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Suppose I have a vector in the covariant basis $\bar e_i$ and a metric $g$ and I want to obtain a unit vector $\bar u_i$ parallel to $\bar e_i$. I would write:

$$\bar e_i = \hat u_i ||\bar e_i|| = \hat u_i \sqrt {\bar e_i \bullet \bar e_i} = \hat u_i \sqrt{g_{ii}} $$

This seems weird because the index $i$ appears a bunch of times but I guess it's okay because it's a free index?

Now suppose I want to calculate the gradient in terms of the vectors $\hat u_i$. We have: $$\mathbf {\nabla} = \vec{e}^j\frac{\partial}{\partial x^j} = g^{ji} \bar{e}_i \frac{\partial}{\partial x^j} = g^{ji}\sqrt{g_{ii}}\hat u_i\frac{\partial}{\partial x^j} $$

Now that's definitely wrong. The dummy index $j$ appears twice, up and down so this is okay, but the dummy index $i$ appears four times! This breaks the rules. I could maybe rewrite it as:

$$\mathbf {\nabla} = \vec{e}^j\frac{\partial}{\partial x^j} = g^{ji} \bar{e}_i \frac{\partial}{\partial x^j} = g^{ji}\sqrt{g_{ik}}\hat u_k\frac{\partial}{\partial x^j} $$

This now satisfies the rules, but it isn't the same equation. I really want the diagonal element $g_{ii}$, not $g_{ik}$. What am I supposed to do?

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I think it goes wrong right from the intent. Einstein notation is great for making sure what you're writing down is independent of your choice of coordinates, but what you want to do is manifestly not coordinate-independent -- it awards a special significance to the particular coordinates you normalize to get the $u_i$s. That's why you cannot get the rules to match -- because the rules are made to match only for things that are coordinate independent. – Henning Makholm Oct 22 '12 at 22:42
More technically, I don't think you have any guarantee that the $\hat u_i$s are going to be a consistent family of tangent space bases for every point on the manifold, in the sense that they won't necessarily arise as the natural basis with respect to any coordinate chart even locally. Given that, it's not clear to me whether "calculate the gradient in terms of the $\hat u_i$s" ought to be a meaningful tensorial operation. – Henning Makholm Oct 22 '12 at 22:47
You might be able to get somewhere if you start out with $n$ general vector fields that just happen to align with your initial distinguished set of coordinates. But still what you want to do may be too nonlinear for the notation to accommodate easily. – Henning Makholm Oct 22 '12 at 22:52
@HenningMakholm Well, rereading the question now, it never states that I must give the answer in Einstein summation, I just assumed I had to. So I guess you're right, it can't be done. I think everything works out if I just use sigmas to denote the sums. – Arthur Oct 23 '12 at 14:12

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