most recent 30 from http://mathoverflow.net 2013-05-24T05:23:10Z http://mathoverflow.net/feeds/question/41488 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/41488/a-question-about-vector-fields Chen 2010-10-08T06:56:12Z 2010-10-09T09:24:35Z

<p>This is a question related to <a href="http://mathoverflow.net/questions/35472/a-question-about-a-one-form-on-riemannian-manifold" rel="nofollow">http://mathoverflow.net/questions/35472/a-question-about-a-one-form-on-riemannian-manifold</a></p> <p>Let M be a Riemannian Manifold, how to construct two vector fields that they didn't have common zeros and perpendicular to each other with the same length only on isolated points. I mean two vector fields $X$ and $Y$, $zero(X)\cap zero(Y)=\varnothing$ and $\langle X(p),Y(p)\rangle=0$ and $|X|=|Y|$ only on isolated points if $X(p)\neq 0,Y(p)\neq 0$. I want to know how to construct the two vector fields? </p> <p>Edit: We assume $X$ and $Y$ are two smooth vector fields. I don't know whether the vector fields exist on any Riemannian Manifold, maybe need some condition.</p> <p>I am sorry, I lost a condition, I need they "perpendicular to each other with the same length only on isolated points", so they can perpendicular on a submanifold. But if they have the same length then only on isolated points. </p>

http://mathoverflow.net/questions/41488/a-question-about-vector-fields/41490#41490 Bruno Martelli 2010-10-08T08:12:39Z 2010-10-08T08:12:39Z

<p>If $\chi(M)\neq 0$ and the zeroes of $X$ and $Y$ are isolated, then you can't find such vector fields. If there were such $X$ and $Y$ with isolated orthogonal points, then the angle between $X$ and $Y$ would be everywhere bigger or smaller than $\pi/2$ except at these isolated points (since $\dim M \geqslant 2$ and $M$ minus these points is connected). </p> <p>Suppose this angle is smaller than $\pi/2$: otherwise replace $Y$ by $-Y$. Then $X$ and $Y$ are never opposite and have distinct zeroes: it follows that $X+Y$ has no zeroes, which is impossible since $\chi (M)\neq 0$.</p> <p>This argument works more generally if $M \setminus $ (the zeroes of $X$ and $Y$) is connected.</p>