Talk:Tangent bundles of bundles (Ex)

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Solution 0.1. There is a short exact sequence of vector bundles over E

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

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

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

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

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

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

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

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

where 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 we get T\CP^{2k+1} \oplus \underline{\mathbb R} = \pi^*(T\Hh P^k \oplus \xi), where \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)).

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

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

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

\to T_{\pi}E \stackrel{i}{\rightarrow} TE \stackrel{j}{\rightarrow} \pi^*TB \to 0$$ where $i$ is the inclusion and $j$ is defined by $j(e,v)=(e, d\pi(v))$. We choose a Riemannian metric on $TE$, then the orthogonal projection to $T_{\pi}E$ gives a splitting of the exact sequence. Which implies that $$ TE \cong \pi^*TB \oplus T_{\pi}E $$ {{endthm}} {{beginthm|Solution}} A necessary condition for $T_{\pi}E$ being the pullback of some bundle over $B$ is that $TF$ is trivial. This is seen by restricting the bundle to a point in $B$. On the other hand, obviously when $E$ is a vector bundle or the bundle $\pi \colon E \to B$ is trivial then $T_{\pi}E$ is the pullback of a vector bundle over $B$. We don't know if this is true in general. {{endthm}} {{beginthm|Solution}} If $\pi \colon E \to B$ is itself a smooth vector bundle, then $\pi^*T_{\pi}E\cong \pi^*E$, therefore $$TE \cong \pi^*TB \oplus T_{\pi}E\cong\pi^*(TB\oplus E).$$ {{endthm}} {{beginthm|Solution}} Denote the associated vector bundle by $\xi$ with projection $\pi_{\xi} \colon E(\xi) \to B$, then $$TE \oplus \underline{\mathbb R} = i^*\pi_{\xi}^*(TB \oplus \xi),$$ where $i \colon E \to E(\xi)$ is the inclusion of the sphere bundle into the vector bundle. {{endthm}} {{beginthm|Solution}} Apply the description of the tangent bundle of sphere bundles in the previous solution to the sphere bundle $S^2 \to \CP^{2k+1} \stackrel{\pi}{\rightarrow} \Hh P^k$ we get $T\CP^{2k+1} \oplus \underline{\mathbb R} = \pi^*(T\Hh P^k \oplus \xi)$, where $\xi$ is the associated vector bundle. Now taking Pontrjagin classes on both sides, we get an identity (since there is no torsion in cohomology) $$p(\CP^{2k+1})=\pi^*(p(\Hh P^k) \cdot p(\xi)).$$ We know that $p(\CP^{2k+1})=(1+x^2)^{2k+2}$ and $p(\xi)=1+p_1(\xi)$, where $p_1(\xi) \in H^4(\Hh P^k)$ is determined by restricting to the case $k=1$ (where $\Hh P^1 = S^4$ whose Pontrjagin class is known to be trivial): $\pi^*p_1(\xi)=p_1(\CP^3)=4x^2$ with $x\in H^2(\CP^{3})$ a generator. Therefore in $H^*(\CP^{2k+1})$ we have the equation $$(1+x^2)^{2k+2}(1+4x^2)^{-1}=\pi^*p(\Hh P^k).$$ It's seen from the Gysin sequence that $\pi^*$ is injective. This is sufficient to determine $p(\Hh P^k)$. {{endthm}}E

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

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

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

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

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

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

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

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

where 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 we get T\CP^{2k+1} \oplus \underline{\mathbb R} = \pi^*(T\Hh P^k \oplus \xi), where \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)).

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

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

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

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