Showing that SO(2,1) is a maximal subgroup of SL(3) We define SO(2,1) as the stabilizer of the matrix diag(1,1,-1) in the action of GL(2) on the space of symmetric matrices, given by $g.M=gMg^t$. 
I know that SO(2) is maximal in SL(2) and I have tried assuming there is $SO(2,1)<H<SL(3)$ and looking at the intersection of H with a copy of SL(2), but it doesn't seem to help.
Could anyone suggest how to approach this problem?
Thanks.
 A: The main step should be looking at the Lie algebras. Via the inclusion $SO(2,1)\hookrightarrow SL(3,\mathbb R)$, the Lie algebra $\mathfrak{sl}(3,\mathbb R)$ becomes a representation of $SO(2,1)$. The Lie algebra $\mathfrak{so}(2,1)$ is an $SO(2,1)$-invariant subspace, which by general results admits an invariant complement. Denoting by $\mathbb R^3$ the standard representation of $SO(2,1)$ it is well known that $\mathfrak{so}(2,1)\cong\Lambda^2\mathbb R^3$ which easily implies that $\mathfrak{sl}(3,\mathbb R)\cong\mathfrak{so}(2,1)\oplus S^2_0(\mathbb R^3)$. Here the second summand is the tracefree part in the symmetric square of $\mathbb R^3$, which is an irreducible represetation of $\mathfrak{so}(2,1)$ and not isomorphic to $\mathfrak{so}(2,1)$. Now if $\mathfrak h\subset\mathfrak{sl}(3,\mathbb R)$ is a Lie subalgebra containing $\mathfrak{so}(2,1)$, then in particular, it is an $\mathfrak{so}(2,1)$-invariant subspace. By standard results in Lie theory, this implies that either $\mathfrak h=\mathfrak{so}(2,1)$ or $\mathfrak h=\mathfrak{sl}(3,\mathbb R)$. Since there is no proper subgroup of $SL(3,\mathbb R)$ with Lie algebra $\mathfrak{sl}(3,\mathbb R)$, we must have $\mathfrak h=\mathfrak{so}(2,1)$. So it only remains to show by elementary means that $SO(2,1)$ is the maximal subgroup of $SL(3,\mathbb R)$ with Lie algebra $\mathfrak{so}(2,1)$. 
