# The Foundations of Relativity II

1. Jan 9, 2006

### Oxymoron

The "The Foundations of Relativity" thread was so helpful for me that I was able to learn a lot very quickly and I thank everyone who made the effort and time to help me out - but in an effort to prevent it from being too long I decided to start the next part in the series in a brand new thread. In this thread, I would like to focus on what I consider the next step in learning the mathematics behind relativity, and that is Differential Geometry.

Now I know there are a lot of good threads on differential geometry on this website, but somehow, the questions asked are not exactly what I needed to know, or they became too involved too quickly (lethe's thread on this topic comes to mind). I will continue to visit the first part with questions on tensors as I come to them.

So lets get this started. The first question obviously is: What is differential geometry? Why is it useful in the study of relativity? And what is a good approach to the topic?

2. Jan 9, 2006

### pervect

Staff Emeritus
The whole of differential geometry is rather a big topic. I think that the next thing you need to understand is probably the precise definition of a manifold.

Wald, in "General Relativity" has a very good mathematically-oriented treatment of manifolds that I would recommend to you. (I wouldn't recommend it to a person who prefers diagrams to math, though).

A manifold is, in the mathematical approach, built on top of a topological space. It is a topological space with a series of "charts", which are homeomorphisms (sp?) from the topological space to $\mathbb{R}^n$ (for a real-valued manifold).

Question: do you already happen to know what a topological space and a homeomorphism are?

The various charts in a manifold have to cover the whole manfold, and when a point is in more than one chart, certain conditions must be met so that the charts join "smoothly" and consistently.

3. Jan 10, 2006

### Oxymoron

I have done a third-year course in topology and I do know what a homeomorphism is, so please by all means, continue.

So you are saying once a topological space has a set of charts whose collection or union of them all covers the space PLUS they must map to $\mathbb{R}^n$, then we say that the space is a manifold? Sounds a bit like joining the notion of compact with Hausdorffness.

4. Jan 10, 2006

### pervect

Staff Emeritus
OK, I have to consult the textbooks to be more precise. This is adapted from Wald's discussion on pg 12 of "General Relativity" but is not an exact quote. Hopefully I haven't screwed anything up in the process of transcribing it.

A n-dimensional, real, $C^\infty$,real manifold M is a topological space M together with a set of homeomorphisms $O_a$ that satisfy the following properties

1) The homeomorphisms $O_a$ each map an open subset of M into an open subset of $\mathbb{R}^n$

2) The homeomorphisms cover M, i.e. every point in M is in the domain of at least one homeomorphism.

These homemorphisms are referred to as charts, or coordinate systems. It is generally assumed that a manifold includes all possible charts, so that one does not worry about manifolds being different because one has an extra chart that another one lacks.

3) The domains of the homeomorphisms may overlap. Let the set of the overlap of the domains of two homeomorphisms $O_a$ and $O_b$ be $X \in M$. Then $O_a(X)$ and $O_b(X)$ must both be open subsets of $\mathbb{R}^n$ Furthermore, $O_b(O_a^{-1})$ is a map from $\mathbb{R}^n \rightarrow\mathbb{R}^n$. This map must be $C^\infty$

(You are free to replace the last requirement with $C^k$ for a $C^k$ manifold).

In GR, we need only consider manifolds which are Hausdorff and paracompact.

Last edited: Jan 10, 2006
5. Jan 10, 2006

### Peterdevis

Maybe it's usefull to explain why scientist has choosen differential geometry to describe GR before running to deep into the math.

1) Einstein (and others) recognised that time was not a parameter (Newtonian view) but a (mathematical) dimension, just like the three spatial one's. So the GR-world is described bij events with four cöordinates put togheter in a topological space (by defening wich events lays in the neighberhood of an other)

2) To compute things you need always numbers. To do this you says that the toplogical space looks local likes $\mathbb{R}^4$. Are simply you make maps. Now you have a manifold (if the maps are smooth continious etc) By doing this you introduce also the calculus of $\mathbb{R}^4$ on the manifold.

3) One of the cornerstones of GR is that laws are the same in al reference frame. So if you have coördinate independent objects for formulating laws, things getting a lot easier. This objects are available in differential geometrie: tensors

6. Jan 10, 2006

### Oxymoron

If one includes all possible charts (therefore including all $C^{\infty}$ charts) then the collection of them is called a maximal atlas? Such an atlas together with the topological manifold becomes a differentiable manifold? Have I got the terminology correct?

Are all manifolds homeomorphic to $\mathbb{R}^n$? Hmmm, maybe not. For example, what if I were to take a straight line. Then this is a manifold. Now what if I take a circle, which is a subset of 2-dimensions, $S^1 \subset \mathbb{R}^2$. The problem comes here with how to map every point of the circle without ambiguity. For example, let

$$U = S^1 \backslash \{(1,0)\}$$

and define a chart (in polar coordinates)

$$\phi(\theta) \,:\, U \rightarrow \mathbb{R}$$

If I have done everything right, then this is a perfectly good chart in polar coordinates, ie. $\phi((x,y)) = \theta$.

Now the image of $U$ under $\phi$ is the interval $(0,2\pi) \subset \mathbb{R}$. But what about the point $(1,0)$? What can I do with this? I thought every point in the manifold must be the domain of a homeomorphism to the real line? I dont see that my chart can handle every point on the circle. If I disconnect it by removing a point then $U = S^1 \backslash \{(1,0)\}$ is homeomorphic to the real line, and finding charts is easy.

However, more than one chart easily covers every point on the circle. Could I say that every 1-dimensional manifold is homeomorphic and requires 1 chart - EXCEPT for the circle?

Last edited: Jan 10, 2006
7. Jan 10, 2006

### robphy

In addition to discussions of the mathematical formalism, there are some good PHYSICAL motivations of the mathematics in the literature.

The OP might enjoy this set of notes by Malament: http://www.lps.uci.edu/home/fac-staff/faculty/malament/FndsofGR.html
http://www.lps.uci.edu/home/fac-staff/faculty/malament/geometryspacetime.html
and
"General Relativity for Mathematicians" by Sachs/Wu
http://www.worldcatlibraries.org/wcpa/top3mset/d370b1b1409a9cf1.html
https://www.amazon.com/exec/obidos/ASIN/038790218X
and
"The Large-Scale Structure of Space-Time" by Hawking/Ellis
http://www.worldcatlibraries.org/wcpa/top3mset/b85cface6f53067b.html
https://www.amazon.com/exec/obidos/ISBN=0521099064
and
"Structure of Space-Time" by Penrose and "Topics on Space-Times" by Lichnerowicz, both in Battelle Rencontres (eds. C M DeWitt, J A Wheeler)
http://www.worldcatlibraries.org/wcpa/top3mset/350016a3492b9289.html
and
"Global structures of spacetimes" by Geroch/Horowitz, in General Relativity, an Einstein Centenary Survey (eds. Hawking, S., and Israel, W.)
http://www.worldcatlibraries.org/wcpa/top3mset/27ed62c8845ce767.html
https://www.amazon.com/exec/obidos/tg/detail/-/0521222850?v=glance
and
"The Geometry of Free Fall and Light Propagation" by Ehlers/Pirani/Schild in General Relativity (ed. O'Raifeartaigh)
http://www.worldcatlibraries.org/wcpa/top3mset/0322f584cc7109ae.html
and
"The nature and structure of space-time" by Ehlers and "Theory of Gravitation" by Trautman, both in The Physicist's Conception of Nature (J. Mehra ed.), Reidel, Dordrecht, 1973
http://www.worldcatlibraries.org/wcpa/top3mset/fae6224329183984.html
https://www.amazon.com/exec/obidos/ASIN/9027703450

Last edited by a moderator: Apr 21, 2017
8. Jan 10, 2006

### George Jones

Staff Emeritus
No!!! A counterexample is your circle.

Again, no. Every point in the manifold S^1 is contained in an open neighbourhood that is homeomorphic to an *open* *subset* of R.

It is impossible for one chart to cover the all of the circle S^1. S^1 is a compact topological space, and the continouous image of a compact set is also compact. Therefore the whole circle S^1 cannot be homeomorphic to an open subset of R, or to alll of R, both of which are open (and thus not compact) subsets of R.

It takes at least 2 charts to cover all of S^1.

Regards,
George

9. Jan 10, 2006

### Oxymoron

Thanks for that. It is always a good thing for me to know the physical consequences of what Im studying in mathematics. I'll keep those points in mind.

Ah, very good. I forget about that.

Exactly what I wanted to hear.

Wow, thanks for the resources Robphy.

OMG. That first link has EXACTLY the same title as my threads!! Hmmm . Anyway, it'll probably take me all day to look through your resources. I'll ask some more questions later.

10. Jan 10, 2006

### Hurkyl

Staff Emeritus
I'm going to say some of the same things, but in a different way.

A manifold is merely something that is locally homeomorphic to Euclidean space. (Some people allow the different connected components of a manifold to be locally homeomorphic to Euclidean spaces of different dimension)

From there, we would then like to say that a differentiable manifold is merely something that is locally diffeomorphic to Euclidean space. However, this raises a problem: what the heck is a differentiable map?!?!

But let's ignore that for the moment.

In Euclidean space, we like to talk about things like functions, and vector fields.

That's easy enough for a manifold, but we can do something even more interesting!

When considering functions, it is often useful to think of their graphs. For example, a function R->C is simply a subset of RxC for which there is exactly one point on each vertical line.

Let's call these vertical lines fibers. (Or fibres) We then say that RxC is the trivial fiber bundle over the base space R with fiber C.

A section of a fiber bundle simply assigns to each point in the base space a single element of the corresponding fiber. So, the notion of a function R->C coincides with the notion of a section of the fiber bundle RxC. (We usually like to talk about continuous sections, which corresponds to a continuous function)

Now, let's consider real-valued functions of the circle S^1.

It's easy to picture the trivial bundle S^1 x R: it's simply a cylinder (or annulus, if you prefer). But it may strike you that there's a more interesting way to form a fiber bundle: you could "twist" the fibers as you go around the circle.

In other words, the Möbius strip is a fiber bundle over S^1, with fiber isomorphic to R!

The salient feature is that any point of S^1 has a neighborhood U in which the corresponding collection of fibers together look like the trivial fiber bundle UxF.

The difference between these two fiber bundles is interesting! one way to tell them is that any continuous section of the Möbius strip must have a zero, whereas the same is not true of the cylinder.

It turns out, I think, that these are the only two fiber bundles over S^1 with fiber R. (Another interesting thing to study!)

If we recall in our work with Euclidean space, one of the important aspects of differentiability is that of tangent vectors. Or more generally, things like vector fields that assign a vector to each point in space. So, this suggests we might like to talk about a vector bundle (a fiber bundle whose fibers are vector spaces) which we'll call the tangent bundle.

TR^n is simply the trivial bundle R^n x R^n. (The first component is the base point, and the second component is the tangent vector)

Then, we can relate all of the tangent bundles for Euclidean spaces. For example, if we have a differentiable map f:R² -> R³, we have the induced map on the tangent bundles $f_*:T\mathbb{R}^2 \rightarrow T\mathbb{R}^3$ given by:

$$f_*(x, v) := (f(x), (Df)(x) v)$$

(Df)(x) v is, of course, simply the directional derivative along v.

We define a differentiable manifold to be a vector bundle that is compatable with all of this structure we've placed on the Euclidean spaces.

(This is the "abstract nonsense" approach -- most introductions will actually construct such beasts, and then later prove that it's compatable with this structure)

So now that we have a differentiable structure, we probably want to do calculus! Thus, differential geometry!

11. Jan 12, 2006

### pervect

Staff Emeritus
OK, so far we have defined tensors, and given a precise defintion of a manifold, I think the next step is to say a few more words about the tangent space of a manifold, which is a vector space.

The physicist's way of looking at it is to draw a sphere to represent the manifold, to draw a plane tangent to the sphere at some point, and then say that for "small enough" displacements, the displacements all lie essentially in the tangent plane, and that one can define infinitesimal displacements and infinitesimal distances on the manifold by working in the tangent plane, which is simply a subset of $\mathbb{R}^n$ and has standard defintions of distances and vectors.

The picture of a sphere is a visual aid for a manifold of dimension 2 - the actual manifold can be of an arbitrary dimension.

The mathematician's way of looking at it is to say that the tangent space actually represents derivative operators.

The mathematican's approach also implies that we never directly define distances, rather we define the rate at which distances change, and integrate where needed.

This approach gets rid of infinitesimals, and replaces them with derivatives, which are easier to deal with by pure mathematical formalism and axioms.

It's important to realize that every point on the manifold has its very own tangent space, regardless of which approach one takes. The tangent space will always have the same dimension, but different points have different "tangent planes", and there is at this point in the development no natural way of taking a vector at one point on the manifold and comparing it to a vector at a different point on the manifold.

The next step after this is to talk about derivative operators. The tangent space defines the derivatives of real-valued functions on the manifold, but suppose we want to take the derivative of a vector field defined on the manifold, or more generally, a tensor field? To do this, we need a derivative operator, sometimes known as the covariant derivative. The symbol for this is $\nabla_a$. The derivative operator operates on a tensor of rank (i,j) and generates a tensor of rank (i,j+1). If applied to a scalar - a tensor of rank (0,0), it generates a tensor of rank (0,1)

In order to take the derivative of a vector field (or of a more general tensor) we need to be able to compare vectors (or tensors) at different points on the manifold. As you recall, we earlier remarked that these vectors lived in different vector spaces, and that there wasn't a natural way of comparing them.

There are many different ways as to how to handle this.

The first approach is to use fibre bundles, as Hurkyl did. Unfortunately, i don't know much about this approach. This involved definining a connection.

Another approach is to first define a metric on the manifold. Given that a metric has been defined, there is then a geometric construction known as Schild's ladder that allows one to "parallel transport" vectors from one point to another.

The detailed description of Schild's ladder in its full original form is somewhat involved (it is discussed, for instance, in MTW's Gravitation), but the basic idea is very simple. Because we have a metric, we can measure the length of the sides of an infinitesimal quadrilateral. We can say that when opposite sides of the quadrilateal have equal lengths, we have a "parallelogram". This parallelogram defines the idea that the opposite sides of the parallelogram are parallel. This notion of "parallel" allows us to parallel transport vectors.

Thus given a metric, Schild's ladder allows us to define "parallel transport" via a geometrical construction. We can then define the derivative of a tensor in basically the normal way

$$\nabla_x T = \lim_{\Delta x \to 0}$$ (T(x+$\Delta x$) parallel transported to x -T(x) )/ $\Delta x$

A more abstract approach to the issue is to consider the general notion of derivative operators. Derivative operators operate on tensors, and generate a tensor of one higher rank. A derivative operator applied to a scalar generates a vector, and we require that the action of a derivative operator applied to a scalar is compatible with our notion of a tangent space.

Five properties are sufficient to completely define a derivative operator

1) linearity
2) Leibnitz rule $\nabla_a (T^b T^c) = T^b (\nabla_a T^c) + (\nabla_a T^b) T^c$
3) Commutativity with contraction (I'm not quite sure why this is needed, frankly)
4) The derivative operator applied to a scalar function must generate a tangent vector as previously defined
5) Torsion free $\nabla_a \nabla_b f = \nabla_b \nabla_a f$ (This is a bit of a mystery to me, too) Actually it's not always necessary that derivative operators always be torsion free, but they are always torsion free in GR. Some other theories of gravity have derivative operators that are not torsion free.

Derivative operators also define parallel transport, via the equation

$$t^a \nabla_a v^b =0$$, where $t^a$ is the the tangent vector for the curve along which the vector is being parallel transported.

There is a unique derivative operator that is compatible with a given metric.

This metric-compatible derivative operator is the one that preserves the dot product of two vectors when both vectors are parallel transported along a curve.

So there are many different routes, but they all lead to the same place, the defintion of the covariant derivative.

Some older books will represent covariant derivatives with a semicolon, i.e an alternate notation is

$$\nabla_a T^b = T^{b;a}$$

Last edited: Jan 12, 2006
12. Jan 13, 2006

### Oxymoron

When I was doing multivariable calculus we spent some time on directional derivatives. Basically what we were doing was differentiating a surface with respect to some vector (roughly speaking). Ultimately we would then find the direction of the fastest increase of the surface.

Now, just say that we have a surface that is not necessarily flat. At each point on the surface there is a vector which points in the fastest increase of gradient of the surface. Every point has a vector associated with it. If we collect all these vectors, then do we have a gradient vector field?

These surfaces where given by functions, $f(\bold{x}(t)) = f(x^i(t))$ of the form

$$f\,:\,\mathbb{R}^n \rightarrow \mathbb{R}$$

usually something simple like $f(x,y,z) = x^2+y^2+z^2$. Anyway, the point is, now we are more advanced than that and we want to generalize this idea that was proposed to me in first-year. So, is it safe to say that these surfaces we were considering are actually manifolds in three-dimensional Euclidean space, and by taking a directional derivative of a manifold is no different to taking a directional derivative of a surface in three-dimensional Euclidean space?

I want to run my idea of this section by you in my own words to see if you agree.

A directional derivative is merely an operator, or a linear differential operator to be precise. It takes an element from the algebra of differential functions, $f$, operates on it, and spits out a real number. The operating bit is given by a rule:

$$Xf=\frac{df(\bold{x}(t))}{dt}=\frac{\partial f(\bold{x})}{\partial x^i}$$

Now, in three-dimensional Euclidan space is was satisfactory to describe the surfaces by its components, thus a simple derivative with respect to its components was sufficient. But in general n-dimensional mainfolds we have to be more careful. In my understanding, we need an approach that is independant of the coordinates. My problem here is that I see no difference between the directional derivative operator and the regular old grad operator in calculus.

My attempt to understand this may be flawed. I first considered only curves through space. Through our work in tensors I can express my curve with raised indices

$$x^i = x^i(t)$$

To define the tangent to the curve we have the vector

$$\bold{v} = (\bold{\dot{x}}) = (\dot{x}^1,\dot{x}^2,\dots,\dot{x}^n)$$

where $\dot{x}^i = (dx^i/dt)$ at $t = t_0$. But I am told that expressing the tangent vector in terms of its components is no good. Why is this so? Why do we need some sort of coordinate-independent approach?

Continuing...This D.D. operator has two properties:

It is linear: $$X(af+bg) = aXf + bXg$$
It obeys the Leibnitz rule: $$X(fg) = f(x_0)Xg + g(x_0)Xf$$

both properties, to me, seem obvious, especially after having done calculus - it just seems natural that the operator should
satisfy them both.

Now lets go back to our manifold. Pick some point on it, say $p$. If our manifold is smooth then there will be a tangent vector, call it $X_p$. The tangent vector is yet another function, it takes elements from the algebra of differentiable functions at p and spits out real numbers. The italicized part is crucial to defining the tangent vector in my opinion. Thus

$$X_p\,:\, \mathscr{F}_p(M) \rightarrow \mathbb{R}$$

is our formal tangent vector.

Now lets collect all possible tangent vectors at this one point on our manifold. By doing so we have created a new vector space (which is to no suprise since the tangent vectors obey linearity and the Leibnitz rule - plus linear combinations - which pretty much means they form their own vector space). We call this new vector space $T_p(M)$ which means the tangent space at a point p on a manifold.

Last edited: Jan 13, 2006
13. Jan 13, 2006

### Oxymoron

The idea I had in my previous post was to use the ideas given to me in multivariable calculus and realize that they were merely specific parts of the more general idea of differential manifolds. I began reasonably well, but I pretty much got stuck on the following notion (which I did write in my previous post but may have been lost under all my ramblings, so I write it here again for clarity):

I want to define my tangent space as simply the space of all tangent vectors at a point p. But the tangent space is meant to be a space of vectors at p. Do we actually know what the tangent to a curve is? These tangent vectors don't necessarily live inside our manifold, so as yet we have no idea what they do!

If all curves through our point p can be written in terms of its n components, each component relating to the coordinate system we use. Is it true that we want to do away with referring to coordinate systems in our differentiation? Why do we want to develop some sort of coordinate independent approach to our directional derivative operator? What would happen if we were to use a coordinate dependant directional derivative operator? Would it be ambiguous? Would it be meaningless? Perhaps meaningless under certain coordinate transformations?

PS. Hurkyl, I dont mean to be rude by not repsonding to your post, its just that it is still a little ahead of what Ive read. But nevertheless, soon I will be there and I can refer to what you've written.

Last edited: Jan 13, 2006
14. Jan 13, 2006

### George Jones

Staff Emeritus
Yes, we do.

First, what is a curve?

A smooth curve is a smooth mapping $\sigma : I \rightarrow M$, where $I$ is a interval (subset) of $\mathbb{R}$. A curve should not be confused with its image, the points that are traced out in $M$.

Let $f : M \rightarrow \mathbb{R}$ be an arbritrary mapping that lives in $C^{\infty} \left( M \right)$. Then $g : I \rightarrow \mathbb{R}$ defined by $g \left( t \right) = f \left( \sigma \left( t \right) \right)$ is a smooth real-valued function of real variable, and, as such, can be differentiated.

Suppose $p = \sigma \left( 0 \right)$. The curve $\sigma$ defines a derivation (tangent vector) $v$ at $p$ by

$$v \left( f \right) = \left[ \frac{dg}{dt} \left( t \right) \right]_{t=0}.$$

Regards,
George

Last edited: Jan 13, 2006
15. Jan 14, 2006

### pervect

Staff Emeritus
Are you perhaps struggling to see why your two defintions of a tangent space are equivalent?

Suppose we have a manifold M, and an open subset O $\subset M$ which is the domain of some specific chart of the manifold $\phi$.

$\phi$ maps O into $\mathbb{R}^n$
$\phi^{-1}$ maps $\mathbb{R}^n$ into M (actually into a subset O of M).

Now we have some function f : M $\rightarrow \mathbb{R}$

f $\in \mathbb{F}$ where

$\mathbb{F}$ denotes infinitely differentiable maps from M $\rightarrow \mathbb{R}$.

Then
$$\frac{\partial}{\partial x^i} (f(\phi^{-1})) \mid_{\phi(p)}$$

is a derivative operator because it is a map from $f \in \mathbb{F} \rightarrow \mathbb{R}$, and it also passes the Lebnitz rule test and the linearity test.

p $\in$ M, so $\phi(p) \in \mathbb{R}^n$, that's why the evaluation is written that way, the evaluation of the partial occurs in $\mathbb{R}^n$

A simple diagram of O, M and the image of O in $\mathbb{R}^n$ helps a lot if you sketch it out, see for instance Wald pg 15.

Last edited: Jan 14, 2006
16. Jan 14, 2006

### pervect

Staff Emeritus
Something I've been meaning to ask:

Hurkyl, what's a good textbook for the "fibre bundle" approach that you used earlier in the thread?

17. Jan 18, 2006

### George Jones

Staff Emeritus
A couple of reference for bundles that I like, but that don't quite follow Hurkyl's 'abstract nonsense' approach, are The Geometry of Physics by Theodore Frankel, and Modern Differential Geometry for Physicists by Chris J. Isham (All three of his books are great!), though both of these books are somewhat abstract.

I find the first, written by mathemetician, to be somewhat less abstract than the second, written by a physicist! Also, unlike the second reference the first reference has a chapter specifically on relativity, and a chapter on connections on manifolds (as opposed to more general connections on bundles). The first: uses 2-dimensional spaces tangent to 2-dimensional surfaces in 3-dimensions to motivate connections; has an interesting chapter on the Dirac equation that has a section (pun intended) on the equation in curved spacetimes.

The second edition of the second reference starts with an introductory chapter on topology that I find interesting, but that I think isn't crucial to the rest of the book. The second reference is geared more towards gauge theory, but its section on tangent spaces is the best that I have seen. This section does a good job of relating tangents to curves to derivations. The second reference is meant as a set of lecture notes, and thus is thinner than the first book.

Both books cover the geometry of Lie group, and of bundles.

Opinions on books are very subjective; these are 2 books that I like very much.

Regards,
George

18. Jan 18, 2006

### Hurkyl

Staff Emeritus
Spivak's Differential Geometry texts talk about vector bundles. (It introduces them before the tangent bundle, I think) I'm not qualified to say if it's a good reference, though. (Though it was once recommended to me, because I was interested in the subject at the time)

Spivak gave three separate definitions of the tangent bundle, and somewhere along the line, proved that the tangent bundle is "essentially unique".

The essential feature of the abstract nonsense approach is that you simply want things to behave correctly for Euclidean space.

You want $T\mathbb{R}^n$ to be (isomorphic to) the trivial bundle $\mathbb{R}^n \times \mathbb{R}^n$.

You want a differentiable map $f:\mathbb{R}^m \rightarrow \mathbb{R}^n$ to induce the correct map on the tangent bundles: $f_*(x, v) = (f(x), (df)(x) \cdot v)$.

(I think there's another thing we really want to happen, but I can't remember what it is... maybe it was that if f is the identity map on any manifold M, then $f_*$ is the identity on TM)

The abstract nonsense theorem is that there is an "essentially unique" way to extend this notion to all differentiable manifolds in a "reasonable" way.

What "reasonable" means should be obvious -- in abstract nonsense, we give the concept a name: the maps $M \rightarrow TM$ and $f \rightarrow f_*$ comprise a functor.

"essentially unique" means that any two different ways to do this must be naturally isomorphic. Loosely speaking, it simply means that if T and T' are two different ways of doing this, then $TM \cong T'M$ for any differentiable manifold M.

The three constructions given by Spivak, intuitively speaking, were:

(1) Tangent vectors are defined by coordinates that transform properly.
(2) Tangent vectors are defined by a curve to which they're tangent.
(3) Tangent vectors are defined as linear differential operators.

The abstract nonsense tells us, in one fell swoop, that all of these constructions (and any others we may imagine) are isomorphic -- which justifies us treating them all as merely different ways of writing tangent vectors.

I really like thinking in this way, since it's more "operational" -- the tangent bundle concept is merely the extension of the obvious notion for Euclidean space. We don't have to worry about what the things "really are" at all.

Last edited: Jan 18, 2006