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Suppose that G is a connected Lie group such that the center of G is trivial.

Question:

Is it true that G is isomorphic (as Lie group) to a closed subgroup of a Linear group GL(n,R) for some natural number n. Or maybe there is an evident counterexample ?

Ofra
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  • Welcome to MSE. Questions like "Here is the task. Solve it for me!" are poorly received on this site. Therefore try to improve your question with an edit. Improving could consist of providing some context concerning your task or by adding what you have tried so far and where did you struggle :) – mrtaurho Nov 24 '18 at 17:43
  • @mrtaurho I was wondering if we can think about some abstract Lie group as Lie subgroup of GL(n,R) for some n. That is my motivation. – Ofra Nov 24 '18 at 18:45

1 Answers1

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The adjoint representation of $G$ realizes it as a closed subgroup of $\mathrm{GL}(\mathfrak{g})$, where $\mathfrak{g}$ is the Lie algebra of $G$ (in general, the kernel of the adjoint representation consists of elements of $G$ that commute with the connected component of $G$ that contains the identity).

Stephen
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  • Naive question: why the image of G by the adjoint representation is closed in GL(g)? – Ofra Nov 24 '18 at 19:01
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    @Ofra It's not such a naive question! In general the image of a Lie group by a representation is not necessarily closed. However, in this case it follows from the facts that (i) the exponential map is a local diffeomorphism from $\mathfrak{g}$ onto $G$, (ii) the exponential is compatible with the adjoint representation, and (iii) a subgroup that is closed in some neighborhood of the identity is closed. – Stephen Nov 24 '18 at 19:15