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Is a metric and a metric tensor the same thing?

Hi all, I am new to differential geometry, but I am reading about metrics and metric tensors now. My understanding is that metric tensors tell you how to find the length between 2 infinitesimally close points, while a metric provides a distance measure between any 2 points. My question is: are the 2 concepts equivalent or otherwise connected? If you have a metric, i.e. a distance measure, can you always find a metric tensor? What about vice versa? Save me!
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A metric tensor is an object defined on a differentiable manifold. Specifically, a metric tensor is a map which, at each point *x* in the manifold, takes in two tangent vectors and spits out a number. For a fixed point *x*, this map behaves like the familiar dot product from linear algebra. It can be used to define the length of tangent vectors and the angle between tangent vectors. Integration of the metric tensor over a curve in the manifold can then give the length of that curve. In this way, some authors talk about the metric tensor as giving a way to measure the distance between infinitesimally close points.

A metric tensor, however, is *not* a metric. A metric on a set *X* (called a metric space) is a map that takes two points in X and spits out a number that is interpreted as the distance between those two points. This map must satisfy certain properties to be a metric. In particular, the "distance" between any two points must be 0 (if the points are the same) or strictly positive (if the points are not the same).

A metric tensor is not a metric because, for one, the metric tensor is really not just one map, but rather a map that depends on the point in the manifold. The map varies from one tangent space to another. But, more important, even when restricting the metric tensor to one particular tangent space, there is no requirement for that map to be a metric. In particular, the "length" the metric tensor assigns to tangent vectors need not be strictly positive for non-zero vectors. In fact, in relativity, the metric tensor does *not* assign positive length to all non-zero vectors.

A metric tensor *can* be used to construct a metric on some manifolds, but the metric tensor must have the property that the length of any non-zero tangent vector is strictly positive. We say that the metric tensor must be positive-definite. Manifolds that have a metric tensor with this property are called Riemannian manifolds. It is then possible to define the distance between any two points in the manifold in such a way that that distance function is a true metric.

How do we do that? By measuring the distance between points along curves. So suppose we have a differentiable manifold with a positive-definite metric tensor and we want to measure the distance between points *x* and *y*. We consider all paths in the manifold that begin at *x* and end at *y*. Each of these paths has a length L. We then simply declare the distance between *x* and *y* is the shortest of all these possible values of *L*. There is some machinery needed here to prove that this minimum length always exists, but once we have that, we have a metric.

The converse is false in general. That is, given a manifold which is also a metric space, the metric (the distance function) is not necessarily derivable from a metric tensor. Why? It turns out that we can prove that in the construction above, the shortest path between *x* and *y* is essentially unique. (There are some extra caveats to this, but they are unimportant.) But there are plenty of metrics we can construct for which there is no such "unique shortest path". For instance, suppose we measure the distance between points in the plane using the "taxicab metric", i.e., the sum of the absolute difference in the x-coordinates and the absolute difference in the y-coordinates. This is the number of "blocks" a taxi has to travel along a grid-like street system to go from *x* to *y*. Then any "staircase path" between *x* and *y* in which all the stairs are parallel to the x- or y-axis is a path of shortest length from *x* to *y*. But this path is far from unique.