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Archive for May, 2011

## Boundary operators

May 12, 2011 8 comments

Consider the vector space of polynomials with coefficients on a field $\mathbb{F},$ $\mathbb{F}[X]$, with the obvious sum of functions and scalar multiplication. For each $n \in \mathbb{N}$, consider the subspaces $\Pi_n$ spanned by polynomials of order $n$,

$\Pi_n = \{ a_0 + a_1 x + \dotsb + a_n x^n : (a_0, a_1, \dotsc, a_n) \in \mathbb{F}^{n+1} \}.$

These subspaces have dimension $n+1.$ Consider now for each $n \in \mathbb{N}$ the maps $\partial_n \colon \Pi_n \to \Pi_{n-1}$ defined in the following way:

$\partial_n \big( a_0 + a_1 x + \dotsb + a_n x^n \big) = \displaystyle{\sum_{k=0}^{n} (-1)^k \sum_{j\neq k} a_j x^{\varphi(j,k)},}$

where $\varphi(j,k) = j$ if $j < k,$ and $\varphi(j,k) = j-1$ otherwise.

Schematically, this can be written as follows

$\begin{pmatrix} a_0 \\ a_1 \\ \vdots \\ a_n \end{pmatrix} \xrightarrow{\partial_n} \begin{pmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{pmatrix} - \begin{pmatrix} a_0 \\ a_2 \\ \vdots \\ a_n \end{pmatrix} + \dotsb + (-1)^k \begin{pmatrix} \vdots \\ a_{k-1} \\ a_{k+1} \\ \vdots \end{pmatrix} + \dotsb + (-1)^n \begin{pmatrix} a_0 \\ a_1 \\ \vdots \\ a_{n-1} \end{pmatrix},$

and it is not hard to prove that these maps are homeomorphisms of vector spaces over $\mathbb{F}.$

Notice this interesting relationship between $\partial_3$ and $\partial_2:$

$\begin{array}{rl} \partial_2(a_0 + a_1 x + a_2 x^2) &= (a_1 + a_2 x) - (a_0 + a_2 x) + (a_0 +a_1 x) \\ &= a_1 + a_1 x \\ \partial_3( a_0 + a_1 x + a_2 x^2 + a_3 x^3) &= ( a_1 + a_2 x + a_3 x^2 ) - ( a_0 + a_2 x + a_3 x^2) \\ &\mbox{} \quad + (a_0 + a_1 x + a_3 x^2) - (a_0 +a_1 x +a_2 x^2) \\ &= (a_1-a_0) + (a_3-a_2)x^2 \end{array}$

The kernel of $\partial_2$ and the image of $\partial_3$ are isomorphic!

$\begin{array}{rl} \ker \partial_2 &= \{ a_0 + a_2 x^2 : (a_0,a_2) \in \mathbb{F}^2 \}. \\ \partial_3\big( \Pi_3 \big) &= \{ b_0 + b_2 x^2 : (b_0, b_2) \in \mathbb{F}^2 \}. \end{array}$

The reader will surely have no trouble to show that this property is satisfied at all levels: $\ker \partial_n = \partial_{n+1} \big( \Pi_{n+1} \big).$ As a consequence, $\partial_n \partial_{n+1} = 0$ for all $n \in \mathbb{N}.$

We say that a family of homomorphisms $\{ \partial_n \colon \Pi_n \to \Pi_{n+1} \}$ are boundary operators if $\partial_n \partial_{n+1} = 0$ for all $n \in \mathbb{N}.$ If this is the case, then trivially $\partial_{n+1} \big( \Pi_{n+1} \big) \subseteq \ker \partial_n.$ The example above is a bit stronger, because of the isomorphism of both subspaces.

So this is the question I pose as today’s challenge:

Describe all boundary operators $\big\{ \partial_n\colon \Pi_n \to \Pi_{n-1} \big\} \big( \partial_n\partial_{n+1} = 0 \big)$

Include a precise relationship between kernels and images of consecutive maps.

$\pi\colon \mathbb{N} \cup \{ 0 \} \to \big( \mathbb{N} \cup \{ 0 \} \big)^2.$
We will accomplish this by creating the corresponding map (and its inverse), that takes each natural number $z$ and drops it at a location in the lattice, as the following diagram suggests: