Differentiable vectors and unitary representations of Fréchet–Lie supergroups

Abstract

A locally convex Lie group G has the Trotter property if, for every $x_1, x_2 \in \mathfrak{g }$,

$$\begin{aligned} \exp _G(t(x_1 + x_2))=\lim _{n \rightarrow \infty } \left(\exp _G\left(\frac{t}{n}x_1\right)\exp _G\left(\frac{t}{n}x_2\right)\right)^n \end{aligned}$$

holds uniformly on compact subsets of $\mathbb{R }$. All locally exponential Lie groups have this property, but also groups of automorphisms of principal bundles over compact smooth manifolds. A key result of the present article is that, if G has the Trotter property, $\pi : G \rightarrow {\mathrm{GL}}(V)$ is a continuous representation of G on a locally convex space, and $v \in V$ is a vector such that $\overline{\mathtt{d}\pi }(x)v :=\frac{d}{dt}|_{t=0} \pi (\exp _G(tx))v$ exists for every $x \in \mathfrak{g }$, then the map $\mathfrak{g }\rightarrow V,x \mapsto \overline{\mathtt{d}\pi }(x)v$ is linear. Using this result we conclude that, for a representation of a locally exponential Fréchet–Lie group G on a metrizable locally convex space, the space of $\mathcal{C }^{k}$-vectors coincides with the common domain of the k-fold products of the operators $\overline{\mathtt{d}\pi }(x)$. For unitary representations on Hilbert spaces, the assumption of local exponentiality can be weakened to the Trotter property. As an application, we show that for smooth (resp., analytic) unitary representations of Fréchet–Lie supergroups $(G,\mathfrak{g })$ where G has the Trotter property, the common domain of the operators of $\mathfrak{g }=\mathfrak{g }_{\overline{0}}\oplus \mathfrak{g }_{\overline{1}}$ can always be extended to the space of smooth (resp., analytic) vectors for G.

Appendices

Appendix A: Some results on Lie groups of maps

The results of this appendix will be used in Appendix below. Let E be a locally convex space, M be a smooth finite dimensional manifold (possibly with boundary) and K be a Lie group (possibly infinite dimensional) with Lie algebra $\mathfrak k$. In the following we write $\Omega ^1_{\mathcal{C }^{r}}(M,E)$ for the space of E-valued 1-forms on M defining $\mathcal{C }^{r}$-functions $TM \rightarrow E$. The space of E-valued smooth forms will be denoted by $\Omega ^1(M,E)$.

We endow $\Omega ^1_{\mathcal{C }^{r}}(M,E)$ with the topology induced by the embedding

$$\begin{aligned} \Omega ^1_{\mathcal{C }^{r}}(M,E)\hookrightarrow \mathcal{C }^{r}(TM,E), \end{aligned}$$

where $TM$ is the tangent bundle and $\mathcal{C }^{r}(TM,E)$ is endowed with the compact open $\mathcal{C }^{r}$-topology, so that $\Omega ^1_{\mathcal{C }^{r}}(M,E)$ is a closed subspace of $\mathcal{C }^{r}(TM,E)$. The space $\Omega ^1(M,E)$ is endowed with the topology induced by the diagonal embedding

$$\begin{aligned} \Omega ^1(M,E)\hookrightarrow \prod _{r=1}^\infty \Omega ^1_{\mathcal{C }^{r}}(M,E). \end{aligned}$$

Lemma 8.1

Let M be a compact smooth manifold (possibly with boundary) and K a Lie group with Lie algebra $\mathfrak{k }$. Then, for each $r \in \mathbb{N }_0 \cup \{\infty \}$, the action of the Lie group $\mathcal{C }^{r}(M,K)$ on $\Omega ^1_{\mathcal{C }^{r}}(M,\mathfrak{k })$ by $(g,\alpha ) \mapsto \mathrm{Ad}(g)\alpha$ is smooth.

Proof

Assume that $d=\dim M$. Every covering $(U_i)_{i \in I}$ of M by compact submanifolds with boundary, which are diffeomorphic to d-dimensional balls and whose interiors define an atlas, yields an embedding

$$\begin{aligned} \Omega ^1_{\mathcal{C }^{r}}(M,\mathfrak{k }) \hookrightarrow \prod _{i \in I} \Omega ^1_{\mathcal{C }^{r}}(U_i,\mathfrak{k }) \cong \prod _{i \in I} \mathcal{C }^{r}(U_i,\mathfrak{k })^d. \end{aligned}$$

Therefore it suffices to show that the action of $\mathcal{C }^{r}(M,K)$ on each space $\mathcal{C }^{r}(U_i,\mathfrak{k })$, given by

$$\begin{aligned} (g,f) \mapsto \mathrm{Ad}(g)f=\sigma _{\mathrm{Ad}} \circ (g,f) \end{aligned}$$

is smooth. This action factors through the Lie group morphisms

$$\begin{aligned} \mathcal{C }^{r}(M,K) \rightarrow \mathcal{C }^{r}(U_i,K), \end{aligned}$$

and for the Lie groups $G:=\mathcal{C }^{r}(U_i,K)$, it coincides with the adjoint action on $\mathbf{L}(G) \cong \mathcal{C }^{r}(U_i,\mathfrak{k })$, which is smooth. This proves the lemma. $\square$

Lemma 8.2

([12]) Let $E_1$ and $E_2$ be locally convex spaces, $U_1 \subseteq E_1$ open, M a compact smooth manifold (possibly with boundary), and $\varphi : U_1 \rightarrow E_2$ be a smooth map. Then, for each $r \in \mathbb{N }\cup \{\infty \}$, the map

$$\begin{aligned} \varphi _* : \mathcal{C }^{r}(M,U_1) \rightarrow \mathcal{C }^{r}(M,E_2), \quad f \mapsto \varphi \circ f \end{aligned}$$

is smooth.

Lemma 8.3

Let U be an open subset of a locally convex space $E,\,F$ a locally convex space, M a compact manifold (possibly with boundary), $r \in \mathbb{N }\cup \{ \infty \}$, and $\alpha \in \Omega ^1(U,F)$. Then the map

$$\begin{aligned} \mathcal{C }^{r}(M,U) \rightarrow \Omega ^1_{\mathcal{C }^{r-1}}(M,F), \quad f \mapsto f^*\alpha \end{aligned}$$

is smooth if $\mathcal{C }^{r}(M,U)$ is considered as an open subset of $\mathcal{C }^{r}(M,E)$.

Proof

Let $\pi : TM \rightarrow M$ denote the bundle projection. Then both components of the map

$$\begin{aligned} \mathcal{C }^{r}(M,U)&\rightarrow \mathcal{C }^{r-1}(TM,TU) \cong \mathcal{C }^{r-1}(TM,U) \times \mathcal{C }^{r-1}(TM,E), \\ f&\mapsto Tf=(f \circ \pi ,\mathtt{d}f) \end{aligned}$$

are restrictions of continuous linear maps, hence smooth. Since $f^*\alpha =\alpha \circ Tf$, smoothness of $\alpha$ and Lemma 8.2 imply that

$$\begin{aligned} \alpha _* : \mathcal{C }^{r-1}(TM,TU) \rightarrow \mathcal{C }^{r-1}(TM, F), \quad h \mapsto \alpha \circ h \end{aligned}$$

is smooth, from which the assertion follows (recall that we topologize $\Omega ^1_{\mathcal{C }^{r-1}}(M,F)$ as a closed subspace of $\mathcal{C }^{r-1}(TM,F)$). $\square$

Proposition 8.4

For any Lie group K with Lie algebra $\mathfrak{k }$, any compact manifold M (possibly with boundary) and any $r \in \mathbb{N }\cup \{\infty \}$, the left logarithmic derivative

$$\begin{aligned} \delta : \mathcal{C }^{r}(M,K) \rightarrow \Omega ^1_{\mathcal{C }^{r-1}}(M,\mathfrak{k }) \end{aligned}$$

is a smooth map with respect to the Lie group structure on $\mathcal{C }^{r}(M,K)$, and

$$\begin{aligned} T_\mathbf{1}(\delta )=\mathtt{d}: \mathcal{C }^{r}(M,\mathfrak{k }) \rightarrow \Omega ^1_{\mathcal{C }^{r-1}}(M,\mathfrak{k }), \quad \xi \mapsto \mathtt{d}\xi . \end{aligned}$$

(28)

Proof

From Lemma 8.1, we already know that the action of the Lie group $G:=\mathcal{C }^{r}(M,K)$ on $\Omega ^1_{\mathcal{C }^{r}}(M,\mathfrak{k })$ by $f\cdot \alpha :=\mathrm{Ad}(f)\alpha$ is smooth. Since the inclusion map $\mathcal{C }^{r}(M,K) \rightarrow \mathcal{C }^{r-1}(M,K)$ is a smooth morphism of Lie groups, the action of G on on $\Omega ^1_{\mathcal{C }^{r-1}}(M,\mathfrak{k })$ is also smooth.

The product rule $\delta (\eta \gamma )=\delta (\gamma )+\mathrm{Ad}(\gamma )^{-1}\delta (\eta )$ means that $\delta$ is a right crossed homomorphism for the smooth action of $\mathcal{C }^{r}(M,K)$ on $\Omega ^1_{\mathcal{C }^{r-1}}(M,\mathfrak{k })$. It therefore suffices to verify its smoothness in a neighborhood of the identity.

Let $(\varphi _K,U_K)$ be a chart of an identity neighborhood of K with $T_\mathbf{1}(\varphi _K)={\mathrm{id}}_\mathfrak{k }$, so that $(\varphi _G,U_G)$ with

$$\begin{aligned} U_G:=\lfloor M,U_K \rfloor :=\{\gamma \in G : \gamma (M) \subseteq U_K \},\quad \varphi _G(\gamma ):=\varphi _K \circ \gamma \end{aligned}$$

is a chart of an identity neighborhood of the Lie group G. If $\kappa _K \in \Omega ^1(K,\mathfrak{k })$ denotes the left Maurer–Cartan form of K, then we have a map

$$\begin{aligned} \mathcal{C }^{r}(M, \varphi _K(U_K))&\rightarrow \Omega ^1_{\mathcal{C }^{r-1}}(M,\mathfrak{k }), \\ \xi&\mapsto \delta (\varphi _K^{-1} \circ \xi )=(\varphi _K^{-1} \circ \xi )^*\kappa _K = \xi ^*(\varphi _K^{-1})^*\kappa _K \end{aligned}$$

whose smoothness follows from Lemma 8.3.

Set $\beta := (\varphi _K^{-1})^*\kappa _K$. Then $\beta _0 = {\mathrm{id}}_\mathfrak{k }$ and (28) follows from the fact that for every $m \in M$ and $v \in T_m(M)$, we have

$$\begin{aligned} \left.\frac{d}{dt}\right|_{t=0} (t\xi )^*\beta v=\left.\frac{d}{dt}\right|_{t=0} \beta _{t\xi (m)}(t \mathtt{d}\xi (m)v)=\lim _{t \rightarrow 0} \beta _{t\xi (m)}\mathtt{d}\xi (m)v=\beta _0 \mathtt{d}\xi (m)v=\mathtt{d}\xi (m)v. \end{aligned}$$

$\square$

Appendix B: $\mathcal{C }^k$-regularity is an extension property

In this appendix we generalize the result that regularity of Lie groups is an extension property to the stronger notion of $\mathcal{C }^k$-regularity for $k \in \mathbb{N }_0$.

Throughout this section $I:=[0,1]$. If $\gamma :(-\varepsilon ,\varepsilon )\rightarrow G$ is a $\mathcal{C }^1$ curve, then the left logarithmic derivative $\delta (\gamma ):(-\varepsilon ,\varepsilon )\rightarrow \mathfrak{g }$ is defined by

$$\begin{aligned} \delta (\gamma )_t:=\mathtt{d}\ell _{\gamma (t)^{-1}}(\gamma (t))\gamma ^{\prime }(t) \end{aligned}$$

where $\ell _g:G\rightarrow G$ denotes the left translation $\ell _g(x):=gx$.

Definition 9.1

Let $k \in \mathbb{N }_0 \cup \{\infty \}$. A Lie group G with Lie algebra $\mathfrak{g }$ is called ${\mathcal{C }^k}$ -regular, if for each $\xi \in {\mathcal{C }^k}(I,\mathfrak{g })$, the initial value problem

$$\begin{aligned} \gamma (0)=\mathbf{1},\quad \delta (\gamma )=\xi \end{aligned}$$

(29)

has a solution $\gamma _\xi$, which is then contained in $\mathcal{C }^{k+1}(I,G)$, and the corresponding evolution map

$$\begin{aligned} {\mathrm{evol}}_G : {\mathcal{C }^k}(I,\mathfrak{g }) \rightarrow G, \quad \xi \mapsto \gamma _\xi (1) \end{aligned}$$

is smooth. The solutions of (29) are unique whenever they exist (cf. [18]). If G is ${\mathcal{C }^k}$-regular, we write

$$\begin{aligned} \mathrm{Evol}_G : {\mathcal{C }^k}(I,\mathfrak{g }) \rightarrow \mathcal{C }^{k+1}(I,G),\quad \xi \mapsto \gamma _\xi \end{aligned}$$

for the corresponding map on the level of Lie group-valued curves. This map is also smooth (cf. [10, Thm. A]). The group G is called regular if it is $C^\infty$-regular.

Remark 9.2

• (a) Any regular Lie group G has a smooth exponential function

  $$\begin{aligned} \exp _G:\mathfrak{g }\rightarrow G\quad \text{ by}\quad \exp _G(x):=\gamma _x(1), \end{aligned}$$

  where $x \in \mathfrak{g }$ is considered as a constant function $I \rightarrow \mathfrak{g }$. As a restriction of the smooth function ${\mathrm{evol}}_G$ to the topological subspace $\mathfrak{g }\subseteq C^\infty (I,\mathfrak{g })$ of constant functions, the exponential function is smooth.

• (b) For $k \le r$, the ${\mathcal{C }^k}$-regularity of a Lie group G implies its $\mathcal{C }^{r}$-regularity because the inclusion map $\mathcal{C }^{r}(I,\mathfrak{g }) \rightarrow {\mathcal{C }^k}(I,\mathfrak{g })$ is continuous linear, hence smooth.

Lemma 9.3

Let G be a Lie group with Lie algebra $\mathfrak{g }$. Then the prescription

$$\begin{aligned} \xi *\gamma :=\delta (\gamma )+\mathrm{Ad}(\gamma )^{-1}\xi \end{aligned}$$

defines a smooth affine right action of the group $\mathcal{C }^{k+1}(I,G)$ on ${\mathcal{C }^k}(I,\mathfrak{g })$.

Proof

That we have an action follows from

$$\begin{aligned} (\xi *\gamma _1)*\gamma _2=\delta (\gamma _2)+\mathrm{Ad}(\gamma _2)^{-1}(\delta (\gamma _1)+\mathrm{Ad}(\gamma _1)^{-1}\xi ) \\ =\delta (\gamma _1\gamma _2)+\mathrm{Ad}(\gamma _1 \gamma _2)^{-1}\xi =\xi *(\gamma _1 \gamma _2). \end{aligned}$$

Since I is a compact manifold with boundary, the smoothness of the action follows from the smoothness of $\delta$ (see Proposition 8.4) and Lemma 8.1 (note that we can identify $\Omega _{\mathcal{C }^k}^{1}(I,\mathfrak{g })$ with $\mathcal{C }^k(I,\mathfrak{g })$). $\square$

Remark 9.4

If $\xi =\delta (\eta )$ for some smooth function $\,h:M \rightarrow G$, then the Product Rule implies that $\delta (\eta \gamma )=\xi *\gamma ,$ so that the action from above corresponds to the right multiplication action on the level of group-valued functions.

Lemma 9.5

(Local regularity criterion) Let G be a Lie group with Lie algebra $\mathfrak{g }$ and $k \in \mathbb{N }_0 \cup \{\infty \}$. Suppose that (29) has a solution for each $\xi$ in an open 0-neighborhood $U \subseteq {\mathcal{C }^k}(I,\mathfrak{g })$. Then it has a solution for each $\xi \in {\mathcal{C }^k}(I,\mathfrak{g })$. If the evolution map

$$\begin{aligned} {\mathrm{evol}}_G:{\mathcal{C }^k}(I,\mathfrak{g }) \rightarrow G, \quad \xi \mapsto \gamma _\xi (1) \end{aligned}$$

is smooth in U, then it is smooth on all of ${\mathcal{C }^k}(I,\mathfrak{g })$.

Proof

Let $\xi \in \mathcal{C }^k(I,\mathfrak{g })$. For $n \in \mathbb{N }$ and $i \in \{0,\ldots , n-1\}$ we define

$$\begin{aligned} \xi _i^n \in \mathcal{C }^k(I,\mathfrak{g }) \quad \text{ by}\quad \xi _i^n(t):=\frac{1}{n}\xi \left(\frac{i+t}{n}\right). \end{aligned}$$

Then for each $m \le k$ we have $\frac{d^m}{dt^m}\xi _i^n(t)=\frac{1}{n^{m+1}}\xi ^{(m)}\left(\frac{i+t}{n}\right)$. We conclude that, for $n \rightarrow \infty$, the sequence $\frac{d^m}{dt^m}\xi _i^n$ tends to 0 in $\mathcal{C }^0(I,\mathfrak{g })$, uniformly in i, and hence that $\xi _i^n$ tends to 0 in $\mathcal{C }^k(I,\mathfrak{g })$, uniformly in i. In particular, there exists some $N \in \mathbb{N }$ for which $\xi _i^N \in U$ for $i = 0,1,\ldots , N-1$.

We define a path $\gamma _\xi : I \rightarrow G$ by

$$\begin{aligned} \gamma _\xi (t):=\gamma _{\xi _0^N}(1)\cdots \gamma _{\xi _{i-1}^N}(1) \gamma _{\xi _i^N}(nt-i) \quad \text{ for}\;\frac{i}{n} \le t \le \frac{i+1}{n} \end{aligned}$$

and observe that $\delta (\gamma _\xi )$ exists on all of I and equals $\xi$. We now put

$$\begin{aligned} \mathrm{Evol}_G(\xi ):=\gamma _\xi \quad \text{ and}\;{\mathrm{evol}}_G(\xi ):=\gamma _\xi (1). \end{aligned}$$

Since, for each i, the assignment $\xi \mapsto \xi _i^N$ is linear and continuous, there exists an open neighborhood V of $\xi$ such that $\eta _i^N \in U$ holds for each $\eta \in V$ and $i =0,1,\ldots , N-1$. It suffices to show that ${\mathrm{evol}}_G$ is smooth on V, but this follows from the fact that

$$\begin{aligned} {\mathrm{evol}}_G(\eta )={\mathrm{evol}}_G(\eta _0^N) \cdots {\mathrm{evol}}_G(\eta _{N-1}^N) \end{aligned}$$

is a product of N smooth functions. $\square$

For the definition of an initial Lie subgroup see [18, Def. II.6.1].

Proposition 9.6

Let G be a ${\mathcal{C }^k}$-regular Lie group with Lie algebra $\mathfrak{g }$ and $H \le G$ an initial Lie subgroup with Lie algebra $\mathfrak{h }\subseteq \mathfrak{g }$, for which there exists an open identity neighborhood $U \subseteq G$ and a smooth function $f: U \rightarrow E$ into some locally convex space E, such that f is constant on $U \cap gH$ for each $g \in U$, and $H \cap U= f^{-1}(0) \cap U$. Then H is $\mathcal{C }^k$-regular.

Proof

The ${\mathcal{C }^k}$-regularity of G implies the existence of a smooth evolution map

$$\begin{aligned} {\mathrm{evol}}_H : {\mathcal{C }^k}(I,\mathfrak{h }) \rightarrow G, \quad \xi \mapsto \gamma _\xi (1), \end{aligned}$$

and since H is initial, it suffices to see that the range of this map lies in H.

If $\xi \in {\mathcal{C }^k}(I,\mathfrak{h })$ such that ${\mathrm{im}}(\gamma _\xi ) \subseteq U$, then for every $t\in I$,

$$\begin{aligned} (f \circ \gamma _\xi )^{\prime }(t)=\mathtt{d}f({\gamma _\xi (t)})\gamma _\xi ^{\prime }(t)=0 \end{aligned}$$

because $\gamma _\xi ^{\prime }(t)=\mathtt{d}\ell _{\gamma _\xi (t)}(\mathbf{1})\xi (t)$ is the derivative of a curve in the set $U \cap \gamma _\xi (t)H$, on which f is constant. Therefore $f \circ \gamma _\xi$ is constant, which leads to ${\mathrm{im}}(\gamma _\xi )\subseteq f^{-1}(f(\mathbf{1}))=f^{-1}(0)=U \cap H$.

If $\xi \in {\mathcal{C }^k}(I,\mathfrak{h })$ is arbitrary, we apply the preceding argument to the curves $t\mapsto \gamma _\xi (t_0)^{-1} \gamma _\xi (t)$ on sufficiently small intervals $[t_0,t_0+\varepsilon ]$ and see that ${\mathrm{im}}(\gamma _\xi )$ is contained in H. $\square$

Theorem 9.7

($\mathcal{C }^k$-regularity is an extension property) Let $q : \widehat{G} \rightarrow G$ be an extension of the Lie group G by the Lie group N and $k \in \mathbb{N }_0$. Then $\widehat{G}$ is ${\mathcal{C }^k}$-regular if and only if N and G are ${\mathcal{C }^k}$-regular.

Proof

Step 1. We assume that G and N are $\mathcal{C }^k$-regular and show that this implies the $\mathcal{C }^k$-regularity of $\widehat{G}$. Since G is $\mathcal{C }^k$-regular, the evolution map

$$\begin{aligned} \mathrm{Evol}_G : {\mathcal{C }^k}(I,\mathfrak{g }) \rightarrow \mathcal{C }^{k+1}_*(I,G) \end{aligned}$$

is smooth [10, Thm A].

Let $U_G \subseteq G$ be an open $\mathbf{1}$-neighborhood for which we have a smooth section $\sigma : U_G \rightarrow \widehat{G}$ with $\sigma (\mathbf{1}_G)=\mathbf{1}_{\widehat{G}}$ and

$$\begin{aligned} p : {\mathcal{C }^k}(I,\widehat{\mathfrak{g }}) \rightarrow {\mathcal{C }^k}(I,\mathfrak{g }), \quad \xi \mapsto q_\mathfrak{g }\circ \xi \end{aligned}$$

be the projection map. Then $V:=p^{-1}(\mathrm{Evol}_G^{-1}(\mathcal{C }^{k+1}(I,U_G)))$ is an open 0-neighborhood in ${\mathcal{C }^k}(I,\widehat{\mathfrak{g }})$. Further, by Lemma 8.2 the map

$$\begin{aligned} \Phi :V \rightarrow \mathcal{C }^{k+1}_*(I,\widehat{G}),\quad \xi \mapsto \sigma \circ \mathrm{Evol}_G(q_\mathfrak{g }\circ \xi ) \end{aligned}$$

is smooth. For $\xi \in {\mathcal{C }^k}(I,\widehat{\mathfrak{g }})$, we find

$$\begin{aligned}&q_\mathfrak{g }(\xi *\Phi (\xi )^{-1})=q_\mathfrak{g }(\mathrm{Ad}(\Phi (\xi ))(\xi -\delta (\Phi (\xi )))) \\&\quad =\mathrm{Ad}(q_G(\Phi (\xi )))(q_\mathfrak{g }\circ \xi -\delta (q_G(\Phi (\xi ))))=\mathrm{Ad}(q_G(\Phi (\xi )))(q_\mathfrak{g }\circ \xi -q_\mathfrak{g }\circ \xi )=0. \end{aligned}$$

This means that $\xi * \Phi (\xi )^{-1} \in {\mathcal{C }^k}(I,\mathfrak{n })$. Now Lemma 9.3, applied to the action of $\mathcal{C }^{k+1}(I,\widehat{G})$ on $\Omega ^1_{{\mathcal{C }^k}}(I,\widehat{\mathfrak{g }}) \cong {\mathcal{C }^k}(I,\widehat{\mathfrak{g }})$, shows that it depends smoothly on $\xi$. We thus obtain a smooth map

$$\begin{aligned} \widehat{E} : V \rightarrow \widehat{G}, \quad \widehat{E}(\xi ):={\mathrm{evol}}_N(\xi *\Phi (\xi )^{-1})\cdot \Phi (\xi )(1). \end{aligned}$$

The curve $\gamma :=\mathrm{Evol}_N(\xi *\Phi (\xi )^{-1})\Phi (\xi )$ satisfies

$$\begin{aligned} \delta (\gamma )=\delta (\Phi (\xi ))+\mathrm{Ad}(\Phi (\xi ))^{-1}(\delta (\Phi (\xi )^{-1}) + \mathrm{Ad}(\Phi (\xi ))\xi )=\xi . \end{aligned}$$

Therefore $\widehat{E}$ is a smooth local evolution map for $\widehat{G}$, and Lemma 9.5 implies that $\widehat{G}$ is $\mathcal{C }^k$-regular.

Step 2. Conversely, we show that the ${\mathcal{C }^k}$-regularity of $\widehat{G}$ implies the ${\mathcal{C }^k}$-regularity of N and G. To see that N is ${\mathcal{C }^k}$-regular, we choose a chart $(\varphi _G,U_G)$ of G and consider the map

$$\begin{aligned} f:=\varphi _G \circ q:q^{-1}(U_G) \rightarrow \varphi _G(U_G), \end{aligned}$$

which is constant on the left cosets of N, lying in this set. Therefore Proposition 9.6 implies that N is ${\mathcal{C }^k}$-regular because it is a submanifold of $\widehat{G}$, hence in particular an initial submanifold.

To see that G is ${\mathcal{C }^k}$-regular, we first choose a continuous linear section $\sigma : \mathfrak{g }\rightarrow \widehat{\mathfrak{g }}$, which induces a continuous linear section

$$\begin{aligned} \sigma _*:{\mathcal{C }^k}(I,\mathfrak{g }) \rightarrow {\mathcal{C }^k}(I,\widehat{\mathfrak{g }}), \quad \xi \mapsto \sigma \circ \xi . \end{aligned}$$

Then, for each $\xi \in {\mathcal{C }^k}([0,1],\mathfrak{g })$, the curve $\gamma _\xi :=q \circ \mathrm{Evol}_{\widehat{G}}(\sigma \circ \xi )$ satisfies

$$\begin{aligned} \delta (\gamma _\xi )=\mathbf{L}(q) \circ \sigma \circ \xi =\xi , \end{aligned}$$

so that ${\mathrm{evol}}_G=q \circ {\mathrm{evol}}_{\widehat{G}} \circ \sigma$ is a composition of smooth maps, hence smooth. $\square$