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On page 54 in his book Introduction to Smooth Manifolds, John Lee says the following:

A linear map $v: C^\infty (M) \rightarrow \Bbb{R}$ is called a derivation at p if it satisfies

\begin{equation} v(fg) = f(p)vg + g(p)vf \end{equation} for all $f,g \in C^\infty (M)$.

My Question:

Why is this useful for defining tangent vectors? This just looks to me like a fancy way of characterizing a set of smooth, linear maps who also happen to satisfy the product rule. What relevance does that have to tangent vectors?

Stan Shunpike
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1 Answers1

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The initial definition of tangent vectors as point-derivations on a manifold is often favored because it is considered somewhat cleaner than other popular definitions, such as using the velocity vectors of smooth curves. This definition has more of a simple, algebraic flavor.

At a point $p$ in the linear space $\mathbb{R}^n$, we have an intuitive notion of the tangent space as a set of directions which we can use to travel, a vector space we can identify with another copy of $\mathbb{R}^n$ itself. Call this vector space $T_p$. Now let $\mathcal{D}_p$ be the set of point-derivations at $p$, and let $\phi$ be the mapping associating a direction in $T_p$ to a directional derivative. Every directional derivative is actually a point-derivation, and we find out the map $\phi$ is actually an isomorphism of vector spaces.

When we switch over to smooth manifolds, there is no longer such a nice linear structure on the space that makes it easy to think of the tangent space as we did before. Thus, we use the more general definition concept of derivations to define the tangent space.

I am not sure what to do to convince you of the importance of the product rule - it is a natural property to be associated with the derivative, and says things like the term $(vf)(vg)$ doesn't matter when looking at $v(fg)$, corresponding with the linear nature of the derivative.

user137769
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