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
<br /> \left(<br /> \begin{array}{cccc}<br /> 1 &amp; &amp; &amp; \\<br /> &amp; 1 &amp; &amp; \\<br /> &amp; &amp; \lambda/(q_3&#039;)^* &amp; \\<br /> &amp; &amp; &amp; A&#039; <br /> \end{array}<br /> \right)<br />
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&#039; \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&#039;||}^{4(n-1)-1}, there is a matrix A&#039; \in Sp(n-1) with A&#039;(q_3&#039;, ..., q&#039;_{n+1}) = (\lambda, 0, ..., 0), \lambda \in \mathbb R. (Then |\lambda| = ||q&#039;||.)

Then I can just take A(v_1, ..., v_{n+1}) := (v_1, v_2, A&#039;(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!