\partial x^j \partial x^i}\,$ .

Notice that this is no longer a first order differential operator (not that we expected it to be), but consider the fact that second partial derivatives commute. Then,

$[\vec{W},\vec{V}] f\,$	$\equiv \vec{W}\left(\vec{V}(f)\right) - \vec{V}\left(\vec{W}(f)\right)\,$ ,
	$= W^j \frac{\partial V^i}{\partial x^j}\frac{\partial f}{\partial x^i} + W^j V^i \frac{\partial^2f}{\partial x^j \partial x^i} - V^j \frac{\partial W^i}{\partial x^j}\frac{\partial f}{\partial x^i} - V^j W^i \frac{\partial^2f}{\partial x^j \partial x^i}\,$ ,
	$= \left(W^j \frac{\partial V^i}{\partial x^j} - V^j \frac{\partial W^i}{\partial x^j}\right) \frac{\partial f}{\partial x^i} \,$ .

Since $f\,$ is arbitrary, we obtain a new first order differential operator (and vector field):

$[\vec{W},\vec{V}] = \left(W^j \frac{\partial V^i}{\partial x^j} - V^j \frac{\partial W^i}{\partial x^j}\right) \frac{\partial}{\partial x^i}\,$ .

The vector field $[\vec{W},\vec{V}]\,$ is known as the Lie bracket of $\vec{W}\,$ and $\vec{V}\,$ . It is the Lie derivative of $\vec{V}\,$ in the direction $\vec{W}\,$ , and can be written as $\mathcal{L}_\vec{W}\vec{V}\,$ .

Finally, if $\lambda\,$ and $\mu\,$ are two coordinates of a coordinate chart, e.g. $y^1 =\lambda\,$ and $y^2 =\mu\,$ , then

$[\vec{W},\vec{V}] f = \frac{\partial^2 f}{\partial y^2 \partial y^1} - \frac{\partial^2 f}{\partial y^1 \partial y^2} = 0\,$ .

Therefore a coordinate basis has the property that

$[\vec{e}_i, \vec{e}_j] = 0\,$ ,

for all $i\,$ and $j\,$ . Although it would appear that this property should hold for general tangent vectors $\frac{d}{d\lambda}\,$ and $\frac{d}{d\mu}\,$ , the fact that second partial derivatives commute is the statement that an infinitesimal displacement $\Delta x^i\,$ along the coordinate curve $x^i\,$ , holding $x^j\,$ for some $j\neq i\,$ fixed, followed by an infinitesimal displacement $\Delta x^j\,$ along the coordinate curve $x^j\,$ , holding $x^i\,$ for $i\neq j\,$ fixed, is equivalent to the infinitesimal displacement in the reverse order. In general, it is always not possible to vary $\lambda\,$ while keeping $\mu\,$ fixed, so that the total derivatives $\frac{d}{d\lambda}\,$ and $\frac{d}{d\mu}\,$ do not commute. In such a case $\lambda\,$ and $\mu\,$ are not good coordinates.