# What is the meaning of non-coordinate basis?

Integral curves of non-coordinate (anholonomic) basis vectors also exist, they just don't form a coordinate system.

This might be a bit difficult to swallow, but the heart of the issue in a coordinate system, the coordinates are independent.

Here's a direct example:

Consider polar coordinates in $\mathbb{R}^2$. These are given by $$x=r\cos\varphi \\ y=r\sin\varphi.$$ The coordinate basis vectors are $$\partial_r=\cos\varphi\partial_x+\sin\varphi\partial_y \\ \partial_\varphi=-r\sin\varphi\partial_x+r\cos\varphi\partial_y.$$

These are orthogonal, but not orthonormal. We can also norm these vectors and get $$\hat{e}_r=\partial_r=\cos\varphi\partial_x+\sin\varphi\partial_y \\ \hat{e}_\varphi=\frac{1}{r}\partial_\varphi=-\sin\varphi\partial_x+\cos\varphi\partial_y.$$ The first set is holonomic, the second isn't, you can calculate $[\hat{e}_r,\hat{e}_\varphi]$ to ascertain this.

To try to interpret what it means for the integral curves of the anholonomic set not forming coordinates, consider that for the holonomic polar coordinate vectors, the $\partial_\varphi$ vector is longer, the further you are away from the origin. This is expected, $\varphi$ is an angular coordinate, the same angular displacement corresponds to larger and larger actual displacements the further you are away from the origin.

Consider now the integral curves of the set $\hat{e}_r,\hat{e}_\varphi$ instead. It is visually clear that the "paths" corresponding to the integral curves are the same, BUT the parametrization for the $\varphi$-curves are different. The vector field $\hat{e}_\varphi$ has the same length everywhere, so all "$\hat{\varphi}$" curves have the same velocity.

Imagine you are at the point $(r_0,\varphi_0)$. You move a parameter $\bar{\varphi}$ along the integral curves of $\hat{e}_\varphi$. Since the integral curves are path-length parametrized (the length of $\hat{e}_\varphi$ is 1 after all), you move a distance of $\bar{\varphi}$, and now you are at $(r_0,\varphi_0+\frac{1}{r_0}\bar{\varphi})$ (remember that I am measuring points in the original "holonomic" coordinates and that the real displacement corresponding to a coordinate displacement $\varphi$ is $\bar{\varphi}=r_0\varphi$, since we are at radius $r_0$).

Now we move radially to $2r_0$, so our coordinates are $(2r_0,\varphi_0+\frac{1}{r_0}\bar{\varphi})$. After this, we move along the integral curves of $\hat{e}_\varphi$ a parameter value of $-\bar{\varphi}$. This is once again, "actual displacement", and its value in $\varphi$-coordinates is $-\frac{1}{2r_0}\bar{\varphi}$, since we are now at radius $2r_0$.

Our new position is $(2r_0,\varphi_0+\frac{1}{r_0}\bar{\varphi}-\frac{1}{2r_0}\bar{\varphi})=(2r_0,\varphi_0+\frac{1}{2r_0}\bar{\varphi})$. Now we move back radially from $2r_0$ to $r_0$, and end up at $(r_0,\varphi_0+\frac{1}{2r_0}\bar{\varphi})$.

What we notice is that we did not do a loop at all, we had a net displacement of $\frac{1}{2r_0}\bar{\varphi}$ in the angular direction.

Now, if there actually did exist an $(r,\hat{\varphi})$ coordinate system attached to the basis $(\hat{e}_r,\hat{e}_\varphi)$, then in this coordinate system our path would have been $$(r_0,\hat{\varphi}_0)\mapsto (r_0,\hat{\varphi}_0+\bar{\varphi})\mapsto (2r_0,\hat{\varphi}_0+\bar{\varphi}) \mapsto (2r_0,\hat{\varphi}_0+\bar{\varphi}-\bar{\varphi})=(2r_0,\hat{\varphi}_0)\mapsto (r_0,\hat{\varphi}_0),$$ so we would have made a "coordinate loop", yet we would have had a net displacement. The "anholonomic coordinates" are not unambigous, because doing a finite loop, we ended up in a different point, yet both points have the same coordinates.

Clearly a coordinate system could never work this way.

I am not sure what your indices mean, but usually a non coordinate basis will be written as, $e^\mu_a(x^\nu)$ such that $g_{\mu\nu}e^\mu_ae^\nu_b = \eta_{ab}$. The thing that classifies a basis as a coordinate basis is a vanishing Lie bracket (see here for a similar question).