Why don't the Christoffel symbols transform as a tensor?

Let $$(P, \pi, M)$$ be a principal $$G$$ -bundle. Given $$A \in L(G),$$ we define the vector field $$X^{A} \in \Gamma(T P)$$ by \begin{aligned} X_{p}^{A}: \mathcal{C}^{\infty}(P) & \stackrel{\sim}{\longrightarrow} \mathbb{R} \\ f & \mapsto[f(p\cdot\exp (t A))]^{\prime}(0), \end{aligned} where the derivative is to be taken with respect to $$t$$. Here $$\Gamma$$ is the set of section in the tangent bundle $$TP$$ of $$(P, \pi, M)$$ and $$L(G)$$ is the lie algebra of the group $$G$$.

We also define connection one-form on the principal bundle $$(P, \pi, M)$$ as a smooth Lie-algebra valued one-form, i.e. a smooth map $$\omega: \Gamma(T P) \stackrel{\sim}{\rightarrow} T_{e} G$$ such that

i) $$\omega\left(X^{A}\right)=A$$;

ii) $$\left((\triangleleft g)^{*} \omega\right)(X)=\left(\operatorname{Ad}_{g^{-1}}\right)_{*}(\omega(X))$$.

Let $$\sigma : U\subset M\to \pi^{-1}(U)$$ be a section in $$(P, \pi, M)$$ the Yang–Mills field is defined as $$\sigma^*\omega:=A(x)=\sum_{\mu=1}^{m} \sum_{a=1}^{\operatorname{dim} G} A_{\mu}^{a}(x) E_{a}\left(d x^{\mu}\right)_{x},$$ where $$\left\{E_{1}, E_{2}, \ldots, E_{\operatorname{dim} G}\right\}$$ is a basis set for $$L(G)$$. If $$\sigma_2=g(x)\sigma$$ is another section and $$G$$ is a matrix group then we have $$\sigma^*\omega_2:=A_2(x)=g(x)^{-1} A_{\mu}(x) g(x)+g(x)^{-1} \partial_{\mu}g(x). \tag 1$$

Let $$LM$$ be a frame bundle. Any chart $$(U,x)$$ of a smooth manifold $$M$$ induces a local section $$\sigma: U\to LM$$ of the frame bundle of $$M$$ by $$\sigma(m):=\bigg(\frac{\partial}{\partial x^1},\ldots,\frac{\partial}{\partial x^m} \bigg)\in L_mM \tag2$$ we define $$\Gamma:= \sigma^* \omega.$$ and if $$(U',x')$$ is another chart with $$U \cap U' \neq \emptyset$$ and that $$\sigma_2(m):=\bigg(\frac{\partial}{\partial x'^1},\ldots,\frac{\partial}{\partial x'^m} \bigg)\in L_mM \tag3$$ One can show by $$(1)$$ that $$\Gamma_{\mu \delta}^{\prime}{ }^{\epsilon}(x)=\sum_{\alpha, \rho=1}^{m} \frac{\partial x^{\alpha}}{\partial x^{\prime \mu}} \frac{\partial x^{\rho}}{\partial x^{\prime \delta}} \frac{\partial x^{\prime \epsilon}}{\partial x^{\chi}} \Gamma_{\alpha \rho}^{\chi}(x)+\sum_{\lambda=1}^{m} \frac{\partial x^{\prime \epsilon}}{\partial x^{\lambda}} \frac{\partial^{2} x^{\lambda}}{\partial x^{\prime \mu} \partial x^{\prime \delta}}.\tag 4$$

As we see expression $$(4)$$ does not come from a coordinate transformation but from a change of section $$\sigma \rightarrow \sigma_2$$.
Now for charts $$(U,x)$$ and $$(U,x')$$ if one makes a coordinate transformation we would have $$\sigma(x) \rightarrow \sigma'(x')=\bigg(\frac{\partial x'^{\nu}}{\partial x^{1}}\frac{\partial}{\partial x'^{\nu}},\ldots,\frac{\partial x'^{\mu}}{\partial x^{m}}\frac{\partial}{\partial x'^{\mu}} \bigg)= \bigg(\frac{\partial}{\partial x^1},\ldots,\frac{\partial}{\partial x^m} \bigg)=\sigma(x) \tag 5$$ and so $$\Gamma(x)=\Gamma'(x')$$ and since only the indices $$\mu$$ depends on the coordinates we should have $$\Gamma_{\mu \delta}^{\prime}{ }^{\epsilon}=\frac{\partial x'^{\nu}}{\partial x^{\mu}}\Gamma_{\nu \delta}{ }^{\epsilon}.$$

Am I wrong?

• Try applying the tensor transformation rule for a rank (1,2) tensor and see what you get Commented Jan 17, 2021 at 15:30
• Commented Jan 28, 2022 at 19:23

Let $$M$$ be a smooth manifold and $$(P,\pi,M,G)$$ a principal $$G$$-bundle over $$M$$. Let $$\omega$$ be a connection one-form and let $$\sigma : U\subset M\to \pi^{-1}(U)$$ be a local section which locally trivializes the bundle by means of $$h:U\times G\to \pi^{-1}(U)$$:

$$h(x,g)=\sigma(x)\cdot g.\tag{1}$$

With $$h$$ we can pullback $$\omega$$ to understand how it is represented on the locally trivial neighborhood. What you must do is evaluate $$h^\ast \omega$$. This is done in Theorem 6.1 of "Modern Differential Geometry for Physicists" by C. J. Isham, and the result is, after identifying $$T_{(x,g)}(U\times G)\simeq T_x U\oplus T_gG$$: $$(h^\ast \omega)_{(x,g)}(\alpha,\beta)={\rm Ad}_{g^{-1}}((\sigma^\ast\omega)_x(\alpha))+\Xi_g(\beta)\tag{2},$$

where $$g\mapsto {\rm Ad}_{g}$$ is the adjoint representation of $$G$$ on the Lie algebra $$\mathfrak{g}$$ and $$\Xi$$ is the Maurer-Cartan form. Observe that a connection one-form is locally specified by a $$\mathfrak{g}$$-valued one form in $$U$$, but there is also the Maurer-Cartan part, this will play a role in a second.

Now, $$\sigma^\ast\omega$$ is a $$\mathfrak{g}$$-valued one-form, and it transforms as such under changes of coordinates on $$U$$, but that is not the point. The point is that if you pick a distinct section $$\widetilde{\sigma}:V\subset M\to \pi^{-1}(V)$$ with $$U\cap V\neq\emptyset$$ you get a distinct local trivialization $$\widetilde{h}:V\times G\to \pi^{-1}(V)$$ and from (2) it gives you a diferent $$\mathfrak{g}$$-valued one-form $$\widetilde{\sigma}^\ast \omega$$.

If you restrict to $$U\cap V$$ you have two $$\mathfrak{g}$$-valued one forms on this open subset: $$\sigma^\ast\omega$$ and $$\widetilde{\sigma}^\ast\omega$$. Then you would like to relate these two and you find out that the Maurer-Cartan form, as could be expected, plays a role. Indeed, defining $$\Omega:U\cap V\to G$$ by $$\widetilde{\sigma}(x)=\sigma(x)\Omega(x)$$, this is shown in Theorem 6.2 of Isham's book:

$$(\widetilde{\sigma}^\ast\omega)_\mu(x)={\rm{Ad}}_{\Omega(x)^{-1}}((\sigma^\ast\omega)_\mu(x))+(\Omega^\ast\Xi)_\mu(x).\tag{3}$$

As it happens for the transformation law of the local representative of a connection you have an inhomogeneous part, which here we see that corresponds to the Maurer-Cartan form. This is the origin of the inhomogeneous term in the transformation law for the Christoffel symbols.

The point IMHO is that in this analysis we are not talking about the transformation of the components of a single one-form but rather we are talking about how two distinct one-forms are related to one another given that they are distinct representatives of the same fundamental object on the principal bundle.

Finally for the specific version of this formalism with $${\rm GL}(m,\mathbb{R})$$ see section 6.1.4 of Isham's book.

• I will edit my question Commented Jan 17, 2021 at 15:57

I am not seeing how since the only part of $${\Gamma_j^i}_\mu$$ which depends on the coordinates is the index $$\mu$$ which transform as one form.

That's not accurate. You're right that the $$\mu$$ index refers to the manifold coordinates and the $$i$$ and $$j$$ indices refer to the tangent space. However, the tangent space basis vectors also depend on the manifold coordinates. In other words, $$\boldsymbol \Gamma \equiv (\Gamma_1,\ldots,\Gamma_{\mathrm{dim}(M)})$$ is a $$T_e\mathrm{GL}(m,\mathbb R)$$-valued one-form (i.e. a collection of matrices), but $$\big(\Gamma_{\mu}\big)^i_{\ j}$$ are the components of the matrix $$\Gamma_\mu$$ in the local tangent space basis.

The connection defines a parallel transport $$T$$ for tangent vectors. If we transport the basis vector $$\hat e_j$$ an infinitesimal distance $$\epsilon$$ along the $$x^\mu$$ direction, we obtain $$T(x^\mu, \epsilon)\hat e_j = (\mathbb I + \epsilon \Gamma_\mu)\hat e_j$$

If $$\Gamma_\mu$$ vanishes, then that implies that the $$j^{th}$$ tangent vector at the point $$(x^1,\ldots,x^N)$$ is parallel-transported to itself at the point $$(x^1,\ldots,x^\mu+\epsilon,\ldots,x^N)$$ along the specified curve. Loosely, we would say that the basis vectors do not depend on the coordinate $$x^\mu$$.

However, even if this is the case in one choice of basis, there is no reason whatsoever that this should be true in any other basis. If we choose a basis which varies with position (such as the coordinate-induced basis in curvilinear coordinates), then $$\mathbf \Gamma \neq 0$$.

In summary, the non-tensorial transformation properties of $$\Gamma_\mu)^i_{\ j}$$ arise because $$\Gamma_\mu$$ measures the rate of change of the basis vectors along the direction $$x^\mu$$. If you're using the coordinate-induced basis for the tangent bundle, then a change of coordinates induces a corresponding change of basis. The spatial rates of change of those basis vectors are clearly non-tensorial - one can see this by noting that their vanishing in one basis does not imply their vanishing in any other basis. As a result, the components $$(\Gamma_\mu)^i_{\ j}$$ do not transform as a $$(1,2)$$-tensor.

As we see expression (4) does not come from a coordinate transformation but from a change of section $$\sigma\rightarrow\sigma_2$$.

Yes, but the section is induced by the coordinate chart.

Now for charts $$(U,x)$$ and $$(U,x')$$ if make a coordinate transformation we would have $$\sigma(x)=\sigma'(x')$$ [...]

This isn't true. It should be clear that if $$x$$ and $$x'$$ are different charts, then $$\left\{\frac{\partial}{\partial x^\mu}\right\} \neq \left\{\frac{\partial}{\partial x'^\mu}\right\}$$. They may have the same span, but that's different from being the same frame.

If you're using the coordinate-induced basis for your tangent spaces, then a change of coordinates is accompanied by a change in frame which ultimately results in the non-tensorial transformation for the connection coefficients.

It appears that you are trying to change coordinates without changing the frame, e.g. transforming from cartesian to polar coordinates but continuing to use the cartesian basis vectors as the basis for the tangent bundle. In principle this is possible, and the connection coefficients would transform as you say. But when we say that the connection coefficients do not transform as a tensor, we refer to coordinate changes and the corresponding induced change in coordinate frame.

$$\sigma(x) \rightarrow \sigma'(x')=\bigg(\frac{\partial x'^{\nu}}{\partial x^{1}}\frac{\partial}{\partial x'^{\nu}},\ldots,\frac{\partial x'^{\mu}}{\partial x^{m}}\frac{\partial}{\partial x'^{\mu}} \bigg)= \bigg(\frac{\partial}{\partial x^1},\ldots,\frac{\partial}{\partial x^m} \bigg)=\sigma(x)$$

This isn't true. $$\sigma'(x')$$ - the frame induced by the chart $$x'$$ - is simply $$\left(\frac{\partial}{\partial x'^1} ,\ldots,\frac{\partial}{\partial x'^m}\right)$$ by the definition you yourself provide in (3).

• I will edit my question to clarify some aspects Commented Jan 17, 2021 at 16:11
• @ J.Murray i updated my question to show why $\sigma(x)=\sigma'(x')$ Commented Jan 17, 2021 at 20:00
• @amiltonmoreira If you read the end of my answer, I do understand that you are trying to change the coordinates without changing the frame. The non-tensorial transformation of the connection coefficients occurs when we change the coordinates and the corresponding induced section of the frame bundle. Commented Jan 17, 2021 at 20:16
• @amiltonmoreira Yes, that's right. When you see similar transformation rules for vectors - namely $V^\mu \mapsto V'^\mu = \frac{\partial x'^\mu}{\partial x^\nu} V^\nu$ - the same coordinate-induced change of frame is assumed. Commented Jan 17, 2021 at 20:31

The Christoffel symbols are not the components of a tensor. Their transformation laws involve the second derivatives of the parameter transformation : \begin{align} \bar{\Gamma}^k_{ij}&\boldsymbol{=} \left[\Gamma^\gamma_{\alpha\beta}\dfrac{\partial u^\alpha}{\partial\bar{u}^i}\dfrac{\partial u^\beta}{\partial\bar{u}^j}\boldsymbol{+}\dfrac{\partial^2 u^\gamma}{\partial\bar{u}^i\partial\bar{u}^j}\right]\dfrac{\partial \bar{u}^k}{\partial u^\gamma}\boldsymbol{\ne} \Gamma^\gamma_{\alpha\beta}\dfrac{\partial u^\alpha}{\partial\bar{u}^i}\dfrac{\partial u^\beta}{\partial\bar{u}^j}\dfrac{\partial \bar{u}^k}{\partial u^\gamma} \tag{01}\label{01}\\ \bar{\Gamma}_{ijk}&\boldsymbol{=} \left[\Gamma_{\alpha\beta\gamma}\dfrac{\partial u^\alpha}{\partial\bar{u}^i}\dfrac{\partial u^\beta}{\partial\bar{u}^j}\boldsymbol{+}g_{\alpha\gamma}\dfrac{\partial^2 u^\alpha}{\partial\bar{u}^i\partial\bar{u}^j}\right]\dfrac{\partial u^\gamma}{\partial \bar{u}^k}\boldsymbol{\ne} \Gamma_{\alpha\beta\gamma}\dfrac{\partial u^\alpha}{\partial\bar{u}^i}\dfrac{\partial u^\beta}{\partial\bar{u}^j}\dfrac{\partial u^\gamma}{\partial \bar{u}^k} \tag{02}\label{02} \end{align}

The clearest way to address the question is to treat the connection as a functional of vector fields $$u = u^μ ∂_μ$$ and $$v = v^ν ∂_ν$$ and a one-form field $$α = α_ρ dx^ρ$$, but to denote it as $$Γ_{uv}^α$$, expressing its operands as "indices". To get backward compatibility with normal index notation, we can establish the convention that $$(\_)^μ$$ and $$(\_)_ν$$ be treated, respectively, as synonymous with $$(\_)^{dx^μ}$$ and $$(\_)_{∂_ν}$$.

Similarly, we can also do the same with frame bases; e.g. $$u^a = u^{h^a}$$ and $$α_b = α_{e_b}$$, if the indices $$a, b, ⋯$$ are being taken with respect to a frame $$\left(h^a: a = 0, 1, ⋯, n-1\right)$$ and its dual $$\left(e_a: a = 0, 1, ⋯, n - 1\right)$$.

We can also extend this to other functionals, e.g. $$∂_u = u^μ ∂_μ = u, \hspace 1em dx^α = dx^ρ α_ρ = α$$ or even $$δ^α_u = δ^ρ_ν α_ρ u^ν = α_μ u^μ = α_u = u^α = u˩α,$$ where $$(\_)˩(\_)$$ is the contraction operator. As you can see, it is a very versatile and robust language that seamlessly connects the Mathematicians' and Physicists' languages.

Now we can address your issue more succinctly! A functional is "tensorial" if it is linear in all of its indices. Each such index can be referred to as "tensorial", so that for functionals that are not tensorial, some of their indices may nonetheless be. Such is the case with the connection.

Geometrically, the connection is defined in terms of a "covariant derivative operator", which is normally written $$∇_u v$$ as a functional of two vector fields $$u$$ and $$v$$. The connection, itself, may be defined in terms of it by: $$Γ^α_{uv} = \left(∇_u v\right)˩α.$$

I won't lay out the properties postulated for the covariant derivative operator, but only illustrate them in use, by running through this example. First, we can decompose $$u$$ in indexed component form: $$∇_u v = ∇_{u^μ ∂_μ} v = u^μ ∇_{∂_μ} v = u^μ ∇_μ v.$$ Linearity in $$u$$ is postulated for $$∇_u$$. Second, we can decompose $$v$$ similarly: \begin{align} ∇_μ v &= ∇_μ \left(v^ν ∂_ν\right) \\ &= \left(∇_μ v^ν\right) ∂_ν + v^ν \left(∇_μ ∂_ν\right) \\ &= \left(∂_μ v^ν\right) ∂_ν + v^ν \left(∇_μ ∂_ν\right). \end{align} Thus \begin{align} ∇_u v &= u^μ \left(\left(∂_μ v^ν\right) ∂_ν + v^ν \left(∇_μ ∂_ν\right)\right) \\ &= \left(u^μ ∂_μ v^ν\right) ∂_ν + u^μ v^ν \left(∇_μ ∂_ν\right) \\ &= \left(∂_u v^ν\right) ∂_ν + u^μ v^ν \left(∇_μ ∂_ν\right). \end{align} The Leibnitz rule applies to the covariant derivative operator and, for scalars, it reduces to an ordinary derivative operator. Thus $$∇_u v$$ and $$∇_μ v$$ are not linear in $$v$$, but only affine. Technically, when it comes to functions: "linear, with offset" is actually referred to as "affine", not "linear". Thus $$f(x) = mx + b$$ (where $$m$$ and $$b$$ are constant coefficients) is not linear, but affine! Rather, it is only $$f(x) = mx$$ that is linear.

Finally, upon contraction with $$α = α_ρ dx^ρ$$, we get: \begin{align} Γ^α_{uv} &= (∇_u v)˩α \\ &= \left(∂_u v^ν\right) ∂_ν˩α + u^μ v^ν \left(∇_μ ∂_ν\right)˩\left(α_ρ dx^ρ\right) \\ &= \left(∂_u v^ν\right) α_ν + u^μ v^ν \left(∇_μ ∂_ν˩dx^ρ\right) α_ρ \\ &= \left(∂_u v^ν\right) α_ν + u^μ v^ν Γ^ρ_{μν} α_ρ, \end{align} using the bilinearity of the contraction operator. Therefore, $$Γ^α_{uv}$$ is tensorial in $$u$$ and $$α$$, but only affine in $$v$$. Its decomposition is: $$Γ^α_{uv} = \left(∂_u v^ν\right) α_ν + u^μ v^ν Γ^ρ_{μν} α_ρ.$$

In particular, the transform between the version indexed by a coordinate frame and the version indexed by a frame basis $$h^a = h^a_μ dx^μ$$ and its dual basis $$e_b = e_b^ν ∂_ν$$ is given by: $$Γ^c_{ab} = \left(∂_a e_b^ν\right) h^c_ν + e_a^μ e_b^ν Γ^ρ_{μν} h^c_ρ.$$

Things get a little cleaner if you write the connection as a one-form, instead, by defining $${Γ^c}_b = Γ^c_{(dx)b} = dx^μ Γ^c_{μb}.$$ Then we can write: $${Γ^α}_v = d\left(v^ν\right) α_ν + v^ν {Γ^ρ}_ν α_ρ.$$ This shows the affine nature of the connection in the $$v$$ index a bit more cleanly.