Talk:Tangent bundles of bundles (Ex)

Solution 0.1. There is a short exact sequence of vector bundles over $E$$\newcommand{\Z}{\mathbb{Z}} \newcommand{\R}{\mathbb{R}} \newcommand{\C}{\mathbb{C}} \newcommand{\N}{\mathbb{N}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\F}{\mathbb{F}} \newcommand{\bZ}{\mathbb{Z}} \newcommand{\bR}{\mathbb{R}} \newcommand{\bC}{\mathbb{C}} \newcommand{\bH}{\mathbb{H}} \newcommand{\bQ}{\mathbb{Q}} \newcommand{\bF}{\mathbb{F}} \newcommand{\bN}{\mathbb{N}} \DeclareMathOperator\id{id} % identity map \DeclareMathOperator\Sq{Sq} % Steenrod squares \DeclareMathOperator\Homeo{Homeo} % group of homeomorphisms of a topoloical space \DeclareMathOperator\Diff{Diff} % group of diffeomorphisms of a smooth manifold \DeclareMathOperator\SDiff{SDiff} % diffeomorphism under some constraint \DeclareMathOperator\Hom{Hom} % homomrphism group \DeclareMathOperator\End{End} % endomorphism group \DeclareMathOperator\Aut{Aut} % automorphism group \DeclareMathOperator\Inn{Inn} % inner automorphisms \DeclareMathOperator\Out{Out} % outer automorphism group \DeclareMathOperator\vol{vol} % volume \newcommand{\GL}{\text{GL}} % general linear group \newcommand{\SL}{\text{SL}} % special linear group \newcommand{\SO}{\text{SO}} % special orthogonal group \newcommand{\O}{\text{O}} % orthogonal group \newcommand{\SU}{\text{SU}} % special unitary group \newcommand{\Spin}{\text{Spin}} % Spin group \newcommand{\RP}{\Rr\mathrm P} % real projective space \newcommand{\CP}{\Cc\mathrm P} % complex projective space \newcommand{\HP}{\Hh\mathrm P} % quaternionic projective space \newcommand{\Top}{\mathrm{Top}} % topological category \newcommand{\PL}{\mathrm{PL}} % piecewise linear category \newcommand{\Cat}{\mathrm{Cat}} % any category \newcommand{\KS}{\text{KS}} % Kirby-Siebenmann class \newcommand{\Hud}{\text{Hud}} % Hudson torus \newcommand{\Ker}{\text{Ker}} % Kernel \newcommand{\underbar}{\underline} %Classifying Spaces for Families of Subgroups \newcommand{\textup}{\text} \newcommand{\sp}{^}E$

$\displaystyle 0 \to T_{\pi}E \stackrel{i}{\rightarrow} TE \stackrel{j}{\rightarrow} \pi^*TB \to 0$ $$\displaystyle 0 \to T_{\pi}E \stackrel{i}{\rightarrow} TE \stackrel{j}{\rightarrow} \pi^*TB \to 0$$

where $i$$i$ is the inclusion and $j$$j$ is defined by $j(e,v)=(e, d\pi(v))$$j(e,v)=(e, d\pi(v))$. We choose a Riemannian metric on $TE$$TE$, then the orthogonal projection to $T_{\pi}E$$T_{\pi}E$ gives a splitting of the exact sequence. Which implies that

$\displaystyle TE \cong \pi^*TB \oplus T_{\pi}E$ $$\displaystyle TE \cong \pi^*TB \oplus T_{\pi}E$$

Solution 0.2. A necessary condition for $T_{\pi}E$$T_{\pi}E$ being the pullback of some bundle over $B$$B$ is that $TF$$TF$ is trivial. This is seen by restricting the bundle to a point in $B$$B$. On the other hand, obviously when $E$$E$ is a vector bundle or the bundle $\pi \colon E \to B$$\pi \colon E \to B$ is trivial then $T_{\pi}E$$T_{\pi}E$ is the pullback of a vector bundle over $B$$B$. We don't know if this is true in general.

Solution 0.3. If $\pi \colon E \to B$$\pi \colon E \to B$ is itself a smooth vector bundle, then $\pi^*T_{\pi}E\cong \pi^*E$$\pi^*T_{\pi}E\cong \pi^*E$, therefore

$\displaystyle TE \cong \pi^*TB \oplus T_{\pi}E\cong\pi^*(TB\oplus E).$ $$\displaystyle TE \cong \pi^*TB \oplus T_{\pi}E\cong\pi^*(TB\oplus E).$$

Solution 0.4. Denote the associated vector bundle by $\xi$$\xi$ with projection $\pi_{\xi} \colon E(\xi) \to B$$\pi_{\xi} \colon E(\xi) \to B$, then

$\displaystyle TE \oplus \underline{\mathbb R} = i^*\pi_{\xi}^*(TB \oplus \xi),$ $$\displaystyle TE \oplus \underline{\mathbb R} = i^*\pi_{\xi}^*(TB \oplus \xi),$$

where $i \colon E \to E(\xi)$$i \colon E \to E(\xi)$ is the inclusion of the sphere bundle into the vector bundle.

Solution 0.5. Apply the description of the tangent bundle of sphere bundles in the previous solution to the sphere bundle $S^2 \to \CP^{2k+1} \stackrel{\pi}{\rightarrow} \Hh P^k$$S^2 \to \CP^{2k+1} \stackrel{\pi}{\rightarrow} \Hh P^k$ we get $T\CP^{2k+1} \oplus \underline{\mathbb R} = \pi^*(T\Hh P^k \oplus \xi)$$T\CP^{2k+1} \oplus \underline{\mathbb R} = \pi^*(T\Hh P^k \oplus \xi)$, where $\xi$$\xi$ is the associated vector bundle. Now taking Pontrjagin classes on both sides, we get an identity (since there is no torsion in cohomology)

$\displaystyle p(\CP^{2k+1})=\pi^*(p(\Hh P^k) \cdot p(\xi)).$ $$\displaystyle p(\CP^{2k+1})=\pi^*(p(\Hh P^k) \cdot p(\xi)).$$

We know that $p(\CP^{2k+1})=(1+x^2)^{2k+2}$$p(\CP^{2k+1})=(1+x^2)^{2k+2}$ and $p(\xi)=1+p_1(\xi)$$p(\xi)=1+p_1(\xi)$, where $p_1(\xi) \in H^4(\Hh P^k)$$p_1(\xi) \in H^4(\Hh P^k)$ is determined by restricting to the case $k=1$$k=1$ (where $\Hh P^1 = S^4$$\Hh P^1 = S^4$ whose Pontrjagin class is known to be trivial): $\pi^*p_1(\xi)=p_1(\CP^3)=4x^2$$\pi^*p_1(\xi)=p_1(\CP^3)=4x^2$ with $x\in H^2(\CP^{3})$$x\in H^2(\CP^{3})$ a generator. Therefore in $H^*(\CP^{2k+1})$$H^*(\CP^{2k+1})$ we have the equation

$\displaystyle (1+x^2)^{2k+2}(1+4x^2)^{-1}=\pi^*p(\Hh P^k).$ $$\displaystyle (1+x^2)^{2k+2}(1+4x^2)^{-1}=\pi^*p(\Hh P^k).$$

It's seen from the Gysin sequence that $\pi^*$$\pi^*$ is injective. This is sufficient to determine $p(\Hh P^k)$$p(\Hh P^k)$.