Tensor #35. Other Notions of Duality텐서 #35. 또 다른 쌍대

These are my notes on XylyXylyX’s What is Tensor lectures. This post covers lecture 36 of the playlist, Other Notions of Duality. In the previous post, Tensor #34, we computed Hodge duals, and this video wraps up the discussion of duality.

The map of dualities so far

The lecture begins with its usual picture. At the top is the space of functions, and below it the spaces $\Lambda^p$ of $p$-forms descend one by one to the top degree $n$. In the same shape there are the spaces of $p$-vectors, also down to degree $n$. The lowest space is one-dimensional, so we call it the top form or top vector.

The Hodge dual is a map from the space of degree $p$ to the space of degree $n-p$. It needs an $n$-dimensional vector space with a metric, and it works because the metric induces metrics on the dual space and on every tensor product space. Between forms and vectors of the same degree, the metric duality, which raises and lowers indices, is the link, and we already know it well.

The duality of this video also goes from degree $p$ to degree $n-p$, even when there is no metric. The Hodge dual can be seen as a two-step composition: this duality followed by the metric duality. There are two reasons to learn this duality separately. One is that many textbooks introduce it before the Hodge dual. The other is that the densitized forms that come later start on this route and use no metric at all, which is the basis for understanding the several ways of visualizing forms.

At this point the lecture adds a word in answer to viewers’ questions. In this subject there is no shortcut that begins from intuition or visualization; you first get the abstract definitions and rules into your hands, and intuition comes afterward.

The volume form that the duality rests on

The metric-free duality rests on an object called the volume form. Once we fix coordinates $x^1, \dots, x^n$ and choose an order, the basis of the top form space is $dx^1 \wedge \cdots \wedge dx^n$. This order is chosen somewhat arbitrarily. Take the same basis at every point of spacetime and multiply it by a function $g(x)$, and we get a field of top forms.

If $g > 0$ everywhere, we regard the spacetime as oriented. A spacetime need not be oriented, and the non-oriented case is tricky, but we lose little by assuming orientation here. The most common choice is $g = 1$, and then this object is called the volume form.

The lecture would also like to call it a density form. The names look reversed because of integration. In an integral this object sits where the differential element is, and looks like a volume element. Viewed as a multilinear map, however, it consumes volume elements and returns real numbers, which is what a density does. Integration is announced as the next topic.

To build the duality we need two volume forms. One is $dx^1 \wedge \cdots \wedge dx^n$ on the form side, and the other is $\partial_1 \wedge \cdots \wedge \partial_n$ on the vector side. The vector side connects to the determinant understood as the volume of the parallelepiped made by the column vectors, so it is natural to call it the volume form. In this post I write the vector-side object as $E$ and tell the story with it.

Defining the duality with the volume form

Pick a degree $p$ and split the indices of $E$ into the first $n-p$ and the last $p$. Writing the first group as $I$ and the last as $J$, the Levi-Civita symbol $\varepsilon^{IJ}$ is the $n$ indices joined together. In Einstein summation the volume form is written like this.

It is written with tensor products, but since $\varepsilon$ is totally antisymmetric it is the same object as a wedge product, and in the wedge basis a factor $1/n!$ appears in front to correct the over-counting. The components $Q_J$ of a $p$-form $Q$ are antisymmetric. The duality feeds $Q$ into the last $p$ slots of $E$. The dual basis produces a delta, the $J$ indices of $E$ are tied to the indices of $Q$, and the first $n-p$ slots remain. The components of the $(n-p)$-vector that comes out are the following.

The factor $1/p!$ in front is a convention. The components carry upper indices because the components of an $(n-p)$-vector are contravariant. This duality never used the metric; what we fixed was only the coordinate order and the orientation, that is, the volume form. Some books use the same star symbol as for the Hodge dual, so you need to tell them apart when reading.

Q. In the sum $\varepsilon^{IJ} Q_J$, doesn’t summing over all $J$ repeat the same terms?
A. It does. $J$ runs over all ordered tuples of $p$ indices, so for each increasing one, $p!$ ordered tuples give the same value, because the Levi-Civita symbol and $Q_J$ change sign together. So with the factor $1/p!$ the result equals the sum over increasing $J$ only, namely $V^{H} = \sum_{J} \varepsilon(H, J) \, Q_J$.

The opposite direction works the same way. To make an $(n-p)$-form from a $p$-vector, we use the form-side volume form $dx^1 \wedge \cdots \wedge dx^n$. The lecture works out one direction and says the other is the same.

The easiest example, the double dual of a function

For $p = 0$, $Q$ is a function $h$ and $J$ is empty. Since $1/0! = 1$, the dual is $h E$. Multiplying the function by the volume form is the dual.

Let us follow the way back. The dual $h E$ is an $n$-vector, with $(n-p) = n$. When going back, this time we feed the Levi-Civita symbol of the form-side volume form into all the slots of $hE$. A factor $1/n!$ appears, and the delta makes the two Levi-Civita symbols multiply against each other.

The sum over complete ordered tuples, $\varepsilon_J \varepsilon^J$, is $n!$. Without dividing by $n!$ the returned value would be $n! \, h$. The prefactor $1/p!$ of the convention was set so that the double dual returns to the start. What comes back is a 0-form, the same as the function.

The formula for going back at a general degree is a part the lecture did not work out. With the same convention, the map from an $(n-p)$-vector to a $p$-form comes out like this. (This is added for understanding and was not covered in the lecture.)

I checked with numbers that $W = Q$ at every degree in 3D and in both signatures. The “going back” line of the toy below is this value.

Q. Why is the dual of a 0-form $h$ just the single object $h E$?
A. Because for $p = 0$ all the indices of $E$ are in the first group, so there are no slots to feed. Nothing is fed, so $E$ stays with the function $h$ multiplied in. At the top degree it is the opposite: every slot is fed, no index is left, and a function comes out.

The route to the Hodge dual

The duality above was built from the volume form $\varepsilon$ alone, and $\varepsilon$ is not a tensor. Its values change under a change of coordinates, so we cannot raise or lower its indices with a metric. With a metric the situation is different. Multiplying by $\sqrt{|g|}$, the square root of the absolute value of the metric determinant, gives the Levi-Civita tensor, which transforms like a true tensor.

With this tensor we can raise and lower indices with the metric freely. So if we choose this tensor as the volume form, we can do the duality above as it is and lower the indices of the result to get an $(n-p)$-form, and this equals the Hodge dual. The lecture puts it in three ways. Adding the metric duality afterward gives the Hodge dual; choosing the volume form well as the Levi-Civita tensor makes a single duality equal to the Hodge dual; and lowering the indices of the Levi-Civita tensor early gives the same. Either way a metric is needed, so this is not a way to sneak into the Hodge dual without a metric. Many textbooks define the Hodge dual this way.

In an orthonormal basis this connection becomes an exact formula. Apply three steps in turn to the $V^H$ obtained above. Raising all the indices of the Levi-Civita tensor multiplies by $\det \eta$, each index of $H$ is lowered with the metric $\eta$, and we multiply by the ordering sign $(-1)^{p(n-p)}$. The ordering sign is needed because $V^H$ is built from $\varepsilon^{HJ}$, while the Hodge dual formula is built from $\varepsilon(I, H)$, so the two groups of indices are in swapped places. (This exact formula is my addition; I checked it against the direct calculation hundreds of times with random components in 3D and in both signatures. The lecture only says that using the Levi-Civita tensor reproduces the Hodge dual.)

The toy below follows this route in 3D. Pick the degree and the signature and enter components, and it shows side by side the step-1 value $V$, the returned $W$, the value after step 2, and the direct value from the Hodge dual formula. If you change the signature, the step-1 values stay put and only the step-2 values change.

Q. Is this duality worth learning separately even though it needs no metric?
A. Yes. It is defined on spaces without a metric, and in situations where we do not want to use one, and the structure of the density used in integration comes out of this route. Knowing that the Hodge dual is this duality plus the metric duality also shows why the metric signature enters the sign of the result only in step 2.

The last form of duality, densitized vectors

The last duality in the lecture does the same job from another direction. Again no metric is used. In this section $E$, unlike in the previous sections, denotes the top form $dx^1 \wedge \cdots \wedge dx^n$ on the form side. In this post we start by picking one $\alpha \in \Lambda^p$.

From $\alpha$ we can make the wedge map $\alpha \wedge$. This map takes a $q$-form and returns a $(p+q)$-form. Restricting to $q = n-p$, it becomes a map that takes an $(n-p)$-form and returns a top form. The lecture calls the space of top forms the space of densitized scalars. This space is one-dimensional, so every element has the form $a E$, and as a vector space under addition and scalar multiples it is the same as the real numbers. The difference is only that the volume element $E$ is multiplied in.

A map of the same shape can be made in another way. Pick an element $\omega$ of the dual space of the $(n-p)$-form space and take its tensor product with the top form $E$. The resulting $\omega \otimes E$ takes an $(n-p)$-form $\beta$ and returns $\omega(\beta) \, E$. Since $\omega(\beta)$ is a real number, the result is again a densitized scalar.

Now, as always, we demand that the two maps be equal. For any $\alpha \in \Lambda^p$ there is an $\omega \otimes E$ with $\alpha \wedge \beta = \omega(\beta) \, E$ for every $\beta$. This $\omega \otimes E$ is the dual of $\alpha$, and since it has the shape of an $(n-p)$-vector multiplied by $E$, it is a densitized $(n-p)$-vector. In the lecture’s words, every $p$-form is dual to a densitized $(n-p)$-vector.

(The next line is my addition. Writing out the components of $\omega$ directly, $\omega^{I}$ has the form $\tfrac{1}{p!} \varepsilon^{JI} \alpha_J$. It comes from the same ingredients as $V^I$ in the previous section and differs only by the ordering sign in $\varepsilon^{JI} = (-1)^{p(n-p)} \varepsilon^{IJ}$. I did not work out the component calculation exactly, so my confidence is low.)

The opposite direction has the same structure. Swap forms and vectors and swap upper and lower indices, and every $p$-vector is dual to an $(n-p)$-form multiplied by the top vector $\partial_1 \wedge \cdots \wedge \partial_n$. This object is the capacity. Volume and capacity are treated as the same idea, and they pair up in that a density acting on a capacity gives a real number. The lecture says this duality has little place in contemporary general relativity and elsewhere, and mentions an unfinished paper that treats it precisely. It matters for visualization, and the next video is said to cover integration and visualization.

Q. How does a densitized vector differ from an ordinary vector?
A. It has one extra factor $E$ multiplied in. As a space it is a tensor product with a one-dimensional space, so the dimension is the same. Under a change of coordinates $E$ changes by a Jacobian determinant factor, so the components transform differently from an ordinary vector, and this is the tensor density from the earlier post.

Summary

What we checked in this post is the following.

• The metric-free duality pairs a $p$-form with an $(n-p)$-vector using only a volume form and the Levi-Civita symbol, as in $V^I = \tfrac{1}{p!} \varepsilon^{IJ} Q_J$.
• The prefactor $1/p!$ is a convention chosen so that the double dual returns to the start, and the function case confirms it as $n!/n! = 1$.
• Choosing the volume form as the Levi-Civita tensor $\sqrt{|g|} \, \varepsilon$ and lowering indices with the metric gives the Hodge dual. This route needs a metric.
• The last duality comes from making $\alpha \wedge$ and $\omega \otimes E$ the same map, and a $p$-form is dual to a densitized $(n-p)$-vector, while a $p$-vector is dual to an $(n-p)$-form carrying a capacity.