# Vertical subspace in HPn (Riemannian submersions)

1. Apr 1, 2012

### Sajet

Hi!

I'm trying to work through a script on Riemannian submersions but I have some problems with one proof in particular (or more likely the general underlying concepts). No worries, this is not about the entire proof, but just one step.

This is about the quaternionic projective space

$\mathbb{H}P^n = \{pS^3 | p \in \mathbb{S}^{4n+3}\}$

as the orbit space under the free isometric group action

$S^3 \times \mathbb{S}^{4n+3} \rightarrow \mathbb{S}^{4n+3}, (g, p) \mapsto pg^{-1}$

The projection $\pi: \mathbb{S}^{4n+3} \rightarrow \mathbb{S}^{4n+3}/S^3 = \mathbb{H}P^n$ should be the associated Riemannian submersion if I understand correctly.

Therefore, we can speak of vertical and horizontal vectors in $T_p \mathbb{S}^{4n+3}$.

Without going into the entire context of the proof, I don't really understand the following:

"Let $v, w$ be horizontal unit vectors in $T_p \mathbb{S}^{4n+3}, p = (1, 0, ..., 0) \in \mathbb H^{n+1}.$ [...] We can regard v, w as vectors in $\mathbb H^{n+1}$ (through the canonical isomorphism). [...] The vertical subspace of $T_p \mathbb{S}^{4n+3}$ in p is generated by $(i, 0, ..., 0), (j, 0, ..., 0), (k, 0, ..., 0).$ Therefore the horizontal vectors v, w are of the form $(0, *, ..., *).$"

I don't understand why the vertical subspace is generated by these three vectors. Why isn't the vector (1, 0, ...) also necessary to generate the subspace?

2. Apr 3, 2012

### Sajet

Ok, I have some new insights that I can use to clarify my question. You can disregard the above. It boils down to this:

$\mathbb S^{4n+3} \subset \mathbb H^{n+1}$ is a manifold. For every $p \in \mathbb S^{4n+3}$ we have a submanifold $pS^3, S^3 \subset \mathbb H$.

What I'm looking for is the tangent space $T_p (pS^3)$ (vertical) and $(T_p (pS^3))^\perp \subset T_p \mathbb S^{4n+3}$ (horizontal) respectively.

$T_p (pS^3)$ seems to be generated by the vectors $ip, jp, kp$

I would be very grateful if someone could explain to me why this is the case.

3. Apr 3, 2012

### quasar987

This sounds like it's a simple computation. The vertical subspace Vp you're after is ker (pi*), where pi* means the derivative of the submersion pi at p. In the context of quotient spaces, Vp is just TpO, the so-called tangent space to the orbit O through p. Consider the map µp:S³-->S^{4n+3} defined by µp(g)=pg-1. Then TpO is just the image of (µp)* (derivative of µp at the identity element of the group). So you only need to compute this to find Vp! To to this, you will may find it helpful instead to consider S³ as acting on the whole H^n, and define instead µp:S³-->H^n. Of course, Vp is still just Im((µp)*), but now you don't have to worry about the unpleasant charts of S^{4n+3}.

4. Apr 3, 2012

### quasar987

More useful is to consider the action of S³ as the restriction of the action of the whole of H\{0}. This allows you to compute TpO as the R-span in H of

$$\dot{\gamma}(0), \ \dot{\sigma}(0), \dot{\tau}(0)$$

where

$$\gamma(t)=1+ti, \ \sigma(t) = 1+tj, \ \tau(t)=1+tk$$

5. Apr 3, 2012

### Sajet

Thank you!

All of your first post makes lots of sense to me. Unfortunately, I don't really see where $\gamma, \sigma, \tau$ come from in your second post and how it connects to the rest.

So, we have the action $\mu_p : S^3 \rightarrow \mathbb H^{n+1} \supset \mathbb S^{4n+3}, g \mapsto pg^{-1}$

I fear this is a very basic concept but I didn't manage to figure out how I can calculate the differential. Maybe this is even what your second post is about, but if so I can't find the connection. Do I need a chart for S³ for that? Or maybe this is why you said I should extend the action of the whole of $H\setminus \{0\}$.

This gives: $\mathbb H\setminus \{0\} \times \mathbb H^{n+1}, g \mapsto pg^{-1}$

But I'm still not sure how to differentiate this. I have never learned about complex or quaternionic differentiation, but maybe this is not even relevant here, I don't know...

The script script I have in front of me makes it sound like it is very trivial that the vertical space is generated by pi, pj, pk...

6. Apr 3, 2012

### quasar987

$\mathbb{H}$ is a real vector space, hence its tangent space at any point p is naturally identified with $\mathbb{H}$ itself (as a real vector space).

The tangent space to $S^3\subset\mathbb{H}$ at 1 is, when considered as a vector subspace of TpH=H, simply the R-span of the quaternions i,j,k. This is because S³={q in H | ||q|| = 1} and TpS³ = {q in H | (p,q)=0}, where ( , ) and || || are just the usual euclidean inner product and norm that you get on H by identifying it with R⁴ (as real vector spaces) via the natural map (1,0,0,0) <--> 1, (0,1,0,0) <--> i, (0,0,1,0) <--> j, (0,0,0,1) <--> k. In particular, just observe that (1,i)=(1,j)=(1,k)=0. Hence, i,j,k belong to T1S³ and they actually span T1S³ for dimensional reasons.

On the other hand, if you have a map F:M-->N between manifolds, and v in TpM a tangent vector, then the derivative of F at p in the direction of v, written F*v is equal to
$$\left.\frac{d}{dt}\right|_{t=0} (F\circ \gamma(t))$$
where $\gamma(t)$ is any smooth path such that $\gamma(0)=p$ and $\dot{\gamma}(0)=v$.

Now that we know that i,j,k span T1S³, we need to compute (µp)*i, (µp)*j and (µp)*k, where µp:S³--> H^n. For this, all we need is 3 curves $\gamma, \ \sigma, \ \tau$ in S³ passing through 1 at t=0 and s.t. $\dot{\gamma}(0)=i$, $\dot{\sigma}(0)=j$, $\dot{\tau}(0)=k$.

This is basically all that I said implicitely in my first post.

Then in the second post I said "Scratch that last part. Instead, consider µp defined on all of H\{0}. This makes our job easier, because now we don't have to take $\gamma, \ \sigma, \ \tau$ in S³, we just have to take them in H\{0}, and there are obvious candidates for such curves, that will make differentiation trivial, namely $\gamma(t)=1+ti, \ \sigma(t) = 1+tj, \ \tau(t)=1+tk$."

7. Apr 3, 2012

### Sajet

Thank you so much! Now I understand :)

You have really helped me out!

8. Apr 3, 2012

### quasar987

That was the plan! :)