# Manifolds of maps

Suppose $M$ is a compact $n$-dimensional manifold without boundary. Let $H^s(M,M)$ denote the Sobolev space on $M$, defined as all maps from $M$ to $M$ whose distributional derivatives up to order $s$ are square integrable. For $s>n/2+1$, let $$\mathcal{D}^s =\{\eta\in H^s(M, M)| \eta ~\text{is bijective and}~ \eta^{-1}\in H^s(M,M)\}.$$ Define right multiplication \begin{align} R_\eta:\mathcal{D}^s \to\mathcal{D}^s \\ \xi\to\xi\circ\eta. \end{align} Why is $R_\eta ~C^{\infty}$ for each $\eta \in \mathcal{D}^s$?

Also, if define left multiplication \begin{align} L_\eta :\mathcal{D}^s \to\mathcal{D}^s \\ \xi \to \eta \circ\xi. \end{align} Why is $L_\eta$ only $C^l$ when $\eta\in \mathcal{D}^{s+l}$?

Essentially what is the difference between $R_\eta ~\text{and} ~L_\eta$?

• If you consider the same operator for Sobolev maps defined on a vector space, right multiplication is linear and thus smooth, whereas left multiplication is not. This might be part of the explanation. – Thomas Richard Sep 27 '17 at 4:04
• Thanks! The conclusion makes sense, since the right multiplication is linear which is naturally smooth, while the regularity of left multiplication depends on the regularity of $\eta$. – Cohen Lu Sep 28 '17 at 1:17

For a closed manifold the space $$H^s (M,N)$$ ($$s$$ bigh enough) carries a Hilbert manifold structure via the usual manifolds of mappings construction (where charts are constructed using a suitable local addition, aka. the Riemannian exponential map, cf. Kriegl, Michor: The convenient setting of global analysis for an explanation of the construction for the manifold of smooth mappings (instead of Sobolev type). The upshot of this construction is however, that there is a canonical identification of the tangent manifold of the manifold of $$H^s$$-mappings: $$TH^s(M,N) \cong H^s (M,TN).$$ Now $$\mathcal{D}^s \subseteq H^s (M,M)$$ inherits its smooth structure from the ambient space and in particular we have an induced identification $$T\mathcal{D}^s \subseteq H^s (M,TM)$$. Under this identification it is not hard to deduce the following identifications $$TR_\eta (f) = f\circ \eta , \qquad TL_\eta (f) = T\eta \circ f, \quad f \in H^s (M,TM).$$ Here of course $$\eta$$ needs to be in $$H^{s+1}$$ for the second formula to make sense, whereas for the first formula we can savely assume that $$\eta \in \mathcal{D}^s$$. Thus whenever $$s< \infty$$ we see that every derivative of the left multiplication looses one order of differentiability of the map $$\eta$$.
Of course the right multiplication is also continuous linear, whence smooth. But to me the above formula makes it explicit how one is loosing orders of differentiability on $$\mathcal{D}^s$$ (which is thus not a Lie group but only a half Lie group).