The goal of this post is to make sense of the notion of the "product integral". Typically this is described with the analogy: a product integral is to products what a Riemann integral is to sums. But it turns out that there are several different ways to even make sense of this analogy.
The Riemann integral
Let $[a,b]$ be an interval. Recall that
- $X$ is a finite subset of $[a,b]$, with $a,b\in X$; we can canonically order the elements of $X$ increasingly as $x_0 = a < x_1 < x_2 < \cdots < x_k = b$.
- $\tau: X\setminus \{a\} \to [a,b]$ satisfies $\tau(x_j) \in [x_{j-1}, x_j]$. For simplicity of notation we will also write $\tau_j$ as a shorthand for $\tau(x_j)$.
Let $\mathcal{P}$ be the set of all tagged partitions of $[a,b]$; we can make $\mathcal{P}$ a directed set with the ordering "by refinement". \[ (X,\tau) \preceq (X', \tau') \iff X \subseteq X' \] Given a function $f:[a,b]\to \mathbb{R}$ and $(X,\tau)\in \mathcal{P}$, we may define the corresponding Riemann sum as \[ \sum_{(X,\tau)} f := \sum_{j = 1}^{|X|-1} f(\tau_j) \cdot |x_j - x_{j-1}| \] As $\mathcal{P}$ is a directed set, this defines a net. The classical Riemann integral can be defined as follows:
Changing the co-domain
First, let's think about how much we may generalize the codomain of the function $f$. In first glance:
- The codomain needs to admit a notion of a finite sum.
- The codomain needs to admit a notion of "multiplication by a non-zero real number" (the $|x_j - x_{j-1}|$ factor in the Riemann sum).
- The codomain needs a topology (so that the convergence of the net of Riemann sums can be defined).
So quite obviously we can extend the notion of Riemann integration to functions $f:[a,b]\to V$ where $V$ is any topological vector space over $\mathbb{R}$.
But can we be more general? Firstly: in vector spaces we have that addition is is commutative, and that the scalar multiplication and addition are compatible. But this is not necessarily required: since our domain $[a,b]$ is ordered and that we can canonically order the partition $X$, we are allowed to use this to order the summation. We may require that the Riemann sums be defined with a specific order in mind. For example, we can write \[ \sum_{(X,\tau)} f := f(\tau_1) \cdot |x_1 - x_0| + f(\tau_2) \cdot |x_2 - x_1| + f(\tau_3) \cdot |x_3 - x_2| + \cdots + f(\tau_k) |x_k - x_{k-1}|. \] The concept of a Riemann sum still makes sense, but of course proving convergence may be more difficult without commutativity or compatibility of addition with multiplication. One could see, however, that in this form that one has to be careful about how one defines derivatives, if one wants an analogue of the fundamental theorem of calculus (order now matters!)
But in fact we can do one even better: why do we need the notion of a multiplication by a non-zero real number? Given that the Riemann sum is not just a sum of the outputs of $f$, but a transformed version of this output, we can make one further abstraction. This leads to the following notion:
Let $[a,b]$ be a non-trivial interval. Let $T$ be a set, and $\mathscr{G}$ be a topological group. Fix a mapping $G:T\times [0,b-a] \to \mathscr{G}$.
Then given a function $f:[a,b]\to T$ and a tagged partition $(X,\tau)$, we may define the $G$-Riemann sum to be \[ \sum^{(G)}_{(X,\tau)} f := G(f(\tau_1), |x_1-x_0|) + G(f(\tau_2), |x_2-x_1|) + \cdots + G(f(\tau_k), |x_k - x_{k-1}|) \] Furthermore, if this net converges in $\mathcal{G}$, we say that $f$ is $G$-Riemann integrable and denote the limit $\int_{[a,b]}^{(G)} f$.
That the function $f$ and its integral may take values in different spaces is in some sense natural. Returning to the idea of the fundamental theorem of calculus, we may regard integration as the "inverse" operation of differentiation. And the theory of differential calculus has been well-developed to the case of manifolds, in which case given a curve $\gamma:[a,b]\to M$, the correct notion of a derivative lives in $\gamma'(x)\in T_{\gamma(x)}M$ the corresponding tangent space. So if our goal is to recover some notion of the fundamental theorem of calculus, we would need to be able to integrate functions taking value in "a tangent space" to get functions taking values in "a space".
Let $\mathscr{G}$ be some Lie group and $\mathfrak{g}$ is Lie algebra, which we interpret as the tangent space of the identity element $e\in \mathscr{G}$.
Then we may set the reconstruction map $G:\mathfrak{g} \times \mathbb{R}\to \mathscr{G}$ to be $G(\zeta, s) = \exp(s \zeta)$, where $\exp$ is the exponential map. Then given a function $f:[a,b]\to \mathfrak{g}$, the Riemann sum can be defined (we will reverse the time ordering here to line up better with convention where the group acts on the left, and use multiplication notation for the Lie group operation) \[ \sum^{(G)}_{(X,\tau)} f := \exp(|x_k-x_{k-1}|f(\tau_k)) \cdot \exp(|x_{k-1}-x_{k-2}|f(\tau_{k-1})) \cdots \exp(|x_1 - x_0|f(\tau_1)) \] In this case, if $f(x) = f_0$ is a constant function, then we have that the $G$-Riemann integral exists and equals $\exp(|b-a| f_0)$.
In fact, the same argument that shows continuous real valued functions on $[a,b]$ are Riemann integrable can be extended to this case. For simplicity let's assume that $\mathscr{G}$ has a bi-invariant Riemannian metric, which induces a bi-invariant distance function $d$. The bi-invariance implies that \[ d(xy, x'y') = d((x')^{-1}x, y' y^{-1}) \leq d((x')^{-1}x, e) + d(e, y' y^{-1}) = d(x,x') + d(y,y'). \] Now, consider the exponential map $\exp:\mathfrak{g}\to G$. Given $\zeta,\eta\in \mathfrak{g}$ we have that \[ d(\exp(\zeta),\exp(\eta)) = d(e, \exp(-\zeta)\exp(\eta)). \] The latter we can estimate by considering the curve \[ s\mapsto \exp(-s\zeta)\exp(s\eta). \] Its derivative is \[ \frac{d}{ds} \exp(-s\zeta)\exp(s\eta) = \exp(-s\zeta)(\eta - \zeta) \exp(s\eta) \] where we abuse notation so that $g\in \mathscr{G}$ is used for the mapping $g: T_x\mathscr{G} \to T_{gx} \mathscr{G}$ induced by the left multiplication (when acting on the left), and similarly on the right. The key however is that bi-invariance of the Riemannian metric implies that $\| \frac{d}{ds} \exp(-s\zeta)\exp(s\eta) \| = \|\eta - \zeta\|$ and so that the Riemannian distance \[ d(\exp(\zeta),\exp(\eta)) \leq \| \eta - \zeta\| \] In other words, the exponential map is $1$-Lipschitz.
Now, given two tagged partitions $(Y,\sigma)$ and $(Y',\sigma')$, both succeeding some tagged partition $(X,\tau)$, and let $f:[a,b]\to \mathfrak{g}$ be continuous (and hence uniformly continuous). By taking $Z = Y\cup Y'$ we can write the Riemann products \[ \sum^{(G)}_{(Y,\sigma)} f = \exp(|z_k - z_{k-1}| \zeta_k) \cdots \exp(|z_1 - z_0| \zeta_0) \] and \[ \sum^{(G)}_{(Y',\sigma')} f = \exp(|z_k - z_{k-1}| \eta_k) \cdots \exp(|z_1 - z_0| \eta_0) \] for some $\zeta$ and $\eta$. By uniform continuity, as long as $X$ is sufficiently finely spaced, we can ensure that $\| \zeta_i - \eta_i\| < \tilde{\epsilon}$ uniformly. Then our argument above shows that the distance between the two Riemann products are also at most $|b-a| \cdot \tilde{\epsilon}$, ensuring that our net of Riemann products is Cauchy.
"Solving Differential Equations"
The original motivation of product integrals was Volterra's attempt at using this to solve differential equations. Let's see how this works formally.
Let $M$ be some manifold, and suppose we are interested in solving a differential equation \[ \dot{x} = F(t,x) \] where $F(t,\cdot)$ is some section of the tangent bundle of $M$ (so a vector field on $M$). What we are interested in is the "solution operator" $G(t,x_0)$ which yields the solution to the differential equation with initial data $x_0$ prescribed at $t = 0$. We can interpret $G$ as a time-dependent curve in the set of automorphisms of $M$. The set of self-maps of $M$ form a group $\mathscr{G}$ under composition. For convenience we can work in the smooth category so these maps are smooth mappings. Now, given a one-parameter family of smooth maps $G(t,x): (a,b)\times M\to M$, we can take its time derivative and get $\frac{d}{dt}G$ is a time-dependent family of sections of the tangent bundle. This mapping is not (pointwise) injective: two distinct families may have the same derivative at one particular time. But let us take a representative: suppose for every smooth vector field $V$ on $M$ we can choose a one-parameter family of smooth maps $G_V(s,x)$ such that $\frac{d}{ds}G_V(0,\cdot) = V$. Then we can try to reconstruct the solution operator with a discrete time approximation. Take $t_0, \ldots, t_k$ an increasing sequence of times with $0 = t_0$ and $T = t_k$, and select $\tau_j \in [t_{j-1}, t_j]$. Then we can build an approximation \[ G(t,\cdot) \approx G_{F(\tau_k)}(t_k - t_{k-1})\circ G_{F(\tau_{k-1})}(t_{k-1}- t_{k-2}) \circ \cdots \circ G_{F(\tau_2)}(t_2 - t_1) \circ G_{F(\tau_1)}(t_1 - t_0). \] This can again be interpreted as a sort of Riemann product.
Comparing to the case of the Lie groups, we see here that in the Lie groups example above we have discovered that $V\in \mathfrak{g}$ the Lie algebra, and that we have taken $G_V(s) = \exp(sV)$ the exponential map. These are the solutions to the case where $F$ is time independent.
To actually apply this to solve differential equations, however, we run into some potential difficulties. One of the biggest ones is as follows:
As we have just explained that a natural choice of the inversion $V\mapsto G_V$ can be taken to be solving the time-independent problem. But in many systems of interest this cannot done for fixed positive time $t > 0$. For example, consider the equation $\dot{x} = x^2$. It is well-known that with initial data $x_0$ the solution is given by \[ x(t) = \frac{x_0}{1 - x_0 t}. \] But this does not extend to a one-parameter family of smooth maps on any time interval! For every $t \neq 0$, there is some choice of initial data $x_0\in \mathbb{R}$ for which the solution blows up exactly at time $t$.
So one interpretation of this procedure is that this gives a formulation of the idea of "variation of constants", whereby if one can solve the autonomous equations $\dot{x} = F(t_0,x)$ for $t_0$ fixed, then one can solve the time-dependent system $\dot{x} = F(t,x)$.
