This might be a dumb question, but are all continuous maps between Lie groups also homomorphisms? I can only seem to think of examples in which they are (i.e., $GL(n,\mathbb{R}) \to \mathbb{R}$ via the determinant, the covering space map from $\mathbb{R} \to S^1$,...). Conversely, a Lie group homomorphism is defined as a homomorphism that is smooth, so what are some examples of homomorphisms between Lie groups that are not smooth?
Continuous map of Lie Groups
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representation-theory
lie-groups
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0Re: first question. So... all continuous maps $\mathbb{R} \to \mathbb{R}$ are of the form $x \mapsto ax$? Re: last question: you can show that measurable homomorphisms between Lie groups are smooth, so you need to seek non-measurable examples. One way would be to take a non-measurable solution to the Cauchy functional equation. – 2012-03-24
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0$S^1$ is a Lie group and there are many continuous maps $S^1 \rightarrow S^1$ which are continuous but no Lie group homomorphisms. Are you sure your formulation expresses your question correctly? – 2012-03-24
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0Take $G$ and $H$ any lie groups and $e \ne h_0 \in H$. Then $G \to H, g \mapsto h_0$ is always continuous but never a homomorphism. – 2012-03-24
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0There exist field isomorphisms of the complex numbers which are not continuous, hence obviously are not smooth. Not sure how to answer the question of how to find continuous non-smooth homomorphism. – 2012-03-24
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2@ThomasAndrews: [Continuous homomorphisms are automatically smooth](http://books.google.com/books?id=gRTDO-wVhj0C&pg=PA49). – 2012-03-24
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2@ThomasAndrews: It's even worse in some cases. For compact Lie groups with finite centers, homomorphisms are automatically continuous, hence automatically smooth. – 2012-03-24