Geometry in calculus

Redirect to:

• Geometry#The metric tensor

We need to develop a way to introduce an arbitrary geometry into the (so far) purely topological setting of cubical complexes.

The need for measuring

Looking back we discover that so far we haven't explicitly used any geometry to develop calculus!

Of course we have relied on the mathematics from calculus courses that may rely on geometry. Let's make a couple of observations nonetheless.

First, we rely on limits that use the absolute value:

$\lim_{x\to a}f(x)=L$ if for any $\epsilon >0$ there is a $\delta >0$ such that $0<|x-a|<\delta$ implies that $|f(x)-L|<\epsilon$.

However, the definition can be easily rewritten without need for measuring:

$\lim_{x\to a}f(x)=L$ if for any open neighborhood $\epsilon$ of $L$ there is an open neighborhood $\delta$ of $a$ such that $x\in\delta$ implies $f(x)\in\epsilon$.

It only requires purely topological ideas such as basis of topology. This also covers the derivative.

Second, do we need geometry for integration? A typical construction of the Riemann integral is as follows. The integral over an interval is the limit of its Riemann sums $R(P)$: $$I=\int_a^b f(x)dx=\lim _{||P||\rightarrow 0} R(P),$$ where $P$ is a partition of the interval $[a,b]$ and $||P||$ is its "mesh" (the largest length among the subintervals in $P$). In other words,

for any $\epsilon >0$ there is $\delta >0$ such that $|I-R(P)|<\epsilon$ whenever $||P||<\delta$.

This can be made purely topological:

for any $\epsilon >0$ there is open cover $\delta$ of $[a,b]$ such that $|I-R(P)|<\epsilon$ whenever all elements of $P$ are covered by the elements of $\delta$.

It is much easier to separate topology from geometry for calculus of discrete differential forms because we have built it from scratch!

For example, it is clear that only the way the cells are attached to each other affects the matrix of the exterior derivative:

It is claar that the sizes or the shapes of the cells are irrelevant.

However, even when those are given, the geometry of the domain remains unspecified:

The two distorted grids can be thought of as two different homeomorphisms of the realization of this cubical complex.

In a calculus course on the topological and algebraic structures to develop the mathematics. There are two main exceptions. One needs inner product and the norm for:

• concavity,
• arc-length and curvature.

In the formulas for the latter, the norm is explicitly present and the example above shows that to introduce the curvature one will need to add more structure to the complex.

The role of geometry in the former is less explicit.

Monotonicity vs concavity

The point where we start to need geometry is when we move from the first to the second derivative.

Shrinking, stretching, deforming the $x$- or the $y$ axes won't change the monotonicity of a function but it will change its concavity.

Here we arbitrarily deformed the $x$-axis:

So, it is sufficient to know the sign of the derivative or the exterior derivative to distinguish between increasing and decreasing behavior. But the latter only uses the topological properties of the cubical complex!

Exercise. To prove the continuous case, consider an orientation preserving diffeomorphism of the $x$-axis and then use the chain rule. What about the discrete case?

Concavity is an important concept that captures the shape of the graph of a function in calculus as well as the acceleration in physics.

The upward concavity of the discrete function below is obvious which reflects the positive acceleration:

Concavity is computed and studied via the second derivative, the rate of change of the rate of change. With $dd=0$ we don't have the second exterior derivative!

As an alternative it is natural to look at the change of change. This can be seen algebraically as the exterior derivative (change) increases $$g(2)-g(1) < g(3)-g(2), g(3)-g(2) < g(4)-g(3), etc.$$ However, does this algebra imply concavity? Not without assuming that the intervals have equal lengths!

Below, on the same, topologically, cubical complex, we have the same discrete function with, of course, the same exterior derivative,

but an opposite concavity!

Metric tensor

A metric tensor is a "parametrized" inner product.

This is what we mean. Given Euclidean space ${\bf R}^n$, a metric tensor (or simply a metric) is a function that associates a number to each location in some region $D$ (the domain) in this space and each pair of vectors (directions) at that location: $$<\cdot,\cdot>(\cdot): D \times {\bf R}^n \times {\bf R}^n \rightarrow {\bf R},$$ $$(x,u,v) \mapsto < u, v >(x) \in {\bf R}$$ that depends continuously on $x$ and satisfies the properties of an inner product at each location $x \in D$:

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

Of course $(x)$ can be suppressed.

The norm now is also location dependent: $$\lVert a \rVert (x) =\sqrt{< a, a >(x)}.$$

This looks very much like the definition of a $2$-form except it's symmetric not antisymmetric!

Just as with the forms, we can use the exponential identity of functions to see that a metric tensor associates an inner product to each location.

With a metric tensor, we can compute the arc-length of a curve $C$ in ${\bf R}^n$: $$l(C)=\int _a^b \lVert p' \rVert dt,$$ if $C$ is parametrized by a differentiable function $p:[a,b] \rightarrow {\bf R}^n$. We can also compute the curvature. Remember that the results will depend on our choice of metric tensor.

What about the discrete case?

We start with a cubical complex $K$ on the square grid in ${\bf R}^n$ and one can simply consider a "discretized metric tensor" as a function that associates a number to each location, i.e., a vertex $A$ in $K$, and each pair of "grid vectors" (directions) $u,v$ at that location: $$<\cdot,\cdot>(\cdot): V(K) \times {\bf Z}^n \times {\bf Z}^n \rightarrow {\bf R},$$ $$(A,v,u) \mapsto < v, u >(A) \in {\bf R},$$ where $V(K)$ is the set of vertices of $K$, that satisfies the properties (1. - 3. above) of an inner product at each location $A$.

This, however, isn't general enough. Indeed, if $n=1$, two edges adjacent to the same vertex can be thought of as vectors $v$ and $-v$. Then, $<v,-v>=-\rVert v \lVert ^2$, which means that they are parallel.

Instead, in the spirit of discrete differential forms, 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 $A$: $$(A,AB,AC) \mapsto < AB, AC >(A) \in {\bf R}.$$

Note: There are no continuity requirements. Also, observe that we only need to use the $0$-skeleton of $K$.

So, we have a function: $$<\cdot,\cdot>(\cdot): T(K) \rightarrow {\bf R},$$ the domain of which $$T(K) = \displaystyle\bigsqcup_{A \in V(K)} T_A(K),$$ is the discrete tangent bundle defined as the disjoint union of all tangent spaces with each simply the $1$-dimensional star of the vertex: $$T_A(K)=\{AB \in K\}.$$

Note: In this sense, $T(K)$ is more like a disk bundle than a vector bundle.

The end result is, again, similar to a certain combination of discrete forms, as we have a number assigned to each edge and to each pair of adjacent edges:

• $a \mapsto < a, a>$;
• $(a,b) \mapsto <a, b >$.

So the metric tensor is given by the table of its values 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:

• $a \mapsto \lVert a \rVert$;
• $(a,b) \mapsto \widehat{ab}$, where

$$\cos\widehat{ab} = \frac{<a, b >}{\lVert a \rVert \lVert b\rVert}.$$ 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: $$< AB , AC >=\lVert AB \rVert\lVert AC \rVert\cos\widehat{BAC}.$$

With a metric tensor, we can compute the arc-length of a discrete curve in ${\bf Z}^n$ as the sum of the lengths of its edges: $$l(\{A, B, C, ..., Z\})=\lVert AB \rVert + \lVert BC \rVert +... + \lVert YZ \rVert.$$ We can also compute the curvature of the curve at a given vertex $B$ as simply the value of the angle $\widehat{ABC}$. Again, the results will depend on our choice of metric tensor.