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Jef
Jef
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Let us first show that a semisimple Lie group with finite center is compactnon-compact if and only if its Lie algebra contains a copy of $sl(2,R)$.

Indeed, if $G$ is compact, the Killing form is definite negative definite, and its restriction to a copy of $sl(2,R)$ would be a definite negative invariant form, but such forms do not exist on $sl(2,R)$. Conversely, if $G$ is not compact, then $Lie(G)$ contains "real roots" for suitable Cartan sub-algebras. Any root vector associated to such a real root will be part of a copy of $sl(2,R)$ inside $Lie(G)$.

Now let $V$ be the adjoint representation of $G$. It induces a faithful representation of $Lie(G)$, hence a non-trivial representation of a fixed copy of $sl(2,R)$ inside $Lie(G)$. So it has an irreducible subrepresentation $(\rho,W)$ which induces a non-trivial, hence faithful, representation of $sl(2,R)$. So the Lie algebra of the closure $\rho(G)$ contains a copy of $sl(2,R)$ and in fact, so does the Lie algebra of the closure of the image of $G$ in $PGL(W)$. Therefore the latter closure is non-compact.

Let us first show that a semisimple Lie group with finite center is compact if and only if its Lie algebra contains a copy of $sl(2,R)$.

Indeed, if $G$ is compact, the Killing form is definite negative, and its restriction to a copy of $sl(2,R)$ would be a definite negative invariant form, but such forms do not exist on $sl(2,R)$. Conversely, if $G$ is not compact, then $Lie(G)$ contains "real roots" for suitable Cartan sub-algebras. Any root vector associated to such a real root will be part of a copy of $sl(2,R)$ inside $Lie(G)$.

Now let $V$ be the adjoint representation of $G$. It induces a faithful representation of $Lie(G)$, hence a non-trivial representation of a fixed copy of $sl(2,R)$ inside $Lie(G)$. So it has an irreducible subrepresentation $(\rho,W)$ which induces a non-trivial, hence faithful, representation of $sl(2,R)$. So the Lie algebra of the closure $\rho(G)$ contains a copy of $sl(2,R)$ and in fact, so does the Lie algebra of the closure of the image of $G$ in $PGL(W)$. Therefore the latter closure is non-compact.

Let us first show that a semisimple Lie group with finite center is non-compact if and only if its Lie algebra contains a copy of $sl(2,R)$.

Indeed, if $G$ is compact, the Killing form is negative definite, and its restriction to a copy of $sl(2,R)$ would be a definite negative invariant form, but such forms do not exist on $sl(2,R)$. Conversely, if $G$ is not compact, then $Lie(G)$ contains "real roots" for suitable Cartan sub-algebras. Any root vector associated to such a real root will be part of a copy of $sl(2,R)$ inside $Lie(G)$.

Now let $V$ be the adjoint representation of $G$. It induces a faithful representation of $Lie(G)$, hence a non-trivial representation of a fixed copy of $sl(2,R)$ inside $Lie(G)$. So it has an irreducible subrepresentation $(\rho,W)$ which induces a non-trivial, hence faithful, representation of $sl(2,R)$. So the Lie algebra of the closure $\rho(G)$ contains a copy of $sl(2,R)$ and in fact, so does the Lie algebra of the closure of the image of $G$ in $PGL(W)$. Therefore the latter closure is non-compact.

Source Link
Jef
Jef
  • 686
  • 6
  • 3

Let us first show that a semisimple Lie group with finite center is compact if and only if its Lie algebra contains a copy of $sl(2,R)$.

Indeed, if $G$ is compact, the Killing form is definite negative, and its restriction to a copy of $sl(2,R)$ would be a definite negative invariant form, but such forms do not exist on $sl(2,R)$. Conversely, if $G$ is not compact, then $Lie(G)$ contains "real roots" for suitable Cartan sub-algebras. Any root vector associated to such a real root will be part of a copy of $sl(2,R)$ inside $Lie(G)$.

Now let $V$ be the adjoint representation of $G$. It induces a faithful representation of $Lie(G)$, hence a non-trivial representation of a fixed copy of $sl(2,R)$ inside $Lie(G)$. So it has an irreducible subrepresentation $(\rho,W)$ which induces a non-trivial, hence faithful, representation of $sl(2,R)$. So the Lie algebra of the closure $\rho(G)$ contains a copy of $sl(2,R)$ and in fact, so does the Lie algebra of the closure of the image of $G$ in $PGL(W)$. Therefore the latter closure is non-compact.