In my lecture notes I have two definitions of the Christoffel symbols. The first is the smooth functions $\Gamma^k_{ij}: U\subseteq M\to\mathbb{R}$ defined for $i,j,k=1,2$ by

$\Gamma^k_{ij}=\dfrac{1}{2}\Sigma_{l=1,2}(g^{-1})^{lk}(\dfrac{\partial g_{jl}}{\partial u_i}+\dfrac{\partial g_{li}}{\partial u_j}-\dfrac{\partial g_{ij}}{\partial u_l})$

The second as $\nabla_{\frac{\partial}{\partial u_i}}(\dfrac{\partial}{\partial u_j})=\Sigma_k \Gamma^k_{ij}\dfrac{\partial}{\partial u_k}$ where $\nabla$ is the Levi-Civita connection.

I see that the first one is a good way of computing the Christoffel symbols, but I have no idea what it means and how it relates to the second definition. Could anybody try to explain how they are related and what exactly these symbols are for?

  • 1
    $\begingroup$ They're equivalent. If I were to try to prove that they were equivalent, I would write down the Kozsul formula in coordinates. $\endgroup$
    – Neal
    Apr 23, 2012 at 14:04
  • 1
    $\begingroup$ There are several equivalent ways to develop the theory, but one way to look at it is that your first equation defines the Christoffel symbols, and your second one then uses now-defined Christoffel symbols to define the Levi-Civita connection. (One then has to do some work to prove that this definition is invariant under coordinate change and satisfies the nice properties we want the connection to have). $\endgroup$ Apr 23, 2012 at 14:05
  • $\begingroup$ @HenningMakholm: I actually think its much better to use the second as a defintion (since this works for connections in general), and then derive the first equation. $\endgroup$ Apr 23, 2012 at 15:34

2 Answers 2


A connection $\nabla$ on $TM$ can be defined abstractly as a map $\nabla: \mathfrak X(M) \times \mathfrak X(M) \to \mathfrak X(M), (X,Y) \mapsto \nabla_X Y$, where $\mathfrak X(M)$ is the set of vector fields on $X$, such that $\nabla$ is $C^\infty(M)$ linear in the $X$ variable, $\mathbb R$ linear in $Y$, and satisfies the Leibniz rule $$ \nabla_X fY = df(X) Y + f\nabla_X Y $$ for smooth functions $f$. If $\partial_{x_i}$ is a local coordinate chart for $TM$ then it is easy to see that $\nabla$ is completely determined by the vector fields $\nabla_{\partial_{x_i}} \partial_{x_j}$. Then $\Gamma_{ij}^k$ is by definition the coordinate functions of $\nabla_{\partial_{x_i}}\partial_{x_j}$. Note that these coefficients do not piece together to form a tensor since a connection is not tensorial (it is not $C^\infty$ linear in $Y$).

Now, by definition, the Levi-Civita connection is the unique connection on $TM$ that has no torsion and is compatible with the metric. From these properties it can be shown that the Christoffel symbols of this connection are given by the first equation you gave.


Let me give a sketch of the answer in the level of the differential geometry of surfaces in $\mathbb{R}^3$. I follow the framework of W.Klingenberg's "A Course of Differential geometry" where you can find all the necessary details, if you wish.

Consider a (regular) surface $S$ parametrized by a smooth immersion $f:U \rightarrow \mathbb{R}^3$. Vectors $f_i := \operatorname{d}{f}(\partial_i)$ form a basis of the tangent plane to surface $S$ at every point $u \in U$ (visualize this as the tangent plane to $S$ passing through the point $p = f(u)$). We can complete this basis to a basis of the tangent space to $\mathbb{R}^3$ at point $p$ if we add a vector orthogonal to the tangent plane $T_p S$. It is convenient to use the unit normal $n = \frac{f_1 \times f_2}{|f_1 \times f_2|}$. The triple $(f_1,f_2,n)$ is called a local frame along $f$. Notice that we can write any vector field $X$ tangent to $S$ as $X = X^i f_i$ (from now on I use the Einstein summation convention).

Let us write $f_{i j} := \frac{\partial{f}}{\partial{u^i}}$. These are some vectors in $\mathbb{R}^3 \equiv T_p{\mathbb{R}^3}$, and they can be expanded in terms of $(f_1,f_2,n)$ as $$ f_{i j} = \Gamma^k_{i j} f_k + II_{i j} n \tag{1} $$ where $\Gamma^k_{i j}$ and $II_{i j}$ are some smooth functions defined on $U$.

Using the fact that $f_{k} \cdot n = 0$ we can get from (1) that $$ f_{i j} \cdot f_k = \Gamma^l_{i j} f_l \cdot f_k = \Gamma^l_{i j} g_{l k} =: \Gamma_{i j k} $$ Immediately we see that $$ \Gamma^k_{i j} = \Gamma^k_{j i} $$ or $$ \Gamma_{i j k} = \Gamma_{j i k} \tag{2} $$ These can be seen as three equations for six indeterminates $\Gamma_{i j k}$ (recall that $i,j,k = 1,2$) One more equation we obtain by differentiating the first fundamental form as follows $$ g_{i j, k}= \frac{\partial}{\partial u^k} g_{i j} = \frac{\partial}{\partial u^k} (f_i \cdot f_j) = f_{i k} \cdot f_j + f_j \cdot f_{j k} = \Gamma_{i k j} + \Gamma_{j k i} \tag{3} $$ Cyclically permuting the indices $i,j,k$ in (3) we obtain the remaining two equations to close the system which is linear. Solving it we find $$ \Gamma_{i j k} = \frac{1}{2}(g_{i k,j} + g_{j k,i} - g_{i j,k}) $$ or, raising index $k$, $$ \Gamma^k_{i j} = \frac{1}{2}g^{k l}(g_{i l,j} + g_{j l,i} - g_{i j,l}) $$

Now we can observe that $\Gamma^k_{i j} f_k$ behaves as a good derivative operation on the tangent vectors (being "good" we formalize by the notion of connection). In particular, it satisfies the product rule, etc.

It makes now sence to introduce the notation $$ \nabla_{f_i}{f_j} := \Gamma^k_{i j} f_k $$ This operation is called the covariant derivative of vector $f_j$ in the direction of vector $f_i$. It can be extended to arbitrary tangent vector fields by (requiring!) linearity.

(A few remarks on the notation. Notice that we do not distinguish between $\partial_i \equiv \frac{\partial}{\partial u^i}$ and $f_i$ (we identify them!), so $\nabla_{f_i}{f_j}$ is just the same thing as $\nabla_{\frac{\partial}{\partial u^i}}{\frac{\partial}{\partial u^j}}$ which we can even write as $\nabla_i \partial_j$ to get a greater simplicity).

Correction. The operation $\nabla_{f_i}{f_j} := \Gamma^k_{i j} f_k$ is defined on coordinate frame and produces a tangent vector again. To extend this operation onto all tangent vectors and make it to behave as a derivation we need to define it in an appropriate way, that is we require that the product rule holds, and extend by linearity. Of course, we also need to ask that $\nabla_{i}{\phi} = \partial_{i}{\phi}$ for any smooth function $\phi:U \rightarrow \mathbb{R}$

  • $\begingroup$ minor typo: 'Using the fact that 𝑓𝑖𝑗⋅𝑛=0', I think you meant $f_k\cdot n =0$. I learned a lot from your answer, thank you. $\endgroup$
    – usr0192
    May 9, 2023 at 15:19
  • $\begingroup$ @usr0192 Thank you for spotting this typo. $\endgroup$ May 9, 2023 at 22:00

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .