Given a finite set \(U\) with \(n\geq 1\) elements, how would you prove that there are as many subsets of \(S\) with even cardinality as there are with odd cardinality? One way is to use the binomial theorem: \[ 0^n = (1-1)^n = \sum_{k=0}^n{n \choose k}(-1)^k = \sum_{\text{$k$ even}}{n \choose k} - \sum_{\text{$k$ odd}}{n \choose k}. \] This identity also holds when \(n=0\), for \(0^0=1\) and there is exactly one subset of the empty set, and its cardinality is even. So that’s a neat proof, but not very combinatorial. For a more combinatorial proof we can use a so called sign-reversing involution.

A function \(f:X\to X\) is an *involution* if it is its own inverse, or, in symbols, \(f(f(x))=x\). Assume that a function \(\epsilon:X\to\{-1,1\}\) is given. We refer to \(\epsilon(x)\) as the *sign* of \(x\). Now, \(f\) is a *sign-reversing involution* if for each non fixed point \(x\) in \(X\) the sign of \(x\) is reversed by \(f\); that is, \(\epsilon(f(x))=-\epsilon(x)\). If \(f\) is a sign-reversing involution of \(X\), then \[\newcommand{Fix}{\mathrm{Fix}}
\sum_{x\in X}\epsilon(x) = \!\!\sum_{x\in \Fix(f)}\!\!\!\epsilon(x),
\] where \(\Fix(f)=\{x\in X: f(x)=x\}\) denotes the set of fixed points of \(f\). This is because, under \(f\), each negative non fixed point is paired with a unique positive non fixed point, and vice versa. For counting purposes we usually also require that all fixed points of \(f\) have the same sign, and then the right hand sum is \(\pm|\Fix(f)|.\)

When counting subsets \(S\) of \(U\) with respect to parity, a natural sign function is \(\epsilon(S) = (-1)^{|S|}\). Assuming \(U\) is equipped with a total order and that \(\hat 0\) denotes the smallest element of \(U\), a sign reversing involution is given by \[ f(S) = \begin{cases} S \setminus \{\hat 0\} & \text{if $\hat 0\in S$}, \\ S \cup \{\hat 0\} & \text{if $\hat 0\notin S $}. \end{cases} \] For instance, if \(U=\{1,2\}\) then \(f(\emptyset)=\{1\}\), \(f(\{1\})=\emptyset\), \(f(\{2\})=\{1,2\}\), and \(f(\{1,2\})=\{2\}\). Note that \(f\) is fixed point free; thus \(\Fix(f)=\emptyset\) and \[ \#\{S\subseteq U: \text{$|S|$ even}\} - \#\{S\subseteq U: \text{$|S|$ odd}\} = |\Fix(f)| = 0. \] Now, that’s a simple and beautiful proof, but is there a more general and “natural” combinatorial proof? E.g. do we have to assume a total order on \(U\) and do we have to mention special elements, such as \(\hat 0\)? Before proceeding to the next paragraph you might want to take a moment and try to come up with such a proof.

Let \(E\) be the combinatorial species of sets, defined by \(E[U]=\{U\}\). Its exponential generating function is \(E(x)=e^x\), and its multiplicative inverse is the virtual species \(E^{-1}\) such that \(E\cdot E^{-1}=1\), where \(1[U]=\{U\}\) if \(U=\emptyset\) and \(1[U]=\emptyset\) otherwise. Note that \[ E^{-1} = (1+E_+)^{-1} = \sum_{k\geq 0} (-1)^k(E_+)^k, \] where \(E_+\) denotes the species of nonempty sets. Thus, \(E^{-1}\) is the species of signed ballots (also called ordered set partitions), where the sign of a ballot \(B_1 B_2\dots B_k\) is \((-1)^k\); that is, the parity of the number of blocks. For instance, \(\{d\}\{a,c,e\}\{b\}\) is a ballot of \(U=\{a,b,c,d,e\}\) and its sign is \((-1)^3=-1\).

The species of subsets \(\newcommand{\Pow}{\mathfrak{P}}\Pow\) is defined by \(\Pow[U]=\{(S,U\setminus S): S\subseteq U\}\). Note that \[ (E \cdot E)[U] = \bigcup_{S\subseteq U} E[S] \times E[U\setminus S] = \bigcup_{S\subseteq U} \{S\} \times \{U\setminus S\} = \Pow[U]. \] That is, \(\Pow=E^2\) and its exponential generating function is \[ \Pow(x)=E(x)^2 = e^{2x}=\sum_{n\geq 0}2^n \frac{x^n}{n!}. \] Further, counting subsets with respect to the sign \((-1)^{|S|}\) we get \[E(x)E(-x) = e^{x}e^{-x} = e^0 = 1. \] Thus, the species interpretation of there being as many subsets of even cardinality as of odd cardinality is \(E\cdot E^{-1}=1\), where the \(1\) on the right hand side stems from the case when the underlying set is empty. We will give a combinatorial proof of this species identity using a sign-reversing involution. The objects of \(E\cdot E^{-1}\) are pairs \((S, \beta)\) such that \(S\subseteq U\) and \(\beta=B_1 B_2\dots B_k\) is a signed ballot of \(U\setminus S\). For example, \((E\cdot E^{-1})[\{a,b,c\}]\) consists of the pairs \[ \begin{array}{c|c} \text{positive pairs} & \text{negative pairs} \\ (\{a,b,c\}, \emptyset) & (\emptyset,\{a,b,c\}) \\ (\emptyset, \{a\}\{b,c\}) & (\{a\}, \{b,c\}) \\ (\emptyset, \{b\}\{a,c\}) & (\{b\}, \{a,c\}) \\ (\emptyset, \{c\}\{a,b\}) & (\{c\}, \{a,b\}) \\ (\emptyset, \{a,b\}\{c\}) & (\{a,b\}, \{c\}) \\ (\emptyset, \{a,c\}\{b\}) & (\{a,c\}, \{b\}) \\ (\emptyset, \{b,c\}\{a\}) & (\{b,c\}, \{a\}) \\ (\{a\}, \{b\}\{c\}) & (\emptyset, \{a\}\{b\}\{c\}) \\ (\{a\}, \{c\}\{b\}) & (\emptyset, \{a\}\{c\}\{b\}) \\ (\{b\}, \{a\}\{c\}) & (\emptyset, \{b\}\{a\}\{c\}) \\ (\{b\}, \{c\}\{a\}) & (\emptyset, \{b\}\{c\}\{a\}) \\ (\{c\}, \{a\}\{b\}) & (\emptyset, \{c\}\{a\}\{b\}) \\ (\{c\}, \{b\}\{a\}) & (\emptyset, \{c\}\{b\}\{a\}) \\ \end{array} \] This suggests the natural sign-reversing involution \[ f(S,B_1 B_2 \dots B_k) = \begin{cases} (B_1, B_2 B_3 \dots B_k) &\text{if $S=\emptyset$,}\\ (\emptyset, SB_1 B_2 \dots B_k) &\text{if $S\neq\emptyset$.} \end{cases} \] The table above is arranged so that pairs on the same row are images of each other under \(f\).

More generally, if \(F\) is a species such that \(|F[\emptyset]|=1\) then the multiplicative inverse, \(F^{-1}\), is the virtual species of lists \(\alpha_1\alpha_2\dots\alpha_k\) of nonempty \(F\)-structures in which the sign is \((-1)^k\). A proof of \(F\cdot F^{-1}=1\) is given by the sign-reversing involution \[ f(\alpha,\alpha_1 \alpha_2 \dots \alpha_k) = \begin{cases} (\alpha_1, \alpha_2 \alpha_3 \dots \alpha_k) &\text{if $\alpha\in F[\emptyset]$,}\\ (\emptyset, \alpha \alpha_1 \alpha_2 \dots \alpha_k) &\text{if $\alpha\notin F[\emptyset]$.} \end{cases} \] It has exactly one fixed point, namely \((\emptyset, \emptyset)\).

For more examples of the use of sign-reversing involutions the reader might want to have a look at my paper with Stuart Hannah.

Finally I’d like to thank my friend Bjarki for suggesting that I write this post.