I'm study Differential Geometry and I'm thinking about a pharese that my professor saying in a class: "Talking about manifolds, working in the domain of a parametrization is better than working in contradomain of a parametrization, i.e, directly on manifold". I not understand what his did mean. For instance, in some books of literature(cf Do Carmo, Differential Geometry of Curves and Surfaces), the Gauss Map is defined on manifold, i.e, let $M \subset R^3$, the Gauss map is defined by $$ N: M \to S^2 $$ In other books, for exemple in Kühnel, Differential Geometry, the Gauss Map is defined on domain of parametrization: Let $f: U \to R^3$ a surface element (a parametrization), the Gauss $$ \nu: U \to S^2 $$ is defined by the formula $$ \nu(u,v) = \frac{f_u \times f_v}{|f_u \times f_v |} $$

I observe that, the definition of Do Carmo, is more geometric than the definition of Kühnel, the formula for Gauss map in Coordinates in Do Carmo, depends of points on Surface, $$ N(p) = \frac{f_u \times f_v}{|f_u \times f_v |}(p) $$ Discussion

Considering that, $f: U \to R^3$, is a parameterization, that is, if $M$ is a manifold in $R^3$, $f$ is a homeomorphism of $U$ on $M$, whose derivative of $f$ is injective, then, study in $M$ , wouldn't that be the same as studying in $U$(topologycally speaking)? The only thing I can see right away is that studying in $M$ ​​seems to be more "Geometric" than studying in $U$. Now, studying in $U$ seems to be independent of the parameterization (or of the chart), that is, it seems that it is something more general. Another thing I realized is about the definition of a tangent plane to a manifold. For example, the tangent plane is defined to be a vector space that contains all vectors tangent to a surface. But, let's analyze the concept of "tangent vector". In some books, the tangent vector to a manifold is defined to be a tangent vector to a curve $a: I \to U$, such that $a'(t)$ is a vector whose property is to be tangent to a at $t$ when it is translated by $a(t)$, that is, the vector $v = a'(t) +a(t)$ is tangent(geometrically speaking) to the curve $a$ at $t$. In this way, if $f$ is the parameterization of the surface, the derivative of the curve $f(a(t))$ is tangent to the surface in the same sense that I just explained in the case of the curve $a$. Where am I going with this? Well, my question is the following: if $a'$ and $b'$ are tangent vectors to the curves $a: I \to U$ and $b: J \to U$, such that $\{a', b'\}$ span the tangent space $T_pU$, ​​where $p$ is a point of $U$, we know that $Df$ is an isomorphism of vector spaces, so $Df(TpU)$ is the tangent space to $M$, that is, it is isomorphic to $T_{f(p)}M$. Thus, it seems to me that nothing is lost, from an algebraic point of view, working on the variety (in the co-domain) than working on the domain.

The conclusion I reached: It seems to me, that there are no topological, or algebraic differences, working in the codomain or in the domain. The only clear difference is that when working in the codomain, things become more geometric than when working in the domain.

Could clarify for me, this sentence? Perhaps an example will make the difference between what you told me clearer for me.

Something came to my mind, in the case, let's say, "the philosophy" of analysis on manifods , I mean, when dealing with an abstract manifold, throughout the course, we always worked in $R^n$, we took the information obtained in $R^n$ to the variety across the charts, then we take notes and bring to manifold again.

  • $\begingroup$ In my opinion, sometimes it is easier to work in local coordinates (i.e. in $\mathbb R^n$), and sometimes it is easier to work coordinate free (i.e. on the manifold), there is not a best choice which always works better. $\endgroup$ Nov 25, 2022 at 16:30

1 Answer 1


There is two ways to address your question:

  • it depends on the person,
  • it depends on the context.

Instead of saying "working in the domain" or "working in the codomain", I will use the more conventional formulations "working without coordinates" and "working with coordinates".

The first way is inherently personal and cultural.

If you are from an Anglo-Saxon country, you will be more likely to learn differential geometry with a large use of coordinates, both from your professors and from textbooks. It seems that historically, this is the way it is taught in the US. On the contrary, I have learnt differential geometry from a coordinate-free point of view, which is surely due to the influence of the French school of differential geometry. I personally have a preference for this latter approach regarding definitions and proofs: I find it easier to understand, to feel, and I confess, I believe coordinate-free proofs are more elegant. An example that I find meaningful is the following: while it takes me a few seconds to understand the equality $\Delta f= - \mathrm{trace}(\nabla^2f)$ and work with it, it takes me a few minutes to understand the equality $\Delta f = -\frac{1}{\sqrt{|g|}}\partial_i\left(\sqrt{|g|} g^{ij}\partial_jf\right)$ (which I can't remember by heart, I had to check online if it was the right formula).

But I know this is my own view and my own taste. Any debate on which is the best is a philosophical debate, driven by highly personal preferences and cultural tastes. Anyone who will tell you the contrary is, I think, a bit arrogant.

The second way is more interesting.

Formally speaking, working with or without coordinates is equivalent, since charts are diffeomorphisms. This means that, formally, for any proof in coordinates, there should exist somewhere a coordinate-free proof of the same statement, and vice-versa. Sometimes, one way is shorter, easier, more obvious, or more elegant than the other. Here are a few examples: consider the two following objects. $$ \begin{array}{|c||c|c|} \hline & \text{Coordinate-free expression} & \text{Expression in coordinates}\\ \hline \\ \text{Equation of geodesics} & \dfrac{\nabla}{dt}\gamma'=0 & \ddot x^k + \Gamma^k_{ij}(x)\dot x^i \dot x^j=0\\ \\ \hline \\ \text{Hessian of a function} & \nabla (df) & \dfrac{\partial^2f}{\partial x^i\partial x^j} - \Gamma^k_{ij}(x)\dfrac{\partial f}{\partial x^k}\\ \\ \hline \end{array} $$ Assume that you are asked to show the following questions:

  1. Show that the integral curves of a vector field $X$ satisfying $\|X\|^2=1$ are geodesics.
  2. Show that for a point $p\in M$ and a tangent vector $v\in T_pM$, there exists a unique maximal geodesic $\gamma$ with $\gamma(0) = p$ and $\gamma'(0)=v$.
  3. Show that the Hessian of a function $f$ is symmetric.
  4. Show that the Hessian of a function $f$ is a tensor.

Then formally, this is equivalent to tackle these questions in a coordinate-free way or in coordinates. However, I think a lot of people will agree with me that addressing questions $1$ and $4$ is easier without coordinates:

  • for the first question, differentiating $\|X\|^2=1$ in the direction of any vector field $Y$, after playing a bit with the Lie derivative, brings $\nabla_XX\perp Y$, and therefore $\nabla_XX=0$.
  • for the fourth question, it is clear from the coordinate-free definition. Due to the appearance of a second-order derivative and of Christoffel symbols (which are not tensors), it is hard to tell directly from the expression in coordinates.

Meanwhile, the two other questions are easily answered thanks to the use of coordinates:

  • the second statement is due to the existence and uniqueness of solution to second order ODE in $\Bbb R^n$.
  • the third question is immediately addressed in coordinates thanks to Schwarz's lemma in $\Bbb R^n$, and the symmetry of the Christoffel symbols in the lower indexes.

Both paradigms have their pros and cons. It has been a few centuries since we know calculus in $\Bbb R^n$, and this is a huge pro for the in-coordinates approach. But sometimes, pure calculus is just what it is: a succession of functional equalities that does not really tell anything before being interpreted in an intrinsic way.

I would like to mention that some area in differential geometry are naturally more inclined to be studied with one or with the other paradigm. The analysis of the Ricci-flow is one area where the in-coordinates approach is really powerful: in order to obtain analytical estimates, it is suitable to express objects in coordinates and use analytic inequalities. On the contrary, contact geometry is an area of differential geometry where the use of coordinates is often not a good idea, since a contact distribution is never tangent to any coordinates.

  • 2
    $\begingroup$ The formula for $\Delta f$ can be derived by integrations by parts: for $u, v$ compactly supported in a coordinate patch in the interior of $M$, $$(\text{grad } u, \text{grad } v) = \int_{M}g^{ij}u_{x_i}v_{x_j}\sqrt{|g|}\,dx = -\int_{M}\frac{1}{\sqrt{|g|}}(\sqrt{|g|}g^{ij}u_{x_i})_{x_j}v\sqrt{|g|}\,dx = -(\Delta u, v).$$ $\endgroup$
    – Mason
    Nov 25, 2022 at 20:57
  • $\begingroup$ @Mason Thanks for the tips, I appreciate. Never thought it this way. It would take me longer to remember than the coordinate-free expression though $\endgroup$
    – Didier
    Nov 25, 2022 at 21:16

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .