Talk:Tangent bundles of bundles (Ex)
(Created page with "<wikitex>; {{beginthm|Solution}} There is a short exact sequence of vector bundles over $E$ $$0 \to T_{\pi}E \stackrel{i}{\rightarrow} TE \stackrel{j}{\rightarrow} \pi^*TB \to...") |
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{{beginthm|Solution}} | {{beginthm|Solution}} | ||
− | If $\pi \colon E \to B$ is itself a smooth vector bundle, then $\pi^*T_{\pi}E\cong \pi^*E$, therefore $TE \cong \pi^*TB \oplus T_{\pi}E\cong\pi^*(TB\oplus E) | + | If $\pi \colon E \to B$ is itself a smooth vector bundle, then $\pi^*T_{\pi}E\cong \pi^*E$, therefore |
+ | $$TE \cong \pi^*TB \oplus T_{\pi}E\cong\pi^*(TB\oplus E).$$ | ||
{{endthm}} | {{endthm}} | ||
{{beginthm|Solution}} | {{beginthm|Solution}} | ||
− | Denote the associated vector bundle by $\xi$ with projection $\pi_{\xi} \colon E(\xi) \to B$, then $TE \oplus \underline{\mathbb R} = i^*\pi_{\xi}^*(TB \oplus \xi) | + | Denote the associated vector bundle by $\xi$ with projection $\pi_{\xi} \colon E(\xi) \to B$, then |
+ | $$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)).$$ | ||
+ | |||
+ | We know that $p(\CP^{2k+1})=(1+x^2)^{2k+2}$ and $p(\xi)=1+p_1(\xi)$, where $p_1(\xi) \in H^4(\Hh P^k)$ is determined by restricting to the case $k=1$ (where $\Hh P^1 = S^4$ whose Pontrjagin class is known to be trivial): $\pi^*p_1(\xi)=4x^2 \in H^4(\CP^{2k+1})$. Therefore in $H^*(\CP^{2k+1})$ we have the equation | ||
+ | $$(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}} | {{endthm}} |
Revision as of 22:27, 29 May 2012
Solution 0.1. There is a short exact sequence of vector bundles over
where is the inclusion and is defined by . We choose a Riemannian metric on , then the orthogonal projection to gives a splitting of the exact sequence. Which implies that
Solution 0.2. A necessary condition for being the pullback of some bundle over is that is trivial. This is seen by restricting the bundle to a point in . On the other hand, obviously when is a vector bundle or the bundle is trivial then is the pullback of a vector bundle over . We don't know if this is true in general.
Solution 0.3. If is itself a smooth vector bundle, then , therefore
Solution 0.4. Denote the associated vector bundle by with projection , then
where 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 we get , where is the associated vector bundle. Now taking Pontrjagin classes on both sides, we get an identity (since there is no torsion in cohomology)
We know that and , where is determined by restricting to the case (where whose Pontrjagin class is known to be trivial): . Therefore in we have the equation
It's seen from the Gysin sequence that is injective. This is sufficient to determine .
where is the inclusion and is defined by . We choose a Riemannian metric on , then the orthogonal projection to gives a splitting of the exact sequence. Which implies that
Solution 0.2. A necessary condition for being the pullback of some bundle over is that is trivial. This is seen by restricting the bundle to a point in . On the other hand, obviously when is a vector bundle or the bundle is trivial then is the pullback of a vector bundle over . We don't know if this is true in general.
Solution 0.3. If is itself a smooth vector bundle, then , therefore
Solution 0.4. Denote the associated vector bundle by with projection , then
where 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 we get , where is the associated vector bundle. Now taking Pontrjagin classes on both sides, we get an identity (since there is no torsion in cohomology)
We know that and , where is determined by restricting to the case (where whose Pontrjagin class is known to be trivial): . Therefore in we have the equation
It's seen from the Gysin sequence that is injective. This is sufficient to determine .