Given a permutation $\sigma \in P_n$, let $I(\sigma)$ be the number of inversions in $\sigma$, i.e. the number of pairs $(i, j)$ with $i<j$ and $\sigma(j)<\sigma(i)$. For every $\sigma \in P_n$ the signum (or signature) of $\sigma$ is defined by $\varepsilon_\sigma = (-1)^{I(\sigma)}$


Consider the product $$V_n = \prod_{i<j}(j-i)$$ For every $\sigma \in P_n$ define $$\sigma(V_n) = \prod_{i<j}[\sigma(j)-\sigma(i)]$$ Since $\sigma$ is a bijection, every factor of $V_n$ occurs precisely once in $\sigma(V_n)$, up to a possible change in sign. Consequently we have $$\sigma(V_n) = (-1)^{I(\sigma)}V_n=\varepsilon_{\sigma}V_n$$ Given $\rho, \sigma \in P_n$ we have similarly $\rho\sigma(V_n) = \varepsilon_{\rho}\sigma(V_n)$. Consequently, $$\varepsilon_{\rho\sigma}V_n=\rho\sigma(V_n) = \varepsilon_{\rho}\sigma(V_n)=\varepsilon_{\rho}\varepsilon_{\sigma}V_n$$ whence, since $V_n \neq 0$, we obtain $\varepsilon_{\rho\sigma} = \varepsilon_{\rho}\varepsilon_\sigma$

I've googled for the different proofs of this theorem, but they involve cycles, parity, etc - way simpler concepts in terms of group theory. And I do understand them.

However, I found this very proof in a book on linear algebra, in "determinants" chapter. And I found myself completely messed up with this proof.

Here are my questions:

  1. What does $I(\sigma)$ show(what is it's meaning)? What is it's domain? What is it's range(meaning)? It's a function $I:n \times n \to \mathbb N \cup \{0\}$, but what is the meaning of this function?

  2. if $\sigma \in P_n$, then range of $\sigma$ is 1..n. And sigma is already defined to be one of $P_n$. Now if we define $\sigma(V_n) = \prod_{i<j}{[\sigma(j)-\sigma(i)]}$, we may get values of $\sigma > n$. So why already defined $\sigma$ is being altered?

  3. What "factors" of $V_n$ do occur precisely once in $\sigma(V_n)$? And where: in domain or in range? And what "change of sign" is mentioned in "up to a possible change of sign"?

I'm asking these questions because I completely can't get any logic path between cause and effect in this argument & it seems using not properly defined notions... Maybe the author is using too cryptic/ambiguous notation, it's just not formal enough for me to understand it without clarifications/tutor. But this book contains "basic linear algebra" in it's title, and I guess it may be used for self study. At least I studied more than 135 of 200 pages without any assistance & this is the very first theorem I'm lost with.

  1. $I(\sigma)$ just count the number of inversions in $\sigma$. For example, let $$\sigma=\begin{pmatrix}1&2&3&4&5&6\\ 3&4&6&2&5&1 \end{pmatrix}.$$ Then $(1,4)$ is an inversion in $\sigma$ since $\sigma(1)>\sigma(4).$ It can be checked that all inversions in $\sigma$ are $$(1,4),(1,6),(2,4)(2,6),(3,4),(3,5)(3,6),(4,6),(5,6).$$ Hence $I(\sigma)=9$.
    The domain of $I$ is $P_n$, while the range is $\{0,1,\dots,\binom{n}{2}\}$.

  2. Strictly speaking, for every $\sigma\in P_n$, $\sigma$ acts on $V_n$ by the rule given. So the function $\sigma$ is not altered, but we want to observe how $\sigma$ affects $V_n$. To avoid confusion, you can check that some books defined $$V_n = \prod_{i<j}(x_j-x_i)$$ and $$\sigma(V_n)=\prod_{i<j}(x_{\sigma(j)}-x_{\sigma(i)})$$

  3. Here I give example by considering $P_3$. Let $\sigma=(123)$. Then $V_n=(2-1)(3-1)(3-2)$ and $\sigma(V_n)=(3-2)(1-2)(1-3)$. You can see that the factors $(2-1),(3-1),(3-2)$ all occurs exactly once in $\sigma(V_n)$ but the sign of $(2-1),(3-1)$ are changed to $-(2-1),-(3-1)$ in $\sigma(V_n)$.

You may refer the book Introduction to Group Theory by Walter Ledermann, pages 133-135 for a proof of this result using similar methods.

  • $\begingroup$ If $P_n$ is the domain of $I$, now I see what the "number of inversions in $\sigma$" means! And as for (2) and (3), what does $x_j$ notation mean? And is $V_n$ a scalar?(it seems to be, from notation). Also, $\sigma$ takes a scalar as an argument. And can we define $\sigma(a*b)$? As what happens in $\sigma(V_n)=\prod_{i<j}(...)$ seems nonsense to me, as $\sigma$ has to be defined over product AND be linear... Just mess. So until we prove this fact, this equation is somewhat doubtful, unless I understand wrong the notation itself... $\endgroup$ – Oleksandr Khryplyvenko Oct 18 '20 at 21:25
  • $\begingroup$ @OleksandrKhryplyvenko $x_j$ are indeterminates. In this case, $V_n$ is a polynomial in $n$ variables. We want to observe that after $\sigma$ "acts on" (or affect) $V_n$, how many factors change their sign. $\endgroup$ – Alan Wang Oct 19 '20 at 1:05
  • $\begingroup$ Now I see, the notation of $\sigma(V_n)$ itself is informal. Strictly speaking, we don't evaluate $\sigma(V_n)$, we just write it down for 'handy usage', and we just evaluate the $\prod_{i<j}(x_{\sigma(j)}- x_{\sigma(i)})$ expression. The last part for me to understand is "Since 𝜎 is a bijection, every factor of 𝑉𝑛 occurs precisely once in 𝜎(𝑉𝑛), up to a possible change in sign". It's not obvious. I'll try to prove it myself $\endgroup$ – Oleksandr Khryplyvenko Oct 19 '20 at 12:12
  • $\begingroup$ I referred to the book you mentioned, but the explanation is priceless: "Clearly, if the indeterminates are subjected to a permutation $\alpha$, the function $\Delta$ either remains unchanged or else is multiplied by -1" Clearly!!! Unfortunately it's not "Clearly" for me why... $\endgroup$ – Oleksandr Khryplyvenko Oct 19 '20 at 13:45
  • $\begingroup$ @OleksandrKhryplyvenko They tried to skip the proof for that part because the proof is tedious. The main point is for $\sigma\in P_n$, $\sigma(V_n)$ must be $V_n$ or $-V_n$. $\endgroup$ – Alan Wang Oct 19 '20 at 13:50

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