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These representation theory notes leave the following claim to the reader:

Recall that a derivation on an algebra is a map $d$ such that $d(ab)=d(a)b+ad(b)$. If $A$ is a finite-dimensional algebra over $\Bbb{R}$ or $\Bbb{C}$, $\gamma:\Bbb{R}\to \operatorname{Aut}_\Bbb{R}(A)$ is differentiable in some neighborhood of $0$, and $\gamma(0)=\operatorname{id}_A$, then $\gamma'(0):A\to A$ is a derivation.

This sounds a lot like the two pictures of the tangent space to a manifold at some point $p$ we get by considering smooth paths through $p$ and derivations of the algebra of germs of smooth real-valued functions at $p$, but that correspondence doesn't appear to me to actually go through.

Specifically, I think that the path of automorphisms of $\Bbb{R}$ given by multiplication by $e^t$ is a counterexample: it's smooth and the identity at $0$, but its derivative at $0$ is $1$, while $1(ab)\neq 1(a)b+a1(b)=2ab$ when $ab\neq 0$. Am I missing something?

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up vote 1 down vote accepted

Multiplication by $e^t$ isn't an algebra automorphism, let alone a map of algebras. In your example, $\mathbb{R} = \mathcal{O}(*)$ is the ring of functions of a point, which has a zero-dimensional tangent space -- i.e., you should expect there to be no possible derivations of $\mathbb{R}$ to itself save the zero map.

Your analogy should really be that $\text{Aut}(A)$ is like the group of diffeomorphisms of the smooth manifold; this finite-dimensional $A$ is playing the role of the ring of smooth functions. And, as you intuit, given a path of diffeomorphisms which passes through the identity at time 0, the derivative at time 0 should yield an element of the Lie algebra of the group of diffeomorphisms -- a.k.a., a vector field on the smooth manifold. This is exactly what a derivation (from the ring of smooth functions to itself) is.

Rigorously, just evaluating the derivative tells us that on any element $f, g \in A$, $$ \gamma'(0)(fg) = \lim_{t \to 0} { \gamma_t(fg) - fg \over t} = \lim_{t \to 0} { \gamma_t f \gamma_t g - \gamma_t f \cdot g + \gamma_t f \cdot g - fg \over t} $$ $$ = \lim_{t \to 0} {\gamma_t f (\gamma_t g - g) \over t} + \lim_{t \to 0} g{(\gamma_t f - f) \over t} $$ $$ = f \gamma'(0)(g) + g \gamma'(0)(f). $$

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