First of all, my question lies on: Differentiable, real, n-dimensional Manifolds and in the context of differential geometry for General Relativity. Also, my level of academic mathematical language do not cover fibre bundles or more complex structures than the intuitive notion of tangent and cotangent bundles.

So, the author in $[1]$ said explicitly the following stroked in red:

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Then, the whole well-know fact that Christoffel symbols aren't tensors has sinked into a whirlpool of confusion. This whirlpool of confusion is due to the classical tensor analysis realization; which we check that the christoffel symbols in fact do not transform like a tensor object. Furthermore other authors have pointed out this fact of a true tensor nature, of the Christoffel symbols, before: $[2]$,$[3]$. The thing is, if they form a components of a $(1,1)-tensor$ $[1]$, then there must to be such abstract object which is the "pure" tensor:

$$\Gamma = \Gamma^{a}_{b} dx^{b} \otimes \frac{\partial}{\partial x^{a}} \tag{1}$$

Well, the fact is, concerning the realization of $[1]$ when he said:

one for each basis vector $\vec{e}_{\nu}$

this motivated me to write a notation for this phrase:

$$ \Gamma^{a}_{(\cdot)b} \tag{2}$$

Where the dot means that, when we put a basis vector $\frac{\partial}{\partial x^{j}}$ we get on return $[2]$:

$$ \Gamma^{a}_{(\cdot)b}\Big(\frac{\partial}{\partial x^{j}}\Big)\equiv \Gamma^{a}_{jb} =: dx^{a}\Big( \nabla_{\frac{\partial}{\partial x^{b}}}\frac{\partial}{\partial x^{j}} \Big) \tag{3}$$

On the other hand, a mathematical fact is that, a Christoffel Symbol can be calculated precisely as $(3)$. So, if we omit the basis vector $j$, then we have, in fact, an operator:

$$\Gamma^{a}_{(\cdot)b}: \mathfrak{X}(M) \to C^{\infty}(M) $$

$$ \Gamma^{a}_{(\cdot)b}=: dx^{a}\Big( \nabla_{\frac{\partial}{\partial x^{b}}} (\cdot)\Big) \tag{4}$$

Or, more gently:

$$\Gamma^{a}_{(\cdot)b}: T_{p}M \to \mathbb{R} $$.

Now, for the tensor transformation law I tried to prove and I get, nicely, a result which is the true tensor law:

Consider then the symbols in a coordinate chart $C'$.

$$ \Gamma'^{a'}_{(\cdot)b'} = \Gamma'^{a'}_{b'} \tag{5}$$

Where in $(5)$ is just a change of notation. Then we perfom a change of coordinates to another chart $C \to C'$

Then the Symbols transforms as:

$$\Gamma'^{a'}_{b'} = dx'^{a'}\Big( \nabla_{\frac{\partial}{\partial x'^{b'}}} (\cdot)\Big) = \frac{\partial x'^{a'}}{\partial x^{c}}dx^{c}\Bigg( \nabla_{\frac{\partial x^{d}}{\partial x^{b'}}\frac{\partial}{\partial x^{d}}} (\cdot)\Bigg) = $$

$$ = \frac{\partial x'^{a'}}{\partial x^{c}}\frac{\partial x^{d}}{\partial x^{b'}}dx^{c}\Bigg( \nabla_{\frac{\partial}{\partial x^{d}}} (\cdot)\Bigg) =\frac{\partial x'^{a'}}{\partial x^{c}}\frac{\partial x^{d}}{\partial x^{b'}} \Gamma^{c}_{d} \tag{6} $$

So indeed the symbols transforms like a tensor, then with an abuse of notation, we can say that "Christoffel symbols" transforms like a tensor. The subtle fact is: for every basis vector we have an Christoffel Symbols; therefore the whole symbol $(3)$ do not transform indeed.

But since we have the $\Gamma^{a}_{b}$, and it's tensor nature, we can therefore say that they form components of the Christoffel Tensor. Then, we can indeed conclude it's abstract form $(1)$.

My question is: Is the tensor law expressed in $(6)$ totally correct? Or, in other words, the operator realization given by $(4)$ makes sense?

$$ * * * $$

$[1]$BERTSCHINGER.B. Introduction to Tensor Calculus for General Relativity. link: http://web.mit.edu/edbert/GR/gr1.pdf pages 20-21.

$[2]$ CHRUSCIEL.P.T. Elements of General Relativity. Birkhauser. pages 16-19.

$[3]$ WALD.R. General Relativity. pages 29-33

  • $\begingroup$ at first glance this has a not so far resemblance to some what is called “connection forms”... I will keep you informed. Meanwhile, let us see if some big shark would drop here some of its geekiness $\endgroup$
    – janmarqz
    Commented May 16, 2020 at 15:57
  • $\begingroup$ @M.N.Raia: Your conclusion is correct. Are you familiar with the notion of affine connection (say, as presented in en.wikipedia.org/wiki/Affine_connection) and the relation between $C^{\infty}(M)$-linearity and being tensorial? $\endgroup$
    – levap
    Commented May 16, 2020 at 16:02

1 Answer 1


It is known that the $$\omega^a{}_b=\Gamma^a{}_{sb}dx^s,$$ define a collection of 1-forms. They are dubbed the holonomic (that is, they're depending on coordinates) connection forms, and clearly they satisfy (abbreviating $\partial_j=\frac{\partial}{\partial x^j}$) \begin{eqnarray*} \omega^a{}_b(\partial_j)&=&\Gamma^a{}_{sb}dx^s(\partial_j),\\ &=&\Gamma^a{}_{sb}\delta^s_j,\\ &=&\Gamma^a{}_{jb},\\ &=&\Gamma^a_{(\cdot) b}(\partial_j). \end{eqnarray*} Hence one gets $\Gamma^a_{(\cdot) b}=\omega^a{}_b$

  • 1
    $\begingroup$ So, in fact, they are the connection one forms. $\endgroup$
    – M.N.Raia
    Commented May 16, 2020 at 19:20
  • 1
    $\begingroup$ yes!... the interesting thing about your method is that the contraction $\omega^a{}_b\otimes\frac{\partial }{\partial x^a}$ comes form the tensor product $\omega^a{}_b\otimes\frac{\partial }{\partial x^c}$ which is a $(1,1)$ tensor $\endgroup$
    – janmarqz
    Commented May 16, 2020 at 19:25
  • 2
    $\begingroup$ @M.N.Raia Yes, for every $a$ and every $b$, the object $\Gamma^a_{(\cdot)b}$ is a tensor, but one should be aware that this assignment is not tensorial in $a$ and $b$. I.e., in $\Gamma^a_{ib}$, fixing $a$ and $b$ and varying the index $i$ you get the components of a tensor, but $a$ and $b$ are not tensor indices, they are more like labels. $\endgroup$ Commented May 21, 2020 at 6:35
  • $\begingroup$ @M.N.Raia I shoulda call a $(1,2)$ and $(0,1)$ the contraction $\endgroup$
    – janmarqz
    Commented Jul 6, 2023 at 19:33

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