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# Tangent bundles

### From Mathematics Is A Science

## 1 Tangent spaces

Suppose we want to find the *work* of a constant force along a straight path. As we know,
$$\text{ work } = \text{ force }\cdot \text{ distance}.$$
This simple formula only works if we carefully take into account the *direction* of motion relative to the direction of the force $F$. For example, if you move forward and then back, the work breaks into two parts and they may cancel each other. The idea is that the work $W$ may be positive or negative and we should speak of the *displacement* $D$ rather than the distance. We then amend the formula:
$$W = F \cdot D.$$

Now, in the context of discrete calculus, the displacement $D$ may be given by a single oriented edge in ${\mathbb R}$, or a combination of edges. It is a $1$-*chain*. Furthermore, the force $F$ defines $W$ as a linear function of $D$. It is a $1$-*form*!

The need for considering directions becomes clearer when the dimension of the space is $2$ or higher. We use *vectors*. First, as we just saw, the work of the force is $W = \pm F \cdot D$ if $F || D$, and we have the plus sign when the two are collinear. Second, $W = 0$ if $F \perp D$. Therefore, only the *projection* of $F$ on $D$ matters when calculating the work and it is the projection when the length of $D$ is $1$.

Then the work $W$ of force $F$ along vector $D$ is defined to be $$W := \langle F , D \rangle .$$ It is simply a (real-valued) linear function of the displacement.

Our conclusion doesn't change: $D$ is a $1$-chain and $F$ is a $1$-form. Even though this idea allows us to continue our study, the example shows that it is impossible to limit ourselves to cubical complexes. Below, we make a step toward discrete calculus over general cell complexes.

On a plane, the force $F$ may vary from location to location. Then the need to handle the displacement vectors, i.e., directions, arises, separately, at every point. The set of all possible directions at point $A\in V={\bf R}^2$ form a vector space of the same dimension. It is $V_A$, a copy of $V$, attached to each point $A$:

Next, we apply this idea to cell complexes.

First, what is the set of all possible directions on a *graph*? We've come to understand the edges starting from a given vertex as independent directions. That's why we will need as many basis vectors as there are edges, at each point:

Of course, once we start talking about *oriented* cells, we know it's about *chains*, over $R$.

**Definition.** For each vertex $A$ in a cell complex $K$, the *tangent space* at $A$ of $K$ is the set of $1$-chains over ring $R$ generated by the $1$-dimensional star of the vertex $A$:
$$T_A=T_A(K):=< \{AB \in K\} > \subset C_1(K).$$

**Proposition.** The tangent space $T_A(K)$ is a submodule of $C_1(K)$.

**Definition.** A *local* $1$-*form* on $K$ is a collection of linear functions for each of the tangent spaces,
$$\varphi_A: T_A\to R,\ A\in K^{(0)}.$$

The work of a force along edges of a cell complex is an example of such a function. Now, if we have a force in complex $K$ that *varies* from point to point, the work along an edge -- seen as the displacement -- will depend on the location.

**Proposition.** Every $1$-form (cochain) is a local $1$-form.

We **denote** the set of all local $1$-forms on $K$ by $T^1(K)$, so that
$$C^1(K)\subset T^1(K).$$

Let's review the setup. First, we have the *space of locations* $X=K^{(0)}$, the set of all vertices of the cell complex $K$. Second, to each location $A\in X$, we associate the *space of directions* determined by the structure of $K$, specifically, by its edges. Now, while the directions at vertex $A\in K$ are given by the edges adjacent to $A$, we can also think of all $1$-*chains* in the star of $A$ as directions at $A$. They are subject to algebraic operations on chains and, therefore, form a module, $T_A$.

We now combine all the tangent spaces into one total tangent space. It contains all possible directions in each location: each tangent space $T_A$ to every point $A$ in $K$.

**Definition.** The *tangent bundle* of $K$ is the disjoint union of all tangent spaces:
$$T(K):=\bigsqcup _{A\in K} \Big( \{A\} \times T_A \Big).$$

Then a local $1$-form is seen as a function on the tangent bundle, $$\varphi =\{\varphi_A: \ A\in K\}:T(K) \to R,$$ linear on each tangent space, defined by $$\varphi (A,AB):=\varphi_A(AB).$$

In particular, the work associates to every location and every direction at that location, a quantity:
$$\varphi(A,AB)\in R.$$
The total work over a path in the complex is the *line integral* of $\varphi$ over a $1$-chain $a$ in $K$. It is simply the sum of the values of the form at the edges in $a$:
$$\begin{array}{llllll}
a=&A_0A_1+A_1A_2+...+A_{n-1}A_n \Longrightarrow \\
&\displaystyle\int_a \varphi:= \varphi (A_0,A_0A_1)+\varphi (A_1,A_1A_2)+... +\varphi (A_{n-1},A_{n-1}A_n).
\end{array}$$

Let's suppose, once again, that this number $\varphi(A,AB)$ is the work performed by a given force while moving from $A$ to $B$ along $AB$. We know that this work should be the negative of the one carried out while going in the opposite direction; i.e., $$\varphi(A,AB)= -\varphi(B,BA).$$ From the linearity of $\varphi$, it follows $\varphi(A,AB)= \varphi(B,AB)$. Then, the local form $\varphi$ defined separately on each of the stars must have matched values on their overlaps. Therefore, it is well-defined as a linear map on $C_1(K)$.

**Theorem.** A local $1$-form that satisfies the matching condition above is a $1$-cochain of $K$.

This conclusion reveals a potential redundancy in the way we defined the space of all directions as the tangent bundle $T(K)$. We can then postulate that the direction from $A$ to $B$ is the opposite of the direction from $B$ to $A$: $$(A,AB)\sim -(B,BA).$$ This equivalence relation reduces the size of the tangent bundle via the quotient construction. Below, we can see the similarity between this new space and the space of tangents of a curve:

It looks as if the disconnected parts of $T(K)$, the tangent spaces, are glued together.

The fact that this equivalence relation preserves the operations on each tangent space implies the following.

**Theorem.**
$$T(K)/_{\sim}=C_1(K).$$

**Exercise.** Prove the theorem.

Then $1$-forms are local $1$-forms that are well-defined on $C_1(K)$. More precisely, the following diagram commutes for any $f\in C^1(K)$: $$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} \newcommand{\la}[1]{\!\!\!\!\!\!\!\xleftarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\ua}[1]{\left\uparrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{ccccccccccccc} T(K)\ & \ra{f} & R &\\ \da{p} & & || & \\ C_1(K)\ & \ra{f} & R, \end{array}$$ where $p$ is the identification map. This justifies our focus on $1$-forms as $1$-cochains.

**Exercise.** Define the analog of the exterior derivative $d:C^0(K)\to T^1(K)$.

Higher order differential forms are *multi*-linear functions but this topic lies outside the scope of this chapter.

## 2 The derivative of a cell map

Consider the two standard ways to write the derivative of function $f$ at $x=a$:
$$\tfrac{dy}{dx} = f'(a).$$
What we know from calculus is that the left-hand side is *not* a fraction but the equation can be rewritten as if it is:
$$dy = f'(a) dx.$$
The equation represents the relation between the increment of $x$ and that of $y$ -- *in the vicinity* of $a$. This information is written in terms of a new coordinate system, $(dx,dy)$ and the best affine approximation (given by the tangent line) becomes a linear function in this system:

Things are much simpler in the discrete case.

Suppose $X$ and $Y$ are two cell complexes and $f:X\to Y$ is a cell map. Then “in the vicinity of point $a$” becomes “in the star of vertex $A$”:

In fact, we can ignore the algebra of the $x$- and $y$-axis if we think of our equation as a relation between the elements of the *tangent spaces* of $A$ and $f(A)$. If we zoom out on the last picture, this is what we see:

Then the above equation becomes: $$e_Y=f(AB)e_X,$$ where $e_X,e_Y$ are the basis vectors of $T_A(X), T_{f(A)} (Y)$, respectively.

The idea is: our function maps both locations *and* directions. The general case is illustrated below:

A cell map takes vertices to vertices and edges to edges and that's what makes the $0$- and $1$-chain maps possible. Then,

- the locations are taken care of by $f_0:C_0(X)\to C_0(Y)$, and
- the directions are taken care of by $f_1:C_1(X)\to C_1(Y)$.

Suppose also that $f(A)=P$, so that the location is fixed for now. Then the tangent spaces at these vertices are: $$T_A(X):=<\{AB \in X \}>\subset C_1(X), \quad T_P(Y):=<\{PQ \in Y \}>\subset C_1(Y).$$

**Definition.** The *derivative of a cell map* $f$ *at vertex* $A$ is a linear map
$$f'(A):T_A(X) \to T_P(Y)$$
given by
$$f'(A)(AB):=f_1(AB).$$

**Example.** Let's consider cell maps of the “cubical circle” (i.e., ${\bf S}^1$ represented by a $4$-edge cubical complex) to itself, $f: X \to X$:

Given a vertex, we only need to look at what happens to the edges adjacent to it. We assume that the bases are ordered according to their letters, such as $\{AB,BC\}$.

The derivatives of these functions are found below.

*Identity:*
$$\begin{array}{lllllll}
f_0(A) = A, &f_0(B) = B, &f_0(C) = C, &f_0(D) = D, \\
\Longrightarrow & f'(A)(AB)=AB, &f'(A)(AD)=AD.
\end{array}$$
It's the identity map.

*Constant:*
$$\begin{array}{llllllll}
f_0(A) = A, &f_0(B) = A, &f_0(C) = A, &f_0(D) = A, \\
\Longrightarrow & f'(A)(AB)=AA=0, &f'(A)(AD)=AA=0.
\end{array}$$
It's the zero map.

*Vertical flip:*
$$\begin{array}{llllllllll}
f_0(A) = D, &f_0(B) = C, &f_0(C) = B, &f_0(D) = A, \\
\Longrightarrow & f'(A)(AB)=DC, &f'(A)(AD)=DA.
\end{array}$$
The matrix of the derivative is
$$\hspace{.37in}f'(A)=\left[
\begin{array}{ccccccc}
0&1\\
1&0
\end{array}
\right].
\begin{array}{cccc}
\\
\hspace{.37in}\square
\end{array}$$

**Exercise.** Repeat these computations for (a) the rotations; (b) the horizontal flip; (c) the diagonal flip; (d) the diagonal fold. Hint: the value of the derivative varies from point to point.

As this construction is carried out for each vertex in $X$, we have defined a function on the whole tangent bundle.

**Definition.** The *derivative of cell map*
$$f:X\to Y$$
between two cell complexes is the map between their tangent bundles,
$$f':T(X) \to T(Y),$$
given by
$$f'(A,AB):=(f_0(A),f_1(AB)).$$

**Exercise.** In this context, define the directional derivative and prove its main properties.

**Theorem (Properties of the derivative).** For a given vertex and an adjacent edge, the derivative satisfies the following properties: $\\$

$\hspace{5mm}\bullet$ The derivative of a constant is zero in the second component: $$(C)'=(C,0),\ C\in Y.$$ $\hspace{5mm}\bullet$ The derivative of the identity is the identity: $$(\operatorname{Id})'=\operatorname{Id}.$$ $\hspace{5mm}\bullet$ The derivative of the composition is the composition of the derivatives: $$(fg)'=f'g'.$$ $\hspace{5mm}\bullet$ The derivative of the inverse is the inverse of the derivative: $$(f^{-1})'=(f')^{-1}.$$

**Exercise.** Prove the theorem.

**Exercise.** Prove that if $|f|$ is a homeomorphism, then $f'=\operatorname{Id}$.

**Notation:** An alternative notation for the derivative is $Df$. It is also often understood as the *tangent map* of $f$ denoted by $T(f)$.

**Exercise.** Show that $T$ is a functor.

We have used the equivalence relation $$(A,AB)\sim (B,-BA)$$ to glue together the tangent spaces of a cell complex: $$T(K)/_{\sim}=C_1(K).$$

**Theorem.** Suppose $X,Y$ are two cell complexes and $f:X\to Y$ is a cell map. Then the quotient map of the derivative
$$[f']:T(X)/_{\sim} \to T(Y)/_{\sim}$$
is well-defined and coincides with the $1$-chain map of $f$,
$$f_1:C_1(X)\to C_1(Y).$$

**Proof.** Suppose $f_0(A)=P$ and $f_0(B)=Q$. Then we compute:
$$\begin{array}{llllll}
f'(A,AB)&=(f_0(A),f_1(AB))\\
&=(P,PQ)\\
&\sim (Q,-QP)\\
&=(f_0(B),-f_1(BA))\\
&=f'(B,-BA).
\end{array}$$
We have proven the following:
$$(A,AB)\sim (B,-BA) \Longrightarrow f'(A,AB)\sim f'(B,-BA).$$
Therefore, $[f']$ is well-defined. $\blacksquare$

## 3 Chain maps

A cell map can't model jumping diagonally across a square.

The issue is related to one previously discussed: cell map extensions vs. chain map extensions (subsection V.3.10). Recall that in the former case, extensions may require subdivisions of the cell complex. The situation when the domain is $1$-dimensional is transparent:

In the former case, we can create a cell map:
$$g(AB):=XY,$$
by extending its values from vertices to edges. In the latter case, an attempt of cell extension (without subdivisions) fails as there is no *single* edge connecting the two vertices. However, there is a *chain* of edges:
$$g(AB):=XY+YZ.$$

Even though the linearity cannot be assumed, the illustration alone suggests a certain continuity of this new “map”. In fact, we know that chain maps are continuous in the algebraic sense: they preserve boundaries, $$g_0\partial = \partial g_1.$$ The idea is also justified by the meaning of the derivative of a cell map $f$: $$f'\Big(A,[A,A+1] \Big)= \Big(f_0(A),f_1([A,A+1]) \Big).$$ It is nothing but a combination of the $0$- and the $1$-chain maps of $f$...

Suppose we are given $f$, a $0$-form on ${\mathbb R}$. Then we would like to interpret the pair $g=\{f,df\}$ as some chain map defined on $C({\mathbb R})$, the chain complex of time. What is the other chain complex $C$, the chain complex of space? Since these two forms take their values in ring $R$, we can choose $C$ to be the trivial combination of two copies of $R$: $$\partial=\operatorname{Id}:R \to R.$$ Below, we consider a more general setting of $k$-forms.

**Theorem.** Cochains are chain maps, in the following sense: for every $k$-cochain $f$ on $K$, there is a chain map from $C(K)$ to the chain complex $C$ with only one non-zero part, $\operatorname{Id}:C_{k+1}=R \to C_k=R$, as shown in the following commutative diagram:
$$
\newcommand{\ra}[1]{\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!}
\newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.}
%
\begin{array}{r|ccccccccccc}
C(K):& ... & C_{k+2}(K) & \ra{\partial} & C_{k+1}(K) & \ra{\partial} & C_k(K) & \ra{\partial} & C_{k-1}(K) & ... \\
f: & \ & \ \da{0} & & \ \da{df} & & \ \da{f} & & \ \da{0}&\\
C: & ... & 0 & \ra{\partial=0} & R & \ra{\partial=\operatorname{Id}} & R & \ra{\partial=0} &0&...
\end{array}
$$

**Proof.** We need to prove the commutativity of each of these squares. We go diagonally in two ways and demonstrate that the result is the same. We use the duality $d=\partial^*$.

For the first square: $$df \partial =(\operatorname{Id}^{-1}f\partial)\partial =\operatorname{Id}^{-1}f0=0.$$ For the second square: $$f\partial =df=\operatorname{Id}df.$$ The third square (and the rest) is zero. $\blacksquare$