In the previous blog post we’ve discussed a proof of the fundamental fact, which is that the signature of a permutation is well defined: every permutation is either expressible as an even number of transpositions or an odd number of transpositions but not both. The main approach we’ve focused on was an approach through exterior algebra and the so-called analyzing map.

In this post, we’re going to discuss a simpler approach that the author has found which is considerably simpler but uses a lot of the same concepts introduced in the previous blog post, namely that of exterior algebras. The basic idea is that instead of using the analyzing map, we reduce the exterior algebra mod 2.

Let $n$ be a positive integer, $M = \mathbb{Z}^n$, $E_M$ be the exterior algebra over $M$. Let $e_1, \ldots, e_n \in M$ be the standard basis elements of $M$. Recall that to prove that the signature homomorphism $\rho$ is well defined for the symmetric group on $n$ elements $S_n$, it’s enough to show that $e_1 \ldots e_n \neq 0$ in $E_M$. Then we simply define, for a permutation $\sigma \in S_n$, $\rho(\sigma) \in \{ -1, 1 \}$ using the following equation

\[e_{\sigma(1)}\ldots e_{\sigma(n)} = \rho(\sigma) \ e_1 \ldots e_n\]

In the following, we denote by $\mathbb{Z}_2$ the ring of integers mod 2. Define $N = (\mathbb{Z}_2)^n \cong M \otimes \mathbb{Z}_2$. There is an obvious $\mathbb{Z}$-module homomorphism $r : M \to N$ given by coordinate wise reduction mod 2:

\[r(a_1, \ldots, a_n) = (\overline{a_1}, \ldots, \overline{a_n})\]

where for $a \in \mathbb{Z}$, $\overline{a} \in \mathbb{Z}_2$ is $a$ reduced mod 2.

Since $r$ is a homomorphism of $\mathbb{Z}$-modules, it extends naturally to a homomorphism of the corresponding exterior algebras

\[R: E_M \to E_N\]

This is helpful for the following reason. $E_N$ is actually a symmetric algebra. This is because for any $n \in N$, $n = -n$, so the same is true by extension for all elements of $E_N$. Actually, $E_N$ is isomorphic to the quotient polynomial ring $\mathbb{Z}_2[x_1, \ldots, x_n]/(x_1^2,\ldots, x_n^2)$.

To see why this is true, remember the construction of $E_N$. It’s constructed by taking the tensor algebra $T_N$ over $N$ and quotienting by the ideal $I$ generated by all elements of the form $n \otimes n \in T_N$. Consider the subideal $I_1$ of $I$ generated by elements of the form $n_1 \otimes n_2 - n_2 \otimes n_1$. Then $T_N/I_1$ is isomorphic to the polynomial ring $Q = \mathbb{Z}_2[x_1, \ldots, x_n]$ where $x_1, \ldots, x_n$ correspond to the standard basis elements $\overline{e_1}, \ldots, \overline{e_n}$ of $N$. Moreover under this isomorphism the image of the full ideal $I$ in $T_N/I_1$ corresponds to the polynomial ideal $(x_1^2, \ldots, x_n^2)$. Thus in fact

\[E_N \cong T_N / I \cong (T_N / I_1) / (I (T_N / I_1)) \cong \mathbb{Z}_2[x_1, \ldots, x_n] / (x_1^2, \ldots x_n^2)\]

Crucially, these are all algebra homomorphisms. Thus the basic $n$-element $e_1 \ldots e_n \in E_M$ maps under $R$ to

\[R(e_1 \ldots e_n) = R(e_1) \ldots R(e_n) = x_1 \ldots x_n \neq 0 \in \mathbb{Z}_2[x_1, \ldots, x_n] / (x_1^2, \ldots, x_n^2)\]

Thus we also have that $e_1 \ldots e_n \neq 0$ in $E_M$, which was to be proven.