# From Topological to Smooth and Holomorphic Vector Bundles

In the last weeks I have been think of the transition from topological vector bundles to smooth and holomorphic vector bundles. This has resulted in a few questions (with a common thread) as follows: Always $\pi:E \to B$ is a topological (complex) vector bundle over a compact base,

(A) For any given smooth manifold structure on $B$, can there exist more than one differential structure on $E$ giving $\pi:E \to B$ the structure of a smooth vector bundle. If so, what is an example?

(B) Same question as above but replacing smooth by holomorphic.

(C) For a choice of smooth vector bundle structure on $\pi:E \to B$, does the de Rham complex of $E$ have an easy relationship with the de Rham complex of $B$. A (very) naive guess would be that $$\Omega^{\bullet}(E) = \Gamma^{\infty}(E) \otimes_{C^\infty(B)}\Omega^{\bullet}(B),$$ but I can't see that there is a well-defined way to define the differential.

(D) Same question as above but for holomorphic structures and the Dolbeault complex

• I think what you want to hear in relation to (C) is the word "Thom isomorphism" Feb 7, 2013 at 11:53
• For question (A), choose two exotic R^4 that is not diffeomorphism, you can view R^4 is R^3 bundle over R^1 Feb 7, 2013 at 12:18
• Siqi He: In your example there is no reason to expect they are vector bundles. Feb 7, 2013 at 12:30
• For (A), what about taking $B$ to be a point and $E$ to be $\mathbb{R}^4$ with two different smooth structures as in Siqi He's comment? I can't see how these examples can fail to be vector bundles. Feb 7, 2013 at 13:13
• Ah, I see what's wrong with my previous comment. The map $\pi\colon E\to B$ is not a smooth vector bundle when $E$ is exotic $\mathbb{R}^4$, since the trivialization $\pi^{-1}(B)\to B\times \mathbb{R}^4$ is not a diffeomorphism. Feb 7, 2013 at 14:53

The answer to A is no. The topological bundle is determined by a continuous homotopy class of maps into the classifying space. Choosing a compatible smooth structure means picking a smooth map in that class. Any two such choices are smoothly homotopic. B however is true. Look up the Jacobian of a Riemann surface. How are you going to grade in C ? Where are the forms on $E$ of degree higher then the dimension of $B$?

• Thank you a lot for your answer. Then for (C), does that mean that in general there is no relationship between the two de Rham complexes? Feb 7, 2013 at 12:54
• ..... or at least that there is no well-known relationship. Feb 7, 2013 at 13:03
• I realise on reading the other responses that I may have misunderstood C. By de Rham cohomology of $E$ did you mean the de Rham cohomology of $E$ as a manifold or some sort of $E$ valued forms on $B$ cohomology ? Feb 8, 2013 at 2:14
• I mean the de Rham cohomology of $E$ as a Riemannian manifold. I've never seen anyone discuss this, and I was wondering why it's not interesting. Feb 23, 2013 at 16:08

(A) As Michael Murray remarks, the answer is no (though some care is needed in interpreting the remark that "any two such choices [of maps into the classifying space] are smoothly homotopic") since in this case the classifying space is an infinite-dimensional manifold (the Grassmannian of $n$-planes in $\mathbb{R}^\infty$).

(B) An elliptic curve $X$ over $\mathbb{C}$ will admit infinitely many distinct holomorphic structures on the (topologically) trivial line bundle! Choose two (distinct) points $x_0, x_1$; then $\mathcal{O}(x_0-x_1)$ is an example (this is the ideal sheaf of $x_1$ tensored with the dual of the ideal sheaf of $x_0$). One can see that this is distinct from e.g. $\mathcal{O}_X$ because it has no non-zero global sections, whereas $\mathcal{O}_X$ does (constant functions).

(C) You need a flat connection to define this complex, as Liviu Nicolaescu remarks. In this case, let $\mathcal{E}$ be the sheaf of flat sections to $E$. Then the de Rham complex is $$\mathcal{E}\otimes_{\mathbb{C}} \Omega^\bullet(B).$$ The differential is defined to be a derivation, which kills $e\otimes 1$ for any section $e$ to $\mathcal{E}$ (this uniquely determines a differential).

(D) Perhaps I'm confused, but I think Alex's answer is misleading--one can define a Dolbeault complex for any holomorphic vector bundle; this doesn't require a Hermitian metric. If $\mathcal{E}$ is the sheaf of holomorphic sections to your vector bundle, this complex is defined as $$\mathcal{E}\otimes_{\mathcal{O}_X} A^{0, \bullet}(B).$$ Again, the differential is a derivation which kills sections to $\mathcal{E}$. Here $A^{0, \bullet}$ is the complex of smooth $(0, \bullet)$ forms; the complex above should compute the sheaf cohomology of $\mathcal{E}$.

• Thanks Daniel. I was little nervous about that point. I was hoping that because $B$ was fixed I could get away with a finite-dimensional Grassmanian but was't sure if it might change as I varied the maps homotopically? I think though the choice o finite dimensional Grassmanian can be fixed by the dimension of $B$? Feb 8, 2013 at 1:30
• You can indeed bound the dimension of the Grassmannian if $B$ is compact, in terms of e.g. the minimal number of contractible open sets required to cover $B$ (think about how you would embed $E$ in a trivial vector bundle). A bit of work should be able to handle the non-compact case, probably; I'll think about it and write something if I come up with a slick argument. Feb 8, 2013 at 1:41

For (C) you need to fix a flat connection (covariant derivative) $\nabla$ on $E$ so that you have the coboundary condition $(d^\nabla)^2=0$. (This may not always be possible; all the Chern classes would have to be torsion classes.)

In any case, if that were possible you would obtain a holonomy morphism

$$h:\pi_1(B) \to \mathrm{GL}_r(\mathbb{C}),$$

where $r$ is the rank of $E$. You can think of $h$ as defining a local coefficient system or, equivalently, a locally constant sheaf $\mathscr{S}$ on $B$. This sheaf has a simple geometric description: it is the sheaf determined by the presheaf of covariant constant sections of $E$ with respect to the chosen connection $\nabla$.

The cohomology of the DeRham complex determined by $\nabla$ is then the cohomology of the sheaf $\mathscr{S}$.

I am under the impression that some of the answers given may be interpreting question (A) in a manner which does not seem --- at least to me --- entirely consistent with how it is stated. Interpreting question (A) quite strictly, it seems to me that there can be in fact two distinct differentiable structures on the total space of a topological vector bundle which both make it into a smooth vector bundle. In fact, for any non-empty manifold $B$ of dimension greater than zero, and any topological vector bundle $E$ over $B$ of dimension greater than zero, there exist uncountably many distinct smooth structures on $E$ for which $E$ becomes a smooth vector bundle over $B$. I will give a very simple (and detailed) example below. Also, I will work --- out of habit --- with real vector bundles, but the exact same construction with ${\mathbb R}$ replaced by ${\mathbb C}$ works equally well for complex line bundles.

Let $B$ be a manifold, and $E$ a topological vector bundle over $B$ with projection $\textrm{proj}:E\to B$ (as usual, we confuse the vector bundle with its total space). Assume the total space $E$ admits a differentiable structure which makes it into a smooth vector bundle over $B$. Denote this smooth vector bundle (and its total space seen as a smooth manifold) by $E^{(1)}$, to distinguish it from the topological vector bundle $E$. For any continuous map $f:B\to{\mathbb R}\setminus\{0\}$ we can construct the map $H_f:E\to E$ given by $$H_f(x)= f(\textrm{proj}(x))\cdot x$$ In other words, the map $H_f$ is the vector bundle map $E\to E$ which sits over the identity $\textrm{id}_B$ on $B$, and which is multiplication by $f(b)$ on the fibre over $b\in B$.

It is obvious that $H_f:E\to E$ is an isomorphism of topological vector bundles whose inverse is $H_{\frac 1 f}$ (since $f$ is never zero). But it does not give a map of smooth vector bundles $E^{(1)}\to E^{(1)}$ unless $f$ is itself smooth. Now transfer the smooth structure on $E^{(1)}$ via the homeomorphism $H_f$, and denote the new smooth manifold by $E^{(f)}$. More precisely, $E^{(f)}$ is the topological space $E$ equipped with the unique differentiable structure which makes $H_f:E_0\to E^{(f)}$ a diffeomorphism.

Recall that $\textrm{proj}:E^{(1)}\to B$ is a smooth vector bundle. Note also that $H_f$ is both a diffeomorphism $H_f:E^{(1)}\to E^{(f)}$ and an isomorphism of topological vector bundles $H_f:E\to E$. As a consequence, the smooth vector bundle structure on $\textrm{proj}:E^{(1)}\to B$ transfers across $H_f$ to a smooth vector bundle structure on $\textrm{proj}:E^{(f)}\to B$ whose underlying topological vector bundle is $E$. In other words, the topological vector bundle $E$ underlying $\textrm{proj}:E^{(f)}\to B$ is actually a smooth vector bundle when we consider the smooth structure $E^{(f)}$ on the total space.

To summarize, we have two smooth vector bundle structures, $E^{(1)}$ and $E^{(f)}$, on the topological vector bundle $E$ over $B$. [By the way, the notation is self-consistent: observe that for $f=1$, $E^{(f)}$ is just the original $E^{(1)}$.] To answer the question (A) affirmatively, and give an example at the same time, assume:

1. the vector bundle $E$ has positive dimension;
2. $B$ is non-empty and has positive dimension.

Then I claim:

Lemma: The identity function on $E$ is a diffeomorphism (or even just a smooth function) $E^{(f)}\to E^{(1)}$ if and only if $f$ is itself smooth. $\square$

On the one hand, it is easy to check that if $f$ is smooth then $H_f$ is a diffeomorphism $E^{(1)}\to E^{(1)}$ (both it and its inverse $H_{\frac 1 f}$ are smooth). Therefore, the smooth structure on $E^{(f)}$ is by definition the same as the smooth structure on $E^{(1)}$, i.e. the identity is a diffeomorphism $E^{(f)}\to E^{(1)}$. On the other hand, assume that the identity $\textrm{id}_E$ is a smooth function $E^{(f)} \to E^{(1)}$. I will prove that $f$ is itself smooth. Consider the composition of diffeomorphisms $$G : E^{(1)} \overset{H_f}{\longrightarrow} E^{(f)} \overset{\textrm{id}_E}{\longrightarrow} E^{(1)}$$ By definition of $H_f$: $$G(x) = f(\textrm{proj}(x))\cdot x$$ and it is fairly easy to use local trivializations for $E^{(1)}$ to conclude that $f$ is smooth. In fact, if $\varphi:U\times {\mathbb R}^n \to E$ is a (smooth) trivialization of $E^{(1)}$ over the open $U\subset B$, it follows that $$\varphi^{-1}\circ G\circ\varphi(u,v) = (u,f(u)\cdot v)$$ and since $E$ has positive dimension, the restriction of $f$ to $U$ is the last component of the smooth function $$u\longmapsto\varphi^{-1}\circ G\circ\varphi(u,(0,\ldots,0,1))$$ Hence $f$ is smooth.

In conclusion, under condition 1 above, the smooth structure on $E^{(1)}$ is the same as the smooth structure on $E^{(f)}$ if and only if $f$ is smooth. Furthermore, under condition 2 above, we can then find a continuous map $f:B\to{\mathbb R}\setminus\{0\}$ which is not smooth. For such a choice of $f$, $E^{(1)}$ and $E^{(f)}$ are two distinct smooth vector bundle structures on $E$.

In fact, more can be said. If we start with $E^{(f)}$ in place of $E^{(1)}$ and apply the above construction, we can easily describe the result: for any continuous functions $f,g:B\to{\mathbb R}\setminus\{0\}$ it is easy to check that $H_g \circ H_f = H_{f\cdot g}$, which implies $$(E^{(f)})^{(g)} = E^{(f\cdot g)}$$ Applying the preceding lemma, we conclude that the smooth structure on $E^{(f)}$ coincides with the smooth structure on $E^{(g)}$ if and only if $\frac f g$ is smooth. Consequently, the construction $f\mapsto E^{(f)}$ gives a set of smooth vector bundle structures on $E$ which is in bijection with the quotient of abelian groups $$C^0(B,{\mathbb R}\setminus\{0\})/C^\infty(B,{\mathbb R}\setminus\{0\})$$ where the multiplication in each of the groups is given by multiplying functions. It is fairly easy to see that this quotient is uncountable under condition 2 above: by giving for each $b\in B$ a continuous function $f_b:B\to{\mathbb R}\setminus\{0\}$ which is smooth everywhere except at the point $b$, we determine an injection of $B$ into the above quotient of abelian groups.

### Essential uniqueness of the smooth vector bundle structure on $E$

In light of the above, what can be said regarding uniqueness of the smooth vector bundle structure on $E$? Well, as some of the other answers have indicated, one can use approximation of continuous functions by smooth functions to prove the following result from its topological counterpart (i.e. the usual topological classification of vector bundles).

Theorem: Let $B$ be a smooth manifold of dimension $n$. Consider the function $$\theta:[B,\textrm{Gr}(k,{\mathbb R}^{k+l})]^{\textrm{smooth}}\longrightarrow \textrm{Vec}^{\textrm{smooth}}_B$$ (where the domain is the set of smooth homotopy classes of smooth functions from $B$ into the Grassmannian of $k$-dimensional linear subspaces of ${\mathbb R}^{k+l}$, and the target is the set of isomorphism classes of smooth $k$-dimensional vector bundles over $B$) defined by $$\theta([f])=f^\ast(\gamma_{k,k+l})$$ where $\gamma_{k,k+l}$ is the tautological smooth vector bundle over the Grassmannian. Then $\theta$ is a bijection if $l\geq n+2$. $\square$

By using this theorem, its topological counterpart (replace smooth by continuous/topological), and the approximation of continuous functions by smooth functions, one can see that the forgetful map from the set of isomorphism classes of smooth vector bundles over $B$ to the set of isomorphism classes of topological vector bundles over $B$ is a bijection.

We do not really need the above theorem to see that isomorphism classes of smooth vector bundles inject into the isomorphism classes of topological vector bundles (although it is a convenient way to prove surjectivity). Using only the approximation of continuous function by smooth functions and smooth partitions of unity, one can approximate any topological isomorphism between smooth vector bundles by a smooth isomorphism. Moreover, given isomorphisms $\varphi,\psi:E\to E'$ of smooth vector bundles over $B$, any homotopy through topological isomorphisms between $\varphi$ and $\psi$ can be approximated by a homotopy through smooth isomorphisms between $\varphi$ and $\psi$. In particular, given two smooth vector bundle structures $E_1$ and $E_2$ on a topological vector bundle $E$ over $B$, there exists a unique homotopy class of smooth vector bundle isomorphisms $E_1\to E_2$ which, as an isomorphism of topological vector bundles, is homotopic to the identity $\textrm{id}_E:E\to E$.

Continuing in the same manner, it is not too hard to conclude that the homotopy fibres of the map $$\textrm{Iso}^{\textrm{smooth}}_B(E_1,E_2) \longrightarrow \textrm{Iso}^{\textrm{top}}_B(E,E)$$ (between the spaces of isomorphisms of smooth/topological vector bundles over $B$) are weakly contractible. Consequently, the map is a weak equivalence. That appears to be the strongest result we can state, to the best of my current knowledge, and especially in light of my examples above.

• Ricardo: Just to be clear, you do in fact agree that any two ways of endowing a topological vector bundle with a smooth structure give rise to isomorphic smooth bundles, right? I honestly think this is a counterexample to a straw-man constructed by misreading "more than one differential structure." Clearly it is reasonable to mean "isomorphism class of smooth structure" by "differential structure"; it seems a bit strong to say my and Michael Murray's answer are wrong. I take your point that it's worth drawing a distinction between objects and isomorphism classes of objects, but really... Feb 8, 2013 at 5:05
• (cont.) that point could have been made with a bit more generosity. Feb 8, 2013 at 5:07
• @Daniel: I understand you objection, and I apologize if my wording was too strong and unpleasant. I have changed it accordingly, and I hope it reads better now. Nevertheless, I will persevere with my statement, as it seems to me that there exist distinct, non-equivalent differentiable structures on the total space of a vector bundle which make it into a smooth vector bundle. I want to be quite clear that I consider a differentiable structure to be either a maximal smooth atlas, or an equivalence class of smooth atlases. (to be continued) Feb 8, 2013 at 6:04
• What's your definition of equivalent Ricardo ? Feb 8, 2013 at 6:08
• (continuation) I absolutely agree that any two ways of endowing a topological vector bundle with a smooth structure which makes it into a smooth vector bundle give rise to isomorphic smooth vector bundles. However, as I attempt to argue in my answer above, that isomorphism cannot in general be the identity function on the total space. Perhaps I am misunderstanding what you mean. If so, please let me know. Feb 8, 2013 at 6:09

For D) For a holomorphic vectorbundle equipped with hermitian metric $h$ and chern connection $\nabla$ there exists the so called twisted dolbeault-complex: You look only at antiholomorphic forms and extend the antiholomorphic part of the exterior derivative with the antiholomorphic part $\nabla^{0,1}=\bar \partial$ of the chern connection. Then you can easily prove, that this gives you a complex.

Good references are Huybrechts "complex geometry" or Griffiths, Harris "principles of algebraic geometry"

• Don't you need flatness too? Feb 7, 2013 at 16:28
• No, because the vectorbundle E is holomorphic: Denote $\bar \partial$ the antiholomorphic part of the exterior derivative and let $\Omega^{p,q}(E)$ be the (p,q)-Forms with values in E. Set $\bar \partial_E (\sum_i \alpha_i \otimes s_i):=\sum_i \bar \partial(\alpha_i) \otimes s_i$ on a desomposed section. Then because of holomorphicity $\bar \partial f_{ij}=0$, for \$f_{ij} so the operator is welldefined if you change trivialization. (taken from Huybrechts, Complex Geometry, Lemma 2.6.23) Feb 8, 2013 at 8:40