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This is a question related to A question about a one-form on Riemannian manifold

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?

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.

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.

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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).

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 $\chi (M)\neq 0$.

This argument works more generally if $M \setminus $ (the zeroes of $X$ and $Y$) is connected.

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At first I mistook the desired property as being generic among pairs of vector fields; thanks for this illuminating answer. –  Benoît Kloeckner Oct 8 '10 at 8:18
    
Just as you say the angle is smaller than $\pi/2$, why it can't be 0? If at some point $X=Y\neq 0$, replace $Y$ by $-Y$, then $X+(-Y)=0$. We can't change the vector fields only at a point. Because if we do it like this, then the vector field becomes uncontinuous. –  Chen Oct 8 '10 at 13:41
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If the angle at some point is smaller than $\pi/2$, then it is so at every point, and therefore you do not replace $Y$. Otherwise, if the angle at some point is bigger than $\pi/2$, then it is so at every point, and you replace $Y$ by $-Y$ at every point of $M$ (so it remains continuous). –  Bruno Martelli Oct 8 '10 at 13:56
    
Thanks,you are right! I think I lost a condition. I need $X$ and $Y$ not only perpendicular to each other but they have the same length. –  Chen Oct 9 '10 at 6:48
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