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The statement I am familiar with regarding classification of vector bundles is :

Given a paracompact space $X$. The set of isomorphism classes of rank $n$ vector bundles over $X$ is in bijective correspondence with the set $[X,G_n]$ of homotopy classes of maps from $X$ to $G_n$.

I am more or less comfortable with the proof of this result.

I could not guess how some one who has done this for the first time thought about it. Were there some spaces $X$ where it is immediately visible that the vector bundles over $X$ have some relation with maps from $X$ to $G_n$?

Question : How did some one guess about the possibility of vector bundles over $X$ being related to homotopy class of maps $X\rightarrow G_n$?

Pointing out a paper where this result is published definitely be useful if it contains some motivation how did the author(s) thought about this.

Praphulla Koushik
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    One potential motivation (I have no idea if this was the historical motivation). Given a smooth $n$-dimensional manifold $M$, for $N$ large enough, there is an embedding $f : M \to \mathbb{R}^N$. As the tangent bundle of $\mathbb{R}^N$ is trivial, the map $p \mapsto df(T_pM)$ can be identified with a map $M \to \operatorname{Gr}(n, N)$. As $n$ grows, so does $N$, so to deal with all manifolds, one needs to take $N \to \infty$. – Michael Albanese Sep 08 '18 at 16:01
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    Yeah, if I had to guess I would assume that the original motivation was an attempt to generalize the Gauss map beyond hypersurfaces – Denis Nardin Sep 08 '18 at 16:03
  • Another potential motivation: complex line bundles on $X$ are classified by their first Chern class, an element of $H^2(X, \mathbb{Z})$. This cohomology group can be identified with the set of homotopy classes of maps $X\rightarrow K(\mathbb{Z}, 2) \cong \mathbb{CP}^{\infty}$. – K.K. Sep 08 '18 at 16:06
  • @Tony The proof I am aware of Complex line bundles being classified by $H^2(X,\mathbb{Z})$ uses that line bundles are classified by $[X,G_1]$ which is same thing as $H^2(X,\mathbb{Z})$... So it would be circular argument... – Praphulla Koushik Sep 08 '18 at 16:10
  • @MichaelAlbanese I guess you are referring to Whitney's embedding theorem which states that any smooth $m$ dimensional manifold can be embedded in $\mathbb{R}^{2m}$... That is clear.... Tangent bundle of $\mathbb{R}^n$ is trivial.. This is also ok... I did not completely understand your statement after that.... It will be good if you can take your time and make it as an answer... – Praphulla Koushik Sep 08 '18 at 16:15
  • @DenisNardin Do you want to write some more details and make it as an answer... I am not able to see what kind of generalization of gauss map gives this idea.. – Praphulla Koushik Sep 08 '18 at 16:17
  • @PraphullaKoushik I don't think it is circular. The exponential sequence gives a map from the set (group) of isomorphism classes of complex line bundles on $X$ to $H^2(X, \mathbb{Z})$. That this is an isomorphism follows from the fact that the sheaf of continuous functions on $X$ is a fine sheaf. – K.K. Sep 08 '18 at 16:28
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    @PraphullaKoushik: What I said about taking $N$ to infinity as $n$ grows doesn't make sense, ignore that part. – Michael Albanese Sep 08 '18 at 16:31
  • @MichaelAlbanese That comment got 3 upvotes... There might be something which still make some sense.. I do not know what the people who upvoted that were thinking.. please do not delete that comment, may be some one can extend it... – Praphulla Koushik Sep 08 '18 at 16:40
  • @Tony Ok.. I remember seeing that version of proof as well.. Can you make it as an answer, please? – Praphulla Koushik Sep 08 '18 at 16:47
  • @Praphulla He's giving a recipe for constructing the tangent bundle of a submanifold of $\Bbb R^n$ as the pullback, by the Gauss map, of the tautological bundle over $\text{Gr}(n, N)$. Verifying that this gives the tangent bundle is a good exercise. – mme Sep 08 '18 at 18:14
  • @MikeMiller I don’t completely understand your comment.. can you write some more details and make it as answer... – Praphulla Koushik Sep 09 '18 at 01:52

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At Praphulla Koushik's request I am posting my comments above as an answer, with a little extra detail added.

Complex line bundles are classified up to isomorphism by their first Chern class. To see this, consider the long exact sequence of cohomology associated to the exponential sequence

$$0\rightarrow \mathbb{Z}\rightarrow \mathcal{C}_X\xrightarrow{\exp(2\pi i -)} \mathcal{C}^*_X\rightarrow 0$$

where $\mathbb{Z}$ is the constant sheaf on $X$ with values in $ \mathbb{Z}$ and $\mathcal{C}_X$ and $\mathcal{C}^*_X$ are the sheaves of continuous functions and non-vanishing continuous functions on $X$, respectively. Since $\mathcal{C}_X$ is a fine sheaf, the connecting homomorphism $H^1(X, \mathcal{C}_X^*)\rightarrow H^2(X,\mathbb{Z})$ is an isomorphism. Since $H^1(X, \mathcal{C}_X^*)$ is in bijection with the set of isomorphism classes of complex line bundles on $X$, we have a bijection

$$\{\text{isomorphism classes of complex line bundles on}\ X\}\rightarrow H^2(X,\mathbb{Z}).$$

Finally, since $H^2(X, \mathbb{Z})$ is in bijection with homotopy classes of maps $X\rightarrow K(\mathbb{Z},2)\simeq \mathbb{CP}^{\infty}$, we obtain that complex line bundles on $X$ are classified by homotopy classes of maps $X\rightarrow \mathbb{CP}^{\infty}$.

K.K.
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  • Can you give a reference for "Since $H^1(X,\mathcal{C}_X^*)$ is in bijection eith the set of isomorphism classes of complex line bundles over $X$" – Praphulla Koushik Sep 08 '18 at 17:50
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    I don't know a reference for this specific fact off the top of my head, but it follows easily from the cocycle description of line bundles. – K.K. Sep 08 '18 at 18:27
  • :D Yes, that is true... I should not even asked for reference... Given a Line bundle $E\rightarrow X$ we have an open cover ${U_\alpha}$ for $X$ and functions $g_{\alpha\beta}:U_\alpha\cap U_\beta\rightarrow \mathbb{C}^$ that determines the line bundle.. This collection ${g_{\alpha\beta}:U_\alpha\cap U_\beta\rightarrow \mathbb{C}^}$ gives an element in first sheaf cohomology of $X$ with sheaf $\mathcal{C}_X^$.. Similarly, given an element in $H^1(X,\mathcal{C}_X^)$ it gives a collection of such transition functions, which will anyways determine a Line bundle over $X$... – Praphulla Koushik Sep 09 '18 at 01:46
  • Thus, there is a bijection between $H^1(X,\mathcal{C}_X^*)$ and the isomorphism classes of complex line bundles... Even I can not think of any reference where this is mentioned... – Praphulla Koushik Sep 09 '18 at 01:51
  • This will force the question how did some one guess there could be a bijection between $H^2(X,\mathbb{Z})$ and the collection of homotopy classes $X\rightarrow K(\mathbb{Z},2)$?... I see going from vector bundle to $H^2(X,\mathbb{Z})$ as natural but could not guess how one can see possibility of bijection from $H^2(X,\mathbb{Z})$ to $[X,\mathbb{C}\mathbb{P}^\infty]$... – Praphulla Koushik Sep 09 '18 at 02:33
  • Can you see https://mathoverflow.net/questions/310306/outline-of-the-proof-that-cech-cohomology-and-singular-cohomology-on-locally-con question... – Praphulla Koushik Sep 11 '18 at 07:16