1
$\begingroup$

I'm trying to understand equation 6.65b from Knapp's 'Lie groups,' 2ed.


Setup: Let $\mathfrak{g}_0$ be a real semisimple Lie algebra with an involution $\theta$. Let $B$ be a bilinear, symmetric, non-degenerate, invariant, $\theta$-invariant form on $\mathfrak{g}_0$. Assume that $B_\theta(X,Y):= -B(X,\theta Y)$ is positive definite on $\mathfrak{g}_0$. We can decompose $\mathfrak{g}_0=\mathfrak{k}_0\oplus\mathfrak{p}_0$ as $\pm 1$ eigenspaces for $\theta$. Let $\mathfrak{g}$ be the complexification $\mathfrak{g}_0\otimes \mathbb{C}$, and by extension of scalars, extend $B$ and $\theta$ to $\mathfrak{g}$. We extend $B_\theta$ to a Hermitian inner product on $\mathfrak{g}$ by $B_\theta(X,Y):=-B(X,\theta\bar{Y})$.

Say $\mathfrak{h}_0 \subset \mathfrak{g}_0$ is a "Cartan subalgebra". That is, say that $\mathfrak{h}:=\mathfrak{h}_0 \otimes \mathbb{C} \subset \mathfrak{g}$ is a Cartan subalgebra. Assume that $\mathfrak{h}_0$ is $\theta$-stable so that we can write $\mathfrak{h}_0 = \mathfrak{a}_0 \oplus \mathfrak{t}_0$ where the components are in $\mathfrak{p}_0$ and $\mathfrak{k}_0$ respectively. Let $\Delta:= \Delta(\mathfrak{g}, \mathfrak{h}) \subset \text{Hom}_\mathbb{C}(\mathfrak{h}, \mathbb{C})$ be the set of roots. Recall, the roots are positive on $\mathfrak{a}_0 \oplus i\mathfrak{t}_0$ and that $B$ is positive definite here. We can thus transfer to an inner product on $V:=\text{span}_\mathbb{R} \Delta$ via the isomorphism $ \mathfrak{a}_0 \oplus i\mathfrak{t}_0 \to V;\ H \mapsto B(H, \cdot)$. This we denote by $\langle \cdot, \cdot \rangle$ and the norm we denote by $|\cdot|$.

Assume we have a root $\beta \in \Delta$ which is "imaginary". That is, it vanishes on $\mathfrak{a}_0$. This shows that the $\beta$-eigenspace satisfies $\theta \mathfrak{g}_\beta = \mathfrak{g}_\beta$ so that $\mathfrak{g}_\beta$ is contained in either $\mathfrak{p}:= \mathfrak{p}_0 \otimes \mathbb{C}$ or in $\mathfrak{k}:= \mathfrak{k}_0\otimes \mathbb{C}$. Assume that $\beta$ is "non-compact", that is, the eigenspace $\mathfrak{g}_\beta$ is contained in $\mathfrak{p}$.

Take any nonzero $E_\beta \in \mathfrak{g}_\beta$. The fact that $\beta$ is imaginary implies that $\overline{E}_\beta \in \mathfrak{g}_{-\beta}$. We can also assume that $B(E_\beta,\overline{E}_\beta) = 2 |\beta|^{-2}$ since we have $$ 0 < B_\theta(E_\beta, E_\beta) = - B(E_\beta, \theta \overline{E}_\beta) = B(E_\beta, \overline{E}_\beta). $$ If we define $H_\beta$ by the property $\beta(H) = B(H, H_\beta)\ \ \forall H \in \mathfrak{h}$ and $H'_\beta:= 2|\beta|^{-2} H_\beta$, we see that $[E_\beta, \overline{E}_\beta] = H'_\beta$ so that the triple $H'_\beta, E_\beta, \overline{E}_\beta$ generates a subalgebra isomorphic to $\mathfrak{sl}_2(\mathbb{C})$.

Now define the operator $$ c_\beta := \exp ad_{\frac{\pi}{4}\left(\overline{E}_\beta - E_\beta\right)} \in \text{Aut} \left(\mathfrak{g}\right) $$ which can be seen to satisfy \begin{equation}\label{use}\tag{1} c_\beta\left(H'_\beta\right) = E_\beta + \overline{E}_\beta. \end{equation}


Question: How do I see from \eqref{use} the identity \begin{equation}\label{ques}\tag{2} \mathfrak{g}_0 \cap c_\beta(\mathfrak{h}) = \text{ker}\left(\beta|_\mathfrak{h_0}\right) \oplus \mathbb{R} \left(E_\beta + \overline{E}_\beta\right). \end{equation}

Attempt: The definition of $c_\beta$ and the fact that $[\overline{E}_\beta - E_\beta, H] = 0$ for every $H\in \ker(\beta)$ shows that $\ker\left(\beta|_\mathfrak{h_0}\right) \subset \mathfrak{g}_0 \cap c_\beta(\mathfrak{h})$. And equation \eqref{use} shows that $\mathbb{R} \left(E_\beta + \overline{E}_\beta\right) \subset \mathfrak{g}_0 \cap c_\beta(\mathfrak{h})$.

It remains to show L.H.S $\subset$ R.H.S in \eqref{ques} which I'm unable to do. I was trying to show that the (real) dimension of the L.H.S is the same as the (real) dimension of $\mathfrak{h}_0$ which would suffice.

Please send help.

$\endgroup$

1 Answer 1

0
$\begingroup$

Er, I guess one can write $$ \mathfrak{h}_0 = \mathbb{R}(iH_\beta') \oplus \text{ker}(\beta|_{\mathfrak{h}_0})\ \text{ so that }\ \mathfrak{h}= \mathbb{C}(iH'_\beta) \oplus\mathbb{C}\cdot\text{ker}(\beta|_{\mathfrak{h}_0}). $$ Then \begin{equation} \begin{split} \mathfrak{g}_0\cap c_\beta(\mathfrak{h}) &= \mathfrak{g}_0 \cap \left(\mathbb{C}\left(\overline{E}_\beta + E_\beta\right) \oplus \mathbb{C}\cdot \text{ker}(\beta|_{\mathfrak{h}_0})\right) \\ &= \mathfrak{g}_0\cap \mathbb{C}\cdot\left(\left(\overline{E}_\beta + E_\beta\right) \oplus \text{ker}(\beta|_{\mathfrak{h}_0})\right) \\ &= \mathbb{R}\left(\overline{E}_\beta + E_\beta\right) \oplus \text{ker}(\beta|_{\mathfrak{h}_0}). \end{split} \end{equation} wefnawiebv;gwbeiwebg!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.