First variation of an action?

Question

I'm working on a problem and I must compute the first variation of an action. Let $\Omega$ is a 2-form on a semi-Riemannian manifold $M$ and $f$ is a smooth function and $\Gamma$ is an 1-form on $M$. I obtained the following equality \begin{equation} \int_M (\langle\Gamma-\Omega(\nabla h,.),\delta\rangle+f(x)h )dV_g=0 \end{equation} for all $h\in C^\infty (M)$ and all 1-form $\delta$ on $M$. This equality cannot be simpler. $\nabla h$ is gradient of $h$.

What can I deduce frome this equality?

Is this true that $\Omega$ and $f$ and $\Gamma$ must be identical to zero?

What is $\nabla$? Is $M$ supposed to be Riemannian or something? — Vectornaut, Aug 11 '15 at 02:29
@ramandi, could you please edit your question to make it clear that $M$ is a Riemannian manifold, rather than just a smooth one? I realize that it's easy to guess from context that $M$ is supposed to be Riemannian, but I think readers shouldn't have to guess that sort of thing. — Vectornaut, Aug 11 '15 at 04:49
@ Vectornaut: $M$ is a semi-riemannian manifold and $\nabla h=grad (h)$ id the gradient of $h$. — Ramand, Aug 11 '15 at 07:45

Anthony Carapetis · Accepted Answer · 2015-08-11T13:31:54.633

First choose $h=0$ and you get $\int \langle \Gamma, \delta \rangle = 0$ for all $\delta$, so $\Gamma = 0$ by a standard lemma.

Now integrate by parts and you get

$$\int h\left(f - \nabla^i(\Omega_{ij} \delta^j) \right) dV= 0$$

for all $h,\delta$, so $f = \nabla^i (\Omega_{ij}\delta^j)$ for all $\delta$.

If $\Omega$ is non-zero then we can choose a vector field $v$ such that $\theta = \Omega(v,\cdot)$ is a non-zero one-form. Then choosing $\delta = \phi v$ we get

$$ f = \theta_i \nabla^i \phi + \phi \nabla^i \theta_i $$

for every smooth function $\phi$. Choosing a point $p$ where $\theta(p) \ne 0$ and a function $\phi$ with $\phi(p) = 0$ and $d \phi(p) = C \theta(p)$, we get $f(p) = C^2 |\theta(p)|^2$ for all real constants $C$, so $\theta(p) = 0$, a contradiction. So $\Omega$ is zero and thus $f$ must be zero too.

Thank you dear. But i missed one factor. take a look please. — ramandi, Aug 11 '15 at 13:17

First variation of an action?

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