Why is a metric?

Question

I have a question about tensors and metrics:

Let $M=\{(t,x,y,z)\in \mathbb{R}^4: t>-1 \}$ and let $g=(1+t)dtdx+dy^2+dz^2$

Show that g is a metric on $M$.

I did the next, I have the basis $\{ \partial_t , \partial_x,\partial_y,\partial_z\}$, then, $g=\begin{pmatrix} 0 &1+t &0 & 0\\ 0 & 0& 0 &0 \\ 0& 0 & 1 & 0\\ 0& 0 & 0 & 1 \end{pmatrix}$.

And this matrix is not symmetric! So g is not a metric.

Where is my mistake?

Thanks a lot!

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EDIT: Which is it's index? I don't know it...

I think that this basis $\{ \partial_t - \partial_x,\partial_t + \partial_x,\partial_y,\partial_z\}$ is ortonormal and has only one temporal vector, so it index is 1 Thanks!!

Answer 1

As $dtdx =\frac{dtdx+dxdt}{2}$, then $g=\begin{pmatrix} 0 &(1+t)/2 &0 & 0\\ (1+t)/2 & 0& 0 &0 \\ 0& 0 & 1 & 0\\ 0& 0 & 0 & 1 \end{pmatrix}$, so $g$ is symmetric and it has inverse. Then, $g$ is a metric on $M$