Suppose a component of tensor field is described by $B^k=\varepsilon^{kij} \phi_{ij}$. If we define $B^k$ in an Euclidean space then does the rising or lowering of the indices of the Levi-Civita symbol change the sign?

I mean does $B^k=\varepsilon^{kij} \phi_{ij}=\varepsilon_{kij} \phi_{ij}$?

  • $\begingroup$ You have $k$ as a free index on the left and as a dummy index on the right, which isn't allowed. You need $B^i$ on the left. $\endgroup$
    – user11266
    Commented Aug 29, 2012 at 16:18
  • $\begingroup$ oh! My mistake.Corrected it. $\endgroup$
    – aries0152
    Commented Aug 29, 2012 at 16:36

1 Answer 1


First, if you're going to keep proper track of covariant and contravariant components, you should lower the index on $B$ and make sure the dummy indices are always of opposite types: $B_k = \varepsilon_{kij} \phi^{ij}$. The reason we can be sloppy in Euclidean space is because of how trivial the metric can be. We can always consider our equations in the basis in which $g_{ij} = \delta_{ij}$, in which case it's clear there's really no effective difference between upper and lower indices.

About Levi-Civita in general: If we're talking about the symbol (which I'll denote with a tilde) we typically define the lower-indexed one with components that are positive for even permutations and negative for odd ones (opposite for left-handed coordinate systems). The upper-indexed one is multiplied by the sign of the determinant of the metric: $$ \tilde{\varepsilon}^{\mu_1\mu_2\cdots\mu_n} = \mathrm{sgn}(g) \tilde{\varepsilon}_{\mu_1\mu_2\cdots\mu_n}. $$ You can freely raise and lower these indices in Euclidean space, as long as you don't change the handedness of your coordinate system.

Regarding the tensor, its components can generally not only change sign but magnitude as well. This is explained in detail in Sean Carroll's Spacetime and Geometry, for which there is a free online preprint (see in particular pp. 51-52 of Chapter 2). The summary of what he says is that the lower-index tensor (without a tilde) is the symbol (again with a tilde) multiplied by the square root of the determinant of the metric: $$ \varepsilon_{\mu_1\mu_2\cdots\mu_n} = \sqrt{\lvert g \rvert} \tilde{\varepsilon}_{\mu_1\mu_2\cdots\mu_n}. $$ Conversely, $$ \varepsilon^{\mu_1\mu_2\cdots\mu_n} = \frac{1}{\sqrt{\lvert g \rvert}} \tilde{\varepsilon}^{\mu_1\mu_2\cdots\mu_n}. $$ Again, if you have a metric with determinant $1$, you can raise and lower without worry.

  • $\begingroup$ Could you please explain why should the dummy indices are always of opposite types?Why can't we put them in the same plane?If we define them in a plane in a first place, can we be able to raise and lower them in the same time? I mean if I define $B_k=\varepsilon_{kij}\phi_{ij}$ then is $B^k=\varepsilon^{kij}\phi^{ij}$ correct? $\endgroup$
    – aries0152
    Commented Aug 30, 2012 at 5:08
  • 1
    $\begingroup$ You could define them at the same level, of course, but I think, if the metric isn't trivial, you run the risk of combining honest tensors into something that doesn't transform properly. Upstairs/downstairs Einstein notation, when used properly, is very good at catching errors. As to your question, just remember that you raised five indices with the metric, not one. You can switch an upstairs/downstairs pair of dummy indices always, since you will be contracting with $g$ and its inverse, but if they start in the same place, moving them requires two copies of the metric that do not cancel. $\endgroup$
    – user10851
    Commented Aug 30, 2012 at 5:37

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