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Let $M$ be a Riemannian manifold or a Lie group whose corresponding exp map (in corresponding context) is denoted by "exp" which is a map $\exp:TM\to M$

We search for the set $\mathcal{H}$ of all smooth maps $f:M\to M$ which satisfy $$f\circ \exp=\exp \circ Df$$

Is there a natural finite dimensional manifold structure on $\mathcal{H}$? In the case that $M$ is a Lie group, is there a Lie group structure on $\mathcal{H}$? In either case what can be said about the dimension of $\mathcal{H}$?

Example: For Lie group $M=\mathbb{R}$ we get $\mathcal{H}=\{ax+b\mid a,b \in \mathbb{R}\}$ the additive group of all affine linear maps on $\mathbb{R}$, a $2$ dimensional Abelian lie group isomorphic to $\mathbb{R}^2$.

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    $\begingroup$ Yes, it's a Lie group -- the defining equation you mention, that the maps commute with exponentiation this tells you your maps are determined by their behaviour on a basis for one tangent space. So the Lie group is at most $m + m^2$ dimensional. Presumably this is only realized for $\mathbb R^m$. I imagine for a compact Riemann manifold this is the isometry group. $\endgroup$
    – Ryan Budney
    Commented Feb 19, 2020 at 18:15
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    $\begingroup$ @RyanBudney For a Riemannian manifold every constant map satisfies this equation so the last part of your comment is not true. Let me think about the first part of your comment. Please read my question again in particular the example in the last part of my post. $\endgroup$
    – Ali Taghavi
    Commented Feb 20, 2020 at 0:19
  • $\begingroup$ @RyanBudney note that a possible group structure on $\mathcal{H}$ is not probably defined via composition but it possibly comes from the group structure of $G$, as it is the case in the example of the post. $\endgroup$
    – Ali Taghavi
    Commented Feb 20, 2020 at 0:26
  • $\begingroup$ @RyanBudney appart from constant map you can find other maps for example $z\mapsto z^2$ on circle, right? $\endgroup$
    – Ali Taghavi
    Commented Feb 20, 2020 at 0:40
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    $\begingroup$ My Lie group comment refers to the maps where Df is invertible. I suppose I imagined that was an assumption you were making. It's certainly a manifold by the frame bundle argument. When the derivative is invertible and the manifold has no self-covering then it's a Lie group. If you allow degenerate maps then I suppose the question is more open-ended. But I don't see a reason for there to be a group structure when allowing degenerate maps, unless you are in some very special case. $\endgroup$
    – Ryan Budney
    Commented Feb 20, 2020 at 4:40

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Let us begin with the “Riemannian” case. In truth, it doesn’t have much to do with the Riemannian structure at all: $\mathcal{H}$ is the space of all affine maps on $(M,\nabla)$, with the affine connection $\nabla$ given by the Levi-Civita connection.

As you note, this will never have a natural Lie group structure unless $M$ is 0-dimensional, since it will always contain the constant maps. If we make the additional restriction that $f$ be a diffeomorphism, however, we get the group of affine transformations, which has been extensively studied and is well-known to be a Lie group when $M$ has finitely many connected components. For compact Riemannian manifolds, the identity component of the group of affine transformations coincides with the identity component of the isometry group.

Regardless of whether we use diffeomorphisms, it is a standard exercise in differential geometry to prove that affine maps are determined entirely by their value and derivative at a single point, so the dimension of $\mathcal{H}$ will be at most $\dim(M)+\dim(M)^2$.

It isn’t a very cohesive space most of the time, though: it will always contain both a copy of $M$ (the constant maps) and the identity map, and a path between a constant map and the identity would constitute a homotopy, which can only occur for $M$ contractible. Thus, in the generic case, it’ll probably just be $M$ together with an isolated point corresponding to the identity map.

In the Lie groups case, it’s essentially the same. Note that if $f$ satisfies the condition, then so does $\mathrm{L}_g\circ f$. Thus, $f$ is of the form $\mathrm{L}_g\circ f_0$, where $f_0$ fixes a point, which we will predictably choose to be $e$. I think it follows that the space you end up with will be something like $G\times\mathrm{Hom}(G,G)$. Again, this won’t naturally be a Lie group outside of the discrete case, but if we restrict to diffeomorphisms, I think we get the Lie group $G\rtimes\mathrm{Aut}(G)$.

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  • $\begingroup$ Thank you for your answer. Your first paragraph:" $\mathcal{H}$ is the group of affine maps" as you noted in your answer it is not realy the case because of constant maps and also $z\mapsto z^k$ on circle. So O wonder if there are some other kind of pathological maps. Any way the example in the post shows that the identity is not isolated. $\endgroup$
    – Ali Taghavi
    Commented Feb 25, 2020 at 17:01
  • $\begingroup$ What is this dimension for circle torus or spher? BTW what is your definition of affine maps when the map is not at least local diffeomorphism? $\endgroup$
    – Ali Taghavi
    Commented Feb 25, 2020 at 17:13
  • $\begingroup$ For having a Lie structure, the case 0 dimensional is not the only relevant case. Please look at the example $\endgroup$
    – Ali Taghavi
    Commented Feb 25, 2020 at 17:15
  • $\begingroup$ @AliTaghavi I’m not sure what you’re referring to in my first paragraph. Perhaps you misread? Also, I don’t think there is much chance of pathological examples; as far as smooth maps go, affine maps are exceptionally well-behaved. Finally, the example in your post is rather specific in that you have chosen a contractible Lie group, so in that specific case you can find a homotopy from the identity to a constant map. As I said, this will almost never happen. Try, for example, any Lie group that isn’t diffeomorphic to $\mathbb{R}^n$. $\endgroup$
    – Robin Goodfellow
    Commented Feb 25, 2020 at 17:21
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    $\begingroup$ For affine maps, you can probably find what you want to know in Chapter VI (“Transformations”) of Kobayashi and Nomizu’s Foundations of Differential Geometry, Vol 1. For contractibility, you can probably find that in any elementary topology textbook. For the Lie groups version, note that if $f(e)=e$, then $f(\exp(X))=\exp(f_*X)$, so you should probably be able to say $f(g)f(\exp(X))=f(g)\exp(f_*X)=f(g\exp(X))$. $\endgroup$
    – Robin Goodfellow
    Commented Feb 27, 2020 at 21:20

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