Time shift operator

$\def \R {\mathbb R} \def \X {\mathcal X} \def \T {\mathbb T} \def \I {\mathsf I} \def \v {\mathsf v} \def \Diff {\text{Diff}} \def \Exp {\text{Exp}}$ Let $X \colon \T \to \X$ be a smooth (infinitely differentiable) curve on space $\X = \R^d$, indexed by time $\T = \R$. At any time $t \in \T$, and for any time displacement $\delta \in \R$, the value of $X(t+\delta)$ is, by Taylor’s theorem:

$$\begin{align*} X(t+\delta) &= X(t) + \delta \dot X(t) + \frac{\delta^2}{2!} \ddot X(t) + \frac{\delta^3}{3!} \dddot X(t) + \cdots \\ &= \left( \I + \delta \frac{d}{dt} + \frac{\delta^2}{2!} \left(\frac{d}{dt}\right)^2 + \frac{\delta^3}{3!} \left(\frac{d}{dt}\right)^3 + \cdots \right) X(t) \\ &= \left( \sum_{n=0}^\infty \frac{\delta^n}{n!} \left(\frac{d}{dt}\right)^n \right) X(t) \\ &= \exp\left( \delta \frac{d}{dt} \right) X(t) \end{align*}$$

where in the above we have treated the time derivative operator $\frac{d}{dt}$ as a formal variable, and understood multiplication as function composition, so for example $(\frac{d}{dt})^2 = \frac{d}{dt} \circ \frac{d}{dt} = \frac{d^2}{dt^2}$.

This means the time shift operator $F_\delta$, which operates on the space of curves by translating time forward by $\delta \in \R$ (or backward if $\delta < 0$):

$$(F_\delta X)(t) := X(t+\delta)$$

can be written as the exponential of the time derivative:

$$F_\delta = \exp\left(\delta \frac{d}{dt}\right)$$

This is understood formally via Taylor expansion as above, but I think we can also make sense of it as the exponential map (in the sense of Lie group) generated by the (rescaled) time differential $\delta \frac{d}{dt}$, which should lie in the Lie algebra (or the tangent space).

On the other hand, we can also write $X(t+\delta)$ in integral form as

$$X(t+\delta) = X(t) + \int_t^{t+\delta} \dot X(s) ds$$

So it seems we can think of integration as exponentiation, in the sense that the effect of exponentiating the time differential is to integrate the time derivative.

I’m not sure how to say this precisely (this blog post is not very well thought out, but that’s okay because it still helps me to write this down, as it’s something that I have been pondering for a while), but the situation may be more clear if we consider $X$ to be the integral curve of a vector field $\v \colon \X \to \R^d$ (where $\R^d$ is really the tangent space of $\X$), which means $X(t)$ satisfies the differential equation

$$\dot X(t) = \v(X(t))$$

so we can write $X(t)$ in integral form as

$$X(t) = X(0) + \int_0^{t} \v(X(s)) ds$$

In this case, as is standard in the theory of differential equation, we should understand $X(t) = T_t(X(0))$ as the value of the time-$t$ map $T_t \colon \X \to \X$ starting from the initial position $X(0)$ at time $t = 0$. Each $T_t$ is a diffeomorphism on $\X$, at least when the vector field $\v$ is nice enough, and the sequence $(T_t)_{t \in \T}$ forms a one-parameter subgroup connected to the identity $T_0 = \I$ in the diffeomorphism group $\text{Diff}(\X)$. This one-parameter subgroup is generated by the vector field $\v$, which is an element of the Lie algebra (the tangent space of $\Diff(\X)$ at the identity), so we can write

$$T_t = \Exp(t\v)$$

where $\Exp$ is really the exponential map in the Lie group $\Diff(\X)$. So in this case, integrating the vector field $\v$ is indeed the same as taking the exponential map.

Moreover, heuristically (or formally), the differential equation $\dot X(t) = \v(X(t))$ gives the relation “$\frac{d}{dt} = \v$”, so the time shift operator above can also be written as

$$F_\delta = \exp\left(\delta \frac{d}{dt}\right) = \exp(\delta \v)$$

which seems to agree with the exponential map conclusion above.