Cotangent space of the sphere - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T23:21:47Z http://mathoverflow.net/feeds/question/112211 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/112211/cotangent-space-of-the-sphere Cotangent space of the sphere unknown (google) 2012-11-12T19:24:55Z 2012-11-12T19:52:10Z <p>In analyzing the spherical pendulum the cotangent space of the sphere is defined as</p> <p>$T^*S^2 = \lbrace (q,p) \in \mathbb{R}^3 \times \mathbb{R}^3; |q| = 1, q \cdot p = 0 \rbrace$ </p> <p>my problem with this is that I see the right-hand side of the equation as a set of points, whereas I see the left-hand side as a set of linear functions on the tangent space of $S^2$. </p> <p>How can I see them as the same?</p> http://mathoverflow.net/questions/112211/cotangent-space-of-the-sphere/112213#112213 Answer by Simon Rose for Cotangent space of the sphere Simon Rose 2012-11-12T19:52:10Z 2012-11-12T19:52:10Z <p>What you have written down to me seems to be the tangent bundle, not the cotangent bundle (though they are isomorphic, I suppose).</p> <p>Consider the projection from your space (let's call it $X$) to the first copy of <code>$\mathbb{R}^3$</code>. This has image exactly those <code>$q \in \mathbb{R}^3$</code> such that $|q| = 1$ i.e. <code>$S^2$</code>. The fibre over a point $q$ is those vectors <code>$v \in \mathbb{R}^3$</code> which are perpendicular to $q$; that is, the tangent plane to $S^2$ at $q$.</p>