An easy way to to explain the equivalence definitions of tangent spaces? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T09:07:13Z http://mathoverflow.net/feeds/question/88880 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/88880/an-easy-way-to-to-explain-the-equivalence-definitions-of-tangent-spaces An easy way to to explain the equivalence definitions of tangent spaces? ssquidd 2012-02-19T02:19:33Z 2012-02-21T08:48:20Z <p>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 $T M$:</p> <ul> <li>Using equivalent classes of smooth curves through $x$</li> <li>Using derivations near $x$</li> <li>Using cotangent vectors at $x$</li> </ul> <p>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.</p> http://mathoverflow.net/questions/88880/an-easy-way-to-to-explain-the-equivalence-definitions-of-tangent-spaces/88884#88884 Answer by Steven Landsburg for An easy way to to explain the equivalence definitions of tangent spaces? Steven Landsburg 2012-02-19T02:43:53Z 2012-02-19T02:43:53Z <p>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.) </p> http://mathoverflow.net/questions/88880/an-easy-way-to-to-explain-the-equivalence-definitions-of-tangent-spaces/88896#88896 Answer by Nick Salter for An easy way to to explain the equivalence definitions of tangent spaces? Nick Salter 2012-02-19T05:59:00Z 2012-02-19T05:59:00Z <p>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. </p> http://mathoverflow.net/questions/88880/an-easy-way-to-to-explain-the-equivalence-definitions-of-tangent-spaces/88947#88947 Answer by Paul Siegel for An easy way to to explain the equivalence definitions of tangent spaces? Paul Siegel 2012-02-19T18:11:06Z 2012-02-19T18:11:06Z <p>What all three definitions have in common is that they each try to capture the first order behavior of a smooth function on $M$.</p> <ol> <li><p>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).</p></li> <li><p>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 $f$.</p></li> <li><p>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 $M$.</p></li> </ol> <p>This intuition acutally allows us to be a little more explicit about how the relevant identifications are made.</p> <p>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 $T_p M$.</p> <p>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 $(D(x_1), \ldots, D(x_n))$.</p> <p>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 $(I/I^2)^*$.</p> <p>Finally, to go from 3 back to 2, let $\ell \in (I/I^2)^*$ and define a point derivation by $D(f) = \ell(f - f(p) + I^2)$.</p>