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I'm reading a paper on complex semi-simple algebraic group geometry at the moment, but finding the going a bit tough since I'm missing alot of the prerequisites. Firstly, the author refers to a principal embedding of one Lie group into another. I guess that this is an embedding of one group into another that is maximal in some sense. The paper states that in the dual Lie algebra setting this means that ${\frak g}$ is obtained from ${\frak g_0}$ by adding a node to the Dynkin diagram. The family of examples used is that of the embedding of $U_{N-1}$ into $SU_{N}$. I've searched the web but can't find a principal embedding defn, could someone point me in the right direction.

Secondly, if $G_0$ is principal embedded into $G$, is there some result about the quotient $G/G_0$ having a complex structure? This works for example $CP^{N-1} = SU_N/U_{N-1}$.

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What is the paper? I have heard of principal $sl_2$ embedding only... An what is the deal with complex structure since you $G$ is already complex? – Bugs Bunny Nov 18 '10 at 16:58
The paper is more of a privately communicated preprint. He uses complex in the introduction (connected, complex, semi-simple, algebraic group) but then speaks about $SL_2$, so I'm confused .... let's omit complex then because I'm interested in SL_2. – John McCarthy Nov 18 '10 at 17:10
The notion of "principal embedding" isn't familiar to me, though it may be so to Lie group specialists. In the Dynkin diagram context, this might just refer to a semisimple or reductive subgroup whose simple roots are a subset of some given simple system for the big group? (Words like "principal" and "regular" are used a lot in mathematics, thus need a reminder of what is meant in a special setting.) – Jim Humphreys Nov 18 '10 at 17:37
I'm confused by your comment, John. Perhaps you could clarify the situation a bit by specifying what is being embedded into what. Is the author looking at embeddings of SL_2 into a complex semisimple G? – Faisal Nov 18 '10 at 17:50
Seems like more context is needed, but it sounds like Bugs might get you on the right track. Specifically, there's the Jacobson-Morozov theorem about embedding $SL_2$. One place to learn about this stuff is the book by Collingwood and McGovern; another is <a href="…;. (By the way, $SL_2$ is often considered as a complex group -- just the usual definition, with complex entries. If this is your context, there's no trouble with $G/G_0$ being complex.) – Dave Anderson Nov 18 '10 at 17:52
up vote 2 down vote accepted

This doesn't answer your question completely, but at least it's a start.

If $G$ is a complex Lie group, then its quotient $G/H$ by a complex subgroup $H$ is always a complex manifold. In case $G$ is semisimple (and connected), then the compact quotients $G/H$ come from what are called parabolic subgroups $H$. For simplicity, let's assume that $G$ is simple. Then, up to conjugacy, the parabolic subgroups of $G$ lie in one-to-one correspondence with subsets of the nodes of the Dynkin diagram of $\mathfrak{g}$. It's easy to find a description of this bijection in the literature, so I will say no more about it. Let me just mention that a parabolic subgroup corresponding to omitting one node from the Dynkin diagram is called maximal parabolic.

Now suppose that $K$ is a compact real form of the complex semisimple group $G$. Then if the quotient space $G/H$ is compact (i.e., if $H$ is parabolic), a theorem of Montgomery asserts that $K$ acts transitively on $G/H$, and so we obtain an identification $G/H = K/(K\cap H)$. The a priori real manifold $K/(K\cap H)$ is thus endowed with a complex structure.

In summary, we have the following result:

Let $K$ be a compact semisimple Lie group with complexification $G$. Then, for a closed subgroup $S$ of $K$, the homogeneous space $K/S$ is complex if $S = K \cap P$ for some parabolic subgroup $P$ of $G$.

Here's an example. Let $G = \rm{SL}(n,\mathbb{C})$. In terms of the bijection between parabolic subgroups of $G$ and nodes, it's not hard to see that the maximal parabolic corresponding to deleting the $k$th node is the subgroup of $G$ preserving a $k$-dimensional subspace of $\mathbb{C}^n$. In particular, if we let $P$ denote the maximal parabolic corresponding to the deletion of the first node, we get $G/P = \mathbb{CP}^{n-1}$. Now the maximal compact subgroup of $G$ is $K = \rm{SU}(n)$. In an appropriate basis, $P$ is the subgroup of $G$ of the form $$ \begin{pmatrix} \ast & \ast & \cdots & \ast \\ 0 & \ast & \cdots & \ast \\ \vdots & \ast & \cdots & \ast \\ 0 & \ast & \cdots & \ast \end{pmatrix}. $$ Thus $K \cap P = \rm{U}(n-1)$ (to see this quickly, just recall that unitary matrices leave invariant the perp of an invariant subspace), where I'm thinking of $\rm{U}(n-1)$ as sitting in $K$ as $$ \begin{pmatrix} (\det A)^{-1} & 0 \\ 0 & A \end{pmatrix}. $$ So by the result stated above we obtain the identification $\rm{SU}(n)/\rm{U}(n-1) = \mathbb{CP}^{n-1}$ given in the OP. Using the other maximal parabolics we can get a similar description for the complex Grassmannian of $k$-planes in $\mathbb{C}^n$.

Here are a couple of useful references:

  1. Wang, Hsien-Chung, Closed manifolds with homogeneous complex structure. Amer. J. Math. 76 (1954). 1–32.

  2. Wolf, Joseph A, The action of a real semisimple group on a complex flag manifold. I. Orbit structure and holomorphic arc components. Bull. Amer. Math. Soc. 75 (1969), 1121–1237.

Both these papers address the issue of finding complex structures on compact homogeneous spaces (so their results apply in particular to homogeneous spaces coming from compact Lie groups). There's a standing assumption of simply connectedness in Wang, but if I remember correctly, there is no such restriction in Wolf's paper.

Edit: Here's some more information that might be helpful. Let $K$ and $G$ be as above. A parabolic subgroup $P$ of $G$ admits "Levi decompositions" $P=LU$, where $U$ is the unipotent radical of $P$ and $L$ is a reductive group (called a Levi factor). In particular, every Levi factor $L$ has a compact real form $L_{\mathbb{R}}$. The different Levi factors of $P$ are all conjugate. There is a choice that is compatible with the choice of maximal compact $K$, and in this case we have $P \cap K = L_{\mathbb{R}}$.

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By the way, I should mention that I interpreted $\mathfrak{g}$ and $\mathfrak{g}_0$ in your question to be the complexifications of the Lie algebras of $G$ and $G_0$. So if $G$ is semisimple, then saying that $G_0$ is principally embedded in $G$ amounts to saying that $\mathfrak{g}_0$ is a maximal parabolic subalgebra of $\mathfrak{g}$. – Faisal Nov 18 '10 at 19:58
Are maximal parabolic subalgebras unique? – John McCarthy Nov 18 '10 at 21:12
No. There's one conjugacy class of maximal parabolics for each node in the Dynkin diagram. I said "the" maximal parabolic in my answer above because I had a specific one in mind (namely the one whose intersection with $K$ gives us a copy of $\rm{U}(n-1)$). I'll edit my answer to fix this. – Faisal Nov 19 '10 at 0:01
In the case of $SU_N$ does a quotient by one of these maximal parabolic subgroups always give $CP^{N-1}$? – John McCarthy Nov 19 '10 at 16:47
You get a Grassmannian. – Faisal Nov 19 '10 at 18:39

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