
# Vector Fields and Finite Machines

Let $Q$ be a quiver. $\dot{Q}$ is the units of $Q$, which satisfy $u^{*}=u={}^*u$.

By a rally on $Q$ we mean a map $r:\dot{Q}\rightarrow Q$ satisfying the equation ${}^*r(u)=u$ for all $u\in\dot{Q}$. This means that if we are at $u$, then $r(u)$ is an arrow from $u$, telling us where to go next, namely to $r(u)^*$. As we'll explore more momentarily, this is discrete analog of the geometric concept of vector field. Now let $Q_R$ be the set of all such rallies.

A machine on $Q$ is a map $\mu:\Sigma\rightarrow Q_R$, where $\Sigma$ could be any set of parameters. It's useful to think of $\Sigma$ as comprising an alphabet, or set of characters. A machine then labels rallies with characters. We'll let $M[\Sigma]$ be the set of all lists of elements of $\Sigma$. This maps, via $mu$, to a list of rallies.

Let $\Omega(Q)$ be the paths of $Q$. A path of $Q$ is defined to be a sequence of arrows of $Q$, which line up tip to tail. Precisely, we mean a $(2k+1)$-tuple of arrows $(a_0, \dots, a_{2k})$ such that $a_{2n}$ is a unit for $0 \le n \le k$, and $a_n^* = {}^*a_{n+1}$ for $0 \le n \lt 2k$, for any $k\ge 0$.

If we pick a unit $u_0$ of $Q$, we can construct a path for each element of $M[\Sigma]$, inductively. For the empty list $()$, we have the trivial path (u_0). If $\omega$ is the path for the list $[\sigma_0, \dots, \sigma_n]$, the for the list $[\sigma_0, \dots, \sigma_{n+1}]$, we create the path as an extension of $\omega$ by an arrow $a = \mu(\sigma_{n+1})(\omega^*)$ and $a^*$. This is the same as the composition of $\omega$ with the path $({}^*a, a, a^*)$ This is precisely the familiar action on a finite state machine, of transitioning from one state, to the state pointed to by the arrow labeled by the next character in a string, in one approach to regular expression recognition.

Instead of a rally, think now about a vector field $X$. For each point $p$, $X(p)$ is a vector. You can think of it as telling you, if you're at $p$, you should next go in the direction pointed out by $X(p)$, at the speed which matches the size of the vector $X(p)$. A differential equation seeks to find an integral curve for the vector field. This means we pick a starting place $p_0$, which is analogous to choosing $u_0$, previously. We want a curve $\eta:T\rightarrow P$, where $T$ is an interval of time, and $P$ is the set of points considered, which could be real numbers, or points in the plane, or points on any smooth manifold. This curve should satisfy $\eta(0) = p_0$, and $\dot{\eta}(t) = X(\eta(t))$, that is the velocity vector of $\eta$ should at all times agree with the current vector specified by the vector field. Such solutions always exist, under somewhat mild conditions on $X$. In many applications, the vector field itself is changing in time. So now $X$ is given by a map from $T$ into the set of all vector fields, analogous to the maps from $\Sigma$ into the rallies of $Q$, or $Q_R$. To complete the analogy, we just need to consider vector fields parametrized by any parameter set $\Sigma$, and then for any path $\sigma:T\rightarrow\Sigma$, we get a corresponding path via the differential equation $\dot{\eta}(t) = X_{\sigma(t)}(\eta(t))$. That these equations have solutions for reasonable choices of $\sigma$ and $X$ is established by a proof that differs little from the usual one, for the special case $\sigma(t) = t$ of time-dependent vector fields.

I'm Andrew Winkler. I hope you've enjoyed exploring these ideas with me. Thanks for your time. If you've got any questions, send an email to 4af502d7148512d4fee9@cloudmailin.net.

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