4-Vectors and four-tensors I want to show that if $\Gamma_{\mu \nu} a^{\mu} a^{\nu}$ is a scalar for any four-vector $a^{\nu}$, then $\Gamma_{\mu \nu}$ is a four-tensor. 
It is $a^{\nu} = g^{\nu \mu} a_{\mu}$, and so I would get 
$\Gamma_{\mu \nu} a^{\mu} a^{\nu}$ = $\Gamma_{\mu \nu} a^{\mu} g^{\nu \mu} a_{\mu}$, where $g^{\nu \mu}$ is the metric tensor. But then I don't know how to go on; I'd be thankful for any help!
Susan 
 A: We have the scalar
$$\phi(x) =\Gamma_{\mu\nu}(x)a^{\mu}(x)a^{\nu}(x)$$
Going to a different coordinate system $x' = x'(x)$ we have
$$a^{\mu}(x') = \frac{dx^{\mu}}{dx'^\alpha} a^{\alpha}(x)$$
Since $\phi(x)$ transforms as a scalar we have $\phi(x'(x)) = \phi(x)$ where
$$\phi(x') = \Gamma_{\mu\nu}(x')a^{\mu}(x')a^{\nu}(x')$$
Using the transformation law for the vector $a$ we get
$$\phi(x') =\Gamma_{\mu\nu}(x')\frac{dx^{\mu}}{dx'^\alpha}\frac{dx^{\nu}}{dx'^\beta} a^{\alpha}(x)a^{\beta}(x) = \Gamma_{\alpha\beta}(x)a^{\alpha}(x)a^{\beta}(x)$$
and it follows that
$$\left(\Gamma_{\mu\nu}(x')\frac{dx^{\mu}}{dx'^\alpha}\frac{dx^{\nu}}{dx'^\beta} - \Gamma_{\alpha\beta}(x)\right)a^{\alpha}(x)a^{\beta}(x) = 0$$
since this holds for all vectors $a$ (this is the linear algebra statement that if $x^TMx = 0$ for all vectors $x$ then $M\equiv 0$) we have
$$\Gamma_{\alpha\beta}(x) = \Gamma_{\mu\nu}(x')\frac{dx^{\mu}}{dx'^\alpha}\frac{dx^{\nu}}{dx'^\beta}$$
which shows that $\Gamma$ transforms as a tensor (and therefore is a tensor).
