Lie subgroups of SU(3) - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T14:47:13Z http://mathoverflow.net/feeds/question/65522 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/65522/lie-subgroups-of-su3 Lie subgroups of SU(3) Alfred Wood 2011-05-20T10:30:17Z 2011-05-24T10:01:14Z <p>Apart from images of representations of subgroups of SU(2), what are the Lie subgroups of SU(3)? Where should I look for a reference?</p> http://mathoverflow.net/questions/65522/lie-subgroups-of-su3/65530#65530 Answer by JosÃ© Figueroa-O'Farrill for Lie subgroups of SU(3) JosÃ© Figueroa-O'Farrill 2011-05-20T11:44:12Z 2011-05-20T12:41:43Z <p>A first approximation to an answer is to determine the Lie subalgebras of $\mathfrak{su}(3)$. Here is their Hasse diagram with edges denoting inclusions. One can work this out iteratively by working out the maximal subalgebras and this follows from Dynkin's work.</p> <p>&nbsp;&nbsp;&nbsp;<img src="http://dl.dropbox.com/u/5096148/su3subalgebras.png" /></p> <p>The notation $\mathfrak{so}(2)_{[\alpha,\beta]}$ means one element of the pencil of one-dimensional subalgebras of $\mathfrak{so}(2)\oplus \mathfrak{so}(2)$ and <code>$\mathfrak{so}(3)_{\text{irr}}$</code> is a subalgebra which acts irreducibly on the 3-dimensional irreducible representation.</p> http://mathoverflow.net/questions/65522/lie-subgroups-of-su3/65834#65834 Answer by Neil Strickland for Lie subgroups of SU(3) Neil Strickland 2011-05-24T10:01:14Z 2011-05-24T10:01:14Z <p>This is essentially just a slower walk through José's diagram.</p> <p>I'll assume you're interested in closed, connected subgroups $G\leq SU(3)$. I'll write $V$ for $\mathbb{C}^3$ regarded as a faithful representation of $G$ via the inclusion in $SU(3)$. If $G$ is abelian then we can decompose $V$ as a sum of three one-dimensional representations, say $V=V_1\oplus V_2\oplus V_3$, and then $$G \leq SU(V)\cap (U(V_1)\times U(V_2)\times U(V_3)) \simeq U(1)\times U(1).$$ All remaining questions about the abelian case are now easy, so I'll assume that $G$ is nonabelian.</p> <p>Now consider the rank of $G$, ie the dimension of its maximal torus. As $G$ is nonabelian and contained in $SU(3)$ the rank must be one or two. </p> <p>If the rank is one, then it is standard that $G$ is isomorphic to $SU(2)$ or $SU(2)/\{\pm I\}\simeq SO(3)$. Let $W_1$ denote the standard two-dimensional representation of $SU(2)$, and let $W_2$ denote the symmetric tensor square of $W_1$, which has dimension $3$. The action of $SU(2)$ on $W_2$ factors through $SO(3)$. The only irreducible representations of $SU(2)$ of dimension $\leq 3$ are $\mathbb{C}$, $W_1$ and $W_2$. It follows that one of the following holds:</p> <ol> <li>$G\simeq SU(2)$, and $V$ corresponds to $\mathbb{C}\oplus W_1$. This means that there is a one-dimensional subspace $L\leq V$ such that $G=\{g\in SU(V):g|_L=1_L\}$.</li> <li>$G\simeq SO(3)$, and $V$ corresponds to $W_2$. This means that there is a real subspace $X\leq V$ with $V=X\oplus iX$, and $G$ is the evident copy of $SO(X)$ in $SU(V)$.</li> </ol> <p>Suppose instead that the rank is two. This means that any maximal torus in $G$ is also a maximal torus in $SU(V)$, so $G$ is a parabolic subgroup. </p> <p>Suppose that $V$ is reducible as a representation of $G$. This implies that there is a one-dimensional subspace $L\leq V$ that is preserved by $G$, and thus that $G$ is contained in the image of the homomorphism $\phi:U(L^\perp)\to SU(V)$ given by $$\phi(g)=g\oplus(\det(g)^{-1}.1_L)$$ As $G$ is connected and nonabelian of rank two, I think it has to be the whole image of $\phi$.</p> <p>Suppose instead that $V$ is irreducible. I think it then follows from the standard story about parabolic subgroups that $G$ is all of $SU(V)$.</p>