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I'm doing some self-studying out of Hughston and Tod's Introduction to General Relativity and I stumbled upon a few problems asking me to solve systems of equations using Levi-Civita and index notation. For example

Solve for $X_i$: $kX_i+\epsilon_{ijk} X_j P_k = Q_i$

Solve for $X_i$ and $Y_i$: \begin{align} aX_i+\epsilon_{ijk}Y_jP_k&=A_i\\ bY_i+\epsilon_{ijk}X_jP_k&=B_i \end{align}

The best thing I could come up with for the first one was $X_i=Q_i/k - \epsilon_{ijk}X_jP_k/k$ but that's not really solving for each $X_i$ since it involves the other components. I tried to do some Google searching but I couldn't find this kind of questions anywhere. What is the technique that the author is trying to teach? And is it actually used in GR?

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So I'm not sure if this is in the spirit of the exercise and this is probably not the most elegant solution but one way to solve your first equation would be to write it as:

$\left(k\delta_{ij}+\varepsilon_{ijk}P_k\right)X_j=Q_i,$

in which case we can call the term in parentheses $M_{ij}$:

$M=\left( \begin{array}{ccc} k & p_3 & -p_2 \\ -p_3 & k & p_1 \\ p_2 & -p_1 & k \\ \end{array} \right).$

This then becomes a simple matrix equation which may be solved by inverting. In particular, using the inverse as a hint, we may try multiplying both sides of the equation by $N$, where:

$N_{ij}=\left(k^2\delta_{ij}+p_ip_j-k\varepsilon_{ij\ell}p_\ell\right)$.

That is, if the original equation is $M_{ij}X_j=Q_i$, writing $N_{ij}M_{jk}X_k=N_{ij}Q_j$. On the left, we have:

$\begin{split}\left(k^2\delta_{ij}+p_ip_j-k\varepsilon_{ij\ell}p_\ell\right)\left(k\delta_{jk}+\varepsilon_{jkm}p_m\right)&=k^3\delta_{ik}+kp_ip_k+k\left(\varepsilon_{ji\ell}\varepsilon_{jkm}\right)p_\ell p_m,\\ &=k^3\delta_{ik}+kp_ip_k+k\left(\delta_{ik}\delta_{\ell m}-\delta_{im}\delta_{\ell k}\right)p_\ell p_m,\\ &=\left(k^3+kp^2\right)\delta_{ik}, \end{split}$

which we see is a multiple of the identity matrix. Thus acting on both sides with this will give a solved equation. I'm not aware of any way we could have deduced the form of $N$ using Levi-Civita identities alone, but it may well be doable.

The second system may be solved similarly (I think) by solving the first equation for $X_i$ in terms of $Y_i$, plugging into the second equation, and obtaining $Y$ using a similar strategy.

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  • $\begingroup$ So you got your $N$ by calculating the inverse of $M$ and then writing it out in index notation? $\endgroup$ – Alex Wang Apr 3 '16 at 23:17
  • $\begingroup$ That's right. I wanted to come up with a systematic way to deduce it directly, but I couldn't think of anything. But as I mentioned in my answer it's very possible I am missing something. $\endgroup$ – commutatertot Apr 4 '16 at 0:44
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    $\begingroup$ I think you basically got it. The book does give the formula for the inverse of a matrix in index notation. $M^{-1}_{jk}=(2\Delta )^{-1}\epsilon_{jpq}\epsilon_{kab}M_{ap}M_{bq}$ where $\Delta$ is the determinant of $M$ given by $1/6\epsilon_{ijk}\epsilon_{lmn}M_{il}M_{jm}M_{kn}$ for a 3x3 matrix. Thanks a lot! $\endgroup$ – Alex Wang Apr 4 '16 at 1:01

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