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What is a metric ? Do a metric depend on the system of coordinates I use ? Does it depend on surfaces (or higher dimensional manifolds. Correct me if I'm wrong using the word) the coordinate frames are chosen on ? Is there any general form for a metric? In this course it's mentioned Gauss pointed it out as restriction for an infinitesimal metric or distance function on surfaces to be $$(dS)^2 = g_{xx} (x,y)(dx)^2 + g_{xy} (x,y)dxdy + g_{yy} (x,y)(dy)^2.$$ And is it that mentioned quadratic form restricted for a metric to look like a Euclidean metric $(ds)^2 = (dx)^2 + (dy)^2$ , is just a belief working in this comment of Gauss or is there any proper logical explanation to this ?

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Do you mean a Riemannian metric? If so, the metric spaces tag should be replaced by the Riemannian geometry tag. – Michael Albanese Jan 21 at 10:23
Riemannian metric is in general a section of $T^*M \otimes T^*M$ so that at each point it is a symmetric non-degenerate 2 form. It is a concept that has a definition that does not depend in any way on coordinate charts. I understand however, that this answer is unlikely to help you, since if you do not know what a metric is you likely do not know what the cotangent bundle is. You may find this. helpful. – s.harp Jan 21 at 10:53
I'm taking a course where discussing about the models of Universe they introduced a function which they called a metric which gives the physical distance between two points in space. Where they told that Gauss gave a general form of a metric on a two dimensional surfaces as $(dS)^2=g_{xx} (x,y)(dx)^2 + g_{xy} (x,y) dxdy +g_{yy} (x,y)(dy)^2$ which reduces to Euclidean form on spaces which can locally or at infinitesimal scales can be treated as Euclidean. – user284090 Jan 21 at 11:27

The word "metric" is used in two completely different senses here.

${\bf 1.}\ $ Firstly a metric on a set $X$ is a distance function $d:\>X\times X\to{\mathbb R}_{\geq0}$ having the properties described in skyking's answer. An example is the metric $$d(x,y):=\sum_{k=1}^n|x_k-y_k|$$ on ${\mathbb R}^n$. Given $x$ and $y$ the distance $d(x,y)$ can be instantly evaluated.

${\bf 2.}\ $ A Riemannian metric on an $m$-dimensional manifold $M$ (say $M:=S^2$) is a law that allows one to compute scalar products of tangent vectors $X$, $Y$ attached at the same point $p\in M$. If $(x_1,\ldots, x_m)$ are local coordinates in the neighborhood of $p$ then this law assumes the form $$\langle X,Y\rangle=\sum_{i, \>k} g_{ik}X_iY_k$$ with a certain positive definite symmetric $m\times m$ matrix $[g_{ik}]$, $g_{ik}=g_{ik}(x_1,\ldots, x_m)$, which transforms in a characteristic way under change of coordinates. Such a Riemannian metric converts each tangent space $T_p$ of $M$ into an $m$-dimensional Euclidean vector space. In particular we can calculate the length of a tangent vector $X\in T_p$, and consequently the length $L(\gamma)$ of curves $\gamma\subset M$. This then finally allows us to establish on $M$ a so called inner metric $d$ in the sense ${\bf 1.}$, which is induced by the given Riemannian metric $[g_{ik}]$: $$d(p,q):=\inf_\gamma\> L(\gamma)\ ,$$ where the $\inf$ ranges over all smooth curves in $M$ that go from $p$ to $q$. This means that in order to determine a distance $d(p,q)$ you have to solve a minimum problem.

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Doesn't sense 1 of metric above only require three properties? I feel like including those properties would improve this answer a lot. – Todd Wilcox Jan 21 at 22:11

$\newcommand{\Reals}{\mathbf{R}}$In the broad sense of your question, a "metric" is a means of quantifying "distance" between points in some "space" (or set) $M$.

One common axiomatic approach leads to the concept of a "topological metric", a function $d$ (for "distance") that assigns a positive real number $d(p, q)$ to each unordered pair of distinct points $p$ and $q$ in $M$, subject to the "triangle inequality", and to the condition $d(p, p) = 0$ for all $p$. This approach is described in multiple answers.

In 1854, Georg Bernhard Riemann gave his habilitation lecture Über die Hypothesen, welche der Geometrie zu Grunde liegen (see Item 13 on the linked page), in which he took an approach leading to the concept of metric in your question.

Suppose $M$ is a space that, near each point, "looks like" the $n$-dimensional Cartesian space. By "looks like", you should imagine that if $p$ is an arbitrary point of $M$, then a sufficiently small neighborhood $U$ of $p$ in $M$:

  1. Is continuously coordinatized by $n$ real-valued functions (i.e., $U$ is homeomorphic to an open subset of $\Reals^{n}$).

  2. There is a well-defined notion of differentiability. Loosely, if $(x^{1}, \dots, x^{n})$ and $(y^{1}, \dots, y^{n})$ are two different sets of coordinates near $p$, then an arbitrary real-valued function $f$ is smooth when expressed in the $x$ coordinates if and only if $f$ is smooth when expressed with respect to the $y$ coordinates. This is equivalent to saying the change of coordinates mapping from $x$ to $y$ is smooth, and invertible with smooth inverse.

Nowadays we call $M$ (together with a fixed notion of "differeitiability") an $n$-dimensional smooth manifold. Examples include Cartesian space itself and any non-empty open subset, spheres of arbitrary dimension (the "unit $n$-sphere" is the set of points at unit distance from the origin in $\Reals^{n+1}$), tori of arbitrary dimension, real or complex projective spaces, and regular level sets of smooth mappings.

Riemann's idea was, in essence, to assume the Pythagorean theorem holds infinitesimally, namely, that if $du$ and $dv$ are orthogonal infinitesimal displacements, then the magnitude of their sum is $ds = \sqrt{du^{2} + dv^{2}}$.

With modern hindsight, Riemann is equipping each "tangent space" of $M$ with an inner product, a real, symmetric, positive-definite bilinear form, now called a Riemannian metric. With respect to a coordinate system $(x^{1}, \dots, x^{n})$, a Riemannian metric and the associated length-squared function look like $$ g = \sum_{i, j} g_{ij}\, dx^{i} \otimes dx^{j},\qquad (ds)^{2} = \sum_{i, j} g_{ij}\, dx^{i} \, dx^{j}. $$ (The Riemannian metric $g_{p}$ at a point $p$ accepts two tangent vectors and returns their inner product; the metric components $(g_{ij})$ are location-dependent matrix entries defining the inner product. The length-squared function is the associated quadratic form, $(ds)^{2}(v) = g(v, v)$.

A Riemannian metric is not the only "distance-inducing geometric structure" a manifold may possess. In addition to general topological metrics, there are, for example, Finsler manifolds, in which each tangent space is equipped with a norm function, and pseudo-Riemannian manifolds, as in general relativity, in which the inner product is not positive-definite.)

Riemannian geometry is the study of Riemannian manifolds, including questions such as

  • When are two Riemannian metrics isometric? (Loosely, when do two different coordinate representations determine "the same geometry"?) Even locally (in a neighborhood of a point) the question is non-trivial. Riemann showed that a Riemannian metric is locally isometric to the Euclidean metric $\sum_{i} (dx^{i})^{2}$ if and only if a certain fourth-order tensor vanishes.

  • Given a tangent vector $v_{p}$ at some point $p$ and a smooth path in $M$ joining $p$ to $q$, is there a "natural" notion of parallel transport along the path, yielding a tangent vector $v_{q}$ at $q$? (The answer is "yes", and the Riemann curvature tensor has the following geometric interpretation. Let $u$ and $v$ be tangent vectors at $p$. Parallel transport around a small (curvilinear) parallelogram with sides (approximately) parallel to $u$, $v$, $-u$, and $-v$ defines an orthogonal transformation of the tangent space $T_{p}M$, and the dependence turns out to be linear in $u$ and $v$. To say the curvature vanishes is to say that parallel transport around a small parallelogram is the identity transformation, as is the case in Euclidean geometry.)

  • Given two points $p$ and $q$ in a Riemannian manifold, does there exist a "shortest path" (technically, a length-minimizing geodesic) from $p$ to $q$? If so, is such a geodesic unique?

Locally, geodesics exist and are automatically length-minimizing, but globally they can fail to exist, or to minimize length, for geometric reasons. For example, if $M$ is the plane $\Reals^{2}$ with the origin removed, and if $g$ is the (restriction of) the Euclidean metric, then a point $p$ and its "opposite" $q = -p$ are not joined by a geodesic in $M$.

A Riemannian manifold is complete if every geodesic is defined for all time. (Loosely, no geodesic "encounters an edge of $M$ in finite time", or "$M$ has no edge".) Even in a complete Riemannian manifold, a sufficiently long geodesic can fail to be length-minimizing: Think of a great circle arc on a sphere subtending more than half of a "full turn".

A connected Riemannian manifold $(M, g)$ becomes a metric space (equipped with a topological metric) if, for points $p$ and $q$ of $M$, we define $d(p, q)$ to be the infimum of the lengths of piecewise-smooth curves from $p$ to $q$. When $M = \Reals^{n}$ and $g$ is the Euclidean (Riemannian) metric, the geodesics are line segments (parametrized at constant speed) and the topological distance reduces to the Euclidean (Pythagorean) distance. In general, however, a topological metric on a smooth manifold $M$ does not arise in this way from a Riemannian metric on $M$.

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Thanks for such a good and simple description. But there's more to do to understand the notations used. – user284090 Jan 21 at 16:25

A metric is a function that is used to define the distance between two points. Normally one denotes it as $d(x,y)$ (being the distance between $x$ and $y$).

In order to have some resemblance with normal distances we're used to one requires it to have certain properties.

First of all the distance between two different points has to be positive. That is if $x\ne y$ we would have $d(x,y)>0$.

Second we require the distance to itself is zero, that is $d(x,x)=0$.

Third we require the triangle inequality to be fulfilled, that is $d(x,z) \le d(x,y) + d(y,z)$.

The concept of metric is not dependent on having a coordinate system at all, so it's independent on whether you add a coordinate system. For example the distance between New York and London is the same regardless of you using longitude+latitude as coordinates or some other coordinate system.

Also it's not required for it to be (based on) a quadratic form, even if it's quite often that it is.

The quadratic form you refer to (in differential geometry) is not the whole story when it comes to metric. The quadratic form is only a local property of the surface, while the actual distance is more of a global property. The distance is then taken to be the length of the shortest path between the points and the quadratic form is then used locally in the definition of path length. So the actual distance need no longer be a quadratic form.

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Can you explain more about the triangle inequality as mentioned above in the answer ? – user284090 Jan 21 at 12:27
@NayanTelrandhe It basically says that the distance between two points is the length of the shortest route. It says this by stating that the distance between $x$ and $z$ gets no shorter if we take the route via pont $y$. It's called triangle because the distance between two corners there is the length of the edge connecting the corners and taking the route via the other corner makes the route no shorter. – skyking Jan 21 at 12:32
The course I'm taking mentions that Gauss pointed that a metric giving distances on infinitesimal scales should look like or be of form $(dS)^2 =g _x_x (x,y)(dx)^2 +g_x_y (x,y)dxdy+ g_y_y (x,y)(dy)^2$ do you have an explanation to this ? – user284090 Jan 21 at 12:33

This is straight out of Greene's and Gamelin's "Introduction to Topology 2nd ed.":

A metric on a set X is a real-valued function d on X $\times$ X that has the following properties:

(1.1) $d(x,y) \geq 0$ $\ \forall \ x,y \in X$

(1.2) $d(x,y) = 0$ if and only if $x = y$

(1.3) $d(x,y) = d(y,x) \ \forall \ x,y \in X$

(1.4) $d(x,z) \leq d(x,y) + d(y,z), \forall \ x,y,z \in X$

The idea of a metric on a set $X$ is an abstract formulation of the notion of distance in Euclidean space. The intuitive interpretation of property (1.4) is particularly suggestive...Consequently (1.4) is referred to as the triangle inequality.

The main idea is that metrics are a way to describe spaces that resemble Euclidean space. Many surfaces are non-Euclidean, but manifolds are locally Euclidean by definition - so they are metrizable. In other words, if you "zoom in" on a point in a manifold, it will "appear to be" a Euclidean space.

For example, a sphere has longitude and latitude as it's coordinates, but if you zoom in, you have basic google-like maps that are flat and have north-south (the y-axis) and east-west (x-axis) as their coordinates.

When you ask "do[es] a metric depend on the system of coordinates" you use, I'm assuming you're asking "Can you change the values of a metric if you transform the basis of the underlying space." The answer is no - changing the basis of the space does not change the structure or shape of a space, but rather the way at which you define points and vectors relative to an origin - all the intrinsic values of the space (length, distance, angles, etc.) remain the same.

For example, say you fix a metric and use our google maps example. If you swap north-south with east-west (maybe you turned somewhere so the map shifts with you), the travel distance is not shorter or longer and angles (ratios between fixed positions) have not changed either.

That said, there are many functions that can be classified as metrics. For example, you are probably most familiar with the standard euclidean metric (i.e. a straight line). However, when it comes to navigating in a city, this straight-line approach isn't the best for communicating actual travel times and distances. Thus, there is something referred to as the manhattan metric in which you zig-zag through the streets of a city.

There are many such metrics, but they all satisfy those four aforementioned properties - this allows us to neatly adapt theorems from calculus (where the euclidean metric is standard) to scenarios where other metrics are more natural (like spheres with longitude and latitude).

Again, one major motivation for defining metric so abstractly is that it allows us to think of spaces with the non-standard Euclidean metric in a manner that allows us to readily apply techniques from calculus.

(Not sure what you mean on the bit about the quadratic function.)

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The course I'm taking mention that Gauss pointed that a metric should look like or be of form $(dS)^2 = g_x_x (x,y)(dx)^2 + g_x_y (x,y)dxdy + g_y_y (x,y)(dy)^2$ they call it generalized form of a metric for tiny scales. – user284090 Jan 21 at 12:55
I've got nothing, sorry. – Dmitri Valentine Jan 21 at 12:59
You can go through the lecture or the transcripts here – user284090 Jan 21 at 13:05

Metric on space $X$ (if $X \neq \emptyset$) is a function $d: X \times X \to \mathbb{R}$ so that

1.) $d(x,y) \geq 0$ and if $d(x,y) = 0 \Leftrightarrow x =y$ $\forall x,y \in X$

2.) $d(x,y) = d(y,x)$ $\forall x,y \in X$

3.) $d(x,y) \leq d(x,z) + d(y,z)$ $\forall x,y,z \in X$

So yes, it does depend on the system of coordinates you use. Metric is specific to X and X is your set of coordinates.

A often encountered metric is the uniform norm.

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"So yes, it does depend on the system of coordinates you use. Metric is specific to X and X is your set of coordinates." Where is a reference to a coordinate system? – s.harp Jan 21 at 10:43
It is now evident that this is not the type of metric the OP was asking about. You may want to delete your answer. – Michael Albanese Jan 21 at 11:40
I suggest that you change your 1st property to $d(x,y)=0 \Leftrightarrow x=y$, since you don't otherwise state the property that $d(x,x)=0\ \ \forall x \in X$. – Scott Jan 22 at 5:10

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