# Subspaces of TCPn THPn

1. Apr 18, 2012

### Sajet

Hi!

I have the following statements in a script on Riemannian submersions:

($\pi$ is the submersion $\mathbb S^{2n+1} \rightarrow \mathbb{CP}^n$ or $\mathbb S^{4n+3} \rightarrow \mathbb{HP}^n$ respectively.)

Regarding a) it is then said: "Let $w \in T\mathbb{CP}^n, \lambda \in \mathbb C$. Let $\bar w$ be a horizontal lift of $w$. Define $\lambda w := \pi_*(\lambda \bar w)$. It is easily checked that this is well-defined."

I thought this was pretty clear. But then in b) they say:

"Let $w \in T\mathbb{HP}^n$, let $\bar w$ be a horizontal lift of w. Define $w\mathbb H := \pi_*(\bar w\mathbb H)$. It is also easily checked that this is well-defined.

Warning: For $\lambda \in \mathbb H$ we cannot set $w\lambda := \pi_*(\bar w\lambda)$ as this is not well-defined."

Now I don't see why exactly the last part is not well-defined. I thought the horizontal lift is unique, therefore $\bar w \lambda$ would be unique and $\pi_*(\bar w\lambda)$ as well.

Or maybe I just don't understand what well-defined means in either case, and why exactly this definition would be viable in a) but not in b).

I'd be very grateful if someone could help me understand this.

2. Apr 18, 2012

### quasar987

The horizontal lift is unique only once a point in the fiber has been chosen.

That is, if p:M--N is a riemannian submersion and y is a point in N with v in TyN a tg vector at y, then to lift v, we need first to chose a point x in p-1(y), and then the horizontal lift of v to TxM is unique.

So they appear to be saying that for complex lambda, $\pi_*(\lambda\overline{w})$ is actually independant of the choice of x to lift too, but not in the case of quaternionic lambda.

But I don't even know what they mean by $\lambda \overline{w}$ in the complex case. Because the tg space of S^{2n+1}, being of odd dimension, cannot support a complex vector space structure...

3. Apr 19, 2012

### Sajet

Wow, I completely ignored the fact that you first have to choose a point in the fiber in order to make the horizontal lift unique...

You're right that $T_p\mathbb S^{2n+1}$ does not carry a complex vector space structure, but the horizontal subspace $T_p^h\mathbb S^{2n+1} = (p\mathbb C)^\perp \cap \mathbb C^{n+1}$ does, and this is enough for this purpose.

Now I'm trying to see exactly why the multiplication is well-defined in a) and not in b).

Ok, I managed to construct a fairly complicated proof for a) but I hope there is an easier way:

I want to give a presentation on this next week and I wouldn't want to make a fool of myself by proving something obvious in such a difficult way.

Do you happen to see if this can be proven much more quickly?

4. Apr 20, 2012

### quasar987

If by "much more quickly" you mean without the need to compute, then I think I do! And as a bonus, we understand what fails in the quaternionic case.

First, a bit of notations. If a group G acts by riemannian isometries on (M,g), write $\pi:M\rightarrow M/G$ for the corresponding riemannian submersion, and write $\theta_g:M\rightarrow M$ for the map $p\mapsto g\cdot p$. Then $\pi=\pi\circ \theta_g$ for any g. Differentiating this relation gives $\pi_*=\pi_*\circ (\theta_g)_*$. Since $(\theta_g)_*$ is an isometric isomorphism, it preverses the horizontal and vertical subspaces. In particular, let p and q=gp be two points of M in the same G-orbit, let $w\in T_{[p]}(M/G)$, and let $\mathrm{Hor}_p(w)$, $\mathrm{Hor}_q(w)$ be the corresponding horizontal lifts of w above p and above q respectively. Then, $(\theta_g)_*(\mathrm{Hor}_p(w))=\mathrm{Hor}_q(w)$.

Now, in the case that interests us, M=S2n+1, G=S1, M/G=CPn, and for a given p in S2n+1, TpS2n+1 is naturally identified with $p^{\perp}\subset\mathbb{C}^{n+1}$. With this identification, $V_pS^{2n+1} = \mathrm{Ker}(\pi_*)=T_p(S^1\cdot p)$ is then naturally identified with $\mathbb{R}ip$, and so $H_pS^{2n+1}=(V_pS^{2n+1})^{\perp}$ is then naturally identified with $(ip)^{\perp}\cap p^{\perp}=\{p,ip\}^{\perp}$. As you noted, this is naturally a complex subspace of $\mathbb{C}^{n+1}$. Moreover, if q=up for some u in S1 are two points in the same orbit, then I hold that $(\theta_u)_*:\{p,ip\}^{\perp}\rightarrow \{q,iq\}^{\perp}$ is just the map multiplication by u itself. To see this is the same trick that I explained to you in an earlier post: extend $\theta_u$ to a map Cn+1-->Cn+1. This is linear, so its derivative is just itself: multiplication by u. Now restrict back to the subspaces that interest you and you get that $(\theta_u)_*:\{p,ip\}^{\perp}\rightarrow \{q,iq\}^{\perp}$ is the map $\mathbf{v}\mapsto u\mathbf{v}$. In particular, it is a C-linear isomorphism: for any $\lambda\in\mathbb{C}$, $(\theta_u)_*(\lambda \mathbf{v})=u\lambda \mathbf{v}=\lambda u\mathbf{v}=\lambda (\theta_u)_*(\mathbf{v})$. This is why it makes sense to define a complex structure on CPn by setting $\lambda w:=\pi_*(\lambda\mathrm{Hor}_p(w))$. Indeed, we have $\pi_*(\lambda\mathrm{Hor}_q(w))=\pi_*(\lambda (\theta_u)_*(\mathrm{Hor}_p(w)))=\pi_*( (\theta_u)_*(\lambda\mathrm{Hor}_p(w)))=\pi_*( \lambda \mathrm{Hor}_p(w))$.

Notice from the above computation that the C-linearity of $(\theta_g)_*$ is equivalent to the commutativity of the multiplication in C. So this is what fails in the quaternionic case! However, given any $u\in S^3\subset \mathbb{H}$ and any $\lambda \in \mathbb{H}$, there exists $\mu \in \mathbb{H}$ such that $(\theta_u)_*(\lambda \mathbf{v})=u\lambda \mathbf{v}=\mu u\mathbf{v}=\mu (\theta_u)_*(\mathbf{v})$. So while it does not make sense to define $\lambda w:=\pi_*(\lambda\mathrm{Hor}_p(w))$ in the quaternionic case, it does make sense to speak of $\mathbb{H}w$ as $\pi_*(\mathbb{H}\mathrm{Hor}_p(w))$ for any p.

5. Apr 20, 2012

### Sajet

Wow, thank you, this helps me tremendously! This is clearly a much better and more insightful way of approaching this.