# Is the raised Levi-Civita symbol a tensor density of weight 1?

In Sean Carroll's GR book, pg 83, between eqs. (2.69-70), the Levi-Civita symbol with raised indices is defined as $$\tilde{\epsilon}^{\mu_1\mu_2...\mu_n}=\text{sgn}(g)\tilde{\epsilon}_{\mu_1 \mu_2...\mu_n}$$ where $$\text{sgn}(g) = \frac{g}{|g|}$$ is the sign function of the metric determinant $$g$$ and $$\tilde{\epsilon}_{\mu_1 \mu_2...\mu_n}$$ is the Levi-Civita symbol we usually use (which is a tensor density of weight $$1$$).

He then said that the Levi-Civita symbol $$\tilde{\epsilon}^{\mu_1\mu_2...\mu_n}$$ is a tensor density of weight $$-1$$.

I tried to prove his statement:

$$$$\tilde{\epsilon}^{\mu_1'\mu_2'...\mu_n'}=\frac{g'}{|g'|}\tilde{\epsilon}_{\mu_1' \mu_2'...\mu_n'} \\ = \frac{|\frac{\partial x^{\mu'}}{\partial x^\mu}| ^{-2} g} {|\frac{\partial x^{\mu'}}{\partial x^\mu}| ^{-2} |g| } | \frac{\partial x^{\mu'}}{\partial x^\mu}| \tilde{\epsilon}_{\mu_1 \mu_2...\mu_n} \frac{\partial x^{\mu_1}}{\partial x^{{\mu}'_1}}\frac{\partial x^{\mu_2}}{\partial x^{\mu_2'}}...\frac{\partial x^{\mu_n}}{\partial x^{\mu_n'}}\\ = |\frac{\partial x^{\mu'}}{\partial x^\mu}|\frac{g}{|g|}\tilde{\epsilon}_{\mu_1 \mu_2...\mu_n} \frac{\partial x^{\mu_1}}{\partial x^{{\mu}'_1}}\frac{\partial x^{\mu_2}}{\partial x^{\mu_2'}}...\frac{\partial x^{\mu_n}}{\partial x^{\mu_n'}}\\ = |\frac{\partial x^{\mu'}}{\partial x^\mu}| \tilde{\epsilon}^{\mu_1\mu_2...\mu_n}\frac{\partial x^{\mu_1}}{\partial x^{{\mu}'_1}}\frac{\partial x^{\mu_2}}{\partial x^{\mu_2'}}...\frac{\partial x^{\mu_n}}{\partial x^{\mu_n'}}$$$$ This shows that $$\tilde{\epsilon}^{\mu_1\mu_2...\mu_n}$$ is a tensory density of weight $$1$$. What am I doing wrong?

• Is there a reason you're writing your coordinates with indices down? – J. Murray Mar 16 at 14:53
• @J.Murray I wrote them wrongly... Should be indices up. Corrected them. – TaeNyFan Mar 16 at 15:30

## 1 Answer

To me, a more conceptually direct derivation goes as follows. The Levi-Civita symbol is a symbol, not a geometrical object. We can denote its value in a coordinate system $$x$$ as $$\tilde \epsilon_{(x)\mu\nu\rho\sigma}$$. On its face, this is silly - the value of $$\tilde \epsilon$$ does not depend on the coordinate system. However, the point here is that we can compare the (trivial) transformation behavior of $$\tilde\epsilon$$ with the transformation behavior we would expect from a $$(0,4)$$-tensor.

As stated, the actual transformation behavior of these components is trivial, i.e. $$\tilde \epsilon_{(x)\mu\nu\rho\sigma}\mapsto \tilde \epsilon_{(y)\mu\nu\rho\sigma} = \tilde \epsilon_{(x)\mu\nu\rho\sigma}$$ If it were a tensor, then it would transform as $$\tilde\epsilon_{(x)\mu\nu\rho\sigma}\mapsto \tilde \epsilon_{(y)\mu\nu\rho\sigma} = \tilde \epsilon_{(x)\alpha\beta\gamma\delta}J^\alpha_{\ \ \mu}J^\beta_{\ \ \nu}J^\gamma_{\ \ \rho}J^\delta_{\ \ \sigma}= J \tilde\epsilon_{(x)\mu\nu\rho\sigma}$$ where $$J^\alpha_{\ \ \beta} \equiv \frac{\partial x^\alpha}{\partial y^\beta}$$ is the Jacobian matrix and $$J$$ is the Jacobian determinant. Therefore, we say that the true transformation behavior is given by

$$\tilde\epsilon_{(x)\mu\nu\rho\sigma}\mapsto \tilde \epsilon_{(y)\mu\nu\rho\sigma} = J^{-1}\tilde \epsilon_{(x)\alpha\beta\gamma\delta}J^\alpha_{\ \ \mu}J^\beta_{\ \ \nu}J^\gamma_{\ \ \rho}J^\delta_{\ \ \sigma}$$

The fact that the power of $$J$$ out front is $$-1$$ means that $$\tilde \epsilon$$ is a tensor density of weight $$-1$$.

We now turn our attention to the symbol with upstairs indices. In this case, we note that the actual transformation behavior is now $$\tilde \epsilon_{(x)}^{\mu\nu\sigma\rho}\mapsto \tilde \epsilon_{(y)}^{\mu\nu\sigma\rho} = \mathrm{sgn}(J)\epsilon_{(x)}^{\mu\nu\sigma\rho}$$ whereas its expected transformation behavior (if it were a tensor) would be $$\tilde \epsilon_{(x)}^{\mu\nu\sigma\rho}\mapsto \tilde \epsilon_{(y)}^{\mu\nu\sigma\rho} =\epsilon_{(x)}^{\alpha\beta\gamma\delta}(J^{-1})^\mu_{\ \ \alpha}(J^{-1})^\nu_{\ \ \beta}(J^{-1})^\rho_{\ \ \gamma}(J^{-1})^\sigma_{\ \ \delta}=J^{-1} \tilde\epsilon_{(x)}^{\mu\nu\rho\sigma}$$

Therefore, we can write that the true transformation behavior is

$$\tilde \epsilon_{(x)}^{\mu\nu\rho\sigma} \mapsto \tilde \epsilon_{(y)}^{\mu\nu\rho\sigma} = \mathrm{sgn}(J) J\epsilon_{(x)}^{\alpha\beta\gamma\delta}(J^{-1})^\mu_{\ \ \alpha}(J^{-1})^\nu_{\ \ \beta}(J^{-1})^\rho_{\ \ \gamma}(J^{-1})^\sigma_{\ \ \delta}$$

The fact that the power of $$J$$ is now $$+1$$ means that the weight is now $$+1$$; the fact that we also have a $$\mathrm{sgn}(J)$$ means that $$\tilde \epsilon^{\mu\nu\rho\sigma}$$ is in fact a pseudotensor density of weight $$+1$$. Carroll does not discuss this because he assumes that the coordinate transformation in question is orientation-preserving, which implies that $$\mathrm{sgn}(J)=+1$$ and the distinction between tensors and pseudotensors (and their respective densities) can be ignored.

In summary, we contrast the actual transformation behavior of the $$\tilde \epsilon$$ symbols with their expected transformation behavior if they were genuine tensors to determine in what way the actual transformation behavior differs. If it transforms like a tensor except for an additional factor of $$J^w$$, we call it a tensor density of weight $$w$$. If it additional picks up a factor of $$\mathrm{sgn}(J)$$, we call it a pseudotensor density.