Is there a method of knowing which of the Christoffel symbols of second kind survive and vanish for a given metric?

Question

I'm trying to solve a problem and the given $g_{ij}$ metric is $$\left[\begin{array}{cc}1&0&0\\0&(x^1)^2&0\\0&0&(x^1 \sin x^2)^2\end{array}\right]$$

The non-zero Christoffel symbols of the second kind, $\Gamma^{i}_{\;\,jk}$, for this problem happen to be:

$i=1$: $\Gamma^{1}_{\;\,22}$, $\Gamma^{1}_{\;\,33},$

$i=2$: $\Gamma^{2}_{\;\,12} =\Gamma^{2}_{\;\,21}$, $\Gamma^{2}_{\;\,33},$

$i=3$: $\Gamma^{3}_{\;\,13} =\Gamma^{3}_{\;\,31}$, $\Gamma^{3}_{\;\,23} =\Gamma^{3}_{\;\,32}$.

My question is, how can one be sure these are the only non-zero ones? It's difficult to check all the possible combinations of $ijk$, especially if there are time constraints.

Answer 1

Christoffel symbols are given by $\Gamma_{ij}^{k}=\frac 1 2 g^{km}(g_{mj,i}+g_{mi,j}-g_{ij,m})$.

I will give a method for diagonal metrics, as the one you have here. I think that for non-diagonal metrics there is no way to simplify significantly the calculation.

We may first notice that for a diagonal metric, only terms with $k=m$ survive. So we may look at Christoffel symbols as $\Gamma_{ij}^{k}=\frac 1 2 g^{kk}(g_{kj,i}+g_{ki,j}-g_{ij,k})$ which simplifies the work since now there is no summation on $m$ - there is only this one term.

Now, because the metric is diagonal, only the Christoffel symbols with two equal indices survive. That is, with $i=j$, $i=k$ or/and $j=k$. Otherwise, all terms in the parentheses vanish. And even if two indices are equal, say $i$ appears twice in the symbol, we need to check whether $g_{ii}$ is dependent on the third index appearing in the symbol. If not, the symbol vanishes. In particular, if $g_{ii}$ is constant, all Christoffel symbols where $i$ appears twice (or more), vanish.

Sidenote (not really important): we notice that if the metric in independent of some variable, a term containing a derivative with respect to it always vanishes, so these indices appear in the symbol only twice or not at all (check that you understand why). In your example, $3$ indeed appears only in pairs, as the metric is independent of $x^3$.

To summarize: in order not to vanish, the Christoffel symbol needs to have two identical indices, where the matching metric element is dependent on the third. (For example, a Christoffel symbol $\Gamma_{ij}^j$ doesn't vanish if and only if $g_{jj}$ is dependent on $x^{i}$). A method of working this out in an efficient way is to write down un-ordered triples of indices fulfilling this, and then let them appear in the Christoffel symbol in all three possible ways.

This way we can quickly decide which Christoffel symbols vanish and which need a calculation.

For example, let's take your case:

After finding which elements in the metric depend on what, we are left with the triples of indices $(2,2,1)$, $(3,3,1)$ and $(3,3,2)$, (since $g_{11}$ is constant, $g_{22}$ depends only $x^1$ and $g_{33}$ depends on $x^1$ and $x^2$) appearing in any possible order in the symbol.

And indeed, as you got, the only Christoffel symbols that don't vanish in your case are those with these triples of indices: $\Gamma_{22}^{1}$, $\Gamma_{21}^{2}=\Gamma_{12}^{2}$, $\Gamma_{33}^{1}$, $\Gamma_{31}^{3}=\Gamma_{13}^{3}$, $\Gamma_{33}^{2}$ and $\Gamma_{32}^{3}=\Gamma_{23}^{3}$.

Answer 2

I suggest the following method of calculation, which does not require any particular theorem, fits to both diagonal and non-diagonal metrics and can be fully automated. Since it relies on all the possible locations of all the obviously non-zero partial derivatives, it provides immediately all non-zero Christoffel symbols instead of searching for the zero symbols.

Assuming that the Christoffel symbols are given by $\quad \Gamma_{ij}^{k}=\frac 1 2 g^{km}(g_{mj,i}+g_{mi,j}-g_{ij,m})\quad $ :

1. list all the non-zero partial derivatives, beginning with the diagonal metrics ; let n be their number,

2. build a table with n+1 rows and 3 columns and write :

   • in row 1 : column 1 : title "NZPD" for "non-zero partial derivatives" ; column 2 : literal expression of the first term inside parentheses $g_{jm,i}$; a column with the second term $g_{im, j}$ is useless since the permutation of the values of the i,j indices yields equal symbols, so column 3 : literal expression $g_{ij, m}$

   • from row 2 up to n+1 in column 1 : successively all partial derivatives given by step 1.

3. in each of the empty cells give to each of the indices of the expression at the head of the column the values of the indices located at the same place at the head of the row,

4. from the value of m get all the possible values for k from the (known) non-zero $g^{km}$,

5. from the current values of k, i and j get the "names" of all the non-zero Christoffel symbols.

The non diagonal metrics lead to obvious redundancies because of previous occurences of the same upper index k, but note pragmatically that they can easier be understood afterwards than predicted through a somewhat sophisticated specific rule …

Here are the results for the earlier discussed metrics $\quad\left[\begin{array}{cc}1&0&0\\0&(x^1)^2&0\\0&0&(x^1 \sin x^2)^2\end{array}\right]\quad $

The list of the non-zero partial derivatives is obvious : $\quad g_{22,1}\quad g_{33,1}\quad g_{33,21}\quad $ and the table gives :

\begin{array}{c|c} {NZPD} & g_{mj,i} & g_{ij,m} \\\hline g_{22,1} & \begin{cases} i=1, \ j=2\\ m=2\Rightarrow k=2 \Rightarrow {\Gamma^2_{12}=\Gamma^2_{21}}\neq 0 \end{cases} &\begin{cases} i=2, \ j=2\\ m=1\Rightarrow k=1 \Rightarrow {\Gamma^1_{22}}\neq 0 \end{cases} \\ \hline g_{33,1} & \begin{cases} i=1, \ j=3\\ m=3\Rightarrow k=3 \Rightarrow {\Gamma^3_{13}=\Gamma^3_{31}}\neq 0 \end{cases} & \begin{cases}i=3, \ j=3\\ m=1\Rightarrow k=1 \Rightarrow{\Gamma^1_{33}}\neq 0 \end{cases} \\ \hline g_{33,2} & \begin{cases} i=3, \ j=3\\ m=3\Rightarrow k=3 \Rightarrow {\Gamma^3_{23}=\Gamma^3_{32}}\neq 0 \end{cases} & \begin{cases}i=3, \ j=3\\ m=2\Rightarrow k=2 \Rightarrow{\Gamma^2_{33}}\neq 0 \end{cases} \end{array} 9 Christoffel symbols have been found, 6 among them are independant because of lower indices symmetries : $$\Gamma^1_{22}\quad \Gamma^1_{33}\quad \Gamma^2_{12}\quad \Gamma^2_{33}\quad \Gamma^3_{13}\quad \Gamma^3_{23}\quad $$

In case of a non-diagonal metric, several values of $k\neq m$ must be taken into account according to every non zero $g^{km}$ and other non-zero Christoffel symbols occur : here are the results with the new metrics $$\left[\begin{array}{cc}1&0&g_{13}(x^2)\\0&(x^1)^2&0\\g_{31}(x^2)&0&(x^1 \sin x^2)^2\end{array}\right]$$

\begin{array}{c|c} NZPD & g_{mj,i} & g_{ij,m} \\\hline g_{22,1} & \begin{cases} i=1, \ j=2\\ m=2\Rightarrow k=2 \Rightarrow {\Gamma^2_{12}=\Gamma^2_{21}}\neq 0 \end{cases} &\begin{cases} i=2, \ j=2\\ m=1\Rightarrow k=1 \Rightarrow {\Gamma^1_{22}}\neq 0 \end{cases} \\ \hline g_{33,1} & \begin{cases}i=1, \ j=3\\m=3\Rightarrow \begin{cases}{\Gamma^1_{13}}={\Gamma^1_{31}}\neq 0 & \text{for k=1} \\{\Gamma^3_{13}={\Gamma^3_{31}}}\neq 0 & \text{for k=3} \end{cases} \end{cases} & \begin{cases} i=3, \ j=3\\ m=1\Rightarrow k=1 \Rightarrow \Gamma^1_{33}\neq 0 \end{cases} \\ \hline g_{33,2} & \begin{cases}i=2, \ j=3\\m=3\Rightarrow \begin{cases}{\Gamma^1_{23}}={\Gamma^1_{32}}\neq 0 & \text{for k=1} \\{\Gamma^3_{23}={\Gamma^3_{32}}}\neq 0 & \text{for k=3} \end{cases} \end{cases} & \begin{cases} i=3, \ j=3\\ m=2\Rightarrow k=2 \Rightarrow \Gamma^2_{33}\neq 0 \end{cases} \\ \hline g_{13,2} & \begin{cases}i=2, \ j=3 \quad (redundant)\\m=1\Rightarrow \begin{cases}{\Gamma^1_{23}}={\Gamma^1_{32}}\neq 0 & \text{for k=1} \\{\Gamma^3_{23}={\Gamma^3_{32}}}\neq 0 & \text{for k=3} \end{cases} \end{cases} & \begin{cases} i=1, \ j=3\\ m=2\Rightarrow k=2 \Rightarrow {\Gamma^2_{13}=\Gamma^2_{31}}\neq 0 \end{cases} \end{array} 15 non redondant Christoffel symbols have been found, among which 9 are independant because of lower indices symmetries : $$\Gamma^1_{13}\quad \Gamma^1_{22}\quad \Gamma^1_{23}\quad \Gamma^1_{33}\quad \Gamma^2_{12}\quad \Gamma^2_{13}\quad \Gamma^2_{33}\quad \Gamma^3_{13}\quad \Gamma^3_{23}\quad$$

Answer 3

CLARIFICATION: In general, computing Christoffel symbols is quite painful and usually can be avoided. Not only do you have to compute every partial derivative of every component of the metric tensor, you also have to compute the inverse tensor. The following is for a diagonal metric tensor only.

Suppose $g_{ij} = a_i\delta_{ij}$. \begin{align*} \partial_kg_{ij} &= \partial_ka_i\delta_{ij}\\ 2g_{kp}\Gamma^p_{ij} &= \partial_ig_{kj} + \partial_jg_{ik} - \partial_kg_{ij}\\ &= \partial_ia_k\delta_{jk} + \partial_ja_k\delta_{ik} - \partial_ka_i\delta_{ij}\\ 2\Gamma^k_{ij} &= a_k^{-1}(\partial_ia_k\delta_{jk} + \partial_ja_k\delta_{ik} - \partial_ka_i\delta_{ij}). \end{align*} This is zero if $i, j, k$ are different from each other. Since the Christoffel symbol is symmetric in $i$ and $j$, it now suffices to consider the following cases:

If $i = j = k$, then $$ 2\Gamma^i_{ii} = a_i^{-1}\partial_ia_i. $$ If $i= j \ne k$, then $$ 2\Gamma^k_{ii} = -a_k^{-1}\partial_ka_i. $$ If $i \ne j = k$, then $$ 2\Gamma^k_{ik} = a_k^{-1}\partial_ia_k. $$