Confusion when dealing with tensors.

Question

I don't understand how tensors work, can someone please explain?

In particular, in the context of Electromagnetism, the dual of the field tensor $F$ is $$(*F)^{\mu\nu}:={1\over 2}\epsilon^{\mu\nu\rho\sigma}F_{\rho \sigma}$$

I am wondering what happens if we take the dual twice. It is given that $$**F=-F$$ But I don't see it trivially.

What I think (probably wrong): I suppose $$(**F)^{ab}={1\over 4} \epsilon^{ab\mu\nu}\epsilon_{\mu\nu\rho\sigma}F^{\rho\sigma}$$

Actually I am very confused as to whether to put the indices upstairs or downstairs.For example, I don't know whether to write the above or $$(**F)^{ab}={1\over 4} \epsilon^{ab\mu\nu}\epsilon_{\mu\nu}.^{\rho\sigma}F_{\rho\sigma}$$ Or in fact other variants, providing that indices to be summed over are such that one's up and one's down.

Help would be appreciated. Thank you.

Answer 1

$\def\a{\alpha} \def\b{\beta} \def\e{\epsilon} \def\d{\delta} \def\m{\mu} \def\n{\nu} \def\r{\rho} \def\s{\sigma}$We have $(*F)^{\m\n} = \frac{1}{2}\e^{\m\n\r\s} F_{\r\s}$, so $(*F)_{\m\n} = \frac{1}{2}\e_{\m\n\r\s} F^{\r\s}.$ Indices are raised and lowered with the Minkowski metric $\eta_{\m\n}$ using the fact that $$\eta^{\m\n}\eta_{\n\r} = \d^\m_\r.$$ Thus, $$\begin{eqnarray*} (**F)^{\m\n} &=& \frac{1}{4} \e^{\m\n\r\s}\e_{\r\s\a\b} F^{\a\b} \\ &=& \frac{1}{4} \e^{\r\s\m\n}\e_{\r\s\a\b} F^{\a\b} \end{eqnarray*}$$ as you've written. To evaluate this we need a result about the Levi-Civita symbol, $$\e^{a_1\cdots a_k a_{k+1}\cdots a_d}\e_{a_1\cdots a_k b_{k+1}\cdots b_d} = (-1)^s k!(d-k)!\ \d^{[a_{k+1}}_{b_{k+1}}\cdots \d^{a_d]}_{b_d},$$ where $(-1)^s = \textrm{sign}\,\det\eta$ and where indices in square brackets are antisymmetrized. In particular, for $d=4$, $k=2$, and $(-1)^s = -1$ we have $$\e^{\r\s\m\n}\e_{\r\s\a\b} = -4 \d_\a^{[\m} \d_\b^{\n]}.$$ Thus, $$\begin{eqnarray*} (**F)^{\m\n} &=& -\d_\a^{[\m} \d_\b^{\n]} F^{\a\b} \\ &=& -\frac{1}{2}(\d_\a^\m \d_\b^\n - \d_\a^\n \d_\b^\m) F^{\a\b} \\ &=& - \frac{1}{2}(F^{\m\n} - F^{\n\m}) \\ &=& -F^{\m\n}, \end{eqnarray*}$$ where we have used the fact that $F$ is antisymmetric.

It is possible to write all this in the language of differential forms in which case we are verifying that the inverse of the Hodge star operation $*$ on 2-forms is $-*$.

Answer 2

There are two concepts here. $F^{ab}$ is dual to $F_{ab}$ in the sense of being contravariant components vs covariant. This isn't denoted with the Hodge star, however. The Hodge star constructs a field that is orthogonal to the original.

The various expressions with different indices upstairs vs downstairs are equivalent.

You will need to evaluate $\epsilon_{abcd} \epsilon^{abef}$ to see -1 come out. In geometric algebra, the -1 is obvious--duality is enforced by contraction with the unit pseudoscalar. This object squares to -1 in 3+1 spacetime, so no more work is needed.

Edit: if you're familiar with clifford algebra, in the guise of Dirac matrices of quantum mechanics, this should be somewhat intelligible. We can interpret the gamma matrices $\gamma_a$ as (contravariant) Cartesian basis vectors for Minkowski space. The algebra they follow by multiplication is the clifford algebra for 3+1 spacetime. Let $\gamma_0 \gamma_0 = -1$ and $\gamma_i \gamma_i = +1$ for $i=1\ldots 3$.

Now then, let's look at the duality operation. Let $G^{cd} = \epsilon^{abcd} F_{ab}$ so that $G = \star F$. We can build up $F$ using the Clifford algebra as

$$F = \frac{1}{2} F_{ab} \gamma^a \gamma^b \implies F_{ab} = F \cdot (\gamma_b \wedge \gamma_a)$$

and $\epsilon$ as

$$\epsilon \equiv \frac{1}{4!} \epsilon^{abcd} \gamma_a \gamma_b \gamma_c \gamma_d = \gamma_0 \gamma_1 \gamma_2 \gamma_3 \implies \epsilon^{abcd} = \epsilon \cdot (\gamma^d \wedge \gamma^c \wedge \gamma^b \wedge \gamma^a)$$

The object $\epsilon$ is called the pseudoscalar (and also sometimes denoted $\gamma_5$), and it represents a unit 4-volume.

Using this notions, we can attack index notation expressions more or less "directly". Let's look at our duality operation:

$$\begin{align*}G^{cd} = G \cdot (\gamma^d \gamma^c) &= \frac{1}{2} F \cdot (\gamma_b \wedge \gamma_a) \epsilon \cdot (\gamma^d \wedge \gamma^c \wedge \gamma^b \wedge \gamma^a) \\ &= \frac{1}{2} [F \cdot (\gamma_b \wedge \gamma_a) \gamma^b \wedge \gamma^a] \epsilon \cdot (\gamma^d \wedge \gamma^c) \\ &= (F \epsilon) \cdot (\gamma^d \wedge \gamma^c)\end{align*}$$

So $G = F \epsilon$, and so $\star G = G \epsilon = F \epsilon \epsilon = -F$. All this follows from the algebraic properties of the $\gamma_a$.

Similarly, proving the proper identity for the Levi-Civita tensors is easy to do with Clifford algebra.

$$\begin{align*} \epsilon_{abcd} \epsilon^{abef} &= \epsilon (\gamma^d \wedge \gamma^c \wedge\gamma^b \wedge \gamma^a) \epsilon (\gamma_f \wedge \gamma_e \wedge \gamma_b \wedge \gamma_a) \\ &= (-2)(-1) (\gamma^d \gamma^c) \cdot ( \gamma_f \gamma_e) \\ &= 2(\eta^c_f \eta^d_e - \eta^c_e \eta^d_f ) \\ &= 4 \eta^{[c}_f \eta^{d]}_e \end{align*} $$