What does it mean for a tensor field to be covariant?

A covariant 2-tensor field is linear over the ring of smooth functions. If $X, Y$ are vector fields and $f, h$ are smooth functions, then $g(fX, hY) = f h g(X,Y)$.

Conditions: $g$ is a covariant 2-tensor field.; $X, Y$ are vector fields.; $f, h$ are smooth functions.

Positive definite means that for any tangent vector $v \in T_pM$, $g_p(v,v) \ge 0$, and $g_p(v,v) = 0$ if and only if $v = 0$. This ensures the metric defines a valid norm.

Conditions: $g$ is a Riemannian metric.; $v \in T_pM$ is a tangent vector.

A Riemannian metric $g$ is symmetric because its value $g_p$ at each point $p$ is an inner product on the tangent space $T_pM$. Inner products are symmetric, so $g_p(v,w) = g_p(w,v)$, which implies $g(X,Y) = g(Y,X)$ for vector fields.

Conditions: $g$ is a Riemannian metric.; $g_p$ is an inner product on $T_pM$ for all $p \in M$.

A Riemannian metric $g$ on a smooth manifold $M$ is a smooth covariant 2-tensor field such that for each point $p \in M$, the value $g_p$ is an inner product on the tangent space $T_pM$. It is symmetric and positive definite.

Conditions: $M$ is a smooth manifold.; $g$ is a smooth covariant 2-tensor field.; For each $p \in M$, $g_p$ is an inner product on $T_pM$.; $g_p(v,v) \ge 0$ for all $v \in T_pM$.; $g_p(v,v) = 0$ if and only if $v = 0$.