In a short talk, I had to explain, to an audience with little knowledge in geometry or algebra, the three different ways one can define the tangent space $T\_x M$ of a smooth manifold $M$ at a point $x \in M$ and more generally the tangent bundle $T M$:


* Using equivalent classes of smooth curves through $x$
* Using derivations near $x$
* Using cotangent vectors at $x$


Just by looking at the definition, it is not at all clear why they should all define the same object. I went through the proof, but judging from their reaction, it was not very meaningful. I wonder if there is any way I can let them "see", with just intuition, that the three definitions are, in certain sense, the same.



An easy way to to explain the equivalence definitions of tangent spaces?
What all three definitions have in common is that they each try to capture the first order behavior of a smooth function on $M$.


1. The derivative of a smooth function $f$ along a curve $\gamma$ with $\gamma(0) = p$ depends on $\gamma$ only insofar as it depends on $\gamma'(0)$, and indeed it recovers the directional derivative of $f$ at $p$ in the direction $\gamma'(0)$. The directional derivatives of $f$ determine the total derivative of $f$ which in turn determines the first order behavior of $f$ (more or less by the definition of the total derivative).
2. Since a derivation $D$ at $p$ sees only the values of a function $f$ and its derivatives at $p$ (not near $p$), we can replace $f$ by a polynomial by Taylor's theorem. By the Leibniz rule, $D(P)$ depends only on the linear part of a polynomial $P$ and hence $D(f)$ depends only on the first order part of $f$.
3. Recall that the cotangent bundle of $M$ at $p$ is the space $I/I^2$ where $I$ is the ideal in $C^\infty(M)$ consisting of functions $f$ such that $f(p) = 0$. If we imagine replacing $C^\infty(M)$ by a polynomial ring then $I$ represents the ideal of polynomials whose lowest order part has degree $1$ and $I^2$ is the ideal of polynomials whose lowest order part has degree $2$. In this case $I/I^2$ is naturally identified with the space of linear polynomials. Thus the cotangent bundle at $p$ is in a sense the space of "first order parts" of smooth functions on $M$.


This intuition acutally allows us to be a little more explicit about how the relevant identifications are made.


It's very easy to go from 1 to 2: if $\gamma$ is a curve in $M$ with $\gamma(0) = p$ then $D(f) = (f \circ \gamma)'(0)$ is a point derivation at $p$ which depends only on the equivalence class of $\gamma$ in $T\_p M$.


To go from 2 back to 1, let $f$ be a smooth function and let $f(x) \sim \sum\_\alpha c\_\alpha x^\alpha$ (multi-index notation) be its Taylor series in a coordinate system centered at $p$. Then for any derivation $D$ at $p$ we have $D(f) = c\_1 D(x\_1) + \ldots + c\_n D(x\_n)$ by the Leibniz rule, so $D$ corresponds to the tangent vector $(D(x\_1), \ldots, D(x\_n))$.


To go from 2 to 3, let $D$ be a derivation at $p$ and observe that $D(f) = 0$ for any $f \in I/I^2$ by Taylor's theorem and the Leibniz rule. Thus $D$ determines a linear functional in $(I/I^2)^\*$.


Finally, to go from 3 back to 2, let $\ell \in (I/I^2)^\*$ and define a point derivation by $D(f) = \ell(f - f(p) + I^2)$.


