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.

  • 1
    $\begingroup$ Maybe part of the conceptual issue is that you can define the "tangent space" of a set $K \subseteq {\mathbb R}^n$ (not a submanifold), but that different notions aren't equivalent. For example, you can try to think of smooth functions on ${\mathbb Q}^n$ as those which are restrictions of smooth functions on ${\mathbb R}^n$, then you will be able to discuss differential operators (as limits of difference quotients), but you won't have smooth curves into the space. On a related note, you might think that the $C^k$ tangent space is different from the $C^{k+1}$ tangent space. $\endgroup$ – Phil Isett Feb 19 '12 at 3:19
  • 6
    $\begingroup$ Why, in a short talk to those without much background, would you be feel compelled to explain these equivalences instead of just using one of them for whatever the main point of the talk was? $\endgroup$ – KConrad Feb 19 '12 at 17:17
  • $\begingroup$ I was talking about homotopy continuation on an algebraic variety. So eventually I must use the last two definitions. But those, I thought, are too abstract, so I wanted to start with something more intuitive, i.e., the first definition. I didn't have to show that they are all equivalent, but that'd be an obvious question to ask. $\endgroup$ – ssquidd Feb 20 '12 at 16:03

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 $D(f) = \ell(f - f(p) + I^2)$.

  • $\begingroup$ This is very good, I did intend to use it in the setting of algebraic varieties, so the use of degree one polynomial feels quite natural. $\endgroup$ – ssquidd Feb 20 '12 at 16:06

In each of the three cases, your definition is capturing the intuition of "directions near x" --- an equivalence class of curves defines a "direction" in which the curves head out from x. A derivation or a linear functional on the cotangent vectors is a directional derivative, hence determined by a choice of direction. (Of course it takes proof that these account for all the derivations, etc. but the intuition is pretty clear.)


Klaus Jänich's undergraduate-level book "Vector Analysis" includes a section showing the equivalence between the three descriptions of the tangent space that you mention. He gives rigorous proofs of everything, and also provides a fair amount of motivation.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.