Example of a Lie group that cannot be represented in matrix form?
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Here (at the beginning) is an example of a local Lie group
https://www.physicsforums.com/insights/journey-manifold-su2mathbbc-part/
However, we have the adjoint representation ##\operatorname{Ad}\, : \,G\longrightarrow \operatorname{GL}(\mathfrak{g})##, and ##\operatorname{Ad}(G)## is a Lie subgroup of ##\operatorname{GL}(\mathfrak{g})##. If it is a monomorphism, we automatically get ##\operatorname{G}\cong \operatorname{Ad}(G)## and have a matrix group.
https://www.physicsforums.com/insights/journey-manifold-su2mathbbc-part/
However, we have the adjoint representation ##\operatorname{Ad}\, : \,G\longrightarrow \operatorname{GL}(\mathfrak{g})##, and ##\operatorname{Ad}(G)## is a Lie subgroup of ##\operatorname{GL}(\mathfrak{g})##. If it is a monomorphism, we automatically get ##\operatorname{G}\cong \operatorname{Ad}(G)## and have a matrix group.
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Your notation for the unitary groups is unconventional.fresh_42 said:Here (at the beginning) is an example of a local Lie group
https://www.physicsforums.com/insights/journey-manifold-su2mathbbc-part/
However, we have the adjoint representation ##\operatorname{Ad}\, : \,G\longrightarrow \operatorname{GL}(\mathfrak{g})##, and ##\operatorname{Ad}(G)## is a Lie subgroup of ##\operatorname{GL}(\mathfrak{g})##. If it is a monomorphism, we automatically get ##\operatorname{G}\cong \operatorname{Ad}(G)## and have a matrix group.
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Here's a description: https://en.wikipedia.org/wiki/Metaplectic_group
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I guess we want to have a faithful representation to call a group a matrix group.WWGD said:Don't you always have a trivial representation sending everything to the identity?
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Are we still talking about Lie groups?WWGD said:How about the Cayley representation then, as a group of permutations?
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Cant every group be described as a permutation group? Unless you want to preserve any other than algebraic properties, it seems it would work, though I don't see how to do it with Lie groups.fresh_42 said:Are we still talking about Lie groups?
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If the group is not finite, these permutations are not going to be matrices.WWGD said:Cant every group be described as a permutation group? Unless you want to preserve any other than algebraic properties, it seems it would work, though I don't see how to do it with Lie groups.
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martinbn said:##\widetilde{SL}_2(\mathbb R)##
martinbn said:Metaplectic
These groups are different. The metaplectic group ##Mp(2)## is not simply connected because it is a double cover of ##Sp(2)=SL_2(\mathbb{R}),## which has fundamental group of ##\mathbb{Z}##.
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Infrared said:These groups are different. The metaplectic group ##Mp(2)## is not simply connected because it is a double cover of ##Sp(2)=SL_2(\mathbb{R}),## which has fundamental group of ##\mathbb{Z}##.
The tilde on top of the group "name" exactly universal cover of that group means. So the two groups are not different.
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The universal cover of a space is simply connected. The metaplectic group is not simply connected. So they are different.dextercioby said:The tilde on top of the group "name" exactly universal cover of that group means. So the two groups are not different.
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Absolutely, it seems my memory betrays me. I stand corrected.
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