Let $M$ and $F$ be differentiable manifolds, and let $F\to E\to M$ be a differentiable fibre bundle over $M$. A trivialising cover $\{(U_i,\phi_i)\,|\,i\in I\}$ of $M$ determines a set $\{t_{ij}:U_{ij}\to\mathrm{Diff}(F)\,|\,i,h\in I\}$ of transition functions satisfying the Čech cocycle conditions. In [Nakahra, Geometry, Topology and Physics, 2003], it is assumed that these transition functions actually take values in some Lie group $G$, and it is simply mentioned that the transition functions are smooth maps. I was pondering:

  • Why are they actually smooth?
  • If one considers the general case that the transition functions take values in $\mathrm{Diff}(F)$, are these functions still smooth? Is this even well-defined, i.e. is $\mathrm{Diff}(F)$ always a manifold?

Any help is much appreciated.


1 Answer 1


You have to be careful when $F$ is not compact, and I don't want to be careful today, so let $F$ be a smooth compact manifold without boundary. Then $\text{Diff}(F)$ is a Frechet manifold: it's locally modeled not on Euclidean spaces, but on topological vector spaces called Frechet spaces. Here the Frechet space that's the tangent space at the identity is precisely $\mathfrak X(F)$, the space of smooth vector fields on $F$.

The point of the topology on this space is that a continuous map $M \to \text{Diff}(F)$ is canonically the same as a continuous family of diffeomorphisms $M \times F \to F$, where each $f_m$ is a diffeomorphism. So the point of the smooth structure is that a smooth map $M \to \text{Diff}(M)$ gives me a smooth family if diffeomorphisms $M \times F \to F$, and vice versa.

Now, when you pick trivializations, they are always meant to be smooth trivializations (as opposed to fiberwise smooth trivializations); and therefore the transition maps $U \times F \to U \times F$ are smooth; composing with the projection map gives us a smooth family of diffeomorphisms $U \times F \to F$. So the cocycle this gives, $U \to \text{Diff}(F)$, is smooth. So indeed the transition maps are smooth. If you only demanded the trivializations be smooth on each fiber, $U \times F \to F$ would only be a continuous family of diffeomorphisms, and $U \to \text{Diff}(F)$ would only be continuous.

The theory of Frechet manifolds is well-developed (see Kriegl-Michor, a convenient setting for analysis), but having not read that book, I tend to try to use it as philosophy (like the above).

  • $\begingroup$ Point taken, thanks for this answer; it clarifies that this touches on stuff I have not seen before. However, in chapter 7 of this book by Michor (mat.univie.ac.at/~michor/gaubook.pdf, via mat.univie.ac.at/~michor/listpubl.html#books), it says that for a smooth manifold $M$ the group $\mathrm{Diff}(F)$ is a submanifold of $C^\infty(M,M)$. How can I see that with this answer? $\endgroup$ Jun 20, 2016 at 20:37
  • $\begingroup$ @B.Pasternak I never told you what the smooth structure was, so you shouldn't see it here! $\text{Diff}(M)$ is an open subset of $C^\infty(M,M)$, so automatically inherits the latter's smooth structure. If it helps, I can tell you what the tangent space $T_f C^\infty(M,N)$ is; it's the space of sections of the pullback bundle $f^*TN$. When $N = M$ and $f = \text{id}$, this is exactly the same as saying it's sections of $TM$, aka vector fields. $\endgroup$
    – user98602
    Jun 20, 2016 at 20:46

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