# On Christoffel symbol and vector fields

Take the defining equation of Christoffel symbols: $$\nabla_{\frac{\partial}{\partial x^\mu}}\frac{\partial}{\partial x^\nu}=\Gamma^{\sigma}_{\mu\nu}\frac{\partial}{\partial x^{\sigma}}$$ Both sides of the above definition are vector fields, in fact, the right side being a linear combination of coordinate vector fields with the coefficients of the combination being precisely the Christoffel symbols $\Gamma^{\sigma}_{\mu\nu}$ that do not transform as tensor. The above fact motivates my following question: If one has a vector field $X$ written out in a chart $x^\mu$ as $X=X^{\mu}\frac{\partial}{\partial x^\mu}$, is it the case that the smooth functions $X^\mu$ on the manifold should always transform as a vector? The right side of the definition above for Christoffel symbols suggest that this claim is not true.

• What makes you say that? The Christoffel symbols are defined at the level of the chart, not the level of the manifold - that's largely the point, that different choices in chart lead to non-trivial differences in the connection coefficient functions. – J. Murray Oct 10 '17 at 4:26
• If I write a vector field $X=f_{i}\frac{\partial}{\partial x^{i}}$, do the smooth functions $f_i$'s on the manifold necessarily transform as vector? – user78032 Oct 10 '17 at 4:34
• No, but that's because you've defined it using a chart already. You would have to make sure that it had the appropriate transformation properties once you expressed it in a different chart. The Christoffel symbols don't. – J. Murray Oct 10 '17 at 4:39
• So, in an appropriate chart I can always make it transform as a vector but in general the $f_i$'s need not transform as a vector, right? Could you please give a reference (Chapter, page etc.) to delve further into it. – user78032 Oct 10 '17 at 4:43
• It should be noted that you haven't actually defined a vector field by expressing it in a chart. A vector field is defined at the manifold level as a "directional derivative" along a smooth curve on the manifold. It can then be expressed in whatever chart you'd like. I don't have any good references on hand, but perhaps somebody else will. – J. Murray Oct 10 '17 at 4:46

If you have some $\{f^\mu\}\subset C^\infty(U)$ where $U$ is some coordinate domain, then $$\tag{1}X=f^\mu \partial_\mu$$ is indeed a vector field in $U$. If with respect to some other coordinate system, we have $$X=g^{\mu'}\partial_{\mu'},$$ then we will of course have $$\tag{2}g^{\mu'}=\frac{\partial x^{\mu'}}{\partial x^\mu} f^\mu.$$ But if $\{f^\mu\}$ has a transformation law other than (2), it does not have to be that $$g^{\mu'}=f^{\mu'}.$$ So basically, if you define a vector field by (1) and then transform, you have to forget whatever auxiliary transformation law the multiplet $\{f^\mu\}$ has.

A geometric object can be defined by its local expression in a particular coordinate system combined with rules governing coordinate change.

So if you treated the lower indices as mere labels, for each choice $n,k$ of $\nu, \kappa$ you could take the right-hand side as the definition of a local vector field

$$Y_{nk} = \Gamma^{\mu}_{nk}\frac{\partial}{\partial x^{\mu}}$$

However, these fields would need to transform under coordinate change according to

$${\tilde Y}^\mu_{nk} = \frac{\partial {\tilde x}^\mu}{\partial x^{\alpha}} \Gamma^{\alpha}_{nk}$$

which in general no longer agrees with the Christoffel symbols expressed in the new coordinates, which are given by

$$\tilde \Gamma^{\mu}_{\nu\kappa} = {\partial \tilde x^\mu \over \partial x^\alpha} \left [ \Gamma^\alpha_{\beta \gamma}{\partial x^\beta \over \partial \tilde x^\nu}{\partial x^\gamma \over \partial \tilde x^\kappa} + {\partial ^2 x^\alpha \over \partial \tilde x^\nu \partial \tilde x^\kappa} \right ]$$

• In $Y = \Gamma^{\mu}_{~\nu\kappa}\partial_{\mu}$ the indices are not balanced, so the extra 2 indices should be ignored. Therefore, you cannot use the transformation formula to conclude anything, because this is made to an object with 3 "active" indices, not to one with 2 dummy ones. – DanielC Oct 10 '17 at 14:55
• The motivation behind posing the Christoffel symbol definition was as follows: If in a chart $(U,x^{i})$ one writes a vector field $$Y\vert_{U}=f_{i}\vert_{U}\frac{\partial}{\partial x^{i}}|_{U}$$, does the smooth functions $f_{i}$'s on $U$, necessarily transform as a contravariant vector? We see from the definition of the Christoffel symbols that the coefficients $f_{i}$'s don't even transform as a tensor, yet the linear combination rendering a vector field in that chart. Am I missing anything? – user78032 Oct 10 '17 at 15:48
• @ Christoph, in the second line above, please correct the typo: should be $\tilde{\Gamma}^{\mu}_{nk}$ instead of $\tilde{Y}^{\mu}_{nk}$. So, you are saying that the object $Y_{nk}$ is not any geometric object, local vector field, in particular, as it behaves inconsistently under change of charts. So, the defining equation of Christoffel symbols only applies to a chart, right? – user78032 Oct 11 '17 at 4:27