Tensor #33. Duality and the Hodge Dual (Part 1)텐서 #33. 쌍대성과 호지 쌍대 (1부)

These are my notes on XylyXylyX’s What is Tensor lectures. This post covers the first half of the video Duality and Hodge Duality, which holds lectures 33 and 34 of the playlist. The video runs over an hour and 40 minutes, so I split it in two. This part covers how the word duality is used and which spaces can be paired. The definition of the Hodge dual starts in the next post, which covers the second half of the video. The tensor densities of the previous post come back at the end of this one.

Two uses of the word dual

Once we fix a vector space $V$, the dual space $V^*$, the collection of all linear maps from $V$ to the real numbers, comes into being automatically.

So far we have used the word dual in this sense. What we call duality in this post is a different use. It points to one chosen map that pairs an element of one space with an element of another space, one to one.

We write the set of all linear maps from $V$ to $W$ as $\Hom(V, W)$. Putting $V^*$ in the place of $W$ gives $\Hom(V, V^*)$, which is also a set that comes into being as soon as we make $V$. But no single map stands out in it. A vector space without an inner product has no canonical map from $V$ to $V^*$, that is, no map fixed without any choice. (One element $\varphi$ of $\Hom(V, V^*)$ fixes a bilinear form by $B(v, w) = \varphi(v)(w)$, so it is the same as one $(0, 2)$ tensor. This explanation is added for understanding and was not covered in the lecture.)

Q. Vector spaces of the same dimension are always isomorphic, so aren’t $V$ and $V^*$ simply the same thing?
A. An isomorphism exists, but it is not fixed uniquely. Choosing a basis lets us build one by $e_i \mapsto e^i$, and a different basis gives a different isomorphism. Canonical means not relying on such a choice. (This explanation is added for understanding and was not covered in the lecture.)

The pairing of vectors and dual vectors fixed by an inner product

Suppose $V$ has an inner product $\langle \cdot, \cdot \rangle$. Put a vector $v$ in the first slot of the inner product and leave the second slot empty, and it becomes a linear map that takes a vector $w$ and gives a real number. This is an element of $V^*$, and we write it $v^\flat$.

The map $v \mapsto v^\flat$ is the one map the inner product picks out of $\Hom(V, V^*)$ in the previous section. If the inner product is nondegenerate, then for $v \neq 0$ there is a $w$ with $\langle v, w \rangle \neq 0$, so $v^\flat \neq 0$, and since the two spaces have the same dimension this map is an isomorphism. The reverse direction $V^* \to V$ also comes from the same inner product. (This explanation is added for understanding and was not covered in the lecture.) In this way, on a $V$ with an inner product, vectors of $V$ and dual vectors of $V^*$ are paired one to one, and that is the first duality.

In the classical notation that does not write the basis, this pairing is the raising and lowering of indices.

$g_{\mu\nu}$ is the $(0, 2)$ tensor that writes the inner product in components, and $g^{\mu\nu}$ is its inverse matrix. Going from a vector to a dual vector we lower the index, and coming back we raise it. Once the inner product is fixed there is nothing to choose, and if we change the inner product the same $v$ goes to a different dual vector. In the toy below, drag the vectors and change the metric. The dual vector is drawn as a family of parallel level lines, and you can also check that $v^\flat(w)$ equals $\langle v, w \rangle$.

Q. Without an inner product, can’t we read a dual vector as a vector?
A. There is just no single way to read it. In general relativity every point of spacetime has an inner product and the whole of them is the metric tensor, so this pairing is always available.

The metric duality that raises and lowers indices

We can do the same for objects with several indices. A $p$-form has completely antisymmetric components and is written like this.

Raising all the indices with the metric gives the components of a $p$-vector.

The same metric is used on every index in the same way, so the antisymmetry is preserved. In this note we call this pairing the metric duality, the name the lecture gives it. It repeats the pairing of the previous section index by index, so there is no new problem to solve. The lecture does not dig further here either.

Counting the dimensions of exterior powers

Now let $V$ be a 4-dimensional real vector space with an inner product, and put the space of $p$-vectors next to the space of $p$-forms. The lecture distinguishes the $p$-vector side with a lower index $\Lambda_p$ and the $p$-form side, made from the dual space, with an upper index $\Lambda^p$. $\Lambda_p$ is the space of wedge products of $p$ vectors of $V$, and $\Lambda^p$ is the space of wedge products of $p$ dual vectors of $V^*$. These spaces are called exterior powers. For $n = 4$ the dimensions are as follows.

• $p = 0$: dimension 1. The space of smooth functions of the spacetime coordinates. The vector side and the form side are the same space.
• $p = 1$: dimension 4. Vectors or dual vectors.
• $p = 2$: dimension 6.
• $p = 3$: dimension 4.
• $p = 4$: dimension 1. The top degree, whose elements are constant multiples of the volume element.

For general $n$ we have $\dim \Lambda^p = \binom{n}{p}$ and $\binom{n}{p} = \binom{n}{n-p}$, so counting from either end gives the same numbers.

Finding pairs of equal dimension

To set up a one-to-one correspondence between two spaces, their dimensions must be equal. So in the list above we take spaces of equal dimension as candidates for pairing. On the $p$-form side, $\Lambda^p$ and $\Lambda^{n-p}$ have equal dimension, and with an inner product there is a one-to-one correspondence joining them canonically. This is the Hodge duality, the most important duality of this series. For $n = 4$ we get $\Lambda^0 \leftrightarrow \Lambda^4$ and $\Lambda^1 \leftrightarrow \Lambda^3$, and $\Lambda^2$ is paired with itself. The $p$-vector side is the same. A self-dual middle space occurs only when $n$ is even. The spaces are the $n + 1$ of $p = 0, \dots, n$, so when $n$ is even there is an odd number and one is left in the middle. When $n$ is odd the two inner spaces pair with each other and none is self-dual.

There are also candidates that cross between the vector side and the form side. A pair of equal dimension such as $\Lambda_1$ and $\Lambda^3$ can be obtained by chaining the Hodge duality and the metric duality. The lecture says it does not yet know whether this is a new duality and moves on. (The chaining is an explanation added for understanding and was not covered in the lecture.)

Here the lecture stresses that the Hodge duality needs an inner product. In general relativity every point of spacetime has an inner product and the whole of them is the metric tensor $g_{\mu\nu}$, which fixes the geometry of spacetime. Requiring an inner product is not an unreasonable condition. It is the center of the theory.

Q. Is equal dimension enough to be a duality?
A. Equal dimension only means a candidate. To accept a duality we must show that a canonical one-to-one correspondence really exists. The lecture also lists the pairs of equal dimension first and then says the existence of a duality must be shown separately.

Densitized exterior powers

There is one more family of spaces with the same dimensions. $\Lambda^4$ has dimension 1, so multiplying it into any tensor product does not change the dimension. So for $p = 0, \dots, 4$ the dimensions of $\Lambda^p \otimes \Lambda^4$ are also $1, 4, 6, 4, 1$. This product moves us to a different tensor product space, so, unlike ordinary tensors, these objects change as tensor densities, and the spaces are called densitized exterior powers. Multiplying in $\Lambda^4$ once more keeps the dimension, so there are doubly and triply densitized ones, going on without end. The lecture draws only the first.

As the lecture has it, candidates for pairing arise among these spaces too, and some of them, unlike the Hodge duality, do not require an inner product. For example, between $\Lambda_p \otimes \Lambda^n$ and $\Lambda^{n-p}$ there is a canonical isomorphism fixed without an inner product, because contracting a $p$-vector into a top-degree form gives an $(n-p)$-form. (This map is an explanation added for understanding and was not covered in the lecture. The lecture only points out the equal dimensions and postpones the concrete maps.)

Summary

What we checked in this post is the following.

• The dual space $V^*$ is the collection of all linear maps, and a duality points to one canonically chosen one-to-one correspondence among them.
• With an inner product, $v \mapsto v^\flat = \langle v, \cdot \rangle$ pairs $V$ and $V^*$ canonically, and in components it is the raising and lowering of indices with the metric.
• In $n$ dimensions $\dim \Lambda^p = \binom{n}{p}$ and $\Lambda^p$ and $\Lambda^{n-p}$ have equal dimension. With an inner product, what joins them is the Hodge duality, and when $n$ is even the middle space is self-dual.
• The densitized exterior powers, with the top degree $\Lambda^n$ multiplied in, have the same dimensions and so are candidates for pairing too.

In the next post we define the Hodge dual among these candidates, using the wedge product and the inner product.