# help understand proof $(\forall \rho, \sigma \in P_n)\ \varepsilon_{\rho\sigma} = \varepsilon_{\rho}\varepsilon_{\sigma}$

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 $$i and $$\sigma(j)<\sigma(i)$$. For every $$\sigma \in P_n$$ the signum (or signature) of $$\sigma$$ is defined by $$\varepsilon_\sigma = (-1)^{I(\sigma)}$$

Proof.

Consider the product $$V_n = \prod_{i For every $$\sigma \in P_n$$ define $$\sigma(V_n) = \prod_{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 $$\sigma(V_n) = \prod_{i, we may get values of $$\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 $$V_n = \prod_{i and $$\sigma(V_n)=\prod_{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.

• 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... – Oleksandr Khryplyvenko Oct 18 '20 at 21:25
• @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. – Alan Wang Oct 19 '20 at 1:05
• 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 – Oleksandr Khryplyvenko Oct 19 '20 at 12:12
• 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... – Oleksandr Khryplyvenko Oct 19 '20 at 13:45
• @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 $-V_n$. – Alan Wang Oct 19 '20 at 13:50