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I was reading a paper by Labourie https://arxiv.org/pdf/math/0401230.pdf and I'm having trouble understanding a proof. First, here's the setup.

  • $G = PSL(n,\mathbb{R})$ and $U$ is the subgroup of diagonal matrices.

  • $P^+, P^-$ are the opposite Borel subgroups (i.e. the upper and lower triangular matrices).

  • Let $M = G/U$ with $m_0$ the class of identity.

and we have

  • $H=PSL(2,\mathbb{R})$ and $A$ is the subgroups of diagonal matrices (in $PSL(2,\mathbb{R}$)).

  • Notice there is a flow on $H$ (the geodesic flow) defined by right multiplication by elements $\begin{pmatrix} e^t & 0\\ 0 &e^{-t} \end{pmatrix}$ in $A$.

  • We have an action of $PSL(2,\mathbb{R})$ on $\mathbb{R}^{n} = \mathrm{Symm}^{n-1}\mathbb{R}^2$. This gives us an embedding of $PSL(2,\mathbb{R})$ into $PSL(n,\mathbb{R})$.

now we can define a map

  • $F:PSL(2,\mathbb{R}) \to M$ with $g\mapsto gU$ (i.e. the left multiplication by the image of the embedding from above).

  • Define $E =F^* TM$ the pullback bundle.

We can equip $E$ with an obvious choice of left-invariant metric

  • First let $q_{m_0}(\cdot,\cdot)$ be an arbitrary metric on the fiber over $m_0\in M = G/U$.

  • For any $gm_0$, define $q_{gm_0}(u,u) = (g^*q_{m_0})(u,u) = q_{m_0}(g_*^{-1}(u), g_*^{-1}(u))$ where $g_∗$ the linear map from $E_{m_0}$ to $E_{gm_0}$ associated to the action of an element g. This metric is left-invariant by construction.

So far I'm OK with the proof. My problem is with the following paragraph:

We finally have a right action of $A$ on $M$ commuting with the left $H$ action, this action of $A$ preserves globally the orbit $F(H)$; the corresponding action of $A$ on $H$ is the geodesic flow, ... . We therefore obtain a right action of $A$ on $E$. If $a$ is an element of $A$ and $q$ a left $H$ invariant metric, $\tilde{q} = a^* q$ is also a $H$-invariant metric completely determined by $q$. By construction, we have $\tilde{q}_{m_0} = Ad(a)q_{m_0}$...

So the questions are

  1. How is this right action by $A$ defined? If it's simply right multiplication by $A$, the image of $A$ under the embedding mentioned above is entirely contained in $U$ and the action on $M=G/U$ will be trivial. I guess it can be a different action but Labourie clearly states that it is the one that corresponds to the geodesic flow on $H=PSL(2,\mathbb{R})$ (which is by right multiplication).
  2. When constructing $\tilde{q}$ is $a^*$ defined in the same manner with $g^*$ in the definition of original $q$ because in that case, I'm not quite sure how he's getting $Ad(a)$ at the end. How is this $\tilde{q}$ constructed exactly?
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  • $\begingroup$ What's wrong with the trivial action? Ideally, the author should have said so, but, alas... $\endgroup$ – Moishe Kohan Aug 19 '20 at 18:55
  • $\begingroup$ @MoisheKohan Correct me if I'm wrong but it rather seems to me the metric in the second question is induced by this action making the action. This is why the trivial action doesn't look quite right... $\endgroup$ – nakanaka2828 Aug 21 '20 at 14:55

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