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My book is Connections, Curvature, and Characteristic Classes by Loring W. Tu (I'll call this Volume 3), a sequel to both Differential Forms in Algebraic Topology by Loring W. Tu and Raoul Bott (Volume 2) and An Introduction to Manifolds by Loring W. Tu (Volume 1).

I refer to Section 27.2 and a comment on one of my (stupid) questions.

The comment by studiosus is:

You can even endow X with a (unique) smooth structure that makes the map f:X→N a diffeomorphism. First pull back the topology of N: declare open subsets of X to be exactly the preimages under f of open subsets of N. Then pull back the smooth structure: if $\{(U_{i},\phi_{i}\}$ is the smooth structure on N, then $\{(f^{-1}(U_{i}),\phi_{i}\circ f)\}$ is a smooth structure on X, and f:X→N becomes a diffeomorphism.

Note: My question is about the frame manifold of a vector space specifically (and not the more general case of a frame bundle of a vector bundle).

Question 1: The text says

Using the bijection $\varphi_V$, we put a manifold structure on $Fr(V)$ in such a way that $\varphi_V$ becomes a diffeomorphism. $\tag{the sentence}$

  • What exactly does the sentence mean, given my understanding as follows?

  • My initial understanding: I initially understood the sentence as a proposition that goes something like "Later on we're going to define a smooth manifold structure on $Fr(V)$. Then one can show that, under the definition given later on, $\varphi_V$ becomes a diffeomorphism."

  • My understanding now: The sentence is actually a definition in the context of some kind of rule: For any bijection $f: A \to B$ of sets $A$ and $B$, we have that if $B$ somehow is given a topology and then given a smooth manifold structure, then we can give $A$ a smooth manifold structure, possibly unique, that makes $f$ into a diffeomorphism. This structure is the one given in the above comment by studiosus (the comment claims uniqueness).

Question 2: If my understanding is correct, then what exactly is the rule? Also, please provide a textbook that gives this rule. I am fairly certain Volumes 1 and 3 don't have any such rule.

  • Guess: I think the rule could be related to how there exists a unique structure such that any smooth manifold subset $X$ becomes a smooth regular/embedded submanifold of a smooth manifold $N$. See link 1, link 2, link 3, link 4, but in those cases we at least have $X$ a subset of $N$, have $X$ with the subspace topology and have that $X$ can become a smooth manifold in the first place.

Question 3: Is there some rule that for any bijection $f: A \to B$ of sets $A$ and $B$, if $B$ can become a group, then there exists a group structure, possibly unique, on $A$ that makes $f$ an isomorphism?

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This is a general idea that if you have a bijection $f:X\to Y$ and one of the sets has some additional structure (a group, ring, topological space, manifold or pretty much anything) then this structure can be "replicated" onto the other set in such a way that $f$ becomes an iso/homeo/diffeo/etc-morphism.

Unfortunately this is a case-by-case work, but let's have a look at some simple examples:

  1. $Y$ is a group with multiplication "$\circ$". Then you can define multiplication "$\cdot$" on $X$ via $x_1\cdot x_2:=f^{-1}\big(f(x_1)\circ f(x_2)\big)$. It turns out that $X$ with this multiplication is a group and $f$ is an isomorphism.
  2. $X$ is a group with multiplication "$\cdot$". So it's the other way around now. Analogously you define multiplication on $Y$ via $y_1\circ y_2:=f\big(f^{-1}(y_1)\cdot f^{-1}(y_2)\big)$.
  3. $Y$ is a topological space. Then we can define topology on $X$ via: $U\subseteq X$ is open if and only if $f(U)$ is. This makes $f$ a homeomorphism.
  4. $X$ is a topological space. Analogously we define $V\subseteq Y$ is open if and only if $f^{-1}(V)$ is. This again makes $f$ a homeomorphism.
  5. $Y$ is a manifold. By 3. we know how $X$ becomes a topological space. And the quoted fragment is a recipe for a manifold structure on $X$. You take a chart on $Y$ and you compose it with $f$ to obtain a chart on $X$. This turns $f$ into a diffeomorphism.
  6. What about for example a vector bundle? Well, this is more complicated, because vector bundle consists of two spaces $X,E$ and a continuous surjection $\pi:E\to X$ between them (with certain properties). So if $f:X\to Y$ is some bijection then we can turn $Y,E$ into a vector bundle by pulling topological structure onto $Y$. The continuous surjection is then $f\circ\pi$. And $f$ together with the identity becomes a vector bundle isomorphism.
  7. What if we again have a vector bundle $\pi:E\to X$ but this time $g:E\to E'$ bijection? Or two bijections: additional $f:X\to Y$? Can you generalize previous points and turn $E',Y$ into an isomorphic vector bundle?

and so on, and so on.

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  • $\begingroup$ Thanks freakish! $\endgroup$
    – user636532
    Oct 19 '19 at 2:15
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    $\begingroup$ (Really minor points: In example 1 and 2, you need an additional use of $f$ of $f^{-1}$. E.g., for 1, it should be $x_1 \cdot x_2 = f^{-1}(f(x_1)\circ f(x_2))$.) $\endgroup$ Oct 19 '19 at 2:19
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    $\begingroup$ @JasonDeVito of course, thanks for noticing. $\endgroup$
    – freakish
    Oct 19 '19 at 6:40

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