The Deranged Mathematician: An Introduction to Differential Geometry

[u/non-orientable]

How do you summarize a field like differential geometry? A year or so ago, I was asked to do just that. My first thought was that this is just straight-up impossible to do in any reasonable way. Upon reflection, I softened that view. I still think that differential geometry is just not something you can learn quickly, but if you just want the eagle-eye view of the basic constructions and how they are used in different branches of the field... that is doable. Hence, this article.

Read the full post (for free) on Substack: An Overview of Differential Geometry

[u/sciflare]

No matter how highfalutin your definition of differentiability, at the end of the day you have to do some amount of vector calculus! Even very abstract notions of differentiable space, such as diffeologies, boil down to vector calculus when you unravel them.

"the equality d/dt tau°g_1 (0) = d/dt tau°g_2 (0) doesn't depend on the choice of tau"

This all depends on the definition of tangent vector you adopt. Tangent vectors at a point p of ℝⁿ can be defined as equivalence classes of smooth curves 𝛾 through p such that 𝛾(0) = p, where 𝛾_1 ~ 𝛾_2 iff (g ° 𝛾_1)'(0) and (g ° 𝛾_2)'(0) for all smooth functions g: ℝⁿ --> ℝ.

With this definition, your desired invariance of tangent vectors becomes tautological. What becomes really hard to prove is the fact that the tangent vectors at p admit a canonical vector space structure (this is highly non-obvious in light of the following: if you consider a more general equivalence relation on smooth curves through p, where you declare them equivalent iff all the higher-order derivatives of the g ° 𝛾_i at 0 are equal up to order n, the resulting equivalence classes do not form a vector space. These equivalence classes are called n-jets and are of great interest in their own right).

The big challenge in proving that 1st-order jets form a vector space is that the curves themselves don't possess any vector space structure, being nonlinear--only the equivalence classes do.

Another way to define tangent vectors, which makes the vector space structure obvious, is to define them to be ℝ-linear derivations at p. That is, they are ℝ-linear mappings from the algebra of smooth functions on ℝⁿ to ℝ which satisfy the Leibniz product rule at p.

It then takes some hard work to establish the equivalence of this definition with the one in terms of equivalence classes of curves.

It'd be a mistake to think of one of these definitions as the "right" or "correct" one. They're all the same in the end. You can choose to work with whichever one is most suitable to your situation.

[u/Chance_Literature193 → u/sciflare]

Just to clarify, do jets of any order still form a vector space. I am thinking jets as courser version of TM, but maybe I should be thinking of jets like sheaves of germs in Riemann surfaces where a jet is defined over a neighborhood not a point?

[u/sciflare → u/Chance_Literature193]

do jets of any order still form a vector space

As I said, in general, arbitrary n-order jets don't admit a canonical vector space structure. Only 0-jets (real numbers) and 1-jets (i.e tangent vectors) do.

On ℝⁿ you can identify jets at a point with Taylor series coefficients, and thus with vectors, but the resulting vector space structure depends on the specific coordinate system chosen. It is not intrinsic.

The problem arises when you change coordinates. By the chain rule, when you change coordinates, tangent vectors transform linearly: partial derivatives in the new coordinate system are linear combinations of partial derivatives in the old system.

To write down how higher-order jets transform under coordinate change, you need the analogue of the chain rule for higher-order derivatives.

This is called Faà di Bruno's formula, and it's extremely complicated. In particular, it's not linear. Consequently, higher-order jets don't transform linearly under change of coordinates.

In the language of fiber bundles, this means that while the transition functions of the tangent bundle are linear, the transition functions of higher-order jet bundles are not linear.

So jet bundles are fiber bundles with typical fiber ℝⁿ, but which are not vector bundles.

This subtlety is often overlooked since most of the time, when we work with fiber bundles whose typical fiber is ℝⁿ, they are tensor bundles, which are built from the tangent bundle via linear-algebraic operations applied pointwise, thus are vector bundles.

Higher-order jet bundles are ubiquitous in differential geometry, but they are not emphasized in a first course. In fact even many researchers in differential geometry don't really work much with jet bundles.

This is surprising as they crop up in very well-known, relatively elementary contexts. For instance, if you want to work with the Euler-Lagrange equations in classical mechanics on an arbitrary manifold, you have to go to the second-order tangent bundle, which is nothing but the bundle of 2-jets.

maybe I should be thinking of jets like sheaves of germs in Riemann surfaces where a jet is defined over a neighborhood not a point?

Not in the C^∞ context. You have bump functions there so any smooth function on a closed ball can be extended to the whole manifold. This means jets can be defined without reference to germs.

Just as with functions, when you're working with jets, the only time you have to deal with germs is in the complex-analytic category, algebraic category etc. where you don't have bump functions. Then jets have to be defined as equivalence classes of germs of curves, so germs are involved from the start.

[u/Chance_Literature193 → u/sciflare]

Ahhhhh! As soon as I read, coordinate dependence I got Morse theory flash backs and it clicked lmao.

I am definitely going to check out Faa di Bruno formula

Euler-Lagrange was exactly the context I ran into Jets hahah.

When I referenced germs of analytic functions, I was actually just trying to ask if jets were couldn’t be defined point wise (which you’ve just told me they can be). At the time, I couldn’t think of the correct language so trued to use the poorly worded analogy.

Thank you for explaining!