## The Christoffel symbol In curved coordinate systems or spaces, derivatives cannot be computed directly without taking into account the geometry of the coordinates or the space itself. Therefore, we introduce the concept of the covariant derivative to properly reflect this, which leads to the introduction of the Christoffel symbols. More on this will be discussed later. Although the discussion here is limited to Euclidean space, the covariant derivative is a critical concept in Riemannian manifolds. >[!Notice] >The Christoffel symbol is not a tensor. See [[General idea on tensor properties]] ### Expressions of Christoffel symbol As stated above, in curvilinear coordinate systems, the basis varies from one to another. Such variation can be described by the partial derivatives $\frac{\partial \mathbf{Z}_i}{\partial Z_j}$, this is $N^{2}$ vectors and each of the basis elements differentiated with respect to each of the coordinates (scalar). Then each of the $N^{2}$ vectors can be decomposed to the covariant basis $\mathbf{Z}_{k}$. These together makes the Christoffel symbol having $N^{3}$ coefficients and three live index. The expression is $\frac{\partial \mathbf{Z}_i}{\partial Z_j} = \Gamma_{ij}^{k} \mathbf{Z}_k$ Now we check some parts of this expression. The differentiation $\frac{\partial \mathbf{Z}_i}{\partial Z_j}$, if we express the covariant basis with position vector, we have $\frac{\partial \mathbf{Z}_i}{\partial Z_j} = \frac{\partial^2 \mathbf{R}}{\partial Z_j \partial Z^i}$ which implies the symmetric property, namely $\frac{\partial \mathbf{Z}_i}{\partial Z_j} = \frac{\partial \mathbf{Z}_j}{\partial Z^i}$ So for the Christoffel symbol, we have $\Gamma_{ij}^{k} = \Gamma_{ji}^{k}$ > [!Note]- > **Origin of the symmetry $\Gamma^k_{ij} = \Gamma^k_{ji}$** > > The symmetry of the Christoffel symbol in its lower indices is not automatic—it is a defining property of the **Levi-Civita connection**, which is the unique connection that satisfies both: > > 1. **Metric compatibility**: $\nabla_k Z_{ij} = 0$, meaning that lengths and angles are preserved under parallel transport; > 2. **Torsion-free condition**: The connection is symmetric in its lower indices, so $\Gamma^k_{ij} = \Gamma^k_{ji}$. (This is equivalent to the general definition $T(X, Y) = \nabla_X Y - \nabla_Y X - [X, Y] = 0$, with Lie bracket.) > > These two conditions uniquely determine the Levi-Civita connection. The torsion-free requirement specifically enforces the symmetry in $\Gamma^k_{ij}$. In the context of a coordinate basis (where $\mathbf{Z}_j = \partial \mathbf{R} / \partial Z^j$), this symmetry also aligns with the commutativity of second partial derivatives: > $ > \frac{\partial^2 \mathbf{R}}{\partial Z^i \partial Z^j} = \frac{\partial^2 \mathbf{R}}{\partial Z^j \partial Z^i} > $ > thereby reinforcing the symmetric structure of $\Gamma^k_{ij}$ when expressed through derivatives of the metric tensor. > In Euclidean space, this symmetry is always satisfied and often trivial. The explicit expression of $\Gamma_{ij}^{k}$ can be obtained by applying a inner product on both sides of $\frac{\partial \mathbf{Z}_i}{\partial Z_j} = \Gamma_{ij}^{k} \mathbf{Z}_k$ with a $\mathbf{Z}^{l}$. Because $\mathbf{Z}^{l}\cdot \mathbf{Z}_{k}=\delta^{l}_{k}$, we have $\mathbf{Z}^l \cdot \frac{\partial \mathbf{Z}_i}{\partial Z_j}= \Gamma_{ij}^{k} \mathbf{Z}_k \cdot \mathbf{Z}^l = \Gamma_{ij}^{k} \delta_k^l = \Gamma_{ij}^{l}$ Rename the index we have $\Gamma_{ij}^{k} = \mathbf{Z}^k \cdot \frac{\partial \mathbf{Z}_i}{\partial Z_j}$ If we consider the contravariant basis, we can first use integration by parts and rewrite $\Gamma_{ij}^{k} = \mathbf{Z}^k \cdot \frac{\partial \mathbf{Z}_i}{\partial Z_j}$, namely $\Gamma_{ij}^{k} = \frac{\partial (\mathbf{Z}^k \cdot \mathbf{Z}_i)}{\partial Z_j} - \frac{\partial \mathbf{Z}^k}{\partial Z_j} \cdot \mathbf{Z}_i$ Since $\mathbf{Z}^k \cdot \mathbf{Z}_i$ gives a $\delta^{k}_{j}$, which is a constant, we then have the latter part only, $\Gamma^k_{ij} = - \frac{\partial \mathbf{Z}^k}{\partial Z^j} \cdot \mathbf{Z}_i$ Based on this we may then find $\frac{\partial \mathbf{Z}^k}{\partial Z^j} = - \Gamma^k_{ij} \mathbf{Z}^i$ >[!Proof]- >The above expression $\frac{\partial \mathbf{Z}^k}{\partial Z^j} = - \Gamma^k_{ij} \mathbf{Z}^i$ can be done by decomposing $\frac{\partial \mathbf{Z}^k}{\partial Z^j}$ to contravariant basis. This is shown in [[Covariant and contravariant basis, matric tensors#Decomposition with respect to a basis by dot product]]. >$\frac{\partial \mathbf{Z}^k}{\partial Z^j} = \left( \frac{\partial\mathbf{Z}^{k}}{\partial Z^{j}} \cdot \mathbf{Z}_{i}\right) \mathbf{Z}^i$ >Then insert what we have, namely $\Gamma^k_{ij} = - \frac{\partial \mathbf{Z}^k}{\partial Z^j} \cdot \mathbf{Z}_i$ to the expression we just obtained, we can see >$\frac{\partial \mathbf{Z}^k}{\partial Z^j} = \left( \frac{\partial\mathbf{Z}^{k}}{\partial Z^{j}} \cdot \mathbf{Z}_{i}\right) \mathbf{Z}^i=-\Gamma^k_{ij} \mathbf{Z}^i$ >[!Notice] >Besides the different sign. The covariant form >$\frac{\partial \mathbf{Z}_i}{\partial Z_j} = \Gamma_{ij}^{k} \mathbf{Z}_k$ >and contravariant form >$\frac{\partial \mathbf{Z}^k}{\partial Z^j} = - \Gamma^k_{ij} \mathbf{Z}^i$ >have the free index (that is commutative) at different position. The previous one is two lower indices and the latter one is a lower and one upper index. **The Christoffel symbol is intrinsic**. As stated in [[Covariant and contravariant basis, matric tensors#Metric tensor and measuring lengths]], this means that it can be expressed with the metric tensors. $\Gamma_{ij}^{k} = \frac{1}{2} Z^{km} \left( \frac{\partial Z_{mi}}{\partial Z_j} + \frac{\partial Z_{mj}}{\partial Z^i} - \frac{\partial Z_{ij}}{\partial Z_m} \right)$ Our previous definition expressed it extrinsically with basis, see the derivation below. In Riemannian geometry, the expression of the Christoffel symbol is this intrinsic form. ^657e8f >[!Proof]- >This is exercise 78, 79, 80 in the book [[Introduction to Tensor Analysis and the Calculus of Moving Surfaces]]. >First we show that >$\frac{\partial Z_{ij}}{\partial Z^{k}} = Z_{li} \Gamma^{l}_{jk} + Z_{lj} \Gamma^{l}_{ik}$ >Rewrite $Z_{ij}$ as covariant basis $\mathbf{Z}_{i}\cdot \mathbf{Z}_{j}$, and rewrite the differentiation as the Christoffel symbol >$\begin{aligned} >\frac{\partial Z_{ij}}{\partial Z^{k}} &= \frac{\partial (\mathbf{Z}_{i}\cdot \mathbf{Z}_{j})}{\partial Z^{k}} \\ &= \frac{\partial \mathbf{Z}_{i}}{\partial Z^{k}}\cdot \mathbf{Z}_{j} + \frac{\partial \mathbf{Z}_{j}}{\partial Z^{k}}\cdot \mathbf{Z}_{i} \\ >&= \Gamma_{ik}^{l} \mathbf{Z}_{l} \cdot \mathbf{Z}_{j} + \Gamma_{jk}^{l} \mathbf{Z}_{i} \cdot \mathbf{Z}_{l} \\ >&= Z_{li} \Gamma^{l}_{jk} + Z_{lj} \Gamma^{l}_{ik} >\end{aligned}$ >Here the index has to be $l$ for both expression because the summation is performed as a whole. >We may denote $Z_{li} \Gamma^{l}_{jk}$ as $\Gamma_{i,jk}$ using [[Index juggling]]. So we have above expression being >$\frac{\partial Z_{ij}}{\partial Z^{k}} = \Gamma_{i,jk} + \Gamma_{j,ik}$ >Now check our target expression, the right side is >$\begin{aligned} >\frac{1}{2} Z^{km} \left( \frac{\partial Z_{mi}}{\partial Z_j} + \frac{\partial Z_{mj}}{\partial Z^i} - \frac{\partial Z_{ij}}{\partial Z_m} \right) &= \frac{1}{2} Z^{km} \left( \Gamma_{m,ij} +\Gamma_{i, mj} + \Gamma_{m,ji} +\Gamma_{j,mi} - (\Gamma_{i,jm}+\Gamma_{j, im}) \right)\\ >&=\frac{1}{2} Z^{km} \left( \Gamma_{m,ij} + \Gamma_{m,ji} \right) \\ >&=\frac{2}{2} Z^{km} \Gamma_{m,ij} \\ >&=Z^{km} \Gamma_{m,ij}\\ >&= Z^{km}Z_{lm} \Gamma_{l,ij}\\ >&= \delta^{k}_{l} \Gamma_{ij}^{l}\\ >&= \Gamma_{ij}^{k} >\end{aligned}$ >So we get the left side, the Christoffel symbol and proved that it is intrinsic. The symmetry property of the Christoffel symbol remains, so $\Gamma_{i,jm}=\Gamma_{i,mj}$. >[!Note]- >A little more flavor on physics. These are exercise 81-84. >For a particle moving along a curve with parametrization with respect to time, and $Z^i \equiv Z^i(t)$. Suppose velocity is $\mathbf{V}(t)$, we can show that the component $V^{i}(t)$ of $\mathbf{V}(t)$ is given by >$V^i(t) = \frac{\mathrm{d} Z^i(t)}{\mathrm{d}t}$ > $\begin{aligned} > V^i(t) &= \mathbf{V} \cdot \mathbf{Z}^{i}\\ > &= \frac{\mathrm{d}\mathbf{R}(t)}{\mathrm{d}Z^{j}}\cdot \mathbf{Z}^{i}\\ > &=\left( \frac{\mathrm{d}\mathbf{R}(t)}{\mathrm{d}{Z}^{j}} \frac{\mathrm{d}{Z}^{j}}{\mathrm{d}t} \right) \cdot \mathbf{Z}^{i}\\ > &=\mathbf{Z}_{j} \frac{\mathrm{d}Z^{j}}{\mathrm{d}t}\cdot \mathbf{Z}^{i}\\ > &=\frac{\mathrm{d}Z^{j}}{\mathrm{d}t} \delta^{i}_{j}\\ > &= \frac{\mathrm{d} Z^i(t)}{\mathrm{d}t} > \end{aligned}$ > This can also be done for acceleration $\mathbf{A}(t)$, the components $A^{i}(t)$ are > $A^i = \frac{\mathrm{d}V^i}{\mathrm{d}t} + \Gamma^i_{jk} V^j V^k$The proof can be first expansion then calculate step by step, or, > $\begin{aligned} > A^{i}(t) &= \mathbf{A}\cdot \mathbf{Z}^{i}\\ > &=\frac{\mathrm{d}\mathbf{V}}{\mathrm{d}t}\cdot \mathbf{Z}^{i}\\ > &=\frac{\mathrm{d}(\mathbf{V}\cdot \mathbf{Z}^{i})}{\mathrm{d}t}-\frac{\mathrm{d}\mathbf{Z}^{i}}{\mathrm{d}t}\cdot \mathbf{V}\\ > &=\frac{\mathrm{d}V^{i}}{dt}-\frac{\mathrm{d}\mathbf{Z}^{i}}{\mathrm{d}Z^{j}} \frac{\mathrm{d}Z^{j}}{\mathrm{d}t} \cdot \mathbf{V}\\ > &=\frac{\mathrm{d}V^{i}}{dt} + \Gamma_{jk}^{i} \frac{\mathrm{d}Z^{j}}{\mathrm{d}t} \mathbf{Z}^{k} \cdot \mathbf{V}\\ > &=\frac{\mathrm{d}V^{i}}{dt} + \Gamma_{jk}^{i} V^{j} \mathbf{Z}^{k} \cdot \mathbf{V}\\ > &=\frac{\mathrm{d}V^{i}}{dt} + \Gamma_{jk}^{i} V^{j} V^{k}\\ > \end{aligned}$ > Here we applied previous result, $\frac{\mathrm{d}Z^{j}}{\mathrm{d}t}=V^{j}$, and replaced $-\frac{\mathrm{d}\mathbf{Z}^{i}}{\mathrm{d}Z^{j}}$ with $\Gamma_{jk}^{i} \mathbf{Z}^{k}$. > Similarly, we can do this further and show the derivative of acceleration, $\mathbf{B}(t) = \mathbf{A}'(t)$, satisfies > $B^i = \frac{\mathrm{d}A^i}{\mathrm{d}t} + \Gamma^i_{jk} A^j V^k$ > With the similar approach, > $\begin{aligned} > B^{i}(t) &= \mathbf{B}\cdot \mathbf{Z}^{i}\\ > &=\frac{\mathrm{d}\mathbf{A}}{\mathrm{d}t}\cdot \mathbf{Z}^{i}\\ > &=\frac{\mathrm{d}(\mathbf{A}\cdot \mathbf{Z}^{i})}{\mathrm{d}t}-\frac{\mathrm{d}\mathbf{Z}^{i}}{\mathrm{d}t}\cdot \mathbf{A}\\ > &=\frac{\mathrm{d}A^{i}}{dt}-\frac{\mathrm{d}\mathbf{Z}^{i}}{\mathrm{d}Z^{j}} \frac{\mathrm{d}Z^{j}}{\mathrm{d}t} \cdot \mathbf{A}\\ > &=\frac{\mathrm{d}A^{i}}{dt} + \Gamma_{jk}^{i} \frac{\mathrm{d}Z^{j}}{\mathrm{d}t} \mathbf{Z}^{k} \cdot \mathbf{A}\\ > &=\frac{\mathrm{d}A^{i}}{dt} + \Gamma_{jk}^{i} V^{j} \mathbf{Z}^{k} \cdot \mathbf{A}\\ > &=\frac{\mathrm{d}A^{i}}{dt} + \Gamma_{jk}^{i} V^{j} a^{k}\\ > \end{aligned}$ > And the general conclusion would be for vector field $\mathbf{U}(t)$ along $Z^i \equiv Z^i(t)$, we have > $\mathbf{U}'(t) = \left( \frac{\mathrm{d}U^i}{\mathrm{d}t} + V^j \Gamma^i_{jk} U^k \right) \mathbf{Z}_i$ > This is equivalent to providing $U'^{i}$ expression, since $\mathbf{U}'=U'^{i}\mathbf{Z}_{i}$, like before, > $\begin{aligned} > U'^{i}(t) &= \frac{\mathrm{d}\mathbf{U}}{\mathrm{d}t}\cdot \mathbf{Z}^{i}\\ > &=\frac{\mathrm{d}(\mathbf{U}\cdot \mathbf{Z}^{i})}{\mathrm{d}t}-\frac{\mathrm{d}\mathbf{Z}^{i}}{\mathrm{d}t}\cdot \mathbf{U}\\ > &=\frac{\mathrm{d}U^{i}}{dt}-\frac{\mathrm{d}\mathbf{Z}^{i}}{\mathrm{d}Z^{j}} \frac{\mathrm{d}Z^{j}}{\mathrm{d}t} \cdot \mathbf{U}\\ > &=\frac{\mathrm{d}U^{i}}{dt} + \Gamma_{jk}^{i} \frac{\mathrm{d}Z^{j}}{\mathrm{d}t} \mathbf{Z}^{k} \cdot \mathbf{U}\\ > &=\frac{\mathrm{d}U^{i}}{dt} + \Gamma_{jk}^{i} V^{j} \mathbf{Z}^{k} \cdot \mathbf{U}\\ > &=\frac{\mathrm{d}U^{i}}{dt} + \Gamma_{jk}^{i} V^{j} U^{k}\\ > \end{aligned}$ > Therefore, if $\mathbf{U}(t)$ is a constant along the curve, we have > $\frac{\mathrm{d}U^i}{\mathrm{d}t} + V^j \Gamma^i_{jk} U^k = 0$ ^876ee2 >[!Notice] >The Christoffel symbol has three indices, the order of index is important (for example, if one wanna assign values). According to our convention, the upper index is the first, while the lower two are the second and third. >For some symbols, like Kronecker delta $\delta_{j}^{i}$, the order does not matter. But for others it is critical to maintain some order by certain convention. One approach is called the dot approach, a dot is added to indicate the index being not the first. For example $A^{i}_{\cdot j}$ for upper index being the first, and $A^{\cdot i}_{j}$ for the lower index being the first. This will be further discussed in [[Index juggling]]. ### Christoffel symbol in coordinates #### Cartesian and affine coordinates $\Gamma^i_{jk} = 0$ It vanishes at all points, this means the covariant basis (as well as the metric tensor) does not vary when changing points. #### Cylindrical coordinates $\Gamma^1_{22} = -r$ $\Gamma^2_{12} = \Gamma^2_{21} = \frac{1}{r}$ #### Spherical coordinates $R=1$, $\Theta=2$, $\Phi=3$, non-zero Christoffel symbols are $\begin{aligned} \Gamma^R_{\Theta \Theta} &= -r \\ \Gamma^R_{\Phi \Phi} &= -r \sin^2 \theta \\ \Gamma^\Theta_{R \Theta} &= \Gamma^\Theta_{\Theta R} = \frac{1}{r} \\ \Gamma^\Theta_{\Phi \Phi} &= - \sin \theta \cos \theta \\ \Gamma^\Phi_{R \Phi} &= \Gamma^\Phi_{\Phi R} = \frac{1}{r} \\ \Gamma^\Phi_{\Theta \Phi} &= \Gamma^\Phi_{\Phi \Theta} = \cot \theta \end{aligned}$