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First let me state a Theorem due to Kazdan and Warner:

``Let M be a compact two dimensional orientable manifold. Let $f: M \rightarrow \mathbb{R}$ be a function that has the same sign as $\chi(M)$, the Euler characteristic of $M$ at some point. Then $M$ admits a metric $g$ such that the Gaussian curvature $K$ is the given function $f$. ''

(I am actually not sure if orientable is necessary). Is anything known about the following question:

``Let M be a compact, orientable manifold. Let $f: M \rightarrow \mathbb{R}$ be a function that has the same sign as $\chi(M)$, the Euler characteristic of $M$ at some point. Then does $M$ admit a metric $g$ such that $$ *e(TM) = f $$ where $e(TM)$ is the Euler class of the Tangent bundle and * is the Hodge star operator?''

Notice that in two dimensions this is precisely the Kazdan Warner theorem, since $$*e(TM) = \frac{K}{2 \pi}$$

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    $\begingroup$ Check this Wikipedia page en.wikipedia.org/wiki/Prescribed_scalar_curvature_problem $\endgroup$ – Liviu Nicolaescu Jan 5 '13 at 15:43
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    $\begingroup$ I am sorry if this is a trivial question or a well known fact, but is the Scalar curvature the same as $*e(TM)$ (maybe upto a constant factor)? $\endgroup$ – Ritwik Jan 5 '13 at 15:54
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    $\begingroup$ I assume that by $\ast e(TM)$ you mean the Gauss--Bonnet integrand. In dimensions at least three, it is different from the scalar curvature. So prescribing scalar curvature is a different question in dimensions at least three. $\endgroup$ – John Pardon Jan 5 '13 at 19:00
  • $\begingroup$ @Ritwik, there is no possibility of such an identity holding. If $*e(TM)$ was some multiple of the scalar curvature $R$, this would imply that $\int R = \int *e(TM) dV = <e(TM),[M]>$, and this is a topological invariant of the (smooth structure) topology of $M$. (see en.wikipedia.org/wiki/…). However, one can show that in dimensions $\geq 3$, all manifolds admit a metric of negative scalar curvature, e.g. $S^3$. So, if the above identity held, we could evaluate $<e(TM),[M]>$ on the standard metric and this other one... $\endgroup$ – Otis Chodosh Jan 5 '13 at 19:09
  • $\begingroup$ ...giving a contradiction. You may be interested in the following MO post: mathoverflow.net/questions/30035/…. $\endgroup$ – Otis Chodosh Jan 5 '13 at 19:10

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