What's $\ln\mathcal{G}$ if $\mathcal{G}:f(x)\mapsto f(x)+1$? I recently learned that $\ln\mathcal{T}=\mathcal{D}$, where $\mathcal{T}:f(x)\mapsto f(x+1)$ and $\mathcal{D}:f(x)\mapsto\frac{d}{dx}f(x)$. I can understand this equation because $\mathcal{T}$ is linear; however, $\mathcal{G}$ is not.
A friend has pointed out that $\mathcal{G}$ is "affine", but this information hasn't helped me figure anything out about its log so far.
So: is $\ln\mathcal{G}$ well-defined, and if so does it have a closed form? And is it multivalued?
 A: Affine operators can be promoted to linear operators in one higher dimension, and I'm going to do this because otherwise I'll get very confused. Namely, let $V$ be whatever space of functions $f(x)$ we're acting on. Then we embed $V$ into $V \oplus \mathbb{R}$ via
$$f(x) \mapsto (f(x), 1)$$
and now we can promote affine operators on $V$ to linear operators on $V \oplus \mathbb{R}$, as follows. Every affine operator has the form $T(f) = A(f) + g$ for some linear operator $A$ and some $g \in V$, and we associate to this affine operator the linear operator
$$T' : (f(x), c) \mapsto (T(f(x)) + c g(x), c).$$
This preserves the embedded copy of $V$ in $V \oplus \mathbb{R}$ and acts on it exactly as $T$ does, as desired. This construction also respects composition. (Crucially, it does not respect addition.)
Now we promote the affine operator $G$ to the one-parameter group of linear operators
$$G'_t : (f(x), c) \mapsto (f(x) + ct, c).$$
Formally every one-parameter group of linear operators can be written $G_t' = \exp(t \ln G_1')$ where $\ln G_1'$ is the generator (in the Lie algebra sense), which we can compute by taking the derivative with respect to $t$ and then setting $t = 0$. This gives
$$\boxed{ \ln G' : (f(x), c) \mapsto (c, 0) }.$$
This is why it was crucial for us to promote to a linear operator: the logarithm does not preserve the embedded copy of $V$! So it doesn't have an interpretation as another affine operator on $V$ at all. But now you can easily check that
$$\exp(t \ln G') = G_t' : (f(x), c) \mapsto (f(x) + ct, c)$$
because $\ln G'$ is nilpotent: only the first two terms of the Taylor series expansion occur and the remaining terms are all zero.
Initially I tried to do this calculation directly in $V$ and got the apparently nonsensical answer that $\ln G$ was the constant affine operator $f(x) \mapsto 1$. This is not consistent with the identity $G_t = \exp(t \ln G)$, at least not if the exponential is being computed in the most naive way. I am actually not quite sure what's going on here; there should be a nice conceptual explanation in terms of the Lie algebra of the affine group but I don't see it quite yet. But in any case, embedding affine operators into linear operators fixes everything.
