I am still trying to get a good grasp on the motivations behind various concepts in Differential Geometry. But I am struggling to come to terms with how certain concepts have this added attribute of being coordinate independent? How does one identify such objects, be it a tangent space or a covariant derivative. How does one go about trying to prove that a certain geometric object is coordinate independent? How is coordinate independence a part of the "geometry" of a given surface or is it?

P.S.: Actually is the concept of a coordinate system part of the intrinsic or extrinsic geometry? I think its the former, but sometimes embedded spaces tend to make me think twice.

Edit: I would appreciate if the covariant derivative could be used as an example.

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    $\begingroup$ In first place, there is a need to parametrize a mere topological space with coordinates to, at least locally, obtain a notion of the usual $\mathbb{R}^n$ differentiabilty. However, one can regard a manifold as an object in its own right without looking at locally to $\mathbb{R}^n$ diffeomorphic neighbourhoods. Therefore, objects need to be globally well defined, i.e. coordinate independent. $\endgroup$ – gofvonx Sep 28 '13 at 19:18

When one thinks intuitively about the geometry of a surface/manifold, coordinates are not involved in the mental picture - you're thinking about directions, lengths, volumes, etc. Thus even if our formalism of differential geometry requires coordinate charts to define everything in terms of, we want the results to be in some sense independent of the particular coordinates we chose. There are some coordinate-dependent objects that appear in DG (e.g. Christoffel symbols), but these are for computational convenience and if they appear in a geometrically meaningful expression they should be paired off with each other so that they are in fact coordinate independent (i.e. Christoffel symbols should only appear when accompanied by partial derivatives as to form the covariant derivative).

Some authors (often physicists) define everything in terms of coordinates and then show that the resulting objects are actually independent of the coordinates used to define them. For example if one defines a vector field $V$ by specifying its components $V^i$ in an arbitrary coordinate system $x^i$, one has to check that the components $V^\mu$ produced in any other coordinate system $\tilde x^\mu$ are related by $$V^i = V^\mu \frac{\partial x^i}{\partial \tilde x^\mu}. \tag{1}$$ Note here that the components $V^i$ are not coordinate-independent, as this equation clearly shows: what is coordinate-independent is the full vector $V^i \partial_i$. The components $V^i$ are often called contravariant instead of invariant, meaning they satisfy the transformation law $(1)$.

Contrasting with this, most modern expositions of differential geometry try to avoid coordinates as much as possible. Now of course at some point you have to get coordinates involved, since they are the very definition of the smooth structure on a manifold; but it may be surprising how quickly you can do away with them. Once you've established that the smoothness of a function is coordinate-independent, you can do everything else in terms of the ring $C^\infty(M)$ of smooth functions, including defining vectors, tensors, metrics, connections, etc. Thus ideally one never has to prove that an object is coordinate independent, since one never got coordinates involved in the definition in the first place.

  • $\begingroup$ Thanks, that is indeed a very insightful answer. But I have some doubts.As you have mentioned yourself, the very notion of a manifold involves defining local coordinates, so if one can do away with them at some stage, then why not go for defining manifolds suitably rather than bringing in local coordinates and all??I hope I am making sense. $\endgroup$ – Vishesh Sep 28 '13 at 4:45
  • $\begingroup$ Also, I keep seeing that isometry invariance is seen as a sign of co-ordinate independence.What does that imply actually?? $\endgroup$ – Vishesh Sep 28 '13 at 4:47
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    $\begingroup$ @Vishesh: I believe it's possible to define manifolds without coordinate charts using sheaf theory; but you still end up with some conditions like "locally isomorphic to $C^\infty(\mathbb{R^n})$". Isometry invariance is indeed an important notion in differential geometry - any geometrically meaningful object should be defined only in terms of the geometric structure (i.e. the metric), so a transformation that preserves the metric should preserve geometric objects. $\endgroup$ – Anthony Carapetis Sep 28 '13 at 5:00
  • $\begingroup$ Sorry for the late reply,so does this mean the Algebraic geometry approach is much more complete?? $\endgroup$ – Vishesh Sep 29 '13 at 6:05

You might be interested in geometric algebra. As Anthony alluded to, physicists are too often obsessed with expressing all geometrical quantities in terms of coordinates and coordinate bases. The more mathematical approach of putting everything in a coordinate-independent way helps, but this is usually done by considering everything as a map. You can absolutely do this, but it also tends to lose the geometric significance of objects. As an example, I spoke with a professor I had for a Riemannian geometry course, and I explained to him the concept that the Riemann tensor is a map from bivectors to bivectors (from oriented planes to oriented planes). He seemed positively surprised by this notion, but in geometric algebra, it is the only way the Riemann tensor is presented (and the only way it needs to be presented).

This is what traditional mathematical presentations lack--the easy ability to deal with geometrical objects beyond vectors, even though such concepts would bring considerable clarity to differential geometry.

One formalism of geometric algebra applied to differential geometry is the "universal" geometric algebra--an infinite-dimensional clifford algebra. This takes away the arbitrary nature of an embedding; a manifold can simply be considered as a set of vectors (not tangent vectors, but vectors in a flat space). And because the points on the manifold are themselves vectors, expressions like $\partial x/\partial x^i$ are no longer hand-waving nonsense but legitimate, well-defined expressions to expose tangent vectors of the manifold (this is actually a major gripe Hestenes has with the identification of $\partial/\partial x^i$ as a vector, partly because the notion of partial derivatives as vectors is incompatible with the clifford algebra structure).

In this view, then, one can always appeal to the basis vectors of the ambient universal GA and identify a geometrical object as having a fixed expression in terms of the those extrinsic basis vectors. That is to say, clearly a tangent vector (or other geometrically significant object) may be expressed in terms of some coordinate basis vectors intrinsic to the manifold, an expression that will change with different choices of coordinates, but the embedding of the manifold in the UGA isn't changing with the choice of coordinates, so the tangent vector cannot be changing. Intrinsically, one would have to resort to the chain rule, as Anthony did, to identify the change in components as following entirely from function composition (from the chain rule).

Of course, from an extrinsic perspective, it's much easier in general to identify geometrically significant quantities. The covariant derivative of a vector field is just the projection of the derivative of the projection, for instance. Intrinsically, it can be harder to visualize.

  • $\begingroup$ Thanks a lot for your reply. Is there any book that deals with tensors and the lot in a geometric manner rather than just as symbols and indices.I am still trying to understand how the covariant derivative is a co-ordinate independent notion. $\endgroup$ – Vishesh Sep 29 '13 at 6:07

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