# Intuition behind Riemannian-metric

What is a Riemannian metric? I have just started reading 'Riemannian Geometry' using primarily do Carmo's text and I've got the idea that to introduce a notion of distance to be able to talk about length of curves on smooth manifolds, angles between curves we introduce something called a 'Riemannian metric'. It's definition on the other hand, seems very unintuitive(Carmo's definition doesn't involve anything tensor related). I guess it involves an inner product in it because we want to measure length of tangent vectors (but why?) and what better way to measure it than using inner product...

To be able to proceed further into the theory and make sure I can make sense of the things, I think it's important I understand the definition very well but after reading it again and again, I'm not sure if I 'get' it. I would love it if someone could give me an explanation for it.

Also, if we want to give a metric structure (to be able to talk about all things distance related) then why not define a metric (the topological one, satisfying positivity, symmetry, triangle inequality) rather than this?

Or maybe I'm confused and Riemannian metric as we've defined is actually that metric(topological one) and together with it, our smooth manifolds becomes a metric space? If that's the case, wouldn't the theory of Riemannian Geometry become somewhat easy as we already know a lot about metric spaces?(I seriously think, that that is not the case)

• You might want to study an undergraduate text on the differential geometry of curves and surfaces to gain insight, examples, and motivation (e.g., for connections) before reading a graduate text. Sep 28, 2020 at 15:56

The idea is to equip the tangent space $$T_p M$$ at $$p$$ to the manifold $$M$$ with an inner product $$\langle -, - \rangle$$ , in such a way that these inner products vary smoothly as $$p$$ varies on $$M$$.

It is then possible to define the length of a curve segment on a $$M$$ and to define the distance between two points on $$M$$.

In terms of the definition of a Riemannian metric, given a smooth $$n$$-dimensional manifold, $$M$$, a Riemannian metric on $$M$$ (or $$TM$$) is a family, $$(\langle -, - \rangle_p )_{p \,\in M}$$, of inner products on each tangent space, $$T_p M$$, such that $$\langle -, - \rangle$$ depends smoothly on $$p$$.

• Huh...you explained the definitions very clearly but I still don't feel very comfortable with it. I guess that may be just because I have just started reading Riemannian-geometry and getting used to will take time, thanks and +1 by the way! Feb 5, 2019 at 19:58
• One question- at some places I see Riemannian-metric defined as you did in terms of inner product family and at other places as $ds^2= \sum_{i=1}^{n} g_{ij} dx_i dx_j$, but they're the same thing, right? Because if we consider co-ordinate system around point $p$ as $(x_1,...x_n)$ then in the definition you gave, Riemannian-metric is infact determined by inner products $g_{ij}$'s where $g_{ij} =<\frac {\partial}{\partial x_i}, \frac{\partial}{\partial x_j} >$. It would be great if you could give more clear explanation for this. Feb 5, 2019 at 20:27
• @clear: This is just linear algebra: If you have $n$-dimensional vector space $V$ with a basis $v_1,...,v_n$ then a symmetric bilinear product $(.,.)$ on $V$ is uniquely determined by its Gram matrix whose entries are $(v_i, v_j)$. Conversely, every symmetric matrix is the Gram matrix of some symmetric bilinear form. Feb 6, 2019 at 23:15
• Oh, yeah. Thanks! Feb 14, 2019 at 5:53
• @MoisheKohan Is the Riemannian metric the Gram matrix of $T_pM$? Sorry if silly question, I'm also learning about this. Mar 23, 2021 at 21:18

I just came across this question as I was searching for answers in a similar direction and thought I'd share a couple of thoughts on this. My knowledge is very limited and I apologize for possible mistakes (in this case, I'd be very happy to be corrected!). In particular, my answers are certainly only partial and there are significantly many people here who can provide significantly more complete answers. Among others, you asked the following:

"we want to measure length of tangent vectors (but why?)": Think of basic calculus, where the length of a curve $$\gamma: [0,1] \to \mathbb{R}^d$$ is calculated as $$\int_0^{1}|\gamma'(t)|dt$$, provided $$\gamma \in C^1([0,1],\mathbb{R}^d)$$. We note that we need to calculate the (norm of the) velocity $$\gamma'(t)$$ at each time $$t$$ to obtain the length of the curve (intuitively: the velocity at time $$t$$ may be thought of as the length of the curve on a small time interval $$[t,t+\epsilon]$$, so we would try to sum the curve's length on all these time intervals. In the limit $$\epsilon \to 0$$, this summation becomes integration). Thus, analogously calculating the length of a (sufficiently smooth) curve $$\gamma: [0,1] \to M$$ in a smooth manifold $$M$$, we also need the derivative $$\gamma'(t)$$ at each $$t$$. However - what should the derivative of this $$M$$-valued path at any given time be? Well, the derivative is just the direction the curve passes through at this instant of time, i.e. at position $$\gamma(t)$$. The collection of all possible such directions is just the tangent space $$T_{\gamma(t)}M$$ at the point $$\gamma(t)$$. Hence, as in the example above, we need a means to measure the length of elements of each tangent space - for which the given metric provides a convenient tool. Then, the formula for the path lengths looks very much alike to the $$\mathbb{R}^d$$-case above. (Note: This general case is in accordance with the example above, as the manifold $$\mathbb{R}^d$$ has tangent space $$\mathbb{R}^d$$ at each point).

"Also, if we want to give a metric structure (to be able to talk about all things distance related) then why not define a metric (the topological one, satisfying positivity, symmetry, triangle inequality) rather than this?" First of all, note that we are NOT defining "a" (that is, one and only one) metric "on M": Instead, we introduce a metric $$g_x$$ on the tangent space $$T_xM$$ for each $$x \in M$$ (since in general, these tangent spaces need not coincide as sets, it is possibly not even meaningful to try to define the same metric, say $$\bar{g}$$, on each tangent space). We do so by introducing an even better object - namely a scalar product $$g_x$$ on each $$T_xM$$, which certainly induces a metric, but has a lot more convenient features (a metric only would not allow the length calculation as pointed out in the above answer, since this formula does require a norm). This induced metric is - put in your words - a "topological" one. Note, however, that this metric is NOT defined on the manifold itself: You do not measure distances of points on the manifold with this metric, but it is defined on the tangent space, which should carefully be separated.

"Or maybe I'm confused and Riemannian metric as we've defined is actually that metric(topological one) and together with it, our smooth manifolds becomes a metric space?" As said in the previous answer, the foremost goal of introducing a Riemannian metric is NOT to metrize the manifold M itself, but its tangent spaces! Please distinguish carefully.

Since you have posted your questions more than 18 months ago, this might be superfluous for you by now. But maybe others will come across this post and find it valuable :)