Tensor #26. P-vectors and P-forms Recap텐서 #26. P-벡터와 P-형식 복습

These are my notes on XylyXylyX’s What is Tensor lectures. This post covers lecture 26 of the playlist, P-vectors and P-forms Recap. We reviewed determinants in the previous post, so this time we look back at the p-forms built so far and place beside them the p-vector, the same construction done with vectors instead of the dual space. It is a recap that fixes the parts of the earlier p-form lecture whose explanation fell short.

Notation: upper and lower indices

Fix one vector space $\V$. We write $\Lambda^p(\V)$ for the space built by wedging $p$ 1-forms, elements of the dual space of $\V$, and $\Lambda_p(\V)$ for the space built by wedging $p$ vectors of $\V$. The position of the index does the distinguishing. In earlier posts I wrote $\Lambda^p \V^*$ with a star, and the lecture corrects that this is not the standard notation. It is the same reason a tensor product space of type $(0, p)$ is written with only $\V$ in it, as in $T^p_q(\V)$, instead of being spelled out with $\V^*$. In this definition $\V$ is the anchor, and the index only says what we pile up from that anchor.

A word of caution about names. A p-vector of degree 4 can be called a 4-vector, but it has nothing to do with the 4-vector of relativity (the vector that bundles energy and momentum). To avoid the confusion the lecture uses a 5-vector as its example.

Q. Is a p-vector the same as the 4-vector of relativity?
A. No. The 4-vector of relativity is an ordinary vector with four components, whereas a p-vector is an element of the exterior power made by wedging $p$ vectors. Only the names overlap.

Simple p-vectors and components

A p-vector is built the same way as a p-form. The wedge $v \wedge w \wedge x \wedge z \wedge q$ of vectors $v, w, x, z, q$ is an element of the fifth exterior power $\Lambda_5(\V)$, and since it is written as one wedge product it is a simple 5-vector. A sum of two such simple ones, like $v \wedge w \wedge x \wedge z \wedge q + a \wedge b \wedge c \wedge d \wedge e$, may not collapse into a single wedge product, and if it does not it is not simple. Usually we study only the simple ones. Everything is linear, so what we learn for simple ones extends to sums, and that sum is the addition inherited from the tensor product space, because $\Lambda_p(\V)$ is a subspace of it.

Components come out the same way. A simple 3-vector expands like this.

The coefficient is the product of the components of $u, v, w$, and the basis is the basis $e_\mu \wedge e_\nu \wedge e_\gamma$ of $\Lambda_3(\V)$. The only difference from a p-form is that the indices have moved up and down. Since a vector can be read as a map acting on elements of the dual space, a p-vector can also be read as a map that takes a p-form and returns a number.

The tower of exterior powers and their dimensions

From one vector space $\V$ we can build tensor product spaces endlessly, one for each $(p, q)$. Exterior powers are different. If $\V$ has dimension $n$, there are only $n + 1$ of them.

From $\Lambda_{n+1}(\V)$ on, everything is 0. Antisymmetry forces this. Many of the interesting facts in physics come out of this finite number of spaces. The lecture says it has not yet explained why these spaces matter, only what they are. The intention is to take the mystery out of the definitions for readers who meet the concept in other books.

Q. Why is $\Lambda_{n+1}(\V)$ zero?
A. With dimension $n$ there are only $n$ basis vectors, so wedging $n + 1$ of them puts at least one basis vector in twice. The wedge product is antisymmetric, so two copies of the same thing give 0. (This explanation is added for understanding and was not covered in the lecture.)

The dimension of each space is a binomial coefficient, the same for p-vectors and p-forms.

We have $\Lambda_0(\V) = \R$, which becomes the smooth functions on spacetime when we treat scalar fields. $\Lambda_1(\V)$ is the vector space $\V$ itself and $\Lambda^1(\V)$ is the dual space. The two ends $\Lambda_0$ and $\Lambda_n$ are both one-dimensional, and $\Lambda_1$ and $\Lambda_{n-1}$ are both $n$-dimensional. This symmetry, where front and back spaces have matching dimensions, underlies the Hodge star later.

Bases for n=3 and n=4

Let us write the bases starting from dimension 3, with numbers 0, 1, 2. In a coordinate basis, put $\partial_\mu$ where $e_\mu$ stands.

The dimensions are $1, 3, 3, 1$, symmetric front to back. For p-forms write $dx^\mu$ in the same place and everything else is the same. When the dimension grows by one to 4, there is one more $e_3$. The second degree grows to 6 with $e_0 \wedge e_3$, $e_1 \wedge e_3$, $e_2 \wedge e_3$ added, the third degree grows to 4 with $e_0 \wedge e_1 \wedge e_3$, $e_0 \wedge e_2 \wedge e_3$, $e_1 \wedge e_2 \wedge e_3$ added, and the fourth degree is the single $e_0 \wedge e_1 \wedge e_2 \wedge e_3$. The dimensions are $1, 4, 6, 4, 1$, again symmetric.

That p-vectors and p-forms have the same structure is more than a matter of counts. The two pair with each other.

(Pairing a p-vector $v_1 \wedge \cdots \wedge v_p$ with a p-form $\alpha^1 \wedge \cdots \wedge \alpha^p$ gives the determinant $\det \left( \alpha^i(v_j) \right)$. The determinant from the previous post is used here. This explanation is added for understanding and was not covered in the lecture.)

The exterior algebra as a direct sum of (n+1)-slot tuples

The exterior algebra that bundles all degrees together is a direct sum.

Being a direct sum it is a vector space, and taking the wedge product as multiplication makes it an algebra. The elements of this set are tuples with $n + 1$ slots. The lecture first says a tuple of dimension $n$, then corrects to $n + 1$ because of the degree-0 slot. The first slot holds an element of degree 0, the second of degree 1, and the last of degree $n$, and everything after degree $n$ is treated as empty and cut off. Addition goes slot by slot. Each slot inherits its addition from its own vector space.

An algebra graded by degree like this is called a graded algebra. The easiest example is polynomials. $x^2$ and $x^3$ cannot be added, but read as tuples with a slot for each degree it becomes polynomial addition, and multiplication works too. The difference is that polynomials have no upper bound on degree, while the exterior algebra ends at $n$.

This structure lets us read expressions that look like elements of different spaces being added. Writing "$a + b$" is shorthand for $(0, a, 0, \ldots) + (0, 0, b, \ldots) = (0, a, b, \ldots)$. This plus sign is defined only inside the frame of the exterior algebra. Whenever forms of different degrees seem to be added, read that place as such a tuple.

The wedge product inside the algebra

The product of two elements is computed by linearity. In an exterior algebra of dimension 2, wedge $(a, b, c)$ and $(d, e, f)$. Here $a, d$ are numbers, $b, e$ are vectors, and $c, f$ are 2-vectors. Expanding as a sum gives nine pairs, and degrees add. A wedge with a number is plain multiplication, and any pair of degree above 2 is 0 because the dimension is 2. What remains collects by degree like this.

The degree-0 slot is the single term $ad$, the degree-1 slot has $ae$ and $db$, and the degree-2 slot has $af$, $b \wedge e$, $dc$. Terms like $b \wedge f$ or $c \wedge e$, whose degree is 3 or more, all disappear. There is no need to expand into a sum: knowing that the product is “linear between slots” is enough.

Three ways to write components

When $Q$ is an element of the $p$-th exterior power, there are three ways to write its components. I write them for p-forms.

The first sums only over lists whose numbers ascend, and these components are called strict components. There is one component per basis element, so there are exactly $\binom{n}{p}$ of them, the number of basis elements. The second runs over all index lists. Swapping two indices flips the sign of both the wedge product and the component, so the same term of the same basis element appears $p!$ times. We multiply by $\frac{1}{p!}$ to correct. It looks wasteful, but writing it this way lets us use the summation convention, which is convenient. The third writes the same thing in the tensor-product basis; since the components are fully antisymmetric, it writes an element of a lower-dimensional subspace with a larger basis, so it is the least efficient.

Q. What happens to the index lists with a repeated number in the second formula?
A. They contribute nothing, because the wedge product is 0. What remains are the lists with distinct numbers, and each is a rearrangement of one ascending list. Each ascending list has $p!$ rearrangements, all giving the same value, which makes $p!$ times as much.

(For example, with $n = 3$ and $p = 2$ the second formula runs over nine index lists. Among them $(0,0)$, $(1,1)$, $(2,2)$ give 0, and the remaining six form three pairs, $(0,1)$ and $(1,0)$, $(0,2)$ and $(2,0)$, $(1,2)$ and $(2,1)$. The two in a pair give equal values, so $\tfrac{1}{2} \cdot 2 = 1$ and the result equals the sum of the strict components. This example is added for understanding and was not covered in the lecture.)

See it yourself in the toy below. Pick the dimension and the degree and switch between p-forms and p-vectors, and only the basis names change while the counts stay the same. Change the strict components with the − + buttons and tap a basis name, and you see the $p!$ index lists that collect on that basis element, with their signs.

How components transform, and the next post

The lecture says it will treat the transformation of components in the next post. The components in the third formula follow the ordinary tensor transformation rule we learned long ago. The strict components do not follow it, and writing the same object in a different way changes the transformation rule, which leads to the concept of tensor densities, it says.

(Read in the standard way, it goes like this. The strict components $Q_{\alpha\beta\cdots\delta}$ are also components of the same tensor $Q$, so under a change of basis they transform by the ordinary rule summed only over ascending index lists. Grouping terms by antisymmetry brings out the $p \times p$ minors of the transformation matrix, and the minors from the previous post are used here. Only the shape of the rule looks different. This explanation is added for understanding and was not covered in the lecture.)

In the next post we look at the transformation of components.