# The Unapologetic Mathematician

## What Does the Bracket Measure? (part 1)

We again pick up the question we posed earlier about what the bracket measures. Clearly it should have something to do with flows and their failure to commute. So we consider vector fields $X$ and $Y$ with flows $\Phi$ and $\Psi$, respectively.

Now, starting with a point $p$ and given some small time displacement $t$ we can flow along $X$ and then $Y$$\left[\Psi_t\circ\Phi_t\right](p)$ — or we can flow along $Y$ and then $X$$\left[\Phi_t\circ\Psi_t\right](p)$ — and get two (potentially different) points in $M$. Unfortunately, we can’t just take their difference as we could when measuring the failure of algebra elements to commute.

However, we do have something rather closely related: the commutator in the sense of a group. That is, we can flow forwards along $X$, then along $Y$, then backwards along $X$ and then backwards along $Y$ to get the point $c(t)$:

$\displaystyle c(t)=\left[\Psi_{-t}\circ\Phi_{-t}\circ\Psi_t\circ\Phi_t\right](p)$

If $\Phi_t$ and $\Psi_t$ were to commute this would just be the constant curve $c(t)=p$, but in general it’s a smooth curve passing through $p$ at $t=0$. The bracket $[X,Y]_p$ will indicate the direction in which $c(t)$ passes through $p$, and something about its speed.

Now, it turns out that the derivative $c'(t)$ always vanishes at $t=0$. In fact, $[X,Y]_p$ is more closely related to the “second derivative” of $c$, though it’s not immediately clear what this means. Indeed, we can define $c'(t)$ for any $t$ in an interval around $0$, but $c'(t)\in\mathcal{T}_{c(t)}M$, and there is in general no good way to compare these vectors for different values of $t$.

Instead, we will prove the following two facts for any smooth $f\in\mathcal{O}U$ — here “smoothness” entails twice differentiability — where $U$ is any neighborhood of $p$:

\displaystyle\begin{aligned}(f\circ c)'(0)&=0\\\frac{1}{2}(f\circ c)''(0)&=[X,Y]_p(f)\end{aligned}

The first indicates that $c'(0)=0\in\mathcal{T}_pM$. Indeed, if $c'(0)\neq0$ then we could surely find some $f$ which changes in the direction it points, which would give a nonzero value to $(f\circ c)'(0)$. The second gives the result we’re really after, as Taylor’s theorem applied to $f\circ c$ shows us that

$\displaystyle[X,Y]_p(f)=\frac{1}{2}(f\circ c)''(0)=\lim\limits_{t\to0}\frac{\left[f\circ c\right](t)-\left[f\circ c\right](0)}{t^2}$

And thus the bracket indeed is the second (and lowest) order term in describing the direction of $c$ as it passes through $p$.