David Roberts smooth fundamental bigroupoid

The smooth fundamental bigroupoid is a smooth version of the topological fundamental bigroupoid, constructed essentially by replacing all continuous maps with smooth ones. This introduces some technicalities, mainly relating to the fact that one can no longer concatentation smooth maps merely if they match up on their common boundary.

First, let us define the points of our space. Let us fix once and for all a manifold $M$ (without boundary). We start with “homotopies of paths”:

Definition

A homotopy of paths in $M$ is a smooth map $\sigma \colon I^2 \to M$ such that $t \mapsto \sigma(t,0)$ and $t \mapsto \sigma(t,1)$ are both constant paths.

The source of $\sigma$ is the path $s \mapsto \sigma(0,s)$ and the target is the path $s \mapsto \sigma(1,s)$.

By a “smooth map” with source $I^2$, we require all derivatives to extend (continuously) to the boundary. Equivalently, the smooth map comes from a morphism in the category of Froelicher spaces. We could express this as simply a smooth map from the quotient of $I^2$ by the two lines, but that is not the same as the space obtained by imagining a “bi-gon” in $\mathbb{R}^2$.

Now we want to establish our equivalence relation.

Definition

We define an equivalence relation on the set of homotopies of paths in $M$ according to the following: $\sigma \simeq \tau$ if there is a smooth map $\Sigma \colon I^3 \to M$ such that:

1. The map $(s,t) \mapsto \Sigma(s,t,0)$ is constant
2. The map $(s,t) \mapsto \Sigma(s,t,1)$ is constant
3. The map $(s,t) \mapsto \Sigma(0,s,t)$ is $\sigma$
4. The map $(s,t) \mapsto \Sigma(1,s,t)$ is $\tau$
5. For each $t$, the map $s \mapsto \Sigma(s,0,t)$ is constant
6. For each $t$, the map $s \mapsto \Sigma(s,1,t)$ is constant

Of course, we need to establish the fact that the above is an equivalence relation. Reflexivity and symmetry are obvious. For transitivity, we need to adjoin two such maps together. We do so by placing the two copies of $I^3$ one on top of the other and defining a map $[0,2] \times I^2 \to I^3$ by $(r,s,t) \mapsto (\phi(r/2),s,t)$ where $\phi \colon I \to I$ is increasing, surjective, and is flat at $1/2$.

Definition

A 2-track is an equivalence class of homotopies of paths under the above relation. We write $\Pi_2(M)$ for the set of all $2$-tracks in $M$.

Now we want to put the structure of a manifold on the space of 2-tracks. We do so following the structure of that of a smooth loop space.

Let $\eta \colon T M \to M$ be a local addition. Let $\sigma$ be a 2-track. As such, it has well-defined source and target paths, say $\partial_0 \sigma$ and $\partial_1 \sigma$. These are smooth paths in $M$ with common end-points. As such, they can be considered to be a smooth map from the smooth space formed by taking two copies of the interval and identifying the endpoints (note that this is not the circle). Let us denote this space by $☽$. Let us write $\partial \sigma \colon ☽ \to M$ for the map which is $\partial_0 \sigma$ on, say, the left half and $\partial_1\sigma$ on the right half.

We then write $\Gamma_☽((\partial \sigma)^* T M)$ for the space of smooth sections of the pull-back of the tangent bundle of $M$ by $\partial \sigma$. We should explain exactly what this means. This is the space of smooth maps $☽ \to T M$ covering $\partial \sigma$. Thus it is the subspace of $\Gamma_I((\partial_0 \sigma)^* T M) \oplus \Gamma_I((\partial_1)^* T M)$ consisting of sections over each piece which agree at the end-points. Thus, in particular, the only requirement at the end-points is that the map be continuous, not smooth.

We now define $\Gamma_☽((\partial \sigma)^* T M) \to \Pi_2(M)$. Let us start with a section, $X$. As this is a section, it defines a map $\tilde{X} \colon ☽ \to T M$. We compose this with the local addition, $\eta \colon T M \to M$, to define a map $\eta\tilde{X} \colon ☽ \to M$. Now we need to fill this in to a $2$-track. We do this by using the diffeomorphisms associated to the local addition.

The local addition defines a smooth map $[0,1] \times T M \to \operatorname{Diff}(M)$. Thus, composing with $\tilde{X}$, we have a smooth map $\psi \colon [0,1] \times ☽ \to \operatorname{Diff}(M)$ (and note that this varies smoothly with $\tilde{X}$). This has the property that $\psi(0,p)$ is always the identity on $M$ and $\psi(1,p)(\partial\sigma(p)) = \eta\tilde{X}(p)$. Thus $\psi(t,p)$ moves the point $\partial\sigma(p)$ to the point $\eta\tilde{X}(p)$, dragging the rest of $M$ along after it like a rubber sheet. So by applying $\psi(1,p)$, we can distort $M$ so that $\partial\sigma$ maps to $\eta\tilde{X}$ (this picture is not quite right, since the distortion is parametrised by the points of $☽$). The trick is now to drag $\sigma$ along with us.

To do this, we need to define a “weight” function on $I^2$ which says how much of the distortion a particular point in $I^2$ feels, and which bit of the boundary of $I^2$ is providing the distortion. More concretely, we want to define a function $\rho \colon I^2 \to I$ which is $1$ on the boundary and we want to define a function $\mu \colon I^2 \supseteq U \to \partial I^2$ with certain properties. We shall then define the 2-track $\tau \colon I^2 \to M$ as

$$\tau(s,t) \coloneqq \psi(\rho(s,t),\mu(s,t))(\sigma(s,t))$$

So we use $\mu$ to tell us which distortion to use and $\rho$ to tell us how much of that distortion to apply. Then we apply this distortion to $\sigma$.

The properties that we want $\tau$ to have tell us what conditions we need on $\rho$ and $\mu$. We want $\tau$ to be smooth and we want $\tau$ to restrict to $\eta \tilde{X}$ on the boundary (via the collapse map $\partial I^2 \to ☽$). We therefore need $\mu$ to induce the same collapse map on $\partial I^2$, thus it must preserve two of the sides pointwise and preserve the other two in totality. This means that $\mu$ cannot be defined on the whole of $\partial I^2$. Therefore $\rho$ must be a bump function on the domain of $\mu$ such that $\rho$ takes the value $1$ on $\partial I^2$.

Let us give a semi-explicit definition of $\mu$. First, we choose a smooth function $\alpha \colon I \to I$ which is increasing and maps $[0,1/4]$ to $0$ and $[3/4,1]$ to $1$. We also choose a smooth function $\beta \colon I \to I$ which is zero on $[0,1/8]$ and $[7/8,1]$ and is $1$ on $[1/4,3/4]$. Then we define

$$\mu(s,t) \coloneqq \big( (1 - \beta(t))s + \beta(t)\alpha(s), \alpha(t)\big)$$

Let us check what happens here. When $t \in [0,1/4] \cup [3/4,1]$ then $\alpha(t)$ is either $0$ or $1$ so $\mu(s,t) \in \partial I^2$. Moreover, if $t \in [0,1/8] \cup [7/8,1]$ then $\beta(t) = 0$ and so the first co-ordinate of $\mu(s,t)$ is just $s$. When $t \in [1/4,3/4]$ then $\beta(t) = 1$ and so the first co-ordinate of $\mu(s,t)$ is $\alpha(s)$. Thus if $s \in [0,1/4] \cup [3/4,1]$ then $\mu(s,t) \in \partial I^2$. Hence $\mu$ takes the square annulus $I^2 \setminus [1/4,3/4]^2$ to $\partial I^2$ and has the desired properties on the boundary. So we restrict $\mu$ to this annulus and choose an appropriate bump function $\rho$.

This then defines our $2$-track $\tau$, and thus our chart map on $\Pi_2(M)$. That the transition functions are diffeomorphisms will follow by a similar argument to that at loop space.

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This argument generalises somewhat. We retain the background space as a smooth manifold, $M$, but generalise the source to a sequentially compact Frölicher space, $N$. We formally declare $N$ to have a “boundary”, $\partial N$. By “formally declare” we are not assuming the existence of a natural boundary of $N$ but simply assert that part of the initial data is a subset of $N$, satisfying some conditions, that we call its boundary. The conditions that we require are as follows:

1. $\partial N$ is a sequentially compact Frölicher space with its inherited structure. Thus $\partial N$ is a closed subspace of $N$.
2. $\partial N$ is a smooth deformation retract of a neighbourhood within $N$. That is, there is a smooth function $\gamma \colon I \to N$ and an open set $U$ containing $\partial N$ such that $\gamma(t,x) = x$ for all $x \in \partial N$, $\gamma(0,x) = x$ for all $x \in N$, and $\gamma(1,x) \in \partial N$ for all $x \in U$.
3. The neighbourhood $U$ above admits a bump function at $\partial N$. That is, there is a smooth function $\rho \colon N \to I$ with support in $U$ such that $\rho(x) = 1$ for $x \in \partial N$.

With these assumptions, we can follow the above, replacing one copy of $I$ by $N$ on all occasions. We start by defining our homotopies; these are smooth deformations of maps from $N$ to $M$ which do not deform $\partial N$. Precisely, we make the following definition.

Definition

A homotopy of $(N,\partial N)$-maps is a smooth map $\sigma \colon I \times N \to M$ such that the induced map $I \times \partial N \to M$ is independent of the $I$-parameter.

From this, we define our equivalence relation.

Definition

We define an equivalence relation on the set of $(N,\partial N)$-maps according to the following rule: $\sigma \simeq \tau$ if there is a smooth map $\Sigma \colon I^2 \times N \to M$ such that:

1. The induced map $I^2 \times \partial N \to M$ is independent of the $I^2$-parameter.
2. The induced map $\{0,1\} \times I \times N \to M$ is $\sigma \amalg \tau$.
3. The induced map $I \times \{0,1\} \times N \to M$ is independent of the $I$-parameter.

That this is an equivalence relation follows exactly as before since the argument uses a smooth reparametrisation in the new $I$-direction.

Definition

An $(N,\partial N)$-track is an equivalence class of homotopies of $(N,\partial N)$-maps under the above relation. We write $\Pi_{(N,\partial N)} (M)$ for the set of all $(N,\partial N)$-tracks in $M$.

The notation is almost the same as that above. If we label $\partial I = \{0,1\}$ as $2$ (which is a reasonable thing to do!) then we are renaming a $2$-track as an $(I,2)$-track.

We now want to put a manifold structure on $\Pi_{(N,\partial N)}(M)$ in the same manner as above. So we fix a local addition $\eta \colon T M \to M$ on $M$ and choose an $(N,\partial N)$-track, $\sigma$. This has well-defined source and target paths, obtained by restricting a choice of map for $\sigma$ to $\{0,1\} \times N$. These are smooth maps $N \to M$ which agree on $\partial N$. As such, they define a smooth map $\partial \sigma \colon N \amalg_{\partial N} N \to M$.

Our model space is then $\Gamma((\partial \sigma)^* T M)$, smooth sections of the pullback of $T M$ along $\partial \sigma$. A point in this space is a smooth map $X \colon N \amalg_{\partial N} N \to T M$ such that $\pi X = \partial \sigma$. We compose $X$ with the local addition $\eta \colon T M \to M$ to obtain a map $\eta X \colon N \amalg_{\partial N} N \to M$. Now we need to fill this in.

Let us choose a representative for $\sigma$, and label it again $\sigma$.

The local addition defines a smooth map $I \times T M \to \operatorname{Diff}(M)$. Thus composing with $X$ we obtain a smooth map $\psi_X \colon I \times (N \amalg_{\partial N} N) \to Diff(M)$ (varying smoothly with $X$). This has the property that $\psi_X(0,x)$ is always the identity on $M$ and $\psi_X(1,x)(\partial \sigma(x)) = \eta X(x)$. Thus $\psi_X(t,x)(\partial \sigma(x))$ is a path from $\partial \sigma(x)$ to $\eta X(x)$ and $\psi_X(t,x)$ is a deformation of $M$ that drags $\partial \sigma(x)$ along that path.

To drag the whole of $\sigma$ along with us, we need a “weight” function on $I \times N$. That is, we need a function $\rho \colon I \times N \to I$ which is $1$ on $\partial(I \times N)$, and we need a function $\mu \colon I \times N \supseteq U \to \partial(I \times N)$. We shall define $\tau \colon I \times N \to M$ as

$$\tau(s,x) \coloneqq \psi_X(\rho(s,x), \mu(s,x))(\sigma(s,x))$$

Thus $\mu$ tells us which distortion to use, $\rho$ tells us how much of that distortion to use, and we then apply that to $\sigma$.

The properties that we want $\tau$ to have tell us what conditions we need on $\rho$ and $\mu$. We want $\tau$ to be smooth and we want it to restrict to $\eta X$ on the boundary (via the obvious collapse map $\partial(I \times N) \to N \amalg_{\partial N} N$). We therefore require $\mu$ to satisfy

1. $\mu(\epsilon,x) = (\epsilon,x)$ for $\epsilon \in \{0,1\}$, $x \in N$
2. $\pi_{\partial N} \mu(t,x) = x$ for $t \in I$, $x \in \partial N$

Since we have more restrictions in the $N$-direction, we start by ensuring that the first condition is met. So choose a smooth function $\alpha \colon I \to I$ which is increasing and maps $[0,1/4]$ to $0$ and $[3/4, 1]$ to $1$. Let $\beta \colon I \to I$ be a smooth function which is zero on $[0,1/8]$ and $[7/8,1]$ and is $1$ on $[1/4,3/4]$. Let $\gamma \colon I \times N \to N$ be a smooth function as in the assumptions on $(N,\partial N)$. Define

$$\mu(t,x) \coloneqq \big(\alpha(t), \gamma(\beta(t),x)\big)$$

Let us check what happens here. When $t \in [0,1/4] \cup [3/4,1]$ then $\alpha(t)$ is either $0$ or $1$ so $\mu(t,x) \in \partial(I \times N)$. Then when $x \in U$ (as in the conditions on $(N,\partial N)$) and $t \in [1/4,3/4]$, $\beta(t) = 1$ so $\gamma(\beta(t),x) \in \partial N$. Hence $\mu$ maps the open set $I \times U \cup ([0,1/4) \cup (3/4,1]) \times N$ into $\partial(I \times N)$.

Now, if $t \in [0,1/8] \cup [7/8,1]$ then $\beta(t) = 0$ and so $\gamma(\beta(t),x) = x$, whence $\mu(\epsilon,x) = (\epsilon,x)$ for $\epsilon \in \{0,1\}$. Then if $x \in \partial N$, $\gamma(s,x) = x$ for all $s \in I$ so $\pi_{\partial N} \mu(t,x) = x$ for $x \in \partial N$.

The function $\rho$ must be a bump function taking value $1$ on $\partial(I \times N)$ and with support on $I \times U \cup ([0,1/4) \cup (3/4,1]) \times N$ into $\partial(I \times N)$. By assumption, there is a smooth function $\rho_N \colon N \to I$ such that $\rho_N(\partial N) \subseteq \{1\}$ and $\rho_N^{-1}(0,1] \subseteq U$. Let $\rho_I \colon I \to I$ be a smooth function such that $\rho_I(\{0,1\}) \subseteq \{1\}$ and $\rho_I^{-1}(0,1] \subseteq [0,1/4] \cup [3/4,1]$. Define $\rho$ by:

$$\rho(t,x) = 1 - (1 - \rho_I(t))(1 - \rho_N(x))$$

Then if $\epsilon \in \{0,1\}$, $\rho(\epsilon,x) = 1$ and if $x \in \partial N$, $\rho(t,x) = 1$. But if $(t,x) \notin I \times U \cup ([0,1/4) \cup (3/4,1]) \times N$ then $\rho_I(t) = 0$ and $\rho_N(x) = 0$ so $\rho(t,x) = 0$. Hence $\rho$ is the required bump function.

Thus we can define our chart map as

$$X \mapsto \big((s,x) \mapsto \psi_X(\rho(s,x), \mu(s,x))(\sigma(s,x))\big)_{\simeq}$$