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This is definitely not my expertise, but here is a stab at it.

A solution to Hilbert's 5th ProblemHilbert's 5th Problem states that given a topological group that is also a manifold there is a unique way to give it the structure of a Lie group.

On the other hand, there are exotic structuresexotic structures on Lie groups (including compact ones).

So take a Lie group $G$ that admits an exotic structure. Let $M$ denote the exotic version of $G$. Then $G$ and $M$ are homeomorphic but not diffeomorphic. However, if $M$ admitted a smooth group operation, then we contradict the uniqueness part of the solution to Hilbert's 5th Problem.

This is definitely not my expertise, but here is a stab at it.

A solution to Hilbert's 5th Problem states that given a topological group that is also a manifold there is a unique way to give it the structure of a Lie group.

On the other hand, there are exotic structures on Lie groups (including compact ones).

So take a Lie group $G$ that admits an exotic structure. Let $M$ denote the exotic version of $G$. Then $G$ and $M$ are homeomorphic but not diffeomorphic. However, if $M$ admitted a smooth group operation, then we contradict the uniqueness part of the solution to Hilbert's 5th Problem.

This is definitely not my expertise, but here is a stab at it.

A solution to Hilbert's 5th Problem states that given a topological group that is also a manifold there is a unique way to give it the structure of a Lie group.

On the other hand, there are exotic structures on Lie groups (including compact ones).

So take a Lie group $G$ that admits an exotic structure. Let $M$ denote the exotic version of $G$. Then $G$ and $M$ are homeomorphic but not diffeomorphic. However, if $M$ admitted a smooth group operation, then we contradict the uniqueness part of the solution to Hilbert's 5th Problem.

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Sean Lawton
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This is definitely not my expertise, but here is a stab at it.

A solution to Hilbert's 5th Problem states that given a topological group that is also a manifold there is a unique way to give it the structure of a Lie group.

On the other hand, there are exotic structures on Lie groups (including compact ones).

So take a Lie group $G$ that admits an exotic structure. Let $M$ denote the exotic version of $G$. Then $G$ and $M$ are homeomorphic but not diffeomorphic. However, if $M$ admitted a smooth group operation, then we contradict the uniqueness partuniqueness part of the solution to Hilbert's 5th Problem.

This is definitely not my expertise, but here is a stab at it.

A solution to Hilbert's 5th Problem states that given a topological group that is also a manifold there is a unique way to give it the structure of a Lie group.

On the other hand, there are exotic structures on Lie groups.

So take a Lie group $G$ that admits an exotic structure. Let $M$ denote the exotic version of $G$. Then $G$ and $M$ are homeomorphic but not diffeomorphic. However, if $M$ admitted a smooth group operation, then we contradict the uniqueness part of the solution to Hilbert's 5th Problem.

This is definitely not my expertise, but here is a stab at it.

A solution to Hilbert's 5th Problem states that given a topological group that is also a manifold there is a unique way to give it the structure of a Lie group.

On the other hand, there are exotic structures on Lie groups (including compact ones).

So take a Lie group $G$ that admits an exotic structure. Let $M$ denote the exotic version of $G$. Then $G$ and $M$ are homeomorphic but not diffeomorphic. However, if $M$ admitted a smooth group operation, then we contradict the uniqueness part of the solution to Hilbert's 5th Problem.

Source Link
Sean Lawton
  • 8.5k
  • 3
  • 46
  • 78

This is definitely not my expertise, but here is a stab at it.

A solution to Hilbert's 5th Problem states that given a topological group that is also a manifold there is a unique way to give it the structure of a Lie group.

On the other hand, there are exotic structures on Lie groups.

So take a Lie group $G$ that admits an exotic structure. Let $M$ denote the exotic version of $G$. Then $G$ and $M$ are homeomorphic but not diffeomorphic. However, if $M$ admitted a smooth group operation, then we contradict the uniqueness part of the solution to Hilbert's 5th Problem.