Questions about the differential

[orion]

In some books, the differential is defined by:

(a) $df(v) = v(f) : T_pM → \mathbb {R}$

while in other books, a more abstract definition is made:

(b) Given two manifolds $M,N$, $v \in T_pM$ and a map $F :M → N$ then $dF$ is defined

$dF(v)(f) = v(f\circ F) : T_pM → T_{F(p)}N$

where $f \in C^∞(N)$.My questions:

(1) How are these two definitions equivalent? And if they are not equivalent how to reconcile them?

(2) I need help understanding (b). Why are there $F$ and $f$? Also, how in what space does $v(F \circ f)$ operate since F is a map from $M$ to $N$ and $v$ belongs to $T_pM$ and $f \in C^∞(N)$?

(3) It would seem that definition (a) is more in line with the idea that a differential is a linear functional on $T_pM$ and hence a covector in a space dual to $T_pM$. How to understand (2) in this context?

 

[WWGD]

To possibly save you some work, in (a), you do start with a map from $\mathbb R^n \rightarrow \mathbb R^m$., both of which are manifolds, your $M,N$ you use below.

in (b), v is a derivation, so it acts in any tangent space. v acts on the tangent space (as a directional derivative) at the tangent space $T_{F(p)}N$ , based at $N$, and it acts on functions $f: N \rightarrow \mathbb R$.

 

[orion]

To possibly save you some work, in (a), you do start with a map from $\mathbb R^n \rightarrow \mathbb R^m$., both of which are manifolds, your $M,N$ you use below.

Thanks for your reply.

I made a mistake in the mapping in (a). It is correct now.

So in (a) what is $F$ and $f$ as defined in (b)?

 

[WWGD]

But please do note that there is a map associated to , or giving rise to , the differential in (a). That will make it easier to compare with the description in (b). Please describe this map, or explain if there is no such map in the definition.

EDIT: I need to leave or a while now, will be back in a few hours.

 

[orion]

Is this correct:

$f :M → \mathbb{R}$
$df: T_pM → T_{f(p)}\mathbb{R}$

but

$T_{f(p)} \mathbb{R} ↔ \mathbb{R}$?

What plays the role of $f$ from definition (b)? Would that be the identity map?

Honestly, I have no idea what I'm doing. I'm just trying to make the maps work out.

Edit: Where does $v$ come into play in all this?

 

[Fredrik]

Everyone uses the df notation for the map defined in (a), but not everyone uses it for the map defined in (b). Your (b) defines a map associated with F called a "pushforward" (because it "pushes" tangent vectors from the point p to the point F(p)). It's usually denoted by something like $F_*|_p$ instead of $\mathrm dF|_p$. I wasn't even aware that some people are using the dF notation until a recent thread where someone asked about it.

I think I found a reason why some people might like the dF notation. (I still prefer $F_*$). $df|_p$ and $F_*|_p$ are both linear maps, so we can compute their https://www.physicsforums.com/threads/matrix-representations-of-linear-transformations.694922/ (one for the domain and one for the codomain). If the manifolds we're dealing with are $\mathbb R^n$ (possibly with different values of n), and we choose the ordered bases associated with the identity maps on these spaces, then in both cases (i.e. $df$ and $F_*$), we end up with the Jacobian matrix of the function.

\begin{align*}
[df|_p]_j &=(df|_p)\left(\frac{\partial}{\partial I^j}\right)_p =\left(\frac{\partial}{\partial I^j}\right)_p f =(f\circ I^{-1})_{,j}(I(p)) =f_{,j}(p)\\
[F_*|_p]^i_j &=\left(F_*|_p\left(\frac{\partial}{\partial I^j}\right)_p\right)^i = \left(F_*|_p\left(\frac{\partial}{\partial I^j}\right)_p\right)(I^i) =\left(\frac{\partial}{\partial I^j}\right)_p (I^i\circ F) =\left(\frac{\partial}{\partial I^j}\right)_p F^i\\
& =(F^i\circ I^{-1})_{,j}(I(p)) =F^i{}_{,j}(p).
\end{align*}

 

[orion]

Is your $I$ the identity map? I'm confused by the step $(F \circ I^{-1})_{,j}(I(p))$ in both your equations. Not the derivative part, but the compositions. How should I understand that? If $M=\mathbb{R^n}$ and $N=\mathbb{R^m}$, then $I$ as you are using it lives where? In $\mathbb{R^n}$ or $\mathbb{R^m}$?

In the books that I've read, they say that they are the same because of the "natural identification" of $T_{f(p)}\mathbb{R}$ with $\mathbb{R}$

Thanks, Fredrik, for your reply. I think it will be immensely helpful. (I just have to understand it completely).

 

[Fredrik]

The equality $\left(\frac{\partial}{\partial I^j}\right)_p f =(f\circ I^{-1})_{,j}(I(p))$ is just the definition of left-hand side. (I think I elaborated a little bit more on this in my first post in your other thread). Yes, $I$ is the identity map. I didn't bother to think about whether its the identity map on $\mathbb R^n$ or on $\mathbb R^m$. I'll think about it now...

OK, in the first calculation, $I$ is the identity map on the manifold that's the domain of f.

In the second, if $F:\mathbb R^n\to\mathbb R^m$ then $I^i$ is the $i$th component of the identity map on $\mathbb R^m$ and $I^j$ is the $j$th component of the identity map on $\mathbb R^n$. It would probably have been less confusing if I had used a different symbol (e.g. J) for the identity map on $\mathbb R^n$.

To understand the second calculation, you need to know that if $v\in T_pM$ and $x:U\to\mathbb R^n$ is a coordinate system such that $p\in U$, we have $v=v(x^i)\left(\frac{\partial}{\partial x^i}\right)_p$. So we can write $v^i=v(x^i)$.

 

[orion]

What I'm not understanding is what is the point of the step $(f \circ I^{-1})_{,j} (I(p))$?

For example, if I were writing that out, it would not occur to me to include that step.

Edit: Forget it. I understand it now.

 

[orion]

No, wait. I'm still having a problem with that step. I don't understand why that step is necessary. I understand the notation from the other thread, but I just don't understand why that step is necessary.

 

[Fredrik]

$\left(\frac{\partial}{\partial I^j}\right)_p$ is defined as the map that takes an arbitrary smooth function f to $(f\circ I^{-1})_{,j}(I(p))$. So I'm just using the definition of the notation. Are you asking why it's defined that way? It's because partial derivatives in calculus are defined using the addition operation on $\mathbb R^n$, but in differential geometry, the domain of the function is a manifold, and most manifolds aren't equipped with an addition operation. So we define the partial derivatives at p of $f:M\to\mathbb R$ with respect to the coordinate system $x:U\to\mathbb R$ by
$$\left(\frac{\partial}{\partial x^i}\right)_p f=(f\circ x^{-1})_{,i}(x(p)).$$

 

[orion]

What I'm asking is why can't we simply write:

$(\partial/\partial I^j)_pF = (\partial F^i/\partial I^j)_p = F^i_{,j}$ ?

Edit: Ok. I accept it as the definition in light of your explanation above.
Edit again: From where did you learn that notation? I haven't seen it anywhere else.

Another question I'm struggling to understand in definition (b) above is what exactly is the role of $f$? Why is that in the definition? I mean, when $f=I$ it's clear, but what about in other cases? I found yet another definition defined in terms of curves ($γ$ and $F \circ γ$) which I understand but neither that definition nor definition (a) above has the additional $f$.

 

[orion]

Here is another proof of the equivalence of (a) and (b) using the identity map $I$:

$df^b(v)(I) = v(I \circ f) = v(f) = df^a(v)$

where $df^a$ and $df^b$ are definitions (a) and (b) respectively.

 

[Fredrik]

Edit again: From where did you learn that notation? I haven't seen it anywhere else.

I first learned it from Spivak, "A comprehensive introduction to differential geometry", vol. 1. Page 35 of the first edition uses the notation
$$\frac{\partial f}{\partial x^i}(p) =\frac{\partial f}{\partial x^i}\bigg|_p =D_i(f\circ x^{-1})(x(p)).$$ Page 39 says that the operator taking f to $\frac{\partial f}{\partial y^i}(p)$ is denoted by $\frac{\partial}{\partial y^i}\big|_p$.

Spivak doesn't use the comma notation for partial derivatives. I think I picked that up from some physics course.

I sometimes use $\big(\frac{\partial}{\partial x^i}\big)_p$ instead of $\frac{\partial}{\partial x^i}\big|_p$. This makes sense because the value at $p$ of a vector field $X$ is denoted by $X_p$, and the map $p\mapsto\frac{\partial}{\partial x^i}\big|_p$ is a vector field that in my opinion should be denoted by $\frac{\partial}{\partial x^i}$.

I see now that Lee is using a different notation, that in my opinion is clearly worse. (Formula (3.8) on page 60 in the second edition). For no good reason, he denotes the coordinate system by $\varphi$, and still denotes the associated partial derivative functional by $\frac{\partial}{\partial x^i}\big|_p$. So his notation doesn't reveal which coordinate system is involved. Why not denote the coordinate system by $x$, or the partial derivative functional by $\frac{\partial}{\partial\varphi^i}\big|_p$ so that the connection is clear?

I noticed something else when I was checking this. Lee used the $F_*$ notation and the "pushforward" terminology in the first edition, but replaced it with the $dF$ notation and the "differential" terminology in the second edition.

Another question I'm struggling to understand in definition (b) above is what exactly is the role of $f$? Why is that in the definition? I mean, when $f=I$ it's clear, but what about in other cases? I found yet another definition defined in terms of curves ($γ$ and $F \circ γ$) which I understand but neither that definition nor definition (a) above has the additional $f$.

Let $M$ be a smooth manifold. Let $p\in M$. Denote the set of all smooth $f:M\to\mathbb R$ by $C^\infty(M)$. The tangent space of $M$ at $p$ is denoted by $T_pM$ and is defined as the vector space of all $v:C^\infty(M)\to\mathbb R$ that satisfy the following two conditions

(1) $v$ is linear.
(2) $v(fg)=v(f)g(p)+f(p)v(g)$ for all $f,g\in C^\infty(M)$.

(By the way, I think my main source for this definition and the basics about tangent vectors was Wald, "General Relativity". Much latter I also read a similar but much more detailed presentation of tangent spaces in Isham, "Modern differential geometry for physicists". I think I've learned a lot of this by putting together pieces from different books).

Edit: I wrote $T_{f(p)}N$ in a few places in the section below. It should have been $T_{F(p)}N$. I have corrected it now.

Now suppose that you want to use a function $F:M\to N$ to map tangent vectors at p to tangent vectors at F(p). We define $F_*|_p:T_pM\to T_{F(p)}N$ by
$$(F_*|_p v)(f)=v(f\circ F)$$ for all $f\in C^\infty(N)$. The way to think here is that if $v\in T_pM$, then $F_*|_p v$ is supposed to be an element of $T_{F(p)}N$, and this is a vector space whose elements are maps from $C^\infty(N)$ into $\mathbb R$. So to specify which element of $T_{F(p)}N$ that $F_*|_p v$ is, we have to specify what real number $(F_*|_p v)(f)$ is for all $f\in C^\infty(N)$.

 

[Fredrik]

Here is another proof of the equivalence of (a) and (b) using the identity map $I$:

$df^b(v)(I) = v(I \circ f) = v(f) = df^a(v)$

where $df^a$ and $df^b$ are definitions (a) and (b) respectively.

The calculation is correct, but I'm not sure what it proves. I guess it proves that the map $v\mapsto df^b(v)(I)$ is $dv^a$. So you have found a simple relationship between $df^a$ and $df^b$, but I don't see why it is a reason to think of them as the same thing, or a reason to use the same notation for them.

 

[mathwonk]

A manifold M can be studied by two natural auxiliary objects, namely maps R-->M of the real numbers into it, and maps M-->R of it into the real numbers, i.e. curves in it, or functions on it.

Secondly we can look at the infinitesimal properties of both of these objects, i.e. the behavior near a given point p of M. So we can study curves through p that have the same velocity vector, and functions defined near p that have the same partial derivatives.

Furthermore, these two approaches are dual to each other, i.e. they admit a natural pairing. namely, given a curve g:R-->M sending 0 to p, and a function f:M-->R defined near p, if we compose we get a function (fog):R-->R whose derivative we can take at 0.

Then to capture just the infinitesimal nature of curves through p, we can identify two such curves g1,g2 if for every function f, the derivatives of (fog1) and (fog2) are equal at 0. Intuitively this means the two curves have the same velocity vector.

Thus we could use this equivalence relation as a means of defining tangent vectors to M at p, I.e. a tangent vector at p is defined as an equivalence class of curves through p that yield the same derivatives for all functions f defined near p.

If we define the tangent space Tp(M) this way, then we could define the dual space to be the formal algebraic dual of this space, namely all linear functionals defined on Tp(M). Then we would see that every function f defined near p gives us such a functional, by taking the derivatives mentioned above. Thus each function f determines a cotangent vector, called df, a linear mapping from Tp(M)-->R.With this definition we also get a definition of the differential or push forward of a mapping of manifolds. Namely if F:M-->N is such a map, we just push curves forward, by composing with F, to define a map F_* on tangent spaces. I.e. a curve g:R-->M goes to the curve F_*(g) = (Fog):R-->N.Then if N = R, and F = f is just a function, we have two definitions, namely df and push forward by f.

Let’s see how they compare. df is a linear map on Tp(M) that takes an equivalence class of curves g through p to their common derivative after composition with f, i.e. to (fog)’(0).

Push forward takes the curve g to the curve (fog). But it takes it to its equivalence class as a curve through f(p). so we want to see that the equivalence class of the curve (fog) is determined by the derivative (fog)’(0). But this follows from the chain rule, i.e. for any further function h, the derivative of the composition ho(fog) will be determined by the value of the derivatives of h and of (fog). I.e. two curves through f(p) having the same derivative at 0, will also have the same derivative after composition with any h.One can also do things in the other direction, assuming as more basic the concept of functions defined near p in M. Thus we could define the cotangent space of M at p or T*p(M) to be the equivalence classes of all functions defined near p, where two functions are equivalent if their compositions with all curves through p have the same derivative.

Then we could define the tangent space Tp(M) as the formal algebraic dual of this, i.e. we could consider a tangent vector to be a linear functional on covectors. Then the natural pairings we have been discussing give us immediately a map from curves to tangent vectors. I.e. given any curve g we get after composition with any function f, a function (fog) whose derivative at zero gives us a number, which is the same for equivalent functions in T*p(M), so our curve g gives us a linear functional on T*p(M), hence a tangent vector.

Finally, the Leibniz rule holds for all these pairings and evaluations, so we are also getting from each curve a real valued mapping on the family of functions, before taking equivalence classes, that obeys the Leibniz rule, i.e. each curve also gives us a “derivation” on functions. So another point of view is to define a tangent vector, not as a linear functional on covectors, i.e. on differentials of functions, but as a derivation on the functions themselves. This is often done in differential geometry and also makes sense in algebraic geometry.

Or you could define both tangent and cotangent spaces separately, Tp(M) as equivalence classes of curves, and T*p(M) as equivalence classes of functions, and then prove that they are formally algebraically dual to one another.

But they are all the same. i.e. you have functions R-->M and functions M-->R. One type gives you tangent vectors and the other gives you cotangent vectors. And you can compose these getting functions R-->R whose derivatives you can take at 0, and this gives a natural pairing of the two concepts. It is up to you what concept you think is more basic, and which one you define to be dual to the other.

 

[orion]

Now suppose that you want to use a function $F:M\to N$ to map tangent vectors at p to tangent vectors at F(p). We define $F_*|_p:T_pM\to T_{f(p)}N$ by
$$(F_*|_p v)(f)=v(f\circ F)$$ for all $f\in C^\infty(N)$. The way to think here is that if $v\in T_pM$, then $F_*|_p v$ is supposed to be an element of $T_{f(p)}N$, and this is a vector space whose elements are maps from $C^\infty(N)$ into $\mathbb R$. So to specify which element of $T_{f(p)}N$ that $F_*|_p v$ is, we have to specify what real number $(F_*|_p v)(f)$ is for all $f\in C^\infty(N)$.

Suppose we have manifolds $M$ and $N$. Suppose $f \in C^\infty: N \rightarrow \mathbb R$ and $F: M \rightarrow N$.
Why is it a push forward of a vector and not a pull back of a curve since the domain of $f \circ F$ is in $M$?

My other question is that since $dF_p$ is a derivation on $f \circ F$, why is it $dF_p$ and not $df_p$? (I know you prefer the notation $F_*$ (push forward) but I've seen quite a few books define the differential this way.) I'm asking this because it seems like $f$ is the function of interest where $F$ is just the map between manifolds.

 

[orion]

Mathwonk, thanks for you detailed reply. I will read it this evening when I'm home.

 

[mathwonk]

you are welcome. here is an other twist on the same stuff. basically we have seen that on a manifold we know what the smooth functions are near each point. moreover, since some of these functions are coordinate functions, these functions tell us everything there is to know about the local structure of the manifold near p. in particular they can be used in various ways to describe the tangent space there. i.e. cotangent vectors are just infinitesimal bits of functions, and vectors are duals of those. here is another way to define infinitesimal bits of functions, i.e.cotangent vectors:

fix a point p on M and let Op(M) or just Op, or just O, be equivalence classes of smooth functions defined near p, where two functions are equivalent if they agree on some neighborhood of p. This is called the "local ring of smooth germs of functions at p". I.e. we are just interested in the behavior of M near p so we only look at functions defined near there and we set them equal if the are the same near p. Moreover we are only interested in first order behavior, or derivatives, so we don't care what their value is, so we just take ones that equal zero. I.e. let mp or just m, be the ideal of those functions in O which vanish at p. This is called the maximal ideal at p, since O is a ring and m is its (only) maximal ideal. (if you know about/care about ring theory.)

Now we are only interested in the first order i.e. derivative behavior of functions at p, so we want to mod out by higher order behavior. I.e. define a smaller ideal m^2 to be all functions in m such that for all curves g through p, the derivative of fog at 0 is zero, i.e. functions ”vanishing (at least) twice at p”. Then m/m^2 is the space of functions vanishing at p, and equated if their difference vanishes twice at p, so this captures only the first order behavior of a function at p. This is another possible definition of the cotangent space T*p(M).

In fact I believe you can show using smooth partial Taylor expansions of smooth functions near p (Lang, Analysis I, page 291, ex.4), that in fact m^2 consists of the square of the ideal m in the sense of ring theory, i.e. a function vanishes at least twice at p if and only if it is (a sum of) products of functions vanishing at least once. This would justify my notation m^2.

This let's us define cotangent spaces in algebraic geometry by taking the ring of polynomials defined near a point, and giving exactly this same definition. Then tangent spaces are defined as the dual of this, so we get away without having access to smooth curves in algebraic geometry. Of course we could also use algebraic derivations of the local ring of polynomial functions defined near p.

If you grasp this general approach, you can read any type of geometry, algebraic, analytic, differential, arithmetic, p adic (? actually i don't know this theory, but what else do they have to work with?), formal, etc... and you can still understand the definitions of tangent vectors etc...

 

[Fredrik]

Suppose we have manifolds $M$ and $N$. Suppose $f \in C^\infty: N \rightarrow \mathbb R$ and $F: M \rightarrow N$.
Why is it a push forward of a vector and not a pull back of a curve since the domain of $f \circ F$ is in $M$?

f isn't a curve, but you could ask why we don't say that F defines a pullback of a smooth functions. The answer is that this terminology is perfectly fine and is used by some people. I think Wald uses it. If I remember correctly, he used the notation $F^*f=f\circ F$. This defines a map $F^*:C^\infty(N)\to C^\infty(M)$, which can be called a "pullback" of smoth functions.

We can also use the pushforward of tangent vectors to define a pullback of cotangent vectors. We define $F^*|_p:T_{F(p)}N^*\to T_pM^*$ by
$$(F^*|_p\omega)(v)=\omega(F_*|_p v)$$ for all $\omega\in T_{F(p)}N^*$ and all $v\in T_pM$.

My other question is that since $dF_p$ is a derivation on $f \circ F$, why is it $dF_p$ and not $df_p$? (I know you prefer the notation $F_*$ (push forward) but I've seen quite a few books define the differential this way.) I'm asking this because it seems like $f$ is the function of interest where $F$ is just the map between manifolds.

This $dF|_p$ is a map from $T_p(M)$ into $T_{F(p)}N$, that's defined using the map F and the point p, but not any specific function f. f is a dummy variable in the definition. We define $dF|_p$ by specifying $dF|_p v$ for each $v\in T_pM$. And since $dF|_pv$ is a map from $C^\infty(N)$ into $\mathbb R$, the way to specify it is to specify the real number $(dF|_pv)(f)$ for each $f\in C^\infty(N)$.

Here's the definition of $dF|_p: T_pM\to T_{F(p)}N$ again: For each $v\in T_pM$, we define $dF|_pv$ by
$$(dF_p v)(f)=v(f\circ F)$$ for all $f\in C^\infty(N)$.

 

[orion]

mathwonk, thanks again for your detailed explanations. So is it correct to say that, given a point $p$ in $M$ and $F: M \rightarrow N$, the cotangent space consists of tangents to functions through $F(p)$ in $N$?

I want to use definition of a tangent vector as a derivation on $M$.

 

[mathwonk]

not tangents to functions, at least that's not what i would call them, but first order approximations to functions, i.e. equivalence classes of functions that behave the same up to first order. to determine that you look at their first order taylor series, or at how they behave when compiosed with curves. of course in some sense you might think of these as tangents to functions, but the name tangents makes people think of tangent vectors instead of cotangent vectors.

i.e. tangent vectors are equivalence classes of curves R-->M, while cotangent vectors are equivalence classes of functions M-->R. I think you got this, and I just quibbled with your way of restating it. But yes, intuitively they are tangents to functions just as vectors are tangents to curves, but I would be careful using this terminology.

You know come to think of it I used to use exactly this terminology when teaching differentials. I.e. I defined what it meant for two functions to be tangent to each other (their graphs are tangent), and then I defined the differential of a function as the unique linear function tangent to the function.So you could begin with functions, then take derivations of functions as tangent vectors, then define two functions to be tangent if all derivations agree on them. Then a cotangent vector is an equivalence class of functions. and it turns out to be the same as an element of the dual space of the space of derivations.

 

[orion]

not tangents to functions, at least that's not what i would call them, but first order approximations to functions, i.e. equivalence classes of functions that behave the same up to first order. to determine that you look at their first order taylor series, or at how they behave when compiosed with curves. of course in some sense you might think of these as tangents to functions, but the name tangents makes people think of tangent vectors instead of cotangent vectors.

I have a question about the differential and approximations because this is where I get confused between $F$ and $f$. In the definition where $dF_p(v)(f) = v(f \circ F)$, $dF_p$ represents a linear approximation to $F$ near $p$? I'm guessing yes because the notation $dF_p$ suggests that.

Edit: I guess if it is an approximation near $p \in M$ then it would have to be because $f$ maps from $N$.

 

[lavinia]

In some books, the differential is defined by:

(a) $df(v) = v(f) : T_pM → \mathbb {R}$

while in other books, a more abstract definition is made:

(b) Given two manifolds $M,N$, $v \in T_pM$ and a map $F :M → N$ then $dF$ is defined

$dF(v)(f) = v(f\circ F) : T_pM → T_{F(p)}N$

where $f \in C^∞(N)$.

You can look at the real numbers in two ways - as a field with multiplication and addition or as a 1 dimensional manifold.

Viewing R as a field then df(v) is just a number in the field.

Viewing R as a manifold df(v) is a tangent vector. Which tangent vector is it? Gven a function g;R ->R then df(v).g = v.(gof).

The key thing to see is that df is a linear map on tangent vectors. df(v) is a tangent vector to the manifold R. In standard coordinates, the tangent vectors to the manifold,R, are all of the form h(x)∂/∂x and df(v) is (v.f)∂/∂x at the point f(x). So the tangent vector way of looking at it is the number way multiplied by ∂/∂x.

This second way of looking at it generalizes to functions that take values in other manifolds. It does not use the field properties of R.

- These ideas also apply when the field is the complex numbers. In this case the functions are complex differentiable and the complex numbers are a one dimensional complex manifold. All tangent vectors to it are of the form h(z)∂/∂z. Between two complex manifolds, df is again a linear map on tangent vectors but now it is linear over the complex numbers.

- For me, it was helpful to think about what the Chain rule really means and how it generalizes from functions on R to multi-variable functions.

The standard statement of the Chain rule for real valued functions is ∂/∂x(fog)(x) = f'(g(x))g'(x).
This can be restated in terms of differentials as

df(dg(∂/∂x)) = ∂/∂x(fog)

dg(∂/∂x) is a tangent vector to R at the point g(x). df(dg(∂/∂x)) is a tangent vector to R at f(g(x)).

You should convince yourself that g'(x) in the manifold way of looking at things is the tangent vector, (∂/∂x.g)∂/∂x at the point g(x). And that (fog)'(x) is the tangent vector (∂/∂x.(fog))∂/∂x at the point f(g(x).

So the Chain rule says that the differential of two composed functions is the composition of their differentials and these at each point are linear functions of tangent vectors. In vector calculus, these linear maps are represented by their Jacobian matrices. The Jacobian is really just the linear map written out with respect to a particular linear basis for the tangent vectors at a point. On a manifold the idea is identical.

- The Leibniz Rule

This is an instructive illustration of differentials and the Chain rule.

Let f and g be functions from the real numbers to the real numbers. Define a function from the plane into the reals by the rule,

F(s,t) = f(s)g(t)

The differential of F is dF =( ∂F/∂s)ds+ (∂F/∂t)dt

This equals( g(t)df/ds)ds + (f(s)dg/dt)dt which can be verified directly by taking partial derivatives.

Now let c(t) map the reals into the plane by c(x) =(x,x)

The Chain rue says that

dFoc(∂/∂x) = dF(dc(∂/∂x)

= dF(∂/∂s + ∂/∂t)

= (( ∂F/∂s)ds+ (∂F/∂t)dt)(∂/∂s + ∂/∂t))

= ( ∂F/∂s)ds(∂/∂s) + (∂F/∂t)dt(∂/∂t) ( since ds(∂/∂t) = dt(∂/∂s) = 0 )

= g(x)f'(x) + f(x)g'(x)

So the Leibniz Rule is an application of the Chain Rule for functions on a field.[/QUOTE]

 

[orion]

I'm getting confused trying to figure out all the maps again so please bear with me.

Let $M,N$ be manifolds and $F: M \rightarrow N$
Then if we have a curve $\gamma$, we have $\gamma : I \subset \mathbb R \rightarrow M$.
$F \circ \gamma$ is the map $I \rightarrow N$ which is the image of $\gamma$ under $F$.

We also have that $dF_p(v)(\phi)$ is a map $T_pM \rightarrow T_{F(p)}N$ where $\phi \in C^\infty (N)$. Correct?
If we have $v \in T_pM$ then $v$ is a map from $f \in C^\infty(M) \rightarrow \mathbb R$ for functions $f$ defined on $M$. Is this correct?

What is the map $N \rightarrow T_{F(p)}N$?

For that matter, what is the map $M \rightarrow T_pM$?

 

[mathwonk]

"In the definition where dFp(v)(f)=v(f∘F), dFp represents a linear approximation to F near p?"

yes, that's right. dFp takes vectors tangent to M to vectors tangent to N. In your notation where vectors are differential operators, it takes differential operators v( _ ) on functions out of M, over to differential operators on functions out of N, namely to v( _ oF).

 

[mathwonk]

In writing math we write what is easiest for us, the author, not for you the reader. tangent vectors are really infinitesimal curves not differential operators, but they can be taken as diff ops for the same reason that the double dual V** of a vector space V is isomorphic to that (finite diml) space V. I.e. saying a tangent vector is a diff op is like saying a vector is a linear operator on linear operators on vectors. Its true but it isn’t very intuitive or easy to keep straight.

So most of your questions are related to the confusion this causes. E.g. there are not too many natural operations in the world, just maps in and out of your space. I.e. as noted earlier, there are curves, or maps from R into your space, called maybe Hom(R, . ), and functions, or maps out of your space to R, or Hom( . ,R). Now the first one is naturally sense preserving and the second one is naturally sense reversing.

I.e. given a curve g:R-->M in M, and a map F:M-->N, we get a curve in N, by composing with F from the left: g goes to F_*(g) = Fog:R-->N. Thus if F:M-->N, then F_* maps Hom(R,M) to Hom(R,N). I.e. F_* (or dF) goes in the same direction as F, i.e. from M to N.

Dually, if we have a function f:N-->R, we get from composing on the right by F, a function F*(f) = foF:M-->R. So if F goes from M to N, then F* goes from Hom(N,R) to Hom(M,R), i.e. it goes backwards, from N to M.

This is preserved by taking infinitesimal pieces, so vectors get pushed forwards, and covectors get pulled back. i.e. if F:M-->N then we get dF:T(M)-->T(N), and F*:T*(N)-->T*(M).

But when we use differential operators things are more confusing. I.e. if we reverse directions twice we get back going in the same direction! So a differential operator is a function on functions, so sort of an element of Hom(Hom( . ,R), R).

Thus if v is a differential operator on functions on M, and F:M-->N maps M to N. then dF(v) is a differential operator on functions on N, taking f in Hom(N,R) to v(foF), i.e. dF(v) = v(foF) = (voF*)(f) = (F*)*(v)(f)! I.e. here dF is defined as (F*)*. I.e. to get something sense preserving we have to do something sense reversing twice!

This is why you had that question about the confusing formula:

dF(v)(f) = v(foF).

So instead of the simple minded principle that to push forward vectors you compose with F on the left and to pull back covectors you compose with F on the right, life is more confusing. Still you do whatever makes sense, and only one thing usually does.By the way, your question about how to relate the behavior of f:M--R thought of as a function, to f thought of as a map of manifolds, also should be asked for g:R-->M thought of as a curve, to g thought of as a map of manifolds. In all cases Lavinia’s advice that the tangent space to R is just R, i.e. a tangent vector to R is just a number, is the way out.

 

[Fredrik]

Then if we have a curve $\gamma$, we have $\gamma : I \subset \mathbb R \rightarrow M$.
$F \circ \gamma$ is the map $I \rightarrow N$ which is the image of $\gamma$ under $F$.

OK, except for a minor terminology issue. I would just call $F\circ\gamma$ the composition of $F$ and $\gamma$. The "image of" terminology is used for subsets of the domain.

We also have that $dF_p(v)(\phi)$ is a map $T_pM \rightarrow T_{F(p)}N$ where $\phi \in C^\infty (N)$. Correct?

We have $dF_p:T_pM\to T_{F(p)}N$. That makes $dF_p(v)$ an element of $T_{F(p)}N$. That makes $dF_p(v)(\phi)$ a real number.

If we have $v \in T_pM$ then $v$ is a map from $f \in C^\infty(M) \rightarrow \mathbb R$ for functions $f$ defined on $M$. Is this correct?

Yes, if you meant what I think. $v$ is a map from $C^\infty(M)$ into $\mathbb R$. So if $f\in C^\infty(M)$ then $v(f)$ is a real number.

What is the map $N \rightarrow T_{F(p)}N$?

For that matter, what is the map $M \rightarrow T_pM$?

I don't think we have discussed maps like that. If M is equipped with a metric, then the "exponential map" at p takes an arbitrary $v\in T_pM$ to $\gamma(1)\in M$, where $\gamma$ is the unique geodesic through p whose tangent vector at p is v. I doubt that the exponential map is always invertible, but I would guess that if it isn't, it probably has a restriction that's invertible.

 

[mathwonk]

Consider the unit sphere M in 3 space. Then certainly the exponential map from the tangent plane at the north pole cannot be invertible as a map to the sphere. However as Fredrik says, the restriction to an open disc in that plane of radius π is a diffeomorphism onto the sphere minus the south pole. If we remove the south pole from the sphere then the exponential map is not even defined off this disc.

 

[lavinia]

For that matter, what is the map $M \rightarrow T_pM$?

Maps from the manifold into the tangent bundle are called vector fields. To each point on the manifold the map associates a tangent vector.

A good exercise is to prove that the space of all tangent vectors to a manifold is itself a manifold. So a vector field is a differentiable map between manifolds.

 

[mathwonk]

be aware Lavinia is talking about something slightly different from a map M-->Tp(M). She is talking about maps M-->T(M), without the fixed subscript p.

I.e. a vector field on M is a map v:M-->T(M) such that for each p in M, v(p) belongs to Tp(M). So the p is varying in a vector field.

 

[orion]

Thanks, everyone. That clears up a lot of issues.

I actually did come across the exponential map in one set of notes I found on the internet, but I put it aside until I get the basics down.

lavinia, I neglected to thank you for your post earlier. Thanks.

 

[lavinia]

I don't think we have discussed maps like that. If M is equipped with a metric, then the "exponential map" at p takes an arbitrary $v\in T_pM$ to $\gamma(1)\in M$, where $\gamma$ is the unique geodesic through p whose tangent vector at p is v. I doubt that the exponential map is always invertible, but I would guess that if it isn't, it probably has a restriction that's invertible.

The exponential map is always locally invertible. Its differential is non-singular at the origin of the tangent space being exponentiated.

The example of the sphere that Mathwonk described shows that the exponential map can have singularities and in general it does. The study of its singularities - so called conjugate points - is the subject of Morse Theory.

In some cases, the exponential map can have no singularities but still not be invertible. This is true for instance of compact flat Riemannian manifolds since on them, the Riemann curvature tensor is identically zero so there are no conjugate points.For instance, for the flat torus, the exponential map is a covering projection.