My course is covering metric tensors in a slap-dash way, so I want to ensure I have understood correctly how they are described.
I hope you can confirm this!
So, I believe that a metric tensor is a $(0,2)$ tensor, $g$, satisfying a certain property. By a $(0,2)$ tensor I mean that it is a map $g : T_pM\times T_pM \mapsto \mathbb{R}$. Where $M$ is some smooth manifold, and $p\in M$.
Now the condition for $g$ to be called a metric tensor is that it must be a inner product.
Now in co-ordinate form we may write $g(X,Y) = X^iY^jg_{ij}$. And so am I right in thinking that g is a metric tensor iff $g_{ii} > 0$ for each $i$?
I.e we could choose any real values $g_{ij}$ and so long as $g_{ii} >0$ for each $i$, then we would have a $(0,2)$ tensor?
I hope you can clarify things for me!
You must also require $g_{ij}=g_{ji}:$ inner products are symmetric. You need bilinearity as well, but let's assume that's what you meant by $g$ being a "map" on $T_p M\times T_p M$.
Now, this is fine as far as it goes, but is actually nothing more or less than the definition of an inner product on a real vector space. What really becomes interesting is when you put a metric on $TM$, rather than just $T_p M$, which is a smoothly varying family of inner products as described above. This is called a (metric) tensor field, and the possibility of putting one on every smooth manifold is the real beginning of Riemannian geometry.