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A smooth manifold $G$ (of dimension $m$) with a group structure given by a multiplication map $\mu: G \times G \to G, \, (g,h) \mapsto gh$ is called a Lie group if both the multiplication $\mu$ and the inversion $\iota:G \to G, \, g \mapsto g^{-1}$ are smooth maps.

It is well-known that smoothness of the multiplication already implies the smoothness of the inversion, which is proven via the inverse function theorem. Smoothness of $\mu$ moreover implies that the left and right multiplication maps $$L_g: G \to G, \, h \mapsto gh \, \text{ and } \, R_g: G \to G, \, h \mapsto hg^{-1} \quad (g \in G)$$ are smooth by pre-composing $\mu$ with suitable smooth maps $G \to G \times G$.

My question is about the converse of the previous statement: Does smoothness of the maps $L_g$ and $R_g$ for all $g \in G$ also imply that $G$ is a Lie group, i.e. that the multiplication $\mu$ is smooth as well? If not, does it at least imply that the multiplication $\mu$ is continuous? If yes, does it also suffice to assume that $L_g$ is smooth for every $g \in G$?

A smooth manifold $G$ (of dimension $m$) with a group structure given by a multiplication map $\mu: G \times G \to G, \, (g,h) \mapsto gh$ is called a Lie group if both the multiplication $\mu$ and the inversion $\iota:G \to G, \, g \mapsto g^{-1}$ are smooth maps.

It is well-known that smoothness of the multiplication already implies the smoothness of the inversion, which is proven via the inverse function theorem. Smoothness of $\mu$ moreover implies that the left and right multiplication maps $$L_g: G \to G, \, h \mapsto gh \, \text{ and } \, R_g: G \to G, \, h \mapsto hg^{-1} \quad (g \in G)$$ are smooth by pre-composing $\mu$ with suitable smooth maps $G \to G \times G$.

My question is about the converse of the previous statement: Does smoothness of the maps $L_g$ and $R_g$ for all $g \in G$ also imply that $G$ is a Lie group, i.e. that the multiplication $\mu$ is smooth as well? If yes, does it also suffice to assume that $L_g$ is smooth for every $g \in G$?

A smooth manifold $G$ (of dimension $m$) with a group structure given by a multiplication map $\mu: G \times G \to G, \, (g,h) \mapsto gh$ is called a Lie group if both the multiplication $\mu$ and the inversion $\iota:G \to G, \, g \mapsto g^{-1}$ are smooth maps (where smooth maps are always assumed to be continuous).

It is well-known that smoothness of the multiplication already implies the smoothness of the inversion, which is proven via the inverse function theorem. Smoothness of $\mu$ moreover implies that the left and right multiplication maps $$L_g: G \to G, \, h \mapsto gh \, \text{ and } \, R_g: G \to G, \, h \mapsto hg^{-1} \quad (g \in G)$$ are smooth by pre-composing $\mu$ with suitable smooth maps $G \to G \times G$.

My question is about the converse of the previous statement: Does smoothness of the maps $L_g$ and $R_g$ for all $g \in G$ also imply that $G$ is a Lie group, i.e. that the multiplication $\mu$ is smooth as well? If not, does it at least imply that the multiplication $\mu$ is continuous? If yes, does it also suffice to assume that $L_g$ is smooth for every $g \in G$?

A smooth manifold $G$ (of dimension $m$) with a group structure given by a multiplication map $\mu: G \times G \to G, \, (g,h) \mapsto gh$ is called a Lie group if both the multiplication $\mu$ and the inversion $\iota:G \to G, \, g \mapsto g^{-1}$ are smooth maps.

It is well-known that smoothness of the multiplication already implies the smoothness of the inversion, which is proven via the inverse function theorem. Smoothness of $\mu$ moreover implies that the left and right multiplication maps $$L_g: G \to G, \, h \mapsto gh \, \text{ and } \, R_g: G \to G, \, h \mapsto hg^{-1} \quad (g \in G)$$ are smooth by pre-composing $\mu$ with suitable smooth maps $G \to G \times G$.

My question is about the converse of the previous statement: Does smoothness of the maps $L_g$ and $R_g$ for all $g \in G$ also imply that $G$ is a Lie group, i.e. that the multiplication $\mu$ is smooth as well? If not, does it at least imply that the multiplication $\mu$ is continuous? If yes, does it also suffice to assume that $L_g$ is smooth for every $g \in G$?

A smooth manifold $G$ (of dimension $m$) with a group structure given by a multiplication map $\mu: G \times G \to G, \, (g,h) \mapsto gh$ is called a Lie group if both the multiplication $\mu$ and the inversion $\iota:G \to G, \, g \mapsto g^{-1}$ are smooth maps.

It is well-known that smoothness of the multiplication already implies the smoothness of the inversion, which is proven via the inverse function theorem. Smoothness of $\mu$ moreover implies that the left and right multiplication maps $$L_g: G \to G, \, h \mapsto gh \, \text{ and } \, R_g: G \to G, \, h \mapsto hg^{-1} \quad (g \in G)$$ are smooth by pre-composing $\mu$ with suitable smooth maps $G \to G \times G$.

My question is about the converse of the previous statement: Does smoothness of the maps $L_g$ and $R_g$ for all $g \in G$ also imply that $G$ is a Lie group, i.e. that the multiplication $\mu$ is smooth as well? If not, does it at least imply that the multiplication $\mu$ is continuous?

A smooth manifold $G$ (of dimension $m$) with a group structure given by a multiplication map $\mu: G \times G \to G, \, (g,h) \mapsto gh$ is called a Lie group if both the multiplication $\mu$ and the inversion $\iota:G \to G, \, g \mapsto g^{-1}$ are smooth maps (where smooth maps are always assumed to be continuous).

It is well-known that smoothness of the multiplication already implies the smoothness of the inversion, which is proven via the inverse function theorem. Smoothness of $\mu$ moreover implies that the left and right multiplication maps $$L_g: G \to G, \, h \mapsto gh \, \text{ and } \, R_g: G \to G, \, h \mapsto hg^{-1} \quad (g \in G)$$ are smooth by pre-composing $\mu$ with suitable smooth maps $G \to G \times G$.

My question is about the converse of the previous statement: Does smoothness of the maps $L_g$ and $R_g$ for all $g \in G$ also imply that $G$ is a Lie group, i.e. that the multiplication $\mu$ is smooth as well? If not, does it at least imply that the multiplication $\mu$ is continuous? If yes, does it also suffice to assume that $L_g$ is smooth for every $g \in G$?