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Community Bot
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Does there exist a (finite dimensional) smooth manifold $M$, such that every Riemannian metric on $M$ has no isometries except the identity?

Of course, such a manifold must not admit a diffeomorphism of finite order.

Since a surface $S$ admits a diffeomorphism of order $n$ iff its mapping class group (MCP) has an element of order $n$ (see herehere), it follows that if $S$ has the above property, then its MCP has only elements of infinite order.

Does there exist a (finite dimensional) smooth manifold $M$, such that every Riemannian metric on $M$ has no isometries except the identity?

Of course, such a manifold must not admit a diffeomorphism of finite order.

Since a surface $S$ admits a diffeomorphism of order $n$ iff its mapping class group (MCP) has an element of order $n$ (see here), it follows that if $S$ has the above property, then its MCP has only elements of infinite order.

Does there exist a (finite dimensional) smooth manifold $M$, such that every Riemannian metric on $M$ has no isometries except the identity?

Of course, such a manifold must not admit a diffeomorphism of finite order.

Since a surface $S$ admits a diffeomorphism of order $n$ iff its mapping class group (MCP) has an element of order $n$ (see here), it follows that if $S$ has the above property, then its MCP has only elements of infinite order.

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Asaf Shachar
Asaf Shachar
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Is there a smooth manifold which admits only rigid metrics?

Does there exist a (finite dimensional) smooth manifold $M$, such that every Riemannian metric on $M$ has no isometries except the identity?

Of course, such a manifold must not admit a diffeomorphism of finite order.

Since a surface $S$ admits a diffeomorphism of order $n$ iff its mapping class group (MCP) has an element of order $n$ (see here), it follows that if $S$ has the above property, then its MCP has only elements of infinite order.