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On the base manifold, the complex projective space ${\mathbb{P}}_1({\mathbb{C}})$ (or ${\mathbf{CP}}_1$), I am interested in knowing how to classify and construct explicit distinct classes of

  1. tautological line bundles $L_{\mathbb{C}}$ over ${\mathbb{P}}_1({\mathbb{C}})$

  2. principle SO(3) bundle over ${\mathbb{P}}_1({\mathbb{C}})$

and how the above two bundles (tautological line bundle and principle SO(3) bundle) have any relations/constraints to

  1. tangent bundle over ${\mathbb{P}}_1({\mathbb{C}})$,

say, through the computations of characteristic classes.

For example, I think we have (the partial info I can obtain, please double check):

$\bullet$ $c_1 (\mathbb{CP}^1) =2 \in H^2(\mathbb{CP}^1, \mathbb{Z})$.

$\bullet$ $w_2(\mathbb{CP}^1)=c_1 (\mathbb{CP}^1) \mod 2=2 \mod 2 =0 \in H^2(\mathbb{CP}^1, \mathbb{Z}_2)$.

$\bullet$ $c_1 (L_{\mathbb C}) = -1 \in H^2(?, \mathbb{Z})$.

$\bullet$ $w_2(L_{\mathbb C})=c_1(L_{\mathbb C}) \mod 2=-1 \mod 2=1$.

Here $c_j$ stands for Chern class, $w_j$ stands for Stieffel Whitney class.

(q1) What are the questions marks above in $H^2(?, \mathbb{Z})$ and $H^2(?, \mathbb{Z}_2)$?

(q2) How do we classify 1. tautological line bundles, 2. principle SO(3) bundle over ${\mathbb{P}}_1({\mathbb{C}})$? How many isomorphism classes there are? How do we construct them? (e.g. relating the associated vector bundles to the principle SO(3) bundle)

(q3) How do we relate the principle SO(3) bundle to the tautological line bundles? (e.g. relating the tautological line bundles to the associated vector bundles, then to the principle SO(3) bundle?)

Here we may write the associated vector bundle $E = 1 \oplus L_{\mathbb{R}} \oplus \dots$ or $E = 1 \oplus L_{\mathbb{C}} \oplus \dots$ in terms of a decomposition into a set of the trivial real vector bundle 1, and the tautological real /or complex line bundle $L_{\mathbb{R}}$ /or $L_{\mathbb{C}}$ (yes ?).

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  • $\begingroup$ What do you mean, classify tautological line bundles? There's only one - it's tautological! $\endgroup$ – KReiser Jul 9 '18 at 23:19
  • $\begingroup$ title changes: Classifying the principle SO(3) bundle, constructed from tautological line bundles $L_{\mathbb{C}}$ $\endgroup$ – wonderich Jul 9 '18 at 23:24
  • $\begingroup$ Still, there is exactly one tautological line bundle. Your q2 and q3 still ask about tautological line bundles, which is nonsense because there's only one such bundle. Unless you're talking about something besides $\mathcal{O}_1$ when you use the term tautological line bundle? $\endgroup$ – KReiser Jul 9 '18 at 23:50
  • $\begingroup$ There is line bundle $L_{\mathbb{C}}$ and trivial vector bundle 1. The principle SO(3) bundle can be constructed/viewed as the associated vector bundle from $L_{\mathbb{C}}$ and $1$, etc. $\endgroup$ – wonderich Jul 10 '18 at 0:30
  • $\begingroup$ For example, the principle SO(3) bundle is 3 dimensions, we may have the associated vector bundle $V=3$ from the trivial vector bundle, or $V=1+L_\mathbb{C}$ from both trivial and the tautological line bundle $L_\mathbb{C}$. $\endgroup$ – wonderich Jul 10 '18 at 0:32
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We have $\mathbb{C}P^1\cong S^2$ so there are are $\pi_2(BU(1))=\pi_2(BS^1)=\pi_1S^1=\mathbb{Z}$ isomorphism classes of complex line bundles over $\mathbb{C}P^1$. The set of all such forms an abelian group under the tensor product and the first Chern class $c_1$ gives an isomorphism of this group with $H^2(S^2;\mathbb{Z})\cong\mathbb{Z}$. You can understand this in another way from the fact that $BS^1\simeq K(\mathbb{Z},2)$ is an Eilenberg-Mac Lane space.

We calculate $c_1(L_\mathbb{C})=\pm 1\in H^2(S^2; \mathbb{Z})\cong\mathbb{Z}$ (sign depending on who you ask) by explicitly constructing the bundle and calculating its transition function, which turns out to be a holomorphic map $U_0\cap U_1\rightarrow S^1\subseteq\mathbb{C}$, where $U_0,U_1$ is the standard cover of $\mathbb{C}P^1$. Thus $L_\mathbb{C}$ is holomorphic.

We infer from the Chern class calculation that the tautological bundle $L_\mathbb{C}$ generates the abelian group of isomorphism classes of complex line bundles over $\mathbb{C}P^1$ under tensor product. That is, any complex Line bundle $E\rightarrow \mathbb{C}P^1$ is isomorphic to some tensor power $\otimes^kL_\mathbb{C}\rightarrow\mathbb{C}P^1$ where $k=\pm c_1(E)$. The fact that $L_\mathbb{C}$ is holomorphic means that any tensor power of it is also holomorphic, as can be seen by considering transition functions. Thus any complex line bundle over $\mathbb{C}P^1$ may be given a holomorphic structure. Moreover, since $H^1(\mathbb{C}P^1;\mathbb{Z})=0$, the holomorphic structure on each line bundle is unique.

Principal $SO(3)$-bundles over $\mathbb{C}P^1\cong S^2$ are classified by $\pi_2BSO(3)\cong\pi_1SO(3)\cong\mathbb{Z}_2$. Thus there are two of them, one of which is the trivial bundle and one of which is not. Since the subgroup inclusion induces a surjection $\mathbb{Z}\cong\pi_1SO(2)\rightarrow \pi_2SO(3)\cong\mathbb{Z}_2$ any $SO(3)$-bundle over $\mathbb{C}P^1$ has a reduction of structure to an $SO(2)\cong S^1$ bundle. This is obvious for the trivial bundle, and for the non-trivial bundle we may view $L_\mathbb{C}$ as an $SO(2)$-bundle by forgetting its complex structure and form the associated bundle $E=L_\mathbb{C}\times_{SO(2)}\mathbb{R}^3$, where $SO(2)\cong S^1$ acts on $L_\mathbb{C}$ by complex multiplication and on $\mathbb{R}^3$ through the subgroup inclusion $SO(2)\hookrightarrow SO(3)$. We find $w_2(E)=c_1(L_\mathbb{C})\mod 2=1\mod (2)$, so this bundle is indeed represents the non-trivial $SO(3)$-bundle.

If you are not comfortable with using classifying spaces and homotopy classes of maps to classify principal (note the spelling) bundles, then you may instead like to use the clutching construction for the suspension $S^2$. We have $U_0,U_1$ being the upper and lower hemispheres, where are open discs with intersection $U_0\cap U_1= S^1\times(-\epsilon,\epsilon)$. Thus the transition function $\phi:U_0\cap U_1\rightarrow S^1$ for some bundle gives a map $S^1\rightarrow S^1$ whose degree is equal to the first Chern class of the bundle. Each bundle gives a transition function with some degree and it is not difficult to see that two bundles give rise to transition functions of the same degree if and only if they are bundle-isomorphic.

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  • $\begingroup$ When you wrote $c_1(L_\mathbb{C})=\pm 1\in H^2S^2\cong\mathbb{Z}$ you mean $c_1(L_\mathbb{C})=\pm 1\in H^2(S^2; \mathbb{Z})\cong\mathbb{Z}$? $\endgroup$ – wonderich Jul 10 '18 at 14:35
  • $\begingroup$ Did you also mean that $w_1(E)=c_1(L_\mathbb{C})\mod 2=1\mod (2)$ or you had typos? Should it be $w_2(E)=c_1(L_\mathbb{C})\mod 2=1\mod (2)$? $\endgroup$ – wonderich Jul 10 '18 at 14:46
  • $\begingroup$ Unless otherwise specified I take integral coefficients for cohomology. Thus $H^2S^2=H^2(S^2;\mathbb{Z})$. And yes, it should be the second Stiefel-Whitney class. I've edited the post. Thanks for pointing those out. $\endgroup$ – Tyrone Jul 10 '18 at 14:55
  • $\begingroup$ thanks, do I see the trivial one has $E=3$ and the nontrivial one has $E=L_{\mathbb C}+1$? $\endgroup$ – wonderich Jul 10 '18 at 15:04
  • $\begingroup$ also a related but in different base manifold, I believe it is still a $\mathbb{Z}_2$-class math.stackexchange.com/q/2846718/79069 $\endgroup$ – wonderich Jul 10 '18 at 15:11

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