Tensor #28. Formal Development of p-forms and p-vectors, Part 2텐서 #28. p-형식과 p-벡터의 정식 전개 2

These are my notes on XylyXylyX’s What is Tensor lectures. This post covers lecture 28 of the playlist, Formal development of p-forms and p-vectors Part 2. Having defined the permutation operator and the antisymmetric projection operator in the previous post, this time we use them to define the wedge product formally and connect it to the wedge we built by hand earlier. The destination is always the same: a wedge of 1-forms (or vectors) is a totally antisymmetric tensor, and it is an element of an exterior power. Below, $\omega_1, \dots, \omega_p \in \V^*$ are 1-forms and the lower number is a name. For p-vectors the analysis is the same with $\V$ in place of $\V^*$, so I write only p-forms.

The definition and two checks

The wedge product is defined as the complete antisymmetrization of the tensor product of the same 1-forms. The $\equiv$ in the formula below marks that the meaning of the symbol is set this way, not that an equality was proved.

The $K_p$ on the right carries a factor $1/p!$, so multiplying by $p!$ leaves the sum over permutations with its signs. In words, we replace the wedges by tensor products, antisymmetrize, and multiply by $p!$. We check that this connects to the old definition with $p = 2$. There are only two permutations, the identity and the swap, so we get this.

For $p = 3$ there are six permutations: the even ones $(123), (231), (312)$ carry $+$ and the odd ones $(213), (132), (321)$ carry $-$.

Swapping any two positions turns each term into a term of opposite sign among the six, so the whole is totally antisymmetric.

Q. Why is the definition multiplied by $p!$ after $K_p$?
A. $K_p$ is an average and carries $1/p!$. Multiplying makes every term’s coefficient $\pm 1$, and the old hand-written form $\omega_1 \otimes \omega_2 - \omega_2 \otimes \omega_1$ comes out unchanged. The $p = 2$ check is the evidence.

The notation as a map

There is also a notation that treats the wedge as a single operation. When the 1-forms are elements of $\V^*$, $\wedge$ is a map from the $p$-fold product of $\V^*$ to the exterior power, taking an ordered list of $p$ dual vectors and returning one element. Antisymmetry is written like this.

Expanding in a basis gives minors

We expand the 1-forms inside the wedge in the coordinate basis. A form with a name number is written $\omega_k = \Omega_{k\mu} \, dx^\mu$. The first index $k$ is a name and is not summed, and only the second index $\mu$ is summed. The capital $\Omega$ is used to tell the name index from the component index by eye, and it is not a transformation matrix. Pulling the coefficients out by linearity gives this.

The sum runs over all values of $\mu$. Now we split the sum in two steps. First pick a set of indices with $i_1 < \cdots < i_p$, then sum over all permutations of that set. Running over the sets and then over the permutations passes through every index row exactly once, so it is the same sum. The basis wedge shuffled by a permutation $\sigma$ differs only by a sign.

Absorbing this sign into the coefficient, the coefficient becomes the following sum, which is exactly the definition of a determinant.

This says the strict-basis components are the $p \times p$ minors obtained by picking $p$ columns from the $p \times n$ component matrix. What we did informally in the previous post, stacking the components and taking minors, is this formula. For example opening $v \wedge w \wedge z$ in four dimensions, the determinant of columns 0, 1, 2 of the $3 \times 4$ matrix that stacks the components of the three vectors in three rows is the coefficient of $dx^0 \wedge dx^1 \wedge dx^2$. Picking columns 0, 1, 3 gives the coefficient of $dx^0 \wedge dx^1 \wedge dx^3$, and four minors appear in all. Since exactly $p$ columns are picked, the minor is always a square matrix.

Q. A general $p$-form may be a sum of terms not of this chained shape, so is this formula enough?
A. It is enough. A general $p$-form is a sum of such simple forms, and since every operation is linear, what holds for simple forms holds for sums.

The wedge adds degrees

For the exterior algebra to be an algebra, the wedge had to take a $p$-form and a $q$-form and give a $(p+q)$-form. Last time this was explained only in words, and now we can check it with formulas. By linearity it is enough to look at the simple forms $\omega_1 \wedge \cdots \wedge \omega_p$ and $\eta_1 \wedge \cdots \wedge \eta_q$. Expanding each in the minor form above and multiplying, the product of two coefficients sits in front of a basis wedge $dx^{I} \wedge dx^{J}$. Here $I$ is an ascending index set of length $p$ and $J$ one of length $q$. If the two sets share an index, this wedge is 0. If they do not, restoring ascending order costs only a sign, so it has the form $\pm dx^{K}$, where $K$ is the ascending union of the two sets, of length $p+q$. This is a strict basis element of the $(p+q)$-forms, so the result is a $(p+q)$-form. The lecture reaches the same conclusion by merging the coefficients of the two expansions into one list of $p+q$ 1-forms and renaming them. A shared index gives 0, so if $p + q > n$ the result is 0. (This explanation counts the basis indices directly and was not covered in the lecture.)

Under this product, the exterior algebra, which bundles the spaces of each degree by a direct sum, becomes a graded algebra. The structure in which the product adds degrees is the reason for the name.

Feeding vectors gives a determinant

A $p$-form is a $(0, p)$ tensor, so it takes $p$ vectors and returns a real number. This value is one of the two rules to remember from this post.

Each 1-form is a map in the dual space taking a vector to a real number, so filling in $\omega_i(v_j)$ for all $i, j$ gives a $p \times p$ real matrix. The value is its determinant. It follows directly from the definition. Unwinding $p! \, K_p$ leaves a sum over permutations whose terms are $\varepsilon(\sigma) \prod_k \omega_k(v_{\sigma(k)})$, and this sum is the determinant. (This explanation unwinds the definition and was not covered in the lecture.) For $p = 2$ it is $\omega_1(v)\omega_2(w) - \omega_1(w)\omega_2(v)$.

In the toy below, fix the components of the 1-forms $\omega_i$ and the vectors $v_j$ and it shows the matrix $[\omega_i(v_j)]$ and its determinant, and computes the same value as the permutation sum of the definition and as the sum of products of minors. Swapping $v_1$ and $v_2$ only flips the sign, and when two vectors are proportional the value is 0.

Q. Why is the value 0 when two vectors are proportional?
A. Two columns of the matrix $[\omega_i(v_j)]$ become proportional, so the determinant is 0. It can also be read as antisymmetry: feeding the same vector twice makes the value the negative of itself, so it is 0.

The induced inner product

If the vector space has an inner product, an inner product is fixed on the dual space, and it in turn fixes one on every tensor product space and on the exterior powers inside them. This is the second rule to remember. The inner product between two simple $p$-forms is the determinant of the matrix whose entries are the inner products between 1-forms.

Whether forms or vectors, it makes sense only between two elements from the same space. Some books write the inner product with a vertical bar like $\langle \cdot | \cdot \rangle$ instead of parentheses and a comma, a notation often seen in quantum mechanics. The lecture says the formula is not hard to prove, but skips the proof and asks us to remember it. These two rules will be used again in the following posts, which treat transformations.