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Trivialisation of Vector Bundles on Stein Spaces

I assume you meant for the fiberwise rank of \$E\$ to be constant, say \$r>0\$ (or at least uniformly bounded above). The answer is "yes" when the Stein space has finite dimension (equivalently, when its analytic irreducible components, all of which must be equidimensional via consideration of their connected normalizations, have uniformly bounded dimension). There is no need to make a "global finite generation" hypothesis (which would be hard to check in practice anyway); dimension finiteness of \$X\$ is entirely sufficient.In fact, for induction purposes it is better to consider more generally \$E\$ that is merely a coherent sheaf for which its fibers \$E(x)\$ at all \$x in X\$ have uniformly bounded dimension, say at most some \$r>0\$. We will show that any such \$E\$ is generated by finitely many global sections. Then when \$E\$ is a vector bundle with fiberwise rank \$le r\$ we can build a finite trivializing cover governed by where various subsets of size \$le r\$ in the global generating set constitute a fiberwise frame. The main content in the argument will be the global theory of analytic irreducible components and their equidimensionality. We may assume \$X\$ is non-empty. Let \$d ge 0\$ be the dimension of \$X\$. If \$d=0\$ then \$X\$ is topologically discrete (with artinian local ring at each point) and everything is clear. Suppose \$d > 0\$, and let \$X_i\$ be the (locally finite) set of analytic irreducible components of \$X\$. For each \$i\$, choose \$x_i in X_i\$ not lying in any other \$X_j\$ (as clearly exists by local finiteness of \$X_i\$ in \$X\$ or many other reasons). Let \$e_j^(i)_1 le j le r\$ be a spanning set of the fiber \$E(x_i)\$. The set \$Z\$ of points \$x_i\$ in \$X\$ is discrete, and with the reduced structure we get a closed immersion \$h:Z hookrightarrow X\$. Let \$s_1, dots, s_r\$ be global sections of \$E\$ such that \$s_j(x_i) = e_j^(i)\$; this exists because of the Stein property of \$X\$ and the surjectivity of the map of coherent sheaves \$\$E

ightarrow h_ast(E|_Z) = bigoplus_i (h_i)_ast(E(x_i))\$\$ where \$h_i: x_i hookrightarrow X\$ is the natural closed immersion (surjectivity uses the discreteness of \$Z\$).Now consider the natural map \$phi:O_X^r

ightarrow E\$ defined by \$(a_j) mapsto sum a_j s_j\$. By design, \$phi\$ is surjective between fibers at all points of \$Z\$, so the coherent sheaf \$F =

mcoker(phi)\$ on \$X\$ has vanishing stalk at each \$z in Z\$. Hence, the coherent ideal sheaf \$

mAnn_O_X(F)\$ has associated closed subspace \$X' subset X\$ whose intersection with each irreducible \$X_i\$ is a proper analytic subspace (as \$x_i

otin X'\$ for all \$i\$). Since each irreducible component of \$X'\$ is contained in one of \$X\$ (by local finiteness considerations with analytic sets) it is therefore clear that the dimension \$d'\$ of \$X'\$ is strictly smaller than \$d\$. (If we are so lucky that \$X'\$ is empty then I suppose we declare \$d' = -infty\$; whatever.)By induction on dimension, \$F\$ viewed as a coherent sheaf on \$X'\$ (with all fibers of dimension at most \$r\$) is generated by a finite set of global sections. These lift to global sections of \$E\$ due to the Stein property of \$X\$, and together with the \$s_j\$'s above constitute a finite global generating set of \$E\$ due to Nakayama's Lemma.QED 1. almost holomorphic line bundles

I think you actually have to define the triple \$(J_L,J_M,

abla)\$ to be an almost holomorphic structure on \$Lto M\$ if there are local nonzero holomorphic sections everywhere, since the definition of 'holomorphic section' depends on \$

abla\$ as well as the choice of \$J_L\$ and \$J_M\$. Also, your choice of \$J_L\$ on the 'symplectic' bundle \$Lto M\$ reduces its structure to an \$mathrmSO(2)\$-structure, so the \$2\$-plane bundle \$L\$ with this structure is just an oriented Euclidean \$2\$-plane bundle. Any two reductions of structure of an oriented \$2\$-plane bundle from \$mathrmGL_(2,mathbbR)\$ to \$mathrmSO(2)\$ are equivalent, so there's really no information in that. It's the \$mathrmSO(2)\$-connection \$

abla\$ and the almost complex structure \$J_M\$ that carry all of the geometry, so you really should be defining your 'almost holomorphic structure in terms of the pair \$(J_M,

abla)\$.Now, when the dimension of \$M\$ is greater than \$2\$ and \$J_M\$ is generic, then there are no nonconstant local holomorphic functions on \$M\$. This means that, if you have two nonvanishing 'holomorphic' sections of \$(L,

abla)\$, then, on overlaps, the ratio of the two (which must be a local holomorphic function), will have to be constant, and hence the bundle \$L\$ will have a flat connection. This will imply that the Euler class of the \$2\$-plane bundle is zero, so that \$L\$ will be a trivial bundle if \$M\$ is simply connected.In special cases, when \$(M, J_M)\$ admits nonconstant local holomorphic functions, you can get nontrivial bundles \$L\$, but that's rare.This is not a complete answer, but it gives you an idea of what to look for and how you might think about modifying your question.

2. Can We understand Vector Bundles as pushouts?

The quotient of an object \$X\$ by an equivalence relation \$Rsubset Xtimes X\$, in any category with enough structure, is defined as the coequalizer of the two projections \$Rstackreltoto X\$. Any coequalizer can be written as the pushout of the corresponding span \$Xleftarrow Rsqcup Xto X\$, so in that sense your vector bundle is a pushout, but that's not a very natural viewpoint.The most natural diagram to use here has, I think, a more complicated shape. Consider the poset structure on the set \$S'=Asqcup Atimes A\$ generated by \$(alpha,beta)leq alpha\$ and \$(alpha,beta)leq beta\$, for each \$(alpha,beta)in Atimes A\$. It is most natural, but immaterial, to also remove the diagonal of \$Atimes A\$ from \$S'\$, yielding finally a poset \$S\$. When \$A\$ contains two points, this construction yields the poset \$alphaleftarrow (alpha,beta)

ightarrow beta\$, so that colimits along \$S\$ for larger index sets \$A\$ are generalized pushouts. Now your vector bundle is the colimit of the \$S\$-indexed diagram of spaces sending \$alphamapsto U_alphatimes mathbbR^n\$, \$(alpha,beta)mapsto p^-1(U_alphacap U_beta)\$, and the comparison maps \$(alpha,beta)leq alpha,(alpha,beta)leq beta\$ to the restrictions of \$h_alpha\$ and \$h_beta\$, respectively. This colimit has exactly the desired effect of identifying \$(x,v)\$ with \$h_beta h_alpha^-1(x,v)\$ when sensible. It might be worth remarking that the same story describes how to describe a manifold as a colimit of its coordinate patches, and for many other situations in which objects are built by gluing together local data. 3. If two smooth manifolds are homeomorphic, then their stable tangent bundles are vector bundle isomorphic

The result you are hoping for is in fact false.In section 9 of Microbundles: Part I, Milnor constructs an open set \$U subset mathbbR^m\$. With its standard smooth structure, the (stable) tangent bundle of \$UtimesmathbbR^k subset mathbbR^mk\$ is trivial, while in Corollary 9. 3, Milnor shows that it admits a smooth structure for which the tangent bundle has a non-zero Pontryagin class. As Pontryagin classes are stable, the stable tangent bundle of the latter manifold is not trivial, and hence not isomorphic to the stable tangent bundle of \$UtimesmathbbR^k\$ with its standard smooth structure

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