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A vector $\vec{\omega}$ in General Relativity is expressed in the language of differential geometry as $$\omega=\omega_r\frac{\partial}{\partial r}+\omega_\theta\frac{\partial}{\partial\theta}$$. However, I want to express this in the usual vector notation as $$\vec{\omega}=\omega(r)\hat{r}+\omega(\theta)\hat{\theta}$$ so as to obtain the magnitude of $\vec{\omega}$ as $\omega=\sqrt{\omega^2(r)+\omega^2(\theta)}$.

Could someone please suggest how to do this?

Richard
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    You should add some context to make your question unambiguous. The vector notation you are referring to does not make sense for abstract manifolds. If your manifold is not an abstract one, i.e, is embedded in $\mathbb R^n$ for some $n$, then the tangent vectors $\frac{\partial}{\partial r},\frac{\partial}{\partial \theta}$ are identified with vectors of $\mathbb R^n$. – Overflowian Jun 08 '22 at 09:57
  • See this answer of mine for some details on how to deal with the normalization of the vectors in polar/cylindrical coordinates. – peek-a-boo Jun 08 '22 at 18:12

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The tangent space at a point $p$ in a manifold consists of the Liebnizian operators (i.e., differential operators) on functions at $p$. By means of a coordinate chart $(x_i)_i$ about $p$, we see that the operators $\frac{\partial}{\partial x_i}$ are such Liebnizian operators, and in fact form a basis for the tangent space.

Now on $\Bbb R^n, \frac{\partial}{\partial x_i}$ is easily seen to be the directional derivative in the direction $\hat x_i$ and for a general vector $\vec v$ with $$\vec v = \sum_i v_i\hat x_i$$ we find that the directional derivative with respect to $\vec v$ is $$\frac{\partial}{\partial \vec v} = \sum_i v_i\frac{\partial}{\partial x_i}$$ Thus we can associate $\vec v$ with its directional derivative $\frac{\partial}{\partial \vec v}$. This same identification is also made for manifolds.

In short, $$\frac{\partial}{\partial r} \equiv \hat r\\\frac{\partial}{\partial \theta} \equiv r\hat \theta$$ Per the comment by peek-a-boo, $\frac{\partial}{\partial \theta}$ is not a unit vector. Its length is $r$. Thus the unit vector $\hat \theta$ is multiplied by $r$.

Paul Sinclair
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  • Thank you for the answer. – Richard Jun 09 '22 at 07:07
  • The final conclusion is misleading, because often (Physics) the hat notation $\hat{r}$ and $\hat{\theta}$ refer to the normalized versions. Of course, $\frac{\partial}{\partial r}$ already has unit norm (with respect to the standard Riemannian metric in the plane) but $\frac{\partial}{\partial \theta}$ has norm $r$. This factor of $r$ is often a source of confusion when comparing formulas from classical vector calculus and differential geometry. So I think it's worth emphasizing lest OP (or future readers) come across such 'discrepancies'. – peek-a-boo Jun 09 '22 at 16:37
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    @peek-a-boo - Thanks. I have overlooked that. I added a correction to the post. – Paul Sinclair Jun 11 '22 at 03:32