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Axioms of calculus

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Introduction

The axiomatic approach allows one to develop and use the same tools in the continuous and discrete cases.

Briefly, calculus is a functor:

  • from topological spaces to a certain kind of chain complexes,

and, further, equipped with inner products.

Suppose $R$ is a ring. Let $\mathcal{Mod}$ (or $R-\mathcal{Mod}$) be the category of modules over $R$ and $Ch(\mathcal{Mod})$ be the category of chain complexes of $R$-modules.

Tangent, cotangent, and tensor bundles

We want to capture the "directions".

If the (multi-)vectors are represented by the chain complex $(C,\partial)$ (the analogue of the tangent bundle), then the covectors are represented by its dual $(C^*,\partial ^*)$ (the analogue of the cotangent bundle). Now, the tensors are represented by "complex" $(T(C),D)$, with $$T(C)=C \otimes C^*,$$ and the grading operator of degree $0$: $$D = \partial \otimes \partial ^*.$$ This is the analogue of the tensor space.

To add geometry we consider a chain complex $(C,\partial )$ with an inner product on $C_1$.

Tangency and inner product

A complex may be quipped with an extra structure. First, in order to capture capture the "directions" in the complex we introduce tangency.

Given a vertex $A$ in $K$, we start with the set of all cells that contain $A$ called the (open) star of vertex $A$, $$St_A(K)=\{a \in K: A \in \partial _1(a) \}$$ $$ = \{AB_i: AB_i \in K^{(1)}\}.$$ We think of the edges that start at $A$ as the tangents.

We define an analogue of tangent spaces of our chain complex $C$ as follows. Given $A\in C_0$, let $$T'_A(C) = \{ a \in Hom([0,1],C): a_0(0) = A \},$$ the set of all parametric curves starting at $A$.

Next, in order to add geometry to chain complex, we introduce a degenerate inner product on $T_A(C)$, for each $A$.

Note: In case of a cell complex, for each location $A$ and each pair of directions at this location, i.e., edges $x=AB,y=AC$ adjacent to $A$, there is a number $< AB, AC >_A \in R$.

More generally, an inner product is a function: $$<\cdot,\cdot>_A: T'_A(C) \rightarrow R$$ that satisfies these properties:

  • 1. $<v,v>_A \geq 0$ for any $v \in T'_A(C)$ -- positive definiteness;
  • 2. $ < u , v >_A = <v,u>_A $ for any $u,v \in T'_A(C)$ -- symmetry;
  • 3. $ < ru +r'u', v >_A=r < u ,v >_A + r' < u' ,v >_A$ for any $u,u',v \in T'_A(C),r,r'\in R$ -- linearity.

We do not assume the common non-degeneracy condition:

  • $< x, y >_A =0$, for all $y \in T'_A(C)$, implies that $x = 0$, or
  • $<v,v>_A=0$ if and only if $v=0$.

The inner product defines a (semi-)norm, the usual way: $$||a||_A = \sqrt{< a,a>}.$$

Under a certain equivalence relation, the space becomes a (non-degenerate) inner product space. We define an equivalence: $$u \sim v \in T'_A(C) \Leftrightarrow ||u-v||_A=0.$$ Then the tangent space of $C$ at $A$ is $$T_A(C)=T'_A(C) / _{\sim}.$$

Further, we define an equivalence on all parametric curves in $C$: $$u \sim v \in Hom([0,1],C) \Leftrightarrow u(0)=v(0)=A\in C_0,||u-v||_A=0.$$ Then the tangent bundle of $C$ is $$T(C)= Hom([0,1],C) / _{\sim}.$$

Hodge-dual complexes

Duality of cubical complexes is defined with respect to some ambient space. That space won't be identified but we are given the ambient dimension $n$.

For any cell complex $K$, we call a cell $a$ of dimension $k=0,1,...,n-1$ a boundary cell of $K$ if $a$ is shared by less than two $n$-cells.

Now, given two cell complexes $K$ and $K^*$, suppose all non-boundary cells of the complementary dimensions are paired up: $$Q_k(K) \leftrightarrow Q_{n-k}(K^*).$$ In this case $K$ and $K^*$ are called dual of each other.

It follows that $n$ is at least as large as the highest dimension of a cell in $K$.

Then the Hodge star operators are linear operators on the chain complexes: $$\star ^k: C_k(K) \rightarrow C_{n-k}(K^*),$$ $$\star ^k: C_k(K^*) \rightarrow C_{n-k}(K),$$ generated by this correspondence if they satisfy: $$\star ^{n-k}\star ^k (a)=c\cdot a,$$ for some $c\in R$, for any non-boundary $k$-cell $a$.

Metric tensor

A metric tensor is a "parametrized" inner product.

We define a discrete metric tensor on $K$ as a function that associates a number to each location, i.e., a vertex $A$ in $K$, and each pair of directions at that location, i.e., edges $x=AB,y=AC$ adjacent to $A$: $$(A,AB,AC) \mapsto < AB, AC >(A) \in R.$$

More precisely, a metric tensor is a function: $$<\cdot,\cdot>(\cdot): T(K) \rightarrow R$$ that satisfies the properties of an inner product at each location $A \in K_0$:

  • 1.
    • $<v,v>(A) \geq 0$ for any $v \in T_A(K)$ -- non-degeneracy.
    • $<v,v>(A)=0$ if and only if $v=0$ -- positive definiteness.
  • 2. $ < u , v >(A) = <v,u>(x) $ for any $u,v \in T_A(K)$ -- symmetry;
  • 3. $ < ru +r'u', v >(A)=r < u ,v >(A) + r' < u' ,v >(A)$ for any $u,u',v \in T_A(K),r,r'\in R$ -- linearity;

(Of course "$(A)$" can be suppressed.)

Note: Observe that of the whole $K$ we only need to use the $1$-skeleton $K_1$.

So the metric tensor is given by the table of its values (in $R$) for each vertex $A$: $$ % \begin{array}{cccccccccc} A & AB & AC & AD &...\\ AB & <AB,AB> & <AB,AC> & <AB,AD> &...\\ AC & <AC,AB> & <AC,AC> & <AC,AD> &...\\ AD & <AD,AB> & <AD,AC> & <AD,AD> &...\\ ...& ... & ... & ... &... \end{array} $$

This data can be used to extract more usable information (may be outside $R$): $$a \mapsto \lVert a \rVert =\sqrt{< a, a >},$$ and $$(a,b) \mapsto \cos\widehat{ab} = \frac{<a, b >}{\lVert a \rVert \lVert b\rVert}.$$ Then we have all the information that we need for measuring: $$ % \begin{array}{cccccccccc} A & AB & AC & AD &...\\ AB & \lVert AB \rVert & \widehat{BAC} & \widehat{BAD} &...\\ AC & \widehat{BAC} & \lVert AC \rVert & \widehat{CAD} &...\\ AD & \widehat{BAD} & \widehat{CAD} & \lVert AD \rVert &...\\ ...& ... & ... & ... &... \end{array} $$

Note: From an arbitrary (positive on the diagonal) collection of numbers in such a table we can reconstruct a metric tensor by using the formula: $$< AB , AC >=\lVert AB \rVert\lVert AC \rVert\cos\widehat{BAC}.$$