Defining a differentiable structure by means of functions.

200 Views Asked by At

I am trying to understand the construction of principal bundles from Kobayashi and Nomizu, and the situation is the following.

Let $M$ be a manifold, $\{ U_\alpha \}_{\alpha \in A}$ an open covering of $M$ and $G$ a Lie group. Define $X_\alpha = U_\alpha \times G$ and $X = \bigcup_{\alpha \in A} X_\alpha$ the disjoint union of the $X_\alpha$. A certain equivalence relation is introduced on $X$ and the resulting quotient space is called $P$, with projection $q:X \to P$ (at this moment, $P$ is just a set). We also have a surjective map $\pi : P \to M$ such that $q_\alpha = q |X_{\alpha} : X_\alpha \to \pi^{-1}(U_\alpha)$ is a bijection for every $\alpha \in A$.

The authors want to give $P$ a differentiable structure, and for this purpose, they write:

"We introduce a differentiable structure in $P$ by requiring that $\pi^{-1}(U_\alpha)$ is an open submanifold of $P$ and that the mapping $q$ induces a diffeomorphism of $X_\alpha$ onto $\pi^{-1}(U_\alpha)$, for every $\alpha \in A$".

I understood the topology of $P$ is the quotient topology induced by $q$. If we postulate a function $f : P \to \mathbb{R}^n$ to be differentiable if and only if $f \circ q_\alpha : U_\alpha \times G \to \mathbb{R}^n$ is differentiable, is that enough to define a differentiable structure in $P$? In other words, just telling what functions are differentiable determines a (the) differentiable structure?

2

There are 2 best solutions below

2
On

I emphisise that $X$ is a differentiable manifold of dimension $n$; and I recall the equivalence relation on $X$: $$ \forall\alpha,\beta\in A,(x,g)\in X_{\alpha},(y,h)\in X_{\beta},\,(x,g)\sim(y,h)\iff x=y\in U_{\alpha}\cap U_{\beta},\psi_{\beta\alpha}(x)g=h $$ where $\psi_{\beta\alpha}:U_{\beta}\cap U_{\alpha}=U_{\beta\alpha}\to G$ are assigned differentiable map that satisfy the cocycle condition. We get: $$ P=X_{\displaystyle/\sim}; $$ there is a natural surjective function: $$ \pi:[(x,g)]_{\sim}\in P\to x\in M, $$ and trivially we consider on $P$ the quotient topology.

In this way, there is a bijection between $X_{\alpha}$ and $\pi^{-1}(U_{\alpha})$ for any $\alpha\in A$; I call this bijection $f_{\alpha}$, and one can prove that $f_{\alpha}$ is a homeomorphism.

By hypothesis $\pi^{-1}(U_{\alpha})$ is locally homeomorphic to $\mathbb{R}^n$ for any $\alpha\in A$; without loss of generality, let $\{(X_{\alpha},g_{\alpha})\}_{\alpha\in A}$ be an atlas of $X$, by hypothesis: $$ \forall\alpha,\beta\in A,\,g_{\beta\displaystyle|X_{\beta\alpha}}\circ g_{\alpha\displaystyle|X_{\alpha\beta}}^{-1}:\mathbb{R}^n\to\mathbb{R}^n $$ are $C^{\infty}$-maps, in particular: $$ g_{\beta\displaystyle|X_{\beta\alpha}}\circ g_{\alpha\displaystyle|X_{\alpha\beta}}^{-1}=g_{\beta\displaystyle|X_{\beta\alpha}}\circ f^{-1}_{\beta\displaystyle|X_{\beta\alpha}}\circ f_{\alpha\displaystyle|X_{\alpha\beta}}\circ g_{\alpha\displaystyle|X_{\alpha\beta}}^{-1} $$ are $C^{\infty}$-maps and therefore $P$ is a smooth manifolds with atlas $\{(\pi^{-1}(U_{\alpha}),g_{\alpha}\circ f^{-1}_{\alpha})\}_{\alpha\in A}$; joint the check that $P$ satisfies the Hausdorff axiom of separation and is basis countable!

After all this, the answer to your first question is yes, but because in this way there is a manifold structure on $P$!

About your second question: I'll take time.

0
On

About your second question:

In other words, just telling what functions are differentiable determines a (the) differentiable structure?

the answer is: depend! X-D

Example: the graphic of the absolute value $|\cdot|$, with the topology inherited from the natural topology of $\mathbb{R}^2$, is not a smooth manifold; but if we induce a smooth structure on this graphic via a homeomorphism with $\mathbb{R}$, it is a smooth curve.

For clarity, if $$ M=\{(x,|x|)\in\mathbb{R}^2\mid x\in\mathbb{R}\},\\ i:M\hookrightarrow\mathbb{R}^2 $$ and $\mathcal{C}^{\infty}_{\mathbb{R}^2}$ is the sheaf of smooth maps on $\mathbb{R}^2$, we get: $$ \forall P\in M,\,\mathcal{C}^{\infty}_{\mathbb{R}^2,P}\simeq i^{-1}\left(\mathcal{C}^{\infty}_{\mathbb{R^2}}\right)_P=\mathcal{C}^{\infty}_{M,P} $$ where the elements of $\mathcal{C}^{\infty}_{\mathbb{R}^2,P}$ are the germs of the smooth maps on $\mathbb{R}^2$ in $P$. In this way, we define the smooth maps on $M$ even if $M$ is not a smooth submanifold of $\mathbb{R}^2$.

Is it all clear?