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.

  • 8
    $\begingroup$ 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 $N \to \infty$. $\endgroup$ – Michael Albanese Sep 8 '18 at 16:01
  • 1
    $\begingroup$ Yeah, if I had to guess I would assume that the original motivation was an attempt to generalize the Gauss map beyond hypersurfaces $\endgroup$ – Denis Nardin Sep 8 '18 at 16:03
  • $\begingroup$ 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 $X\rightarrow K(\mathbb{Z}, 2) \cong \mathbb{CP}^{\infty}$. $\endgroup$ – Tony Sep 8 '18 at 16:06
  • $\begingroup$ @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... $\endgroup$ – Praphulla Koushik Sep 8 '18 at 16:10
  • 1
    $\begingroup$ @PraphullaKoushik: What I said about taking $N$ to infinity as $n$ grows doesn't make sense, ignore that part. $\endgroup$ – Michael Albanese Sep 8 '18 at 16:31

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 $X\rightarrow \mathbb{CP}^{\infty}$.

  • $\begingroup$ 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 $X$" $\endgroup$ – Praphulla Koushik Sep 8 '18 at 17:50
  • 2
    $\begingroup$ 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. $\endgroup$ – Tony Sep 8 '18 at 18:27
  • $\begingroup$ :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 $X$... $\endgroup$ – Praphulla Koushik Sep 9 '18 at 1:46
  • $\begingroup$ 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... $\endgroup$ – Praphulla Koushik Sep 9 '18 at 1:51
  • $\begingroup$ 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 $[X,\mathbb{C}\mathbb{P}^\infty]$... $\endgroup$ – Praphulla Koushik Sep 9 '18 at 2:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.