Isomorphism between tangent bundle of $S^2$ and the kernel of a bundle homomorphism

Question

Let $S^{4n+3} \to \mathbb{H}P^n$ be the standard projection which is a fiber bundle with fiber $S^3$. By the action of $S^1$ on $S^3$ we get a fiber bundle $$ \mathbb{C}P^1 \xrightarrow{\iota} \mathbb{C}P^{2n+1} \xrightarrow{\pi} \mathbb{H}P^n $$ where $\iota$ is a inclusion into one fiber.

Taking the tangent bundle over each space we get the following diagram $\require{AMScd}$ \begin{CD} T\mathbb{C}P^1 @>>> T\mathbb{C}P^{2n+1} @>>> T\mathbb{H}P^n\\ @VVV @VVV @VVV\\ \mathbb{C}P^1 @>\iota>> \mathbb{C}P^{2n+1} @>\pi>> \mathbb{H}P^n \end{CD} Pulling back via $\pi$ and using that a fiber bundle projection is a submersion we get an isomorphism $T\mathbb{C}P^{2n+1} \cong \pi^{*}(T\mathbb{H}P^n) \oplus L_n$ where $L_n$ is the kernel of the bundle homomorphism $T\mathbb{C}P^{2n+1} \twoheadrightarrow \pi^{*}(T\mathbb{H}P^n)$ and it is a two-dimensional real vector bundle. Now we can pull this back via $\iota$ and get a monomorphism $$ T\mathbb{C}P^1 \hookrightarrow \iota^*(T\mathbb{C}P^{2n+1}) \cong \iota^*(\pi^*(T\mathbb{H}P^n) \oplus L_n) \cong \theta^{4n} \oplus \iota^*(L_n) $$ where we use that $\pi \circ \iota$ is a constant map and $\theta$ denotes the trivial bundle.

How can I show that this induces an isomorphism $T\mathbb{C}P^1 \cong \iota^*(L_n)$ (without using the Pontryagin class of $\mathbb{H}P^n$ as I want to calculate them this way)?

Answer 1

Since $\pi\circ\iota$ is constant, we have $d\pi\circ d\iota = 0$, so the image of $d\iota$ is contained in the kernel of $d\pi$, i.e. $(d\iota)(T\mathbb{CP}^1) \subseteq L_n$. Since $\iota$ is an embedding, the map $d\iota : T\mathbb{CP}^1 \to L_n \subset T\mathbb{CP}^{2n+1}$ is injective. As $T\mathbb{CP}^1$ and $L_n$ both have rank $2$, we see that $d\iota$ is a vector bundle isomorphism onto its image which covers $\iota$, so $T\mathbb{CP}^1 \cong \iota^*L_n$ (see Lemma 3.1 of Characteristic Classes by Milnor and Stasheff for example).

By the same argument, if $F \to E \xrightarrow{\pi} B$ is a smooth fiber bundle, then for any inclusion of a fiber $\iota : F \to E$, we have $\iota^*(\ker d\pi) \cong TF$.