# Can we define a specific metric tensor to all manifolds?

First, my question it maybe doesn't make sense at all, and that's because I don't truly understand these things, if it doesn't make sense please explain to me why.

When I say a metric tensor I mean the first fundamental form where every student learns in a differential geometry course. Meaning the standard inner product $$\|w\|=\langle w,w\rangle$$ of an element of the tangent space $$T_pS$$ of a normal surface $$S$$. Can we use this for every n-manifold to calculate length, angles, surface area etc?

what's the point of considering all these other tensors we use in Riemannian geometry since a manifold is locally euclidean can't we just use the standard inner product to do all these things?

• The standard inner product locally in each chart in general do not glue together to form a metric on the manifold. May 13, 2021 at 17:27
• some manifolds doesn't use the Euclidean metric, by example the Minkowski space that is used to represent the general theory of relativity May 13, 2021 at 17:28
• @Masacroso In the Minkowski space don't we use the metric of the inner product $<x,y>=x_1y_1+x_2y_2+x_3y_3 - x_4y_4$ ? There is no essential difference, we just define the inner product, and we already have a metric right ? May 13, 2021 at 17:32
• The metric you describe is defined with respect a chosen set of coordinates. So it is ill-defined on a manifold, where you want the metric to be well-defined, independent of the coordinates used. For example, suppose you start with coordinates $(x,y)$ and set the metric equal to $dx^2 + dy^2$. Now define new coordinates $(x',y')$, where $x = x'\cos y'$ and $y = x'\sin y'$. If you then set the metric to $(dx')^2 + (dy')^2$, you get a metric different from the first one. May 13, 2021 at 17:42
• "Professor Daene don't leave me cliffhanging here, come on that was a fair question, wasn't it?" That's pretty rude. Believe it or not, I do have work to do and don't monitor your post continuously. I suggest you try to read a textbook really carefully before asking more questions. May 13, 2021 at 18:56

$$\newcommand{\R}{\mathbb{R}}$$ Even though you're asking about manifolds in general, it suffices to understand this when the manifold is just an open set $$M \subset \mathbb{R}^n$$. The fact that it is a subset of $$\R^n$$ means there is already a set of coordinate functions $$x^k: M \rightarrow \R$$. However, there can be other coordinate functions $$y^k: M \rightarrow \R$$. For example, if $$M = (0,1) \times (0,2\pi)$$, the standard coordinates would be $$x^1(a,b) = a$$ and $$x^2(a,b) = b$$. Another set of coordinates would be $$y^1(a,b) = a\cos b$$ and $$y^2 = a\sin b$$.

When studying geometry or physics, coordinates are a necessary evil because we need to be able to measure and compare things. However, geometric properties and physical laws should not depend on the coordinates used. For example, physical laws should not depend on whether we are measuring lengths in meters or feet.

The first step towards defining a Riemannian metric is understanding what a tangent space is. The key idea here is that every point $$p \in M$$ has its own tangent space $$T_pM$$. It consists of all possible velocity vectors of curves. If we fix one set of coordinates, then $$T_pM$$ is obviously isomorphic to $$\R^n$$. If we switch to a different set of coordinates, then $$T_pM$$ still isomorphic to $$\R^n$$, but the isomorphism is different from the first one. The crucial observation is that the map $$\R^n \rightarrow T_pM \rightarrow \R^n$$ defined using the two different sets of coordinates is linear. What this allows us to do is to view $$T_pM$$ as an abstract vector space, where a set of coordinates defines an isomorphism $$T_pM \rightarrow \R^n$$. Put differently, a set of coordinates implies a basis of $$T_pM$$, commonly denoted as $$(\partial_1, \dots, \partial_n)$$.

Now recall that a dot or inner product on an abstract vector space $$V$$ is a bilinear function $$g: V \times V \rightarrow \R$$ such that $$g(v,v) \ge 0$$, with equality holding only if $$v = 0$$. A Riemannian metric is simply an inner product defined on each $$T_pM$$. It is important to note that we are not assuming any relationship between the inner products on two different tangent spaces $$T_pM$$ and $$T_qM$$.

Next, recall that, given a basis $$(v_1, \dots, v_n)$$ of $$V$$, an inner product $$g$$ is uniquely determined by the positive definite symmetric matrix $$[g_{ij}]$$, where $$g_{ij} = g(v_i,v_u).$$

So if we choose a set of coordinates, $$(x^1, \dots, x^n)$$ on $$M$$, then, at each point, we have a basis $$(\partial_1, \dots, \partial_n)$$ of $$T_pM$$, which then allows us to write the Riemannian metric at $$p$$ as the matrix $$[g_{ij}(p)]$$, where $$g_{ij}(p) = g(p)(\partial_i, \partial_j)$$ Here's the next important point: This matrix is a matrix of functions that depend on the point on the manifold. On top of that, the matrix depends on which coordinates you used. What's important is that the matrix changes in just the right way, so that, no matter what coordinates you use, the matrix, used with the basis of $$T_pM$$ defined by the coordinates, defines the same abstract inner product on $$T_pM$$.

To study this further, I encourage you to look at fundamental examples such as the standard sphere, hyperbolic space, surfaces of revolution in $$\R^3$$, and graphs of functions in $$\R^n$$.

• Thanks a lot for answering, that was the answer I was hoping to get. May 14, 2021 at 9:23
• Great answer! The only thing I would have added is that if we take a partition of unity then we can define the metric over all of $M$. May 24, 2021 at 18:49

(1) When most people say "an $$n$$-manifold $$M^n$$," what they mean typically mean is an (abstract) smooth $$n$$-manifold (which is virtually the same as an (abstract) differentiable $$n$$-manifold). I don't have time to go through the definition here, but there are many fantastic books on abstract manifolds (e.g., John Lee's Introduction to Smooth Manifolds), as well as a Wikipedia page:

https://en.wikipedia.org/wiki/Differentiable_manifold

The point is that an abstract $$n$$-manifold $$M^n$$ is not necessarily a subset of euclidean space $$\mathbb{R}^N$$. However, if it happens to be the case that your abstract $$n$$-manifold lives in some euclidean space, say $$M^n \subset \mathbb{R}^N$$, then we say that $$M^n$$ is a smooth submanifold of $$\mathbb{R}^N$$. If you've taken a differential geometry course on surfaces in $$\mathbb{R}^3$$, then you've studied the smooth $$2$$-dimensional submanifolds of $$\mathbb{R}^3$$.

(2) Given a smooth submanifold $$M^n \subset \mathbb{R}^N$$, then yes, it is possible to use the standard inner product $$\Vert v \Vert = \sqrt{\langle v,v \rangle}$$ on $$\mathbb{R}^N$$ to give each tangent space $$T_pM \subset \mathbb{R}^N$$ a positive-definite inner product that varies smoothly with $$p \in M$$: $$$$g_p(v,w) := \langle v,w \rangle. \tag{1}$$$$

(3) In general, however, an abstract manifold $$M^n$$ doesn't necessarily lie in a euclidean space $$\mathbb{R}^N$$. Nevertheless, it would be interesting if we could give each tangent space $$T_pM$$ a smoothly varying positive-definite inner product $$g_p$$. A choice of smoothly varying positive-definite inner product on each $$T_pM$$ is called a Riemannian metric (or metric tensor or Riemannian structure). The pair $$(M,g)$$ is then called a Riemannian manifold.

It is a fact (mentioned in Moe's answer) that every abstract $$n$$-manifold can be given (uncountably many!) Riemannian metrics $$g$$. Therefore, every abstract manifold $$M$$ can be made into a Riemannian manifold $$(M,g)$$ in uncountably many ways. If it so happens that $$M^n \subset \mathbb{R}^N$$, then there is a natural choice of Riemannian metric $$g$$ coming from equation (1), but one could in principle give $$M^n$$ other Riemannian metrics (a.k.a. metric tensors) besides this.

• Thanks a lot, that was really helpful ! May 13, 2021 at 19:10
• @Jesse Madnick, great answer! But one thing perhaps would be nice to mention is that indeed, every manifold can be embedded into some Euclidean space. So if M is a manifold, then we can always think $M \subseteq \mathbb{R}^N$ for large enough N. May 15, 2021 at 15:30

Every differentiable manifold admits a Riemannian structure. This is actually fairly intuitive to see: Every differentiable manifold is locally Riemannian, thus sewing up all the local metrics to a global metric gives us the conclusion we seek.

• What is a "Riemannian structure", I am not familiar with the term-definition. May 13, 2021 at 17:54
• @PetrosK Its a smooth assignment to each point $p$ on the manifold $M$ an inner product $g_p$ on $T_pM$. May 13, 2021 at 17:55
• I found this about the inner product $g_p$ [1]: math.stackexchange.com/questions/1850900/… but I can't understand what's the formula of the inner product May 13, 2021 at 18:04