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Normally in differential topology, we consider both manifolds with and without a boundary. However, for a Lie group do we only restrict our attention to manifolds without boundary?

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A Lie group is homogeneous: for every $g,h \in G$ there exists a diffeomorphism $f : G \to G$ taking $g$ to $h$, namely $f(k) = k g^{-1} h$.

But an $n$-dimensional manifold with nonempty boundary is not homogeneous: if $x \in \partial M$ and $y \in \text{int}(M)$ then there is no diffeomorphism $f : M \to M$ such that $f(x)=y$, because the local homology groups $H_n(M,M-y) \approx \mathbb{Z}$ and $H_n(M,M-x) \approx 0$ are not isomorphic, as they would have to be if $f$ existed.

Therefore, yes, by necessity we only restrict our attention to manifolds without boundary when we are studying Lie groups.

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Recall that a Lie group is also a topological group. Therefore, there is a homeomorphism between any two points. Therefore, if one point is internal, all are. Therefore, every point is internal.

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I believe you want to ask can a Lie group be a manifold with boundary ? No, since every point in a Lie group has a neighborhood diffeomorphic to a neighborhood of the identity.

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  • $\begingroup$ Yes, that is my question. Can it be? $\endgroup$
    – z.z
    Oct 10, 2017 at 13:41

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