Isometric diffeomorphism on HPn/S^4n+3

[Sajet]

Hi!

I'm working through this script and I'm not sure if if there is a mistake at one point, or if I'm just thinking wrong.

It is to be shown that for every two unit vectors $v, w \in T \mathbb{HP}^n$ there exists an isometric diffeomorphism $\iota: \mathbb{HP}^n \rightarrow \mathbb{HP}^n$ with $\iota_*(v) = w, (\mathbb{HP}^n = \mathbb S^{4n+3}/S^3)$.

To prove this, the group $Sp(n+1) = \{A \in M(n+1, \mathbb H) | A^*A = I\}$ is used. Its elements operate linearly and isometrically on $\mathbb{S}^{4n+3}$ and therefore induce isometries on $\mathbb{HP}^n$.

The proof starts by first defining two unit vectors $v, w \in T_p \mathbb S^{4n+3}, p := (1, 0, ..., 0) \in \mathbb H^{n+1}.$. Now it says: "It suffices to find $A \in Sp(n+1)$ with $A_*v = A_*w$."

Is this correct? Doesn't it have to say $A_*v = w$? Wouldn't $A_*v = A_*w$ imply that $v = w$ since all matrices in the symplectic group are invertible?

[By the way, it goes on: "If we regard v, w as vectors in $H^{n+1}$ this problem is equivalent to: There is $A \in Sp(n+1), Ap = p, Av = w$. Maybe this helps."]

 

[quasar987]

Seems like a typo to me also.

 

[Sajet]

Thank you!

 

[Sajet]

I have a small follow up question. I hope it is ok that I post here again, as it is closely related to my first post.

Lateron, the curvature of $\mathbb{CP}^n, \mathbb{HP}^n$ is analyzed, namely:

Let $v \in T_{\bar p}\mathbb{HP}^n$ be a vector of length 1. Then for every $w \in (v\mathbb H)^\perp$: $R(w, v)v = w$ (plus the identical statement on CPn)

The proof goes as follows: [Because of the statement in my first post,] we can assume $\bar p = pS^3$ for $p = (1, 0, ..., 0), v = \pi_*(J_p(q))$ where $q = (0, 1, 0, ..)$ and $J_p$ the canonical isomorphism, $\pi: \mathbb S^{4n+3} \rightarrow \mathbb H^n$ the Riemannian submersion.

Let $w \in (v\mathbb H)^\perp$. Then $w = \pi_*(J_p(q'))$ for $q' = (0, 0, q_3', ...)$.

Now it goes on:

Choose $A \in Sp(n+1)$ with $A(p) = p, A(q) = q, A(q') = (0, 0, \lambda, 0, ..., 0), \lambda \in \mathbb R$.

And I don't see why such a matrix A necessarily exists. I mean, because of the statement in my first post, it is clear that there is a matrix A with $A(p) = p, A_*(q) = A(q) = q$. And because it is an isometric diffeomorphism we still have $A(q') \perp q \Rightarrow A(q') = (0, 0, s_3', s_4', ...)$. But I don't see how I can necessarily get $A(q') = (0, 0, \lambda, 0, ...)$

 

[quasar987]

Does it work to simply take A to be block diagonal of the form
$$\left(<br /> \begin{array}{cccc}<br /> 1 & & & \\<br /> & 1 & & \\<br /> & & \lambda/(q_3')^* & \\<br /> & & & A' <br /> \end{array}<br /> \right)$$
for A' any Sp(n-3) matrix?

 

[Sajet]

I don't think this matrix will necessarily be in Sp(n+1). Also, it is not guaranteed that $q_3' \neq 0$.

But I think I know how it works now. Your matrix gave me an idea:

Because $Sp(n-1)$ acts transitively on $\mathbb S_{||q'||}^{4(n-1)-1}$, there is a matrix $A' \in Sp(n-1)$ with $A'(q_3', ..., q'_{n+1}) = (\lambda, 0, ..., 0), \lambda \in \mathbb R$. (Then $|\lambda| = ||q'||$.)

Then I can just take $A(v_1, ..., v_{n+1}) := (v_1, v_2, A'(v_3, ..., v_{n+1}))$. This should be in $Sp(n+1)$ and have the desired properties :)

Thank you for your help!

 

[quasar987]

This certainly does the trick. Good job!