Deane

Agreed. Why did you say this was disingenuous?

That’s exactly the point of affine space. In Euclidean geometry, you never add two points. It makes no sense geometrically. A vector $v$ is simply a line segment with a direction where you designate which endpoint is the start and which is the end. In Euclidean geometry, given a point $p$, you can put the start of the line segment at $p$ and define $p+v$ to be the end of this line segment. All of this is done without any reference to an origin. And points are definitely not line segments and therefore not vectors.

@StevenThomasHatton, what you should do is to show that your definition of affine space satisfies all of the properties in Moretti's definition.

A natural model of $n$-dimensional affine space is an $n$-dimensional hyperplane $A \subset V$, where $V$ is an $(n+1)$-dimensional vector space and $A$ does not contain the origin.

I like to say that an affine space $A$ is like a vector space but you "forget" where the origin is. Affine space is the natural model of a flat geometric space. In such a space, every point looks like any other. There is no special point called the origin. But associated to $A$ is a vector space $V$, where if you do designate a point $p \in A$ to be the "origin", then there is a natural bijection $A \rightarrow V$, where $V$ is the set of position vectors relative to $p$.

You say that affine space is a set of poiints and that $V$ is a vcector space, but you don't say whether $V$ is the affine space of points. Then you say that given a vector space $V$, we designate $\vec{O}$ to be an origin. But by the definition of a vector space, $V$ already has an origin, so you cannot designate another vector (point?) to be an origin..

Here's another way to express this: The concepts and theorems of geometry and physics should not depend on the coordinates or their infinitesimal analogues (namely a basis of the tangent space). If you choose a pointwise isomorphism between the tangent and cotangent space, then it does induce isomorphisms between exterior multivectors and differential forms. But in the end, whatever you do, you'll have to prove that your final result is independent of this isomorphism.

What turns out to be particularly important is to distinguish which aspects of differential geometry do not require a Riemannian metric and which do, all while keeping track of how things depend on coordinates. Physicists in fact understand this. They know that the cross product of two vector fields is not an honest vector field. So they call it a pseudovector field (en.wikipedia.org/wiki/Pseudovector).

I'm going disagree with @TedShifrin. It is possible to do multilinear algebra and differential geometry without explicitly using the concept of a dual vector space or differential forms or the exterior derivative. Many physicists still do. It's not so bad in dimensions 2 and 3, because you can use the cross product and curl. But the calculations are messy and hard to understand This is particularly true if there is no Riemannian metric. Differential geometers didn't adopt this fancy stuff for nothing. It both simplifies calculations and clarifies the geometry.

It’s not wrong to be skeptical. My suggestion as you learn everything carefully, is to try to reformulate everything without using forms. Sometimes you’ll like your version better. But if not, this is a good way to really appreciate the power of using the dual vector space and differential forms

Could you provide a formula for an example of $f$?

Here's another issue: I don't really know what $F$ is. Could you provide an example of an $F$ that appears in your work or whatever you're reading?

I gotta say that if I were to do this calculation, I would follow the advice in my first comment and substitute the special form of the metric only at the end. That avoids all the subtleties you are running into.

@J.G. is correct.

A possible way out of this is to assume the frame is time dependent

Well, since the metric is time dependent there has to be a time derivative of the metric in one of the equations, depending on which one you start with. Your calculation demonstrates this.

Sorry. I don’t think both both equation 1 and the next equation can both be correct, as your calculation shows.

If you use an orthogonal frame, then indices can be raised and lowered freely. Normally if you want to lower an index, you have to multiply by the metric but since the metric is identity matrix, it disappears. This why I often do my calculations in an arbitrary frame. For me this can be less confusing than using an orthogonal one.

Without using any inner product.

There's almost never a need to take wedge products of tangent vectors, especially if you want basis-independent (i.e., coordinate independent) definitions and theorems.

And could you remind me what $\Psi$ is? The definition of $\bigwedge^2 V$ is forced on you once you decide that $\bigwedge^2V^*$ is defined as a subspace of $V^*\otimes V^*$ and want $\bigwedge^2V^* = (\bigwedge^2V)^*$. The ad hoc assumption is that you want this duality. In most areas of differential geometry, we get around this by never using $\Lambda^2V$ at all. We simply assume that differential forms are multilinear functions of the tangent space and work with them that way.

Could we stick to characteristic 0?

So, despite the natural isomorphism $(V^*)^*=V$, in many settings (such as differential geometry), the vector spaces $V$ and $V^*$ do not play symmetric roles. You start with a vector space $V$ that is fundamental (e.g., tangent vectors on a manifold) and then view $V^*$ as being defined as linear functions on $V$ (e.g., cotangent vectors). So it is OK to have $\bigwedge^kV^*$ be a subspace of $(V^*)^{\otimes k}$ but $\bigwedge^kV$ be a quotient space of $V^{\otimes k}$.

This stems from the fact that if $S$ is a subspace of a vector space $X$, $S^*$ is not a subspace of $X^*$.

The asymmetry is the crucial point. In any application of exterior algebras, where you do not want to assume there is an inner product (or nondegenerate quadratic form), it is not possible to define in a basis-independent way $\bigwedge^2V$ as a subspace of $V\otimes V$ and $\bigwedge^2V^*$ as a subspace of $V^*\otimes V^*$ and have $\bigwedge^2V$ and $\bigwedge^2V^*$ be dual to each other (i.e., have a basis-independent pairing).

Ok. But latex isn’t rendered here?

I don't understand your last comment. Maybe there's a typo? If $v, w \in V$, then $v\wedge w \in (\Lambda^2V^*)^*$ is defined as follows: For any $\Theta \in \Lambda^2V^*$, $$(v\wedge w)(\Theta) = \Theta(v,w). $$ This is equivalent to the definition I give in my answer.

@NicholasTodoroff, as I say in the last sentence, these can all be defined as functors between the appropriate categories and therefore satisfy the relevant universal properties. So they are not ad hoc. I chose not to define everything in terms of universal properties, because it's longer and harder to follow. Instead, I provided explicit definitions. The universal properties can be verified easily, for example, using bases for all of the vector spaces.

@NicholasTodoroff, since I define $\Lambda^2V = (\Lambda^2V^*)^*$, the pairing is there by definition.

@NicholasTodoroff, where or what is $\Psi$? Here's a short summary of how I now see things: 1) Define $V^*\otimes V^*$ as space of bilinear functions on $V$ 2) Define $\Lambda^2V^*\subset V^*\otimes V^*$ as subspace of antisymmetric bilinear functions. 3) There is a canonical isomorphism $V^*\otimes V^*=(V\otimes V)^*$. 4) The natural dual of $\Lambda^2V^*$ is $$(\Lambda^2V^*)^* = (V^*\otimes V^*)^*/(\Lambda^2V^*)^\perp=(V\otimes V)/(\Lambda^2V^*)^\perp.$$ Define $S^2V= (\Lambda^2V^*)^\perp$ and $$\Lambda^2V=(\Lambda^2V^*)^* = (V\otimes V)/S^2V.$$

@LSpice, apologies! I I hope it's fixed now.

@LSpice, I added the missing text. I think you have a point with the $\frac{1}{2}$.This is supposed to all work for a (free?) nodule over $\mathbb{Z}$, right? I gotta work out the details on that.

@KritikerderElche, I was referring to the version cited in the first paragraph of the question. It’s indeed not the standard version of stereographic projection.

@PaulFrost, indeed. You can choose the parameterization of the vertical angle to be any monotone map from $[0,1]$ to $[0,\pi]$ to define the map to $U$. That then uniquely determines the map to $U'$.

@PaulFrost, I agree that quaternionic stereographic projection is not radial scaling (even though standard stereographic projection is). But OP is asking about the transition function given in the last line of the second paragraph. That is radial scaling. No quaternionic multiplication or division is used.

According to my calculations, if \begin{align*} \left(\frac{z}{|z|}\sin\pi|z|, \cos\pi|z|\right) &= \left(\frac{z'}{|z'|}\sin\pi(1-|z'|), \cos\pi(1-|z'|)\right), \end{align*} then $|z'|=1-|z|$ and therefore \begin{align*} z' &= \frac{1-|z|}{|z|}z \end{align*}

@PaulFrost, perhaps I'm misreading the question. Is the desired transition function not $$z' = \frac{1-|z|}{|z|}z?$$ That's radial scaling.

\begin{align*} z' &\mapsto \left(\frac{z'}{|z'|}\sin\pi(1-|z'|),\cos\pi(1-|z'|)\right)\‌​end{align*}

Note that unlike stereographic projection, the multiplicative structure of the quaternions is not used at all. As @coiso says, this looks like spherical coordinates to me. Something like \begin{align*} B &\rightarrow U\\ z &\mapsto \left(\frac{z}{|z|}\sin\pi|z|, \cos \pi|z|\right) \end{align*}

Here's are some basic facts you can prove yourself: If a function $g$ is differentiable, then its distributional derivative is equal to its derivative. If $f = g$ a.e., then the distributional derivatives of $f$ and $g$ are equal.

Two things: 1) I agree that the distributional (weak) derivative is at first a strange thing. 2) It's great to think about examples like the one you posted. But always try to compute the answer yourself using just the definitions used and, if necessary, some basic facts you already know. It's often easier than you might think at first.

Well, you seemed to have provided a pretty ironclad proof of that.

Good!. Now remember that $\phi$ is compactly supported and therefore so is $\phi'$. So it suffices to integrate over a compact interval that contains the support of $\phi'$. Then use the fundamental theorem of calculus.

By the way, the fact you're using (but maybe initially forgot) is that if $f = g$ a.e., then $\int f = \int g$.

Yup. Now go back to the definition of the distributional derivative of the function $f$. See if you can figure out what the answer is. If not, just write out the formula and ask again.

I mean, simplify it.

Yes! Now, could you revise your answer to the second integral?

How about the first integral?

Well, could you at least tell me what the answers to my latest questions are, and then I can help guide you further.

You could start with some simpler examples. For example, suppose $$u(x) = \begin{cases} 1 &\text{ if }0 \le x \le 1\\ 0 &\text{ otherwise} \end{cases}. $$ What is $$\int u(x)f(x)\,dx?$$ Next, let $u$ be any compactly supported smooth function. What is $$\int u(x)f(x)\,dx?$$