bijective proof

- binary linear code
- chord diagram
- combinatorial design
- matroid
- permutation
- Latin square
- shuffle
- Young diagram

category: combinatorics

In enumerative combinatorics, a “bijective proof” refers to a basic method of counting the number of structures of a certain type supported on a finite set of underlying points, by analyzing structure in two different ways. It is really a special case of “categorification”: an identity $a = b$ where $a$ and $b$ are natural number expressions is proved by taking underlying cardinalities of a bijection $A \cong B$ between sets of structures, hence the name. (If taking cardinalities is an example of “decategorifying”, then going the other way from $a = b$ to $A \cong B$ would be “categorifying” or promoting an equation to an isomorphism.)

In some sense this is what the theory of species is all about: formalizing the operations adhering to bijective proofs. Often bijective proofs give an impression of elegance or enlightenment: rather than prove natural number or polynomial identities through acts of symbolic manipulation, one *sees* the identity by drawing pictures of structures that the two sides of the identity are counting.

For the moment we’ll just jot down some possible examples for illustrating the method; these are to be filled in later, or to be linked to other nLab articles where the bijective proof is given.

See for now binomial theorem.

For more details, see the Wikipedia article.

Algebraically, one may introduce the $q$-binomial coefficients $\binom{n}{k}_q$ by introducing a $q$-affine plane (or quantum affine plane, to be connected with quantum affine algebra) and its associated (non-commutative) algebra

$k[x, y]_q = k\langle x, y \rangle/(y x = q x y)$

where $q \in k$ is a parameter, and then writing $(x + y)^n = \sum_{k = 0}^n \binom{n}{k}_q x^k y^{n-k}$.

However, an illuminating combinatorial interpretation of the $q$-binomial coefficients is that they count points in a Grassmannian consisting of $k$-dimensional linear subspaces of an $n$-dimensional vector space $\mathbb{F}_q^n$ over a finite field $\mathbb{F}_q$ of cardinality $q$. This interpretation can be used to provide a bijective proof of the $q$-Pascal identity

$\binom{n}{k}_q = \binom{n-1}{k}_q + q^{n-k}\binom{n-1}{k-1}_q.$

For, if we choose a standard affine hyperplane $H = \{x_n = 1\}$ in $\mathbb{F}_q^n$, then any linear $k$-plane $W$ in $\mathbb{F}_q^n$ either intersects $H$ in an affine $(k-1)$-plane in the affine $(n-1)$-space, or intersects $H$ in the empty set, i.e., $W$ is contained in the linear space $\{x_n = 0\}$. The number of $W$ satisfying the latter condition is $\binom{n-1}{k}_q$, and the number of $W$ satisfying the former is in bijection with the number of $(k-1)$-dimensional affine subspaces of $\mathbb{F}_q^{n-1}$, which is $q^{n-1}$ times the number of linear subspaces $\binom{n-1}{k-1}_q$ divided by $q^{k-1}$, the number of translations which stabilize a given affine subspace. (Maybe to be rewritten later.)

Cayley’s theorem, here not to be confused with Cayley’s observation that every group (monoid) is a subgroup (submonoid) of a permutation group (endofunction monoid), says that the number of tree structures on an $n$-element set $\{x_1, \ldots, x_n\}$ is $n^{n-2}$.

There are many proofs of this fact in the literature. One, due to Gilbert Labelle and made popular by André Joyal, proceeds by establishing a bijection between trees equipped with a pair $(x_i, x_j)$ of nodes (allowing $x_i = x_j$) and the number of endofunctions on $X = \{x_1, \ldots, x_n\}$.

On the one hand, such a structure $(T, x_i, x_j)$ on $X$ can be identified with a linearly ordered set (the path $P$ from $x_i$ to $x_j$ in $T$) together with, for each node $y$ along that path, a rooted tree structure $T_y$ consisting of nodes $x \in X$ for which the shortest path in $T$ from $x$ to $P$ ends at $y$. (As a subgraph of $T$, this naturally forms a tree $T_y$ rooted at $y$.) Thus, such a structure $(T, x_i, x_j)$ on $X$ determines and is determined by an equivalence relation $E$ on $X$, a rooted tree structure on each equivalence class, and a linear ordering of the set $X/E$ of equivalence classes.

On the other hand, each endofunction on $X$, say $f: X \to X$, has an eventual image or stable set $stab(f) = \bigcap_{n \geq 0} f^n(X)$, on which $f$ acts by a permutation (i.e., bijectively). For each $y \in stab(f)$, there is a rooted tree $T_y$ which consists of all $x \in X$ for which the set of iterates $x, f(x), f^2(x), \ldots$ first lands in $stab(f)$ at $y$. Thus an endofunction on $X$ determines and is determined by an equivalence relation $E$ on $X$, a rooted tree structure on each equivalence class, and a permutation on the set $X/E$ of equivalence classes.

Since the number of permutations on $X/E$ is equinumerous with the number of linear orderings of $X/E$, we conclude that the set of “bipointed tree” structures on $X$ is equinumerous with the set of endofunctions on $X$. Thus the number of bipointed tree structures $(T, x_i, x_j)$ on $X$ is $n^n$, and therefore the number of tree structures $T$ on $X$ is $n^{n-2}$.

One method for proving polynomial identities (equations $p = q$ between polynomials $p, q \in \mathbb{C}[x_1, \ldots, x_k]$ in several variables) is first to give a bijective proof for the special case in which natural numbers $m_i$ are substituted for the variables $x_i$, and then reason that there are enough natural numbers that the polynomial identity must hold generally.

More formally, two polynomials are equal, $p(x_1, \ldots, x_k) = q(x_1, \ldots, x_k)$, if $p(m_1, \ldots, m_k) = q(m_1, \ldots, m_k)$ for all choices of natural numbers $m_i$. Relatedly, the set $\mathbb{N}^k \subseteq \mathbb{C}^k$ is Zariski-dense (i.e., the closure of $\mathbb{N}^k$ in $\mathbb{C}^k$ equipped with the Zariski topology is all of $\mathbb{C}^k$).

This is intuitively obvious of course. As a public service, we prove a more general result.

Suppose $B$ is an infinite subset of an integral domain $A$. Then a polynomial $p(x_1, \ldots, x_k) \in A[x_1, \ldots, x_k]$ is uniquely determined by its values it takes at arguments $b_1, \ldots, b_k \in B$.

It suffices to show that if $p(b_1, \ldots, b_k) = 0$ for every choice of $b_i \in B$, then $p = 0$. We prove this by induction on $k$. The case $k = 0$ is trivial.

Assuming the assertion as induction hypothesis for case $k$, suppose $p \in A[x_1, \ldots, x_k, x_{k+1}]$ satisfies $p(b_1, \ldots, b_k, b) = 0$ for all choices $b_i \in B$ and $b \in B$. Then by induction, for each fixed $b \in B$ the polynomial $p(x_1, \ldots, x_k, b) \in A[x_1, \ldots, x_k]$ is identically zero. Thus if $F$ is the field of fractions of $A[x_1, \ldots, x_k]$, and if we regard $p$ as an element of the Euclidean domain $F[x_{k+1}]$ under the obvious embedding

$A[x_1, \ldots, x_k, x_{k+1}] \cong A[x_1, \ldots, x_k][x_{k+1}] \hookrightarrow F[x_{k+1}],$

then $p$ has infinitely many roots $b$ in $A$ and thus in $F$, forcing $p$ to be identically zero.

Last revised on August 26, 2018 at 08:15:59. See the history of this page for a list of all contributions to it.