How to prove the covariant derivative cannot be written as an eigendecomposition of the partial derivative?

Question

The Question

How does one prove that Rindler's definition of the covariant derivative of a covariant vector field $\lambda_a$ as \begin{align} \lambda_{a;c} = \lambda_{a,c} - \Gamma^{b}_{\ \ ca} \lambda_{b} \tag{Rindler} \end{align} cannot be written in terms of an eigendecomposition of the partial derivative such as \begin{align} \lambda_{a;c} = \dfrac{\partial y_d}{\partial x_a} \dfrac{\partial }{\partial x_c} \left( \dfrac{\partial x_b}{\partial y_d} \lambda_{b} \right) \tag{linuxfreebird} ? \end{align}

The Definitions

The following symbols are defined by Rindler and interpreted by linuxfreebird:

\begin{align} \Gamma^{b}_{\ \ ca} = g^{bk} \Gamma_{k ca} \tag{Christoffel symbol} \end{align}

are the Christoffel symbols and

\begin{align} \tag{inverse metric tensor} g^{bd} = \dfrac{\partial x_b}{\partial y_k} \dfrac{\partial x_d}{\partial y_k} \end{align}

is the inverse of the metric tensor $g_{bd}$.

The eigendecomposition of a partial derivative is analogous to eigendecomposition of matrix operators. Given a matrix operator $A$, one can express $A$ in terms of its eigenvalue diagonal matrix $\Lambda$ and its eigenvector matrix $V$ such that $A = V\Lambda V^{-1}$.

The Reason for the Question

Taken from Rindler, the Riemann tensor $R^{d}_{\ \ abc}$ is defined using the commutator of the covariant derivative from Eq. (Rindler) as

\begin{align} \lambda_{a;bc} - \lambda_{a;cb} = R^{d}_{\ \ abc} \lambda_{d} \tag{Riemman tensor}. \end{align}

However, if one attempts to calculate the Riemman tensor from Eq. (Riemman tensor) by using the definition of the covariant derivative from Eq. (linuxfreebird), one obtains the following result

\begin{align} \lambda_{a;bc} &=& \dfrac{\partial y_n}{\partial x_a} \dfrac{\partial }{\partial x_b} \left( \dfrac{\partial x_m}{\partial y_n} \dfrac{\partial y_f}{\partial x_m} \dfrac{\partial }{\partial x_c} \left( \dfrac{\partial x_d}{\partial y_f} \lambda_{d} \right) \right) \\&=& \dfrac{\partial y_n}{\partial x_a} \dfrac{\partial^2 }{\partial x_b \partial x_c} \left( \dfrac{\partial x_d}{\partial y_n} \lambda_{d} \right) , \end{align}

which states that the commutator of the covariant derivative is zero, hence the Riemman tensor is zero. This was the inconsistency I discovered from Eq. (linuxfreebird) and proves Eq. (linuxfreebird) cannot be an equivalent representation of Eq. (Rindler). However, I have been unable to prove why Eq. (linuxfreebird) is incorrect. Section (The Supporting Material) provides my formulation of Eq. (linuxfreebird).

The Supporting Material

Below is my original formulation of Eq. (linuxfreebird):

Rindler defines the Christoffel symbol of the first kind as \begin{align} \tag{12} \Gamma_{b ca} = \dfrac{1}{2} \left( \dfrac{\partial g_{bc}}{\partial x_{a}} + \dfrac{\partial g_{ab}}{\partial x_{c}} - \dfrac{\partial g_{ca}}{\partial x_{b}} \right) , \end{align} where \begin{align} \tag{13} g_{ab} = \dfrac{\partial y_k}{\partial x_{a}} \dfrac{\partial y_k}{\partial x_{b}} \end{align} is the metric tensor. The partial derivative of $g_{ab}$ is \begin{align} \tag{14} \dfrac{\partial g_{ab}}{\partial x_{c}} = A_{acb} + A_{bca} , \end{align} where \begin{align} \tag{15} A_{acb} = \dfrac{\partial y_k}{\partial x_{a}} \dfrac{\partial^2 y_k}{\partial x_{c}\partial x_{b}} . \end{align} One can express $\Gamma_{bca}$ in terms of $A_{abc}$ as \begin{align} \tag{16} \Gamma_{b ca} = \dfrac{1}{2} \left( A_{bac} + A_{cab} + A_{acb} + A_{bca} - A_{cba} - A_{abc} \right) = A_{bac} , \end{align} which proves that \begin{align} \tag{17} \Gamma_{b ca} = \dfrac{\partial y_k}{\partial x_{b}} \dfrac{\partial^2 y_k}{\partial x_{c}\partial x_{a}} . \end{align} Rindler defines the Christoffel symbol of the second kind as \begin{align} \tag{18} \Gamma^{b}_{\ ca} = g^{bk} \Gamma_{k ca} , \end{align} where \begin{align} \tag{19} g^{ab} = \dfrac{\partial x_{a}}{\partial y_k} \dfrac{\partial x_{b}}{\partial y_k} \end{align} is the inverse of $g_{ab}$. One can express $\Gamma^{b}_{\ ca}$ in terms of the Jacobian components $\partial x_{a}/\partial y_{k}$ and prove the following equivalent expressions \begin{align} \tag{20} \Gamma^{b}_{\ ca} &=& \left( \dfrac{\partial x_{b}}{\partial y_n} \dfrac{\partial x_{k}}{\partial y_n} \right) \left( \dfrac{\partial y_m}{\partial x_{k}} \dfrac{\partial^2 y_m}{\partial x_{c}\partial x_{a}} \right) \\ &=& \tag{21} \dfrac{\partial x_{b}}{\partial y_n} \left( \dfrac{\partial x_{k}}{\partial y_n} \dfrac{\partial y_m}{\partial x_{k}} \right) \dfrac{\partial^2 y_m}{\partial x_{c}\partial x_{a}} \\ &=& \tag{22} \dfrac{\partial x_{b}}{\partial y_n} \left( \delta_{mn} \right) \dfrac{\partial^2 y_m}{\partial x_{c}\partial x_{a}} \\ &=& \tag{23} \dfrac{\partial x_{b}}{\partial y_m} \dfrac{\partial }{\partial x_{c}} \left( \dfrac{\partial y_m}{\partial x_{a}} \right) \\ &=& \tag{24} \dfrac{\partial }{\partial x_{c}} \left( \dfrac{\partial x_{b}}{\partial y_m} \dfrac{\partial y_m}{\partial x_{a}} \right) - \dfrac{\partial y_m}{\partial x_{a}} \dfrac{\partial }{\partial x_{c}} \left( \dfrac{\partial x_{b}}{\partial y_m} \right) \\ &=& \tag{25} \dfrac{\partial }{\partial x_{c}} \left( \delta_{ab} \right) - \dfrac{\partial y_m}{\partial x_{a}} \dfrac{\partial }{\partial x_{c}} \left( \dfrac{\partial x_{b}}{\partial y_m} \right) \\ &=& \tag{26} - \dfrac{\partial y_m}{\partial x_{a}} \dfrac{\partial }{\partial x_{c}} \left( \dfrac{\partial x_{b}}{\partial y_m} \right) . \end{align} Rindler defines the covariant derivative of a covariant vector field $\lambda_{a}$ as \begin{align} \tag{27} \lambda_{a;c} = \lambda_{a,c} - \Gamma^{b}_{\ ca} \lambda_{b} . \end{align} One can define the covariant derivative in terms of $\partial x_{a}/\partial y_{k}$ as \begin{align} \tag{28} \lambda_{a;c} &=& \dfrac{\partial \lambda_{a} }{\partial x_{c}} + \dfrac{\partial y_m}{\partial x_{a}} \dfrac{\partial }{\partial x_{c}} \left( \dfrac{\partial x_{b}}{\partial y_m} \right) \lambda_{b} \\&=& \tag{29} (\delta_{ab}) \dfrac{\partial \lambda_{b} }{\partial x_{c}} + \dfrac{\partial y_m}{\partial x_{a}} \dfrac{\partial }{\partial x_{c}} \left( \dfrac{\partial x_{b}}{\partial y_m} \right) \lambda_{b} \\&=& \tag{30} \left( \dfrac{\partial y_m}{\partial x_{a}} \dfrac{\partial x_{b}}{\partial y_m} \right) \dfrac{\partial \lambda_{b} }{\partial x_{c}} + \dfrac{\partial y_m}{\partial x_{a}} \dfrac{\partial }{\partial x_{c}} \left( \dfrac{\partial x_{b}}{\partial y_m} \right) \lambda_{b} \\&=& \tag{31} \dfrac{\partial y_m}{\partial x_{a}} \left( \dfrac{\partial x_{b}}{\partial y_m} \dfrac{\partial (\lambda_{b} ) }{\partial x_{c}} + \dfrac{\partial }{\partial x_{c}} \left( \dfrac{\partial x_{b}}{\partial y_m} \right) \lambda_{b} \right) \\&=& \tag{32} \dfrac{\partial y_m}{\partial x_{a}} \dfrac{\partial }{\partial x_{c}} \left( \dfrac{\partial x_{b}}{\partial y_m} \lambda_{b} \right) . \end{align}

Answer 1

Let there be given a manifold $(M,\nabla)$ equipped with a (not necessarily torsionfree) tangent bundle connection $\nabla$.

I got the (possibly faulty) impression from reading the first lines in OP's question formulation (v18) that OP is asking:

Is it possible that the local coordinate expression for the covariant derivative of a co-vector/one-form $\lambda =\lambda_a \mathrm{d}x^a$ (in some local coordinate system $x^a$) could be on the form $$\tag{1} (\nabla_c \lambda)_a~=~\frac{\partial y^d}{\partial x^a} \frac{\partial }{\partial x^c}\left(\frac{\partial x^b}{\partial y^d} \lambda_b \right) $$ for some (invertible, smooth, locally defined) functions $y^d=y^d(x)$?

This is equivalent to asking:

Is it possible that the Christoffel symbols$^1$ (in some local coordinate system $x^a$) could be of the form $$\tag{2} \Gamma^{(x)b}_{ca}~=~\frac{\partial^2 y^d}{\partial x^c\partial x^a} \frac{\partial x^b}{\partial y^d} $$ for some (invertible smooth, smooth, locally defined) functions $y^d=y^d(x)$?

Recall that the Christoffel symbol $\Gamma^{\lambda}_{\mu\nu}$ does not transform as a tensor under a local coordinate transformation $x^a \to y^b=y^b(x)$, but rather with an inhomogeneous term, which is built from the second derivative of the coordinate transformation

$$\tag{3}\frac{\partial y^f}{\partial x^c} \Gamma^{(x)c}_{ab} ~=~\frac{\partial y^d}{\partial x^a}\, \frac{\partial y^e}{\partial x^b}\, \Gamma^{(y)f}_{de} +\frac{\partial^2 y^f}{\partial x^a \partial x^b}. $$

Therefore formulation (2) is equivalent of asking:

Does there exists a local coordinate system $y^d$ such the Christoffel symbols vanish $$\tag{4} \Gamma^{(y)c}_{ab}~=~0? $$

This, in turn, is equivalent to asking

Is $(M,\nabla)$ flat?

This is equivalent to asking

Does there exist a local coordinate system $y^d$ such that the metric tensor $$\tag{5} g^{(y)}_{ab}~=~\eta_{ab} $$ is constant?

This is equivalent to asking

In an arbitrary coordinate system $x^a$, is the metric tensor of the form $$\tag{6} g^{(x)}_{ab}~=~\frac{\partial y^c}{\partial x^a}\eta_{cd} \frac{\partial y^d}{\partial x^b} $$ for some (invertible smooth, smooth, locally defined) functions $y^d=y^d(x)$?

Rereading OP's question formulation, condition (6) seems to be one of OP's assumptions. If we are granted assumption (6), then OP's initial claim (1) holds.

--

$^1$ It is convenient to call $\Gamma^c_{ab}$ Christoffel symbols even if the tangent-space connection $\nabla$ is not torsionfree.

Answer 2

Equation (13) expresses the metric on an embedded hypersurface given by the relations $y^k = y^k(x^a)$. However, the equation for the inverse metric (4-th equation) is in general not correct.

Take for example a hypersurface defined by: $y^1 = x^1$, $y^2 = x^2$, $ y^3 = x^2$. In our case, the partial derivative of $x^2$ with respect to $y^2$ or $y^3$ cannot be defined. For example, in order to differentiate with respect to $y^2$ we must vary $y^2$ while keeping $y^3$ constant. This is impossible to do on the hypersurface.

Update: This is a detailed answer to linuxfreebird comment.

Although I did not receive my education from Rindler's book, the material of the subject of this question appears in many pedagogical introductions of general relativity, in which an $N$ dimensional space time $X$ is isometrically embedded in a larger flat $M$-dimensional flat space $Y$:

$$\begin{matrix} X \rightarrow Y \\ y^k = y^k(x^1, ...., x^N), k = 1, ..., M \end{matrix}$$

The embedding is given as a set of equations defining $X$ as a hypersurface into $Y$. The functions need not be invertible as we want them to describe embeddings of spaces of different dimensions.

Equation (4) is just the relation expressing that the embedding is isometric. This can be seen more explicitly if we write this equation in the form:

$$ g_{a b} =\eta_{kl} \frac{\partial y^k }{\partial x^a} \frac{\partial y^l}{\partial x^b}$$

Where $\eta$ is the flat metric on the manifold $Y$. Now, this equation looks as a metric transformation, i.e., it preserves distances.

This approach is called extrinsic geometry. It can be used to define the various geometrical objects of the space-time $X$ (such as the Levi-Civita connection and the Riemann curvature) in terms of the embedding equations. This approach is correct, because for any space-time $X$ there exists such an embedding for some high enough $M$ (This is called Nash embedding theorem), but it is seldom used in general relativity real work. Most work in general relativity uses intrinsic geometry without resorting to any embedding equations.

The key point is to remember that the $y$s are functions of the $x$s and every time a partial derivative of $x$ with respect to $y$ is used, means that these equations were inverted, but this cannot be done in general, except in the particular case where these equations are invertible.

Now, in the case when the equations are invertible (it necessarily implies that $ M=N$ ), everything is correct, but this case refers only to space-times that can be isometrically embedded into a flat space-time of the same dimension. These space-times are necessarily flat, thus not general enough to cover interesting cases in general relativity.

May be you can profit from this analysis if you continue your exercise and prove that the Riemann tensor computed from your form of the Levi-Civita connection is globally identically zero.

Answer 3

Special thanks to Qmechanic and David Bar Moshe for providing their answers for this post. Combining Qmechanic's and David Bar Moshe's insights, I finally figured out the confusing element in the problem.

The covariant derivative can be written in terms of an eigendecomposition of the partial derivative, the Christoffel symbols can be written in terms of non-invertible extensive variables, and the Riemman tensor is not necessarily equal to zero. Here is my answer:

I have written the covariant derivative \begin{align} \lambda_{a;c} = \dfrac{\partial y^k}{\partial x^a} \dfrac{\partial }{\partial x^c} \left( \dfrac{\partial y_k}{\partial x^p} g^{pb} \lambda_{b} \right) \end{align} in terms of its eigendecomposition and indices "a" and "c" and again \begin{align} \lambda_{b;d} = \dfrac{\partial y^j}{\partial x^b} \dfrac{\partial }{\partial x^d} \left( \dfrac{\partial y_j}{\partial x^q} g^{qn} \lambda_{n} \right) \end{align} with different indices "b" and "d". Combining the covariant derivatives forms the following: \begin{align} \lambda_{a;cd} =\\ \dfrac{\partial y^k}{\partial x^a} \dfrac{\partial }{\partial x^c} \left( \Lambda^{\ \ j}_{k} \dfrac{\partial }{\partial x^d} \left( \dfrac{\partial y_j}{\partial x^q} g^{qn} \lambda_{n} \right) \right) \end{align} where \begin{align} \Delta^{\ \ j}_{k} = \dfrac{\partial y_k}{\partial x^p} g^{pb} \dfrac{\partial y^j}{\partial x^b} \neq \delta_{k}^{\ \ j} \end{align} because \begin{align} g^{pb} \neq \dfrac{\partial x^p}{\partial y^n} \dfrac{\partial x^b}{\partial y_n} \end{align}

is ill-defined.

The Riemann tensor $R_{acd}^{ \ \ \ \ m} $ is created by computing the commutator of the covariant derviatives as follows:

\begin{align} \lambda_{a;[c,d]} = \left( \dfrac{\partial y^k}{\partial x^a} \dfrac{\partial \Delta^{\ \ j}_{k} }{\partial x^c} \dfrac{\partial }{\partial x^d} \left( \dfrac{\partial y_j}{\partial x^p} g^{pm} \right) \right)_{[c,d]} = R_{acd}^{ \ \ \ \ m} \lambda_m . \end{align}

If the number of dimensions of $y$ equals the number of dimensions in $x$, then $\Delta^{\ \ j}_{k} = \delta^{\ \ j}_{k}$, thus $R_{acd}^{ \ \ \ \ m} = 0$.