Let $G_\mathbb{C}$ be a complex simple Lie group and let $V_\lambda$ be its finite dimensional irreducible representation with highest weight $\lambda$. Define $\mathcal{H}\_{\mathbb{C}} \subset V_\lambda$ to be the set of all possible highest weight vectors for all possible choices of Borel subgroups in $G_\mathbb{C}$.

It is well known that the projectivization $\mathbb{P}\mathcal{H}\_\mathbb{C} \subset \mathbb{P}V$ is homogeneous space $G\_\mathbb{C}/P$ where $P$ is a parabolic subgroup easily determined by $\lambda$. As was reminded to me by Sasha and Alex, the action of $G_\mathbb{C}$ is even transitive on $\mathcal{H}\_\mathbb{C}$. It is also known that the space $G\_\mathbb{C} / P$ is in fact a projective algebraic variety cut out by quadrics.

Now consider a real form $G_\mathbb{R}$ and its real irreducible representation $V\_\mathbb{R}$ such that its complexification is $V_\lambda$.

Question 1: Does $G\_\mathbb{R}$ act transitively on the set of real points of $\mathcal{H}\_\mathbb{C}$ and of $\mathbb{P}\mathcal{H}\_\mathbb{C}$?

Consider a real form $G_\mathbb{R}$ with irreducible complex representation $V$ such that $V \simeq V_\lambda$ as a $G\_\mathbb{C}$ representations and let $v_\lambda$ be a highest weight vector.

Question 2: What are the $G\_\mathbb{R}$ orbits of $v_\lambda$ and $[v_\lambda]\in\mathbb{P}V_\lambda$?

Edit: Due to a highly nontrivial amount of confusion on my part the question was edited three times and so some of the answers and comments may seem out of place. I deeply apologize for that.

  • $\begingroup$ Regarding question 1: Fix a Borel subgroup $B$ and corresponding highest weight vector $v_{\lambda}$. Then for any other Borel subgroup $B'$, there exists $g \in G$ such that $gBg^{-1} = B'$. In this case, the highest weight vector $v_{\lambda}'$ associated to $B'$ is simply $gv_{\lambda}$. Therefore $G$ does act transitively on highest weight vectors. It seems to me that the stabilizer would then be the unipotent radical of $B$. $\endgroup$ Nov 9, 2011 at 16:48
  • $\begingroup$ You are right, except that highest weight vectors are only unique up to scale. That's why the classical result talks about projectivization. One needs an element in $G$ that acts as scaling in order to have transitivity on $\mathcal{H}$. $\endgroup$ Nov 9, 2011 at 17:29
  • $\begingroup$ What exactly do you mean by the "real case"? $\endgroup$
    – Faisal
    Nov 9, 2011 at 18:59
  • $\begingroup$ If $v'$ is a highest weight vector for the Borel $B'$, then the elements of $B'$ will act as scalings on $v'$. That's the scalings you need. $\endgroup$
    – Alex
    Nov 9, 2011 at 19:14
  • $\begingroup$ re latest edit: I still don't really understand what you mean by real Lie groups. This highest weight business doesn't work anymore. Are you talking about real forms of complex simple Lie groups? $\endgroup$
    – Faisal
    Nov 9, 2011 at 20:56

2 Answers 2


There seems to be some confusion in the question, so let me try to recap the basic setup. Thus let $G$ be a complex simple Lie group, $V^\lambda$ the irrep of $G$ of highest weight $\lambda$, and $G/P$ the $G$-orbit in $\mathbb P V^\lambda$ through the line spanned by a highest weight vector in $V^\lambda$. In the question it's claimed that if $G_0$ is a real form of $G$ then the action of $G_0$ on $G/P$ is transitive. This is not true. For example, if $G=\operatorname{SL}_2 \mathbb C$ then the action of $G_0 = \operatorname{SL}_2 \mathbb R$ on the sphere $G/P = \mathbb P^1$ has three orbits (two hemispheres and the circle in between them). In general, there will be finitely many $G_0$-orbits in $G/P$. A wealth of information can be found in

Joseph Wolf, 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.

If $G_0$ is a compact real form then there is only one orbit, i.e. $G_0$ acts transitively $G/P$. This gives the familiar "compact picture" $G_0/G_0 \cap P$ of the flag variety. On the other hand, if $G_0$ is noncompact then it's relatively uncommon for its action on $G/P$ to be transitive, but it is possible. Joseph Wolf worked out the list of $G_0$ for which this is the case in

Joseph Wolf, Real groups transitive on complex flag manifolds, Proc. Amer. Math. Soc. 129 (2001), 2483-2487.

  • $\begingroup$ I guess in case of a real group a real representation is considered. $\endgroup$
    – Sasha
    Nov 10, 2011 at 5:40
  • $\begingroup$ I am sorry but I edited my question once again. Your answer solves the second part of my question 2. Thank you! $\endgroup$ Nov 10, 2011 at 11:48

Let $v$ be a fixed highest weight vector and $B$ the corresponding Borel. Then $B$ acts on $v$ via the map $B \to B/U = H \stackrel{\lambda}\to G_m$. Let $K_\lambda$ denote the kernel. Then ${\mathcal H} = G/K_\lambda$.

ADDITION. In the real case the answer depends on the type of group and representation. For example, if you take $SL$ acting on the standard representation $V$, the action on ${\mathcal H} = V \setminus \{0\}$ is transitive, but if you take $SO$ and any representation then the action is not transitive since the length of a vector is preserved. And if $G = SL\times SO$ then if you take the standard representation of $SL$ the action is transitive and if you take any representation of $SO$ it is not.

In general you should again consider the map $H \stackrel{\lambda}\to G_m = {\mathbb R}^\times$. If the map is surjective then the action is transitive. And if the image of the map is contained in $S^1$ (this is always the case if $G$ is compact) then the action is not transitive.

  • $\begingroup$ I am sorry, but I don't understand. What is $U$, $G_m$ and how is the morphism $\lambda$ defined? $\endgroup$ Nov 9, 2011 at 18:51
  • 2
    $\begingroup$ $U$ is the unipotent radical of $B$, $G_m$ is $\mathbb{C}^\times$ (in this case), $H=B/U$ is isomorphic to a maximal torus of $G$, so you can see $\lambda$ as a map from $B/U$ to $ùmathbb{C}^\times$. If $v$ is a highest weight vector for $B$, then $U$ acts trivially on $v$, and $B/U$ acts by multiplication by this $\lambda:B/U\rightarrow\C^\times$, because that is the definition of a highest weight vector. Sasha's message means : 1/ yes, $G$ acts transitively on the set of highest weight vectors 2/ the stabilizer of a highest weight vector $v$ for $B$ is the $K_\lambda$ he or she defined. $\endgroup$
    – Alex
    Nov 9, 2011 at 19:21
  • $\begingroup$ Sorry for the math. I meant "you can see $\lambda$ as a map from $B/U$ to $\mathbb{C}^\times$". $\endgroup$
    – Alex
    Nov 9, 2011 at 19:22
  • $\begingroup$ Ach so... I forgot that the maximal torus for complex groups is not a product of circles but rather of $\mathbb{C}^\times$. Thank you! I will edit my question accordingly. $\endgroup$ Nov 9, 2011 at 20:38

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