Tensor #34. Elementary Hodge Dual Calculations텐서 #34. 호지 쌍대 계산 예

These are my notes on XylyXylyX’s What is Tensor lectures. This post covers lecture 35 of the playlist, Elementary Hodge Dual Calculations. The video of the previous post, lecture 33/34, runs 1 hour 41 minutes, so I split the notes in two. The first half is in the previous post. This post starts from the latter part of that video, the definition of the Hodge dual, and continues with the examples of lecture 35.

The latter part of the previous video: defining the Hodge dual

Suppose an $n$-dimensional vector space has an inner product. It induces an inner product on the dual space, and then an inner product $\ip{\cdot}{\cdot}$ on the space $\Lambda^p$ of $p$-forms. For simple forms it is given by a determinant.

There are two ways to get one real number from a $p$-form $\omega$. The first is the inner product. Choose a reference $p$-form $\beta$ and take $\ip{\omega}{\beta}$. The second is the wedge product. Choose a reference $(n-p)$-form $\alpha$ and form $\omega \wedge \alpha$, which lies in the space of top degree $n$. That space is one-dimensional, so the result is a real multiple of the unit volume form $E = dx^1 \wedge \cdots \wedge dx^n$, that is, of the form $x \, E$, and we get one real number $x$. If we demand that the two ways give the same real number, then $\alpha$ is determined by $\beta$, and this is the Hodge dual.

So $\star$ sends an element of $\Lambda^p$ to an element of $\Lambda^{n-p}$. The two spaces have the same dimension, and $\star$ is an isomorphism that we can use in both directions. Because the top degree is one-dimensional, each of the two methods gives exactly one real number and they can be matched. In higher dimensions we would get several real numbers and this matching would not work.

Q. Where does the metric enter in the definition of the Hodge dual?
A. In the inner product $\ip{\omega}{\beta}$ on the right side. The inner product of the vector space induces the inner product of forms, so if the metric changes, the dual of the same $\beta$ changes. Also, fixing the order in $E$ amounts to fixing an orientation, so we agree to take $dx^1 \wedge \cdots \wedge dx^n$ as the positive orientation.

Three facts follow directly from the definition.

• $\star$ is linear. So once we know the duals of the basis forms, the dual of any form comes from pulling out the components and substituting.
• For the basis $1$ of the 0-forms, $\omega$ is a function, so $\omega \wedge \star 1 = \omega \, E$ and $\star 1 = E$. This holds for any dimension and any metric.
• Putting in $\omega = E$ gives $E \wedge \star E = \ip{E}{E} \, E$, and in an orthonormal basis $\ip{E}{E}$ is the determinant of $\ip{dx^i}{dx^j}$. Its diagonal entries are $\pm 1$, so the value depends on how many are negative. If the metric has $s$ positive eigenvalues, there are $n - s$ negative ones, and $\star E = (-1)^{n-s}$.

In an orthonormal basis, the dual of a basis form is written in one line. Here $j_1, \dots, j_{n-p}$ are the indices not among $i_1, \dots, i_p$.

There is no summation sign, because fixing the $i$ fixes the set of $j$. Whatever order we write the $j$ in, the sign of the wedge product and the sign of $\varepsilon$ change together and the result is the same. The inner products multiplied in are $\pm 1$, so the signature of the metric appears only as a sign. (The reason this formula satisfies the definition is added for understanding. Wedge $dx^{i_1} \wedge \cdots \wedge dx^{i_p}$ onto both sides; since $\varepsilon^2 = 1$ the right side becomes $\ip{dx^I}{dx^I} \, E$. This was not covered in the lecture.)

The dual of 1-forms in 3D Euclidean space

Now the calculations of lecture 35. First is $n = 3$ with a Euclidean metric whose signature is three positives. In the coordinates $x^1, x^2, x^3$, the form $dx^1 \wedge dx^2 \wedge dx^3$ is the positive volume form. All the inner products are $+1$, so only $\varepsilon$ decides the sign.

For $dx^1$ the index set is $\{1\}$ and the remaining indices are $\{2, 3\}$. Any order is allowed, so we take increasing order. Since $\varepsilon_{123} = 1$, we get $\star dx^1 = dx^2 \wedge dx^3$. For $dx^2$ the remaining indices are $\{1, 3\}$, and $\varepsilon_{213}$ is an odd permutation with one swap, so it is $-1$. For $dx^3$, $\varepsilon_{312}$ is an even permutation with two swaps, so it is $+1$.

By linearity an arbitrary 1-form $a = a_1 dx^1 + a_2 dx^2 + a_3 dx^3$ keeps its components.

Elementary textbooks use $x, y, z$ and rewrite $- dx \wedge dz$ as $dz \wedge dx$ by flipping the order, which removes the minus sign. These notes keep the increasing basis and leave the minus sign.

Q. Why does the dual of $dx^2$ carry a minus sign?
A. Because $(2, 1, 3)$ is an odd permutation, one swap away from $(1, 2, 3)$. Since $- dx^1 \wedge dx^3 = dx^3 \wedge dx^1$, it is the same object with the order changed, and the dual is the same whichever way we write it.

The dual of 2-forms and the top degree in 3D

2-forms work the same way. One index remains each time, and the metric factors are $+1$.

This is because $\varepsilon_{132} = -1$ and $\varepsilon_{231} = +1$. An arbitrary 2-form $\alpha = \alpha_{12} \, dx^1 \wedge dx^2 + \alpha_{13} \, dx^1 \wedge dx^3 + \alpha_{23} \, dx^2 \wedge dx^3$ becomes this.

The spaces of 1-forms and 2-forms are both three-dimensional, and $\star$ moves the components of one into the other. This is natural, since the dimensions of degrees $p$ and $n - p$ are always equal. In the top degree, $\star (dx^1 \wedge dx^2 \wedge dx^3) = 1$, which agrees with $(-1)^{n-s} = (-1)^{3-3} = 1$ from the definition.

The dual of 1-forms and 2-forms in 4D Lorentzian space

The second case is $n = 4$ with a Lorentzian metric of signature three positives and one negative. The indices are $0, 1, 2, 3$, with $\ip{dx^0}{dx^0} = -1$ and the others $+1$. The positive volume form is $dx^0 \wedge dx^1 \wedge dx^2 \wedge dx^3$. Now the metric product decides the sign along with $\varepsilon$.

For the 1-form $dx^0$, $\varepsilon_{0123} = +1$ and the metric factor is $-1$, so a minus sign appears. For $dx^2$, the remaining indices are $\{0, 1, 3\}$, $\varepsilon_{2013}$ is an even permutation with two swaps, $+1$, and the factor is $+1$.

For 2-forms both factors appear. For $dx^0 \wedge dx^1$, $\varepsilon_{0123} = +1$ and the metric product is $(-1)(+1) = -1$. For $dx^0 \wedge dx^2$, $\varepsilon_{0213} = -1$ and the metric product is $-1$, so two minus signs meet and give a plus.

In four dimensions the space of 2-forms is six-dimensional and $n - p$ is also 2, so $\star$ sends this space to itself. We have to put parentheses around the wedge of two forms and write $\star (dx^0 \wedge dx^2)$. Without parentheses, $\star dx^0 \wedge dx^2$ does not say whether the dual comes first or the wedge does.

3-forms in 4D and the double dual

3-forms are computed by the same rule.

The first has $\varepsilon_{0123} = +1$ with metric factor $-1$. The second has $\varepsilon_{0132} = -1$ with factor $-1$, which gives a plus. A basis containing $dx^0$ always has metric product $-1$, and one without it has $+1$. That each result is a 1-form, as the dual of a 3-form should be, is checked by $4 - 3 = 1$.

What remains are $\star 1$ and $\star E$. Earlier we had $\star 1 = E = dx^0 \wedge dx^1 \wedge dx^2 \wedge dx^3$ and $\star E = (-1)^{4-3} = -1$. Therefore $\star \star 1 = \star E = -1$. Taking the dual twice may change the sign instead of returning to the start, and the signature decides it. The lecture says the sign depends only on the parity of the signature. In the signature with one positive and three negatives, $n - s = 3$ is odd too, so we get the same $-1$.

The general double dual formula is read off from all the calculations above. On a $p$-form, $\star \star \omega = (-1)^{p(n-p)} (-1)^{n-s} \, \omega$, and I checked it on every basis form in the two spaces of this post. (This formula is added for understanding and was not covered in the lecture.) In 3D Euclidean space $p(3-p)$ is always even and $n - s = 0$, so $\star \star = 1$.

Q. Why does taking the dual twice not always return to the start?
A. Each dual adds a sign from the product of inner products $\ip{dx^i}{dx^i}$ and a sign from the permutation that reorders the indices. Going through twice multiplies all these signs again, and if an odd number of negative inner products remain, a minus sign is left. In Euclidean space there are none.

The toy below shows this calculation step by step for every basis form of the two spaces. Pick the indices and check the permutation sign, the metric product, the result, and the sign of the double dual.

The cross product is one dual

In elementary physics the cross product is an operation that makes one vector out of two 3D vectors. $\mathbf{i} \times \mathbf{j} = \mathbf{k}$, $\mathbf{j} \times \mathbf{k} = \mathbf{i}$, $\mathbf{k} \times \mathbf{i} = \mathbf{j}$. These rules come from the 3D duals computed above. The same formula works on the vector side, and for the basis vectors $e_1, e_2, e_3$ it gives this.

The three equations come from $\varepsilon_{123} = \varepsilon_{231} = \varepsilon_{312} = +1$. So the cross product is the operation that first makes a 2-vector by the wedge product and returns to a vector by the dual.

Q. Why is the cross product defined this way only in three dimensions?
A. Because $u \wedge v$ lies in the degree-2 space and $\star$ sends it to the degree-$(n-2)$ space. To return to vectors we need $n - 2 = 1$, so $n = 3$. In four dimensions, $\star (\partial_0 \wedge \partial_1)$ is again a 2-vector of the form $\partial_2 \wedge \partial_3$. The lecture only mentions in passing that a cross product also exists in seven dimensions, but by an odd construction unlike this one.

Summary

What we checked in this post is the following.

• The Hodge dual is defined by $\omega \wedge \star\beta = \ip{\omega}{\beta} \, E$. The inner product and the wedge product each give one real number, and the top degree being one-dimensional is what lets us match them.
• In an orthonormal basis, $\star dx^I$ is a one-line rule: multiply the permutation sign $\varepsilon_{I,J}$ by the inner product $\pm 1$ for each index of $I$, and put the result on $dx^J$.
• In 3D Euclidean space only the permutation sign decides the sign, while in 4D Lorentzian space a basis containing $dx^0$ picks up the metric’s $-1$.
• Since $\star 1 = E$ and $\star E = (-1)^{n-s}$, in Lorentzian space the double dual can change the sign.
• In 3D, $u \times v = \star (u \wedge v)$, and the cross product returns to a vector only when $n = 3$.

The next post looks at other dualities that can be built even without a metric.