Well-defined map: transvections in symmetric space

[Sajet]

Hi!

I'm trying to understand a proof for the fact that the isometry group of a symmetric space is a Lie group. The proof uses a lemma and I don't see how the lemma works. Here is the statement in question:

Let $M$ be a symmetric space with involutions $s_p$, transvections $\tau_v$ and a point $p_0 \in M$. Then:

$\tau_p(q)$ depends smoothly on $(p, q) \in M \times M$ where $\tau_p$ is $\tau_v$ such that $\tau_v(p_0) = p.$

(Let me give you the definition for $\tau_v$: Let M be a symmetric space and $c:\mathbb R \rightarrow M$ a geodesic with $p = c(0), v = \dot c(0).$ Then for $t \in \mathbb R$ the isometries

$\tau_{tv} = s_{c(t/2)}\circ s_{c(0)}$

are called transvections.)

I don't see how the map $(p, q) \mapsto \tau_p(q)$ is even well-defined. There can be more than one transvection mapping $p_0$ to $p$, and different transvections will in general give different values $\tau_p(q)$.

 

[jvicens]

Hello,

Thank you for your question. I understand that the proof for the isometry group of a symmetric space being a Lie group can be confusing, especially when it involves a lemma. Let me try to explain it in simpler terms.

First, let's define what a symmetric space is. A symmetric space is a smooth manifold where each point has an involution (a transformation that is its own inverse) and this involution preserves the geometry of the space. This means that if you apply the involution to any point on the space, the distance and angles from that point to other points will remain the same.

Now, let's look at the lemma in question. It states that for a symmetric space M with involutions s_p and transvections \tau_v, the map (p, q) \mapsto \tau_p(q) is a smooth function. This means that for any two points p and q on the symmetric space, the value of \tau_p(q) will depend smoothly on the points p and q. In other words, if you move p and q slightly, the value of \tau_p(q) will also change smoothly.

The reason for this is because the transvections are defined in such a way that they always map one specific point (p_0) to another point (p). This means that for any given point p, there is only one transvection that maps p_0 to p. Therefore, the map (p, q) \mapsto \tau_p(q) is well-defined and depends only on the points p and q.

I hope this helps to clarify the lemma and how it relates to the isometry group of a symmetric space being a Lie group. If you have any further questions, please don't hesitate to ask. Thank you for your interest in this topic.