# Question in transpositioning with Einstein summation notation

In the process of getting the formula of Christoffel symbols in the terms of metric tensors, we get

$$\partial_n g_{rm}+\partial_m g_{rn}-\partial_r g_{mn}=2\Gamma ^t_{mn} g_{rt}.$$

The book I have read goes ' multiply both sides of the equation by the inverse metric tensor' and yields

$$\Gamma ^t_{mn}={1\over 2} g^{rt}(\partial_n g_{rm}+\partial_m g_{rn}-\partial_r g_{mn}).$$

But I am confused with dummy variables. In the first equation, r was the real variable and t was a dummy variable, but in the second one, the opposite seems to be true. If we write the right hand side of the first equation without summation, it'll be

$$2(\Gamma ^0_{mn} g_{r0}+\Gamma ^1_{mn} g_{r1}+\Gamma ^2_{mn} g_{r2}+\Gamma ^3_{mn} g_{r3})$$

I will multiply this with $$g^{rt}$$. In this case it's for t=0, 1, 2, 3, so the left hand side is $$(\Gamma ^0_{mn} g_{r0}+\Gamma ^1_{mn} g_{r1}+\Gamma ^2_{mn} g_{r2}+\Gamma ^3_{mn} g_{r3})(g^{r0}+g^{r1}+g^{r2}+g^{r3}) \\ = \Gamma ^0_{mn}+\Gamma ^1_{mn}+\Gamma ^2_{mn}+\Gamma ^3_{mn}$$ since $$g_{rm}g^{rn}=\delta ^n_m$$. (Let's divide both sides with 2)

The left hand side will be $${1\over 2} (g^{r0}+g^{r1}+g^{r2}+g^{r3})(\partial_n g_{rm}+\partial_m g_{rn}-\partial_r g_{mn})$$ and it is not a summation with r because r is not a dummy variable.

So I don't know the way to get to the right result from the last equation. And how the real and dummy variables changed during the calculation.

• Because t is dummy variable you can change it with any letter you want, for example change t in first equation with an other letter like f and try solution Feb 27 at 14:55

No, it is not working like that.

As one disposes of ( as one has 3 different indices r, m, n which are not contracted of which each runs from 0 to 3 -- so 4 different values) $$4\times 4\times 4$$ equations (some of these might be identical due to symmetry relations) one can combines these equations upon multiplication with the metric tensor $$g^{rs}$$ --- i.e. the multiplication with the metric tensor is not just multiplication but also implicates summation:

$$g^{rs}(\partial_n g_{rm} + \partial_m g_{rn} -\partial_r g_{nm})= 2g^{rs}g_{rt}\Gamma^t_{mn}$$

which actually means: $$\sum\limits_{r=0,\ldots, 3} g^{rs}(\partial_n g_{rm} + \partial_m g_{rn} -\partial_r g_{nm})=\sum\limits_{r=0,\ldots, 3} 2 g^{rs}g_{rt} \Gamma^t_{mn}$$

This means that one takes from the $$4\times 4\times 4$$ equations $$4\times 4$$ eqns (for m and n) but with fixed $$r=0$$ multiplying it with $$g^{0s}$$, adds it to the next set of equations for $$r=1$$ multiplying that one with $$g^{1s}$$ and so forth, i.e. also adds the set of eqns for $$r=2$$ weighted by $$g^{2s}$$ and for $$r=3$$ weighted by $$g^{3s}$$. And of course one does it for each $$r$$ on LHS and RHS.

Then one gets: $$\sum\limits_{r=0,\ldots, 3} g^{rs}\frac{1}{2}(\partial_n g_{rm} + \partial_m g_{rn} -\partial_r g_{nm})= \delta^s_t\Gamma^t_{mn} = \Gamma^s_{mn}$$

because

$$\sum\limits_{r=0,\ldots, 3} g^{rs}g_{rt} = \delta^s_t$$

where $$\delta^s_t$$ is the Kronecker symbol. Or applying Einstein's summation convention the summation symbol can be omitted.

$$g^{rs}\frac{1}{2}(\partial_n g_{rm} + \partial_m g_{rn} -\partial_r g_{nm})= \delta^s_t\Gamma^t_{mn} = \Gamma^s_{mn}$$

As the very last step one can replace the index $$s$$ by $$t$$ again. But the name of an index can be chosen (almost) arbitrarily --- better do not use indices which are already used in the equation.

Act on your first equation with $$g^{ar}$$. Then the action on the right-hand side will produce $$2\Gamma^t_{mn} \delta^a_t$$. Contracting with the Kronecker delta and rearranging yields: $$\Gamma^a_{mn} = \frac{1}{2} g^{ar} \left( \partial_n g_{rm} + \partial_m g_{rn} - \partial_r g_{mn}\right)$$ This is identical to the expression in your second line (with the free index $$t$$ replaced with $$a$$, and $$g^{ab} = g^{ba}$$).

You are right about what is free and what is dummy. When you contract with the "inverse metric" you should use a new free index, $$a$$, and the old free index $$r$$. The book then, confusingly, renames the new free index $$t$$ again, even though it was summed over in the calculation.