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Sep 12, 2017 at 6:08 vote accept Ali Taghavi
Sep 11, 2017 at 23:12 answer added Robert Bryant timeline score: 12
Sep 11, 2017 at 20:35 history edited Ali Taghavi CC BY-SA 3.0
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Sep 11, 2017 at 20:33 comment added Ali Taghavi @RobertBryant Thank you. I revise it again.
Sep 11, 2017 at 16:39 comment added Robert Bryant Your question is better now, but why not just say that the sup is to be taken over all $1$-forms $\alpha$ of $g$-length $1$? Thinking of the vector field $X$ doesn't seem to be relevant.
Sep 11, 2017 at 16:28 history edited Ali Taghavi
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Sep 11, 2017 at 16:27 history undeleted Ali Taghavi
Sep 11, 2017 at 14:47 history deleted Ali Taghavi via Vote
Sep 11, 2017 at 13:59 comment added Ali Taghavi @RobertBryant Now I revise my question. Could I clarify my question?
Sep 11, 2017 at 13:41 history edited Ali Taghavi CC BY-SA 3.0
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Sep 11, 2017 at 13:36 comment added Ali Taghavi I am sorry for not paying attention to the Euler characteristic of odd dimensional manifold. Moreover, to every unit length vector field $X$ we associate a i-form $\alpha_X$ with $\alpha_X(Y)=<X,Y>$. Then we take $\sup$ over all possible unit length vector fields $X$.
Sep 11, 2017 at 13:25 comment added Robert Bryant First, the Euler characteristic of any compact $3$-manifold without boundary is $0$, so this is no restriction. Second, if $\alpha(Y) = g(X,Y)$ for all vector fields $Y$, then $\alpha = X^\flat$, so there is only one such $\alpha$ for any given unit vector field $X$. Can you reformulate your question? Did you want to just require that $\alpha(X)=1$, which still allows for some variation in $\alpha$? Now the metric is irrelevant. It only depends on $X$.
Sep 11, 2017 at 13:23 history edited Ali Taghavi CC BY-SA 3.0
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Sep 11, 2017 at 13:15 history asked Ali Taghavi CC BY-SA 3.0