Invariant vectors/eigenvectors of R(., v)v

[Sajet]

I'm afraid I need help again...

First, these two things are shown:

1) Let $v \in T_{\bar p}\mathbb{CP}^n, ||v|| = 1$. Then: $R(w, v)v = w \forall w \in (\mathbb Cv)^\perp$

2) Let $v \in T_{\bar p}\mathbb{HP}^n, ||v|| = 1$. Then: $R(w, v)v = w \forall w \in (v\mathbb H)^\perp$

Afterwards the following is supposed to be proven:

a) $R(iv, v)v = 4iv$ (in the case of $CP^n$)
b) $R(w, v)v = 4w \forall w \in (\mathbb Rv)^\perp\cap(v \mathbb H)$ (in the case of $HP^n$)

Unfortunately, I don't understand the very beginning of the following proof:

"It is already clear that $iv$ is an eigenvector of $R(., v)v$ (meaning $R(iv, v)v = \kappa iv$ for some $\kappa$)"

I've been on this since yesterday but I don't see why this is the case. Does it somehow follow from 1)?

In b) it is basically the same thing (I think) but the script is a little bit more elaborate - so maybe this helps. It reads:

"We have already shown that $(vH)\cap(\mathbb Rv)^\perp$ is an invariant subspace of the endomorphism $R(., v)v$. Let $w \in (vH)\cap (\mathbb Rv)^\perp$ be an eigenvector."

Do these two statements immediately follow from 1) and 2)? I mean 1) basically shows:

$R(., v)v|_{(\mathbb Cv)^\perp} = id_{(\mathbb Cv)^\perp}$

But I can't make the connection to $R(iv, v)v = \kappa iv$...

 

[quasar987]

Do you have an electronic version of the document you're reading?

 

[Sajet]

I do but it is in German.

 

[quasar987]

I don't understand either, sorry. :(

 

[Sajet]

Thank you anyway :)

 

[Sajet]

I might have an idea. At a different point the following theorem is introduced:

For every $\bar p \in \mathbb{CP}^n$ the tangent space $T_{\bar p} \mathbb{CP}^n$ carries the structure of a complex vector space. For $\iota \in U(n+1)$ we have $\bar \iota_*(\lambda v) = \lambda\bar \iota_*(v)$ for all $\lambda \in \mathbb C, v \in T\mathbb{CP}^n$. ($\bar \iota$ is the induced map on $\mathbb{CP}^n$).

Then there is a similar statement about $\mathbb{HP}^n$, namely $\iota \in Sp(n+1) \Rightarrow \bar \iota_*(v\mathbb H) = \bar \iota_*(v)\mathbb H$

Sp(n+1) and U(n+1) are defined as matrices A fulfilling $AA^* = I$.

If the map $B(w) := R(w, v)v$ were in $Sp(n+1), U(n+1)$, then the above theorem might be what the proof is referring to. I figured out that B is a self-adjoint endomorphism, therefore $B = B^*$ (right?). But that doesn't mean $BB^* = I$. For that to be true, it needs to be $B = B^{-1}$, meaning $R(R(w, v)v, v)v = w$. I've worked on this for the last couple of hours but I think this is not even true...

 

[Sajet]

Okay, I've got it now. Not that clear imo (at least for me it wasn't) but this is the explanation for anybody who cares:

$R(., v)v$ is a self-adjoint endomorphism. Therefore the tangent space has an orthonormal basis of eigenvectors and all eigenvalues are real. It follows that iv must be an eigenvector.

Then basically the same applies in the case of HPn.