Tensor #27. Formal Development of p-forms and p-vectors, Part 1텐서 #27. p-형식과 p-벡터의 정식 전개 1

These are my notes on XylyXylyX’s What is Tensor lectures. This post covers lecture 27 of the playlist, Formal development of p-forms and p-vectors Part I. Having reviewed p-vectors and p-forms in the previous post, in this post and the next we rebuild the same objects in a strict order. If we walk the road other books take once, it will be our second time when we meet that book later. This post is the preparation, defining the operators that shuffle indices. The definitions come one after another, but they gather in one place at the end.

Permutations and naming

A permutation $\sigma$ is a one-to-one correspondence that rearranges the numbers $1, \dots, p$ without overlap. $\sigma(1), \sigma(2), \dots, \sigma(p)$ is the same numbers in a changed order, and in this post we count numbers from 1, not 0. A row of indices is written $i_1, i_2, \dots, i_p$ and the shuffled row $i_{\sigma(1)}, \dots, i_{\sigma(p)}$. Subscripts of subscripts appear, which is clumsy, but every notation has a flaw, so we switch between them as the situation calls for.

The number in a vector’s name deserves its own remark. For elements $v_1, v_2, \dots, v_p$ of a vector space $\V$, the lower number is part of the name, not a tensor index. Written in components, as in $v_1 = v_1^{\mu} e_\mu$, the name number and the component index appear together in one expression, and we number the names only because the alphabet runs short.

The permutation operator

Let $A$ be a $(0, p)$ tensor, that is, a tensor whose indices are all covariant. $A$ takes $p$ vectors and returns a real number. For each permutation $\sigma$ we define a new tensor $P_\sigma A$ as follows.

It is the tensor that shuffles the vectors it receives according to $\sigma$ and feeds them to $A$. In general $A$ gives a different value when the order of its inputs changes, so $P_\sigma A$ is a different tensor from $A$.

For example let $p = 2$ and $A = a_{12} \, e^1 \otimes e^2$, with a single component and $a_{11} = a_{21} = a_{22} = 0$. If $\sigma$ is the permutation that swaps 1 and 2, we get this.

So $P_\sigma A = a_{12} \, e^2 \otimes e^1$. Shuffling the inputs and shuffling the order of the basis product give the same result. In components it reads as follows.

In the example, $(P_\sigma A)_{21} = A_{12} = a_{12}$. Shuffling twice is again one shuffle: doing $\pi$ first and $\sigma$ afterwards gives $P_\sigma P_\pi = P_{\sigma\pi}$, where $(\sigma\pi)(k) = \sigma(\pi(k))$. This product is used again and again in the computations below.

Q. Does $P_\sigma$ shuffle the inputs or the component indices?
A. They are the same thing. The definition acts on the inputs, and written in components it happens on the indices. In the example, reading $e^1(v_2) \, e^2(v_1)$ in the other order turns the basis into $e^2 \otimes e^1$, and the component $a_{12}$ moves to the place $21$.

The symmetric projection operator

For a tensor $T$ that is fully covariant (or fully contravariant), applying every possible permutation once and averaging defines the projection operator $S_p$. The lecture first calls it the symmetrizer and then corrects the name to projection operator.

Its components are marked with parentheses. The parentheses signal that this component belongs to $S_p T$, not to $T$ itself.

Let us follow the computation of the component for the index row $1, 3, 1, 2, 5, 6$ in six dimensions. The identity permutation gives $T_{131256}$, swapping only the first two gives $T_{311256}$, and swapping only the second and third gives $T_{113256}$. We add all $6! = 720$ permutations this way and divide by 720. This is an ordinary tensor and not a form, so repeated indices are allowed.

If we compute the rearranged row $3, 1, 1, 2, 5, 6$ the same way, the terms being added are a rearrangement of the same terms, so the value is identical. Only the basis attached to it changes, to $e^3 \otimes e^1 \otimes \cdots$. This is what a symmetric tensor is. On the other hand $1, 3, 3, 2, 5, 6$ is not a rearrangement of the earlier row, so it gives an independent component.

The important property of this operation is that $S_p T$ does not change when a permutation is applied.

This is because when $\pi$ runs over all permutations, $\sigma\pi$ also runs over all permutations once. Note that this does not say $P_\sigma T = T$. What stays unchanged is the symmetric part of $T$. (The name projection can be understood as coming from the fact that applying $S_p$ twice equals applying it once. This follows directly from the formula above and was not covered in the lecture.)

The antisymmetric projection operator

The antisymmetric side is trickier because signs come in. It leads to determinants rather than plain sums. The definition is this.

$\varepsilon(\sigma)$ is the Levi-Civita symbol: $+1$ if $\sigma$ is an even number of swaps and $-1$ if odd. An antisymmetric tensor $Q$ is defined as a tensor that only picks up a sign when a permutation is applied.

What $K_p$ does must agree with this definition. We check two things. First, it must do nothing to an already antisymmetric $Q$.

Since $\varepsilon(\sigma)^2 = 1$ and there are $p!$ permutations, $Q$ is added $p!$ times. The factor $1/p!$ in front is the normalization that makes the result come out as $Q$ itself. Second, starting from any $T$, the result $K_p T$ must be antisymmetric.

The middle equality uses $\varepsilon(\pi) = \varepsilon(\sigma) \, \varepsilon(\sigma\pi)$, and the last uses that $\sigma\pi$ runs through every permutation once. Together the two checks say that $K_p$ is a projection onto the space of antisymmetric tensors. Every element of that space is its own antisymmetric part, so there is no antisymmetric tensor that fails to be the antisymmetric part of some tensor.

Q. Why does $\varepsilon(\pi) = \varepsilon(\sigma) \, \varepsilon(\sigma\pi)$ hold?
A. Signs of permutations multiply. $\varepsilon(\sigma\pi) = \varepsilon(\sigma) \, \varepsilon(\pi)$, because the numbers of swaps add. Multiplying both sides by $\varepsilon(\sigma)$ and using $\varepsilon(\sigma)^2 = 1$ gives the formula above. (This explanation is added for understanding and was not covered in the lecture.)

The toy below fixes the components of a $(0, 2)$ or $(0, 3)$ tensor in three dimensions and computes $P_\sigma T$, $S_p T$ and $K_p T$. Tap a cell to change its value and pick a permutation. For any $T$, $S_p T$ stays put under $P_\sigma$, and $K_p T$ only picks up the sign $\varepsilon(\sigma)$.

Q. Can we write $T$ as the sum of $S_p T$ and $K_p T$?
A. For $p = 2$ yes: $T_{(ij)} + T_{[ij]} = T_{ij}$. From $p = 3$ on, generally not. For $T = e^1 \otimes e^1 \otimes e^2$ a repeated index gives $K_3 T = 0$ but $S_3 T \neq T$, because a part that is neither symmetric nor antisymmetric is left over. (This example is added for understanding and was not covered in the lecture.)

Components of the antisymmetric part

The components of the antisymmetric part are marked with square brackets. The brackets signal a component of $K_p T$, while the parentheses were for $S_p T$.

Here $\delta^{i_1 \dots i_p}_{j_1 \dots j_p}$ is the generalized Kronecker delta we learned earlier. If the upper row of indices is a permutation of the lower row it equals the sign of that permutation, and otherwise it is 0. This symbol picks the sign $\varepsilon$ that was attached to each permutation in the sum, all at once. So only the terms whose upper indices match survive and the rest vanish. Some books put the $1/p!$ inside the delta by definition, so in this post $\delta$ is taken to give only signs and zeros, and the $1/p!$ is written outside.

In practice we choose a row of distinct indices and compute. For example if we set $j = (1, 3, 4, 2, 6)$, these numbers pick the basis element we are computing, and all the $i$ indices are summed over. This is the “sum over all permutations” of the definition written in one line. If a number repeats within $j$, the delta gives 0 and so the component is 0.

One tensor, two bases

The formula above writes the antisymmetric part in the tensor product basis $e^{j_1} \otimes \cdots \otimes e^{j_p}$. $K_p T$ is still an element of the space of $(0, p)$ tensors, but because it is antisymmetric it sits in a smaller subspace, the space of $p$-forms. Its components differ only by sign or are 0, so this basis is full of redundancy. Grouping equal terms lets us rewrite it in the strict basis.

This is the same tensor written in a different basis. The distinction matters in the coming posts, because the transformation rule for tensors we have learned is for components in the tensor product basis, and components in the strict basis change differently.

In this post we defined three operators: the permutation operator $P_\sigma$, the symmetric projection operator $S_p$, and the antisymmetric projection operator $K_p$. In the next post we use them to define the wedge product formally.