Double Sums in Einstein Notation

Question

Suppose that you are given a metric $g_{\mu\nu}$ s.t. the metric is defined as $$ds^2=g_{\mu\nu}dx^{\mu}dx^{\nu}=-e^{2\phi}dt^2+\left(1-\frac{b}{r}\right)^{-1}dr^2+r^2(d\theta^2+\sin^2(\theta)d\phi^2).$$ Given this metric we can calculate the Ricci tensor $R_{\mu\nu}$. The equation for the ricci tensor follows as $$R_{\mu\nu}=\partial_\lambda\Gamma^{\lambda}_{\mu\nu}-\partial_\mu\Gamma^{\lambda}_{\nu\lambda}+\Gamma^{\lambda}_{\lambda\sigma}\Gamma^{\sigma}_{\mu\nu}-\Gamma^{\lambda}_{\mu\sigma}\Gamma^{\sigma}_{\nu\lambda},$$ notice that $\lambda$ and $\sigma$ are dummy indices and are being contracted, while $\mu$ and $\nu$ are our only free indices. Since we are trying to calculate $R_{\mu\nu}$ we must contract on the dummy indices. Contraction yields $$R_{\mu\nu}=\partial_\lambda\Gamma^{\lambda}_{\mu\nu}-\partial_\mu\Gamma^{\lambda}_{\nu\lambda}+\Gamma^{\lambda}_{\lambda\sigma}\Gamma^{\sigma}_{\mu\nu}-\Gamma^{\lambda}_{\mu\sigma}\Gamma^{\sigma}_{\nu\lambda}=\partial_t\Gamma^{t}_{\mu\nu}+\partial_r\Gamma^{r}_{\mu\nu}+\partial_\theta\Gamma^{\theta}_{\mu\nu}+\partial_\phi\Gamma^{\phi}_{\mu\nu}-\partial_\mu\Gamma^{t}_{\nu t}-\partial_\mu\Gamma^{r}_{\nu r}-\partial_\mu\Gamma^{\theta}_{\nu\theta}-\partial_\mu\Gamma^{\phi}_{\nu\phi}+...$$ The problem I am currently having is expanding the double sums $$\Gamma^{\lambda}_{\lambda\sigma}\Gamma^{\sigma}_{\mu\nu}$$ and $$\Gamma^{\lambda}_{\mu\sigma}\Gamma^{\sigma}_{\nu\lambda}.$$ What exactly will these sums look like fully expanded? Im just having trouble distinguishing what values the dummies will take and in what order.

Answer 1

You'll have that $$\Gamma^\lambda_{\lambda \sigma} \Gamma^\sigma_{\mu\nu}= \Gamma^\color{red}{t}_{\color{red}{t}\color{blue}{t}} \Gamma^\color{blue}{t}_{\mu\nu}+\Gamma^\color{red}{t}_{\color{red}{t}\color{blue}{r}} \Gamma^\color{blue}{r}_{\mu\nu}+\ldots + \Gamma^\color{red}{r}_{\color{red}{r}\color{blue}{t}} \Gamma^\color{blue}{t}_{\mu\nu} + \Gamma^\color{red}{r}_{\color{red}{r}\color{blue}{r}} \Gamma^\color{blue}{r}_{\mu\nu}+\ldots$$ and $$\Gamma^\lambda_{\mu\sigma} \Gamma^\sigma_{\nu \lambda} = \Gamma^\color{red}{t}_{\mu\color{blue}{t}}\Gamma^\color{blue}{t}_{\nu\color{red}{t}}+\Gamma^\color{red}{t}_{\mu\color{blue}{r}}\Gamma^\color{blue}{r}_{\nu\color{red}{t}}+\ldots + \Gamma^\color{red}{r}_{\mu\color{blue}{t}}\Gamma^\color{blue}{t}_{\nu\color{red}{r}}+\Gamma^\color{red}{r}_{\mu\color{blue}{r}}\Gamma^\color{blue}{r}_{\nu\color{red}{r}}+\ldots$$

Answer 2

I also used to also find these sorts of things confusing. It helps me to think of the general case: take a vector of $N$ numbers ${a_i}$ and a vector of $M$ numbers ${b_i}.$ So long as $N,M$ are finite, we have $$ \sum_i \sum_j a_i b_j = \sum_j \sum_i a_i b_j \equiv \sum_{i,j} a_i b_j. $$ Where the double sum notation $\sum_{i,j}$ is used because the order of the summation does not matter (note that this is implicit in Einstein notation). You can see this by writing the matrix of numbers $T_{ij} = a_i b_j.$ The double sum is then just adding all the entries in this matrix together, and it's clear that this can be done in any order. This is essentially the discrete version of Fubini's theorem from calculus, and by thinking about the matrix $T_{ij}$ it's clear that if $N,M$ are infinite that this interchange of summation may not necessarily be true in all cases, which suggests the continuous version of Fubini's theorem.

Your specific case is just when the vectors $a,b$ are no longer vectors but objects with more indices, but everything follows the same. Indeed, the benefit of Einstein notation is that given fixed indices $\mu, \nu$ we can just assume everything is a number and the sort of higher-index structure is not important in performing the summation.

Answer 3

The trick is do just do the sums one at a time. For simplicity, I'll open the expression in two spacetime dimensions $t$ and $x$, but it works quite similarly in general.

We have $$ \Gamma^{\lambda}{}_{\mu\sigma} \Gamma^{\sigma}{}_{\nu\lambda} = \sum_{\lambda,\sigma} \Gamma^{\lambda}{}_{\mu\sigma} \Gamma^{\sigma}{}_{\nu\lambda}. $$ Due to associativity and commutativity of addition, we can choose which sum we want to compute first. You don't need to worry about the order in which you perform the sums. Hence, we have, for example, \begin{align} \Gamma^{\lambda}{}_{\mu\sigma} \Gamma^{\sigma}{}_{\nu\lambda} &= \sum_{\lambda} \sum_{\sigma} \Gamma^{\lambda}{}_{\mu\sigma} \Gamma^{\sigma}{}_{\nu\lambda}, \\ &= \sum_{\lambda} \left(\Gamma^{\lambda}{}_{\mu t} \Gamma^{t}{}_{\nu\lambda} + \Gamma^{\lambda}{}_{\mu x} \Gamma^{x}{}_{\nu\lambda}\right), \\ &= \sum_{\lambda} \Gamma^{\lambda}{}_{\mu t} \Gamma^{t}{}_{\nu\lambda} + \sum_{\lambda} \Gamma^{\lambda}{}_{\mu x} \Gamma^{x}{}_{\nu\lambda}, \\ &= \left(\Gamma^{t}{}_{\mu t} \Gamma^{t}{}_{\nu t} + \Gamma^{x}{}_{\mu t} \Gamma^{t}{}_{\nu x}\right) + \left(\Gamma^{t}{}_{\mu x} \Gamma^{x}{}_{\nu t} + \Gamma^{x}{}_{\mu x} \Gamma^{x}{}_{\nu x}\right), \\ &= \Gamma^{t}{}_{\mu t} \Gamma^{t}{}_{\nu t} + \Gamma^{x}{}_{\mu t} \Gamma^{t}{}_{\nu x} + \Gamma^{t}{}_{\mu x} \Gamma^{x}{}_{\nu t} + \Gamma^{x}{}_{\mu x} \Gamma^{x}{}_{\nu x}. \end{align}

A cool practice exercise is to perform the calculation in the opposite order ($\lambda$ first, then $\sigma$) and realize the answers are the same.