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Given a spherical metric $ds^2 = dr^2 + r^2(d\theta^2 + \sin^2(\theta) d\phi^2)$.

Is it correct that $\nabla_\theta \theta = 1$? Or do I need to consider the metric?

blablu
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  • Crossposted to https://physics.stackexchange.com/q/721823/2451 – Qmechanic Aug 07 '22 at 09:55
  • Since $\nabla_{e_i} e_j = \Gamma^k_{ji} e_k$, it suffices to calculate the Christoffel symbols. In this sense, yes; you do need to consider the metric, since the Christoffel symbols are calculated using the metric (and aren't always flat, e.g. in spherical coordinates) – Brevan Ellefsen Aug 07 '22 at 10:17
  • What confuses me here is that $\theta$ is just a scalar function which I assumed would simplify the derivative. E.g. I want to calculate $\nabla_\theta \theta = \nabla_{\frac{\partial}{\partial \theta}} \theta$ and not $\nabla_\theta \vec{\theta}$ – blablu Aug 07 '22 at 10:24
  • From the looks of it you are using the round metric on a given sphere, so we have $ds^2 = d\theta + sin^2(\theta) d \phi^2$. I'm not sure what you are trying to specify by adding the vector arrow over theta... The Covariant derivative $\nabla_X Y$ at a point $p$ depends on the values of $Y$ in an infinitesimal neighborhood of $p$ and not just at $p$, so you already have to extend as a vector field (though the choice of extension won't change the result). You also can't just get an answer of "1", since there has to be a basis vector.... – Brevan Ellefsen Aug 07 '22 at 10:44
  • ... Compare to flat 3-space, where for example we have $\nabla f = f_1 e_1 + f_2 e_2 + f_3 e_3$ (we usually write this as a vector with three complements, each the partial with respect to one of the coordinates. Adding basis vectors is just how we keep track of this over manifolds where there is no longer a global vector space structure on the manifold itself, only on the tangent bundle) – Brevan Ellefsen Aug 07 '22 at 10:46
  • By $\theta$ I mean the function $f:M \to \mathbb{R}$ which assigns the coordinate value $\theta$ to every point $p$. This is clearly a scalar (field). – blablu Aug 07 '22 at 10:54
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    That's not a well-defined global, smooth function on the sphere (think one dimension lower, in polar coordinates... The angle isn't a function because it jumps by $2\pi$). It's instead forced to be something locally defined, so that technically only $d\theta$ makes sense as a smooth object (which you can view either as a vector field or a one form). Indeed, $d\theta$ is not exact, since it's not the exterior derivative of any function. Moving back to the sphere, you have the same problem...theta can be defined locally, and even extended locally, but not globally. You get a vector-field/form. – Brevan Ellefsen Aug 07 '22 at 11:06
  • i.e., it is clear to me that $\nabla_\theta \vec{\theta} = \Gamma^k_{\theta \theta} e_k = -r e_r$ – blablu Aug 07 '22 at 11:06
  • so I can not write $\nabla_\alpha \theta = \delta _\alpha ^\theta$ for example? – blablu Aug 07 '22 at 12:06

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Yes it is right. Here, we have a certain open set $U\subset\Bbb{R}^3$, which is the domain of the spherical coordinate system, and on this domain, we have the spherical coordinate maps $(r,\theta,\phi):U\to (0,\infty)\times (0,\pi)\times (0,2\pi)$, and $r,\theta,\phi$ are each smooth functions so we can consider their exterior derivatives $dr,d\theta,d\phi$, which are thus $1$-forms on $U$. We also have the coordinate-induced vector fields on $U$, $\frac{\partial}{\partial r}, \frac{\partial}{\partial \theta}, \frac{\partial}{\partial \phi}$. Now, \begin{align} \nabla_{\theta}\theta\equiv\nabla_{\frac{\partial}{\partial \theta}}(\theta):=(d\theta)\left(\frac{\partial}{\partial \theta}\right):=\frac{\partial\theta}{\partial\theta}=1. \end{align} The first is simply notation. The second is because $\theta$ being a smooth function means by definition that $\nabla\theta:=d\theta$. The third equality is again a definition (of the exterior derivative acting on functions). The last should be clear.

The only thing I would caution you is to not confuse $\nabla_{\theta}\theta$ (which, again, really means $\nabla_{\frac{\partial}{\partial \theta}}(\theta)$) with $\nabla_{\frac{\partial}{\partial \theta}}\frac{\partial}{\partial\theta}$. The former is a smooth real-valued function on $U$, while the latter is a vector field on $U$.

More generally, on any smooth manifold $M$ with a connection $\nabla$ on the tangent bundle (which is then extended in the usual way to all tensor bundles, in particular to the smooth functions), if you take any coordinate chart $(U,x=(x^1,\dots, x^n))$, then we have $\nabla_{\mu}(x^{\nu})=dx^{\nu}\left(\frac{\partial}{\partial x^{\mu}}\right)=\delta^{\nu}_{\mu}$.

And no this doesn't depend on the metric because a connection $\nabla$ really refers to a connection on the tangent bundle $TM$. We then extend this to all the tensor bundles, and in particular to the space of smooth functions, where by definition we set $\nabla f=df$. At this point, I should again caution you that $\nabla f$ does not refer to the gradient vector field you may have seen in multivariable calculus courses; here, $\nabla f=df$ is a $1$-form/covector field. The usual gradient vector field is something which depends on the metric tensor, and is defined as $\text{grad}_g(f):=g^{\sharp}(df)=g^{\sharp}(\nabla f)$ (see this for a related question).

peek-a-boo
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