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  Vector bundles where the rank can only be defined locally?

+ 1 like - 0 dislike

As I understand it, a real or complex vector bundle of rank $k$ consists of

  •   a manifold $X$ which is the base space of the vector bundle $E$
  •   a  bundle projection $\pi: E \to X$
  •   for every $x \in X$ the \fibre $\pi^{-1}(x)\in F$ has the structure of a $R^n$- or $C^n$-space

Locally, $E$ is a product space in the following sense:

For every point $x\in X$ there is a suitable  environment $U \subset X$ and $F_U\subset F$, a natural number $k$, and a mapping (a diffeomorphism) $\phi_U: U \times F_U \to \pi^{-1}(U)$ such that for all $x\in U$,

  • $(\pi \circ \phi_U)(x,v) = x$ for all vectors $v \in F_U$
  • The map $v \mapsto \phi_U(x,v)$ from $F_U$ to $\pi^{-1}(U)$
    is linear and bijective.

The pair $(U,\phi_U)$ is called a local trivialization.

If the rank of the fibre $F$ is globally $k$, $E$ is a vector bundle of rank $k$}. If $k=1$ we have a
line bundle.

My question now is:

Do there exist any mathematical structures that look like a vector bundle for which the rank $k$ can not be defined globally, for example because it varies over the base space $X$?

And if such mathematical structures exist, what are they useful for in theoretical physics?

asked Nov 27, 2017 in Mathematics by Dilaton (6,240 points) [ no revision ]

2 Answers

+ 4 like - 0 dislike

Yes. Let me answer with an example. Consider a vector bundle $E \to X$ for some smooth topological space $X$. Then one can compute the Yang-Mills functional and find "topological" solutions of the ASD equation $F^{+}=0$. This is a differential equation whose solutions represent instantons, i.e. local connections $A \in \Omega^1(X,End(E))$ obeying the ASF equation. These instantons have a moduli space (i.e. the set of all solutions of the ASD equation modulo the gauge group). The resulting space can be very singular and often non-Hausdorff because there exist instantons that run off to infinity that become point-like and thus singular exactly because there the bundle changes rank. So in order to obtain one nice moduli space one can consider various ways to cure these singularities, one of which is called Gieseker-Maruyama compactification where we add the singular connections (reducible connections or abelian "instantons") to the moduli space. This corresponds to the moduli space of coherent sheaves over $X$. Thus a generalization of vector bundles can be given by coherent torsion free sheaves which intuitively can be thought as vector bundles of rank $r$ except some points of $X$ where the rank suddenly jumps. A good reference is Le Potier's lectures on vector bundles. The usefulness to physics appears in whole bunch of examples. Vafa-Witten theory computes the Euler characteristic of exactly this moduli space. In general instantons correspond to torsion free sheaves, while vector bundles correspond to locally free sheaves. (Topological) D-branes can be modelled using the bounded derived category of coherent sheaves. In general, I would say that using sheaves allows us to use the power of algebraic geometry for exact computations. 

answered Nov 27, 2017 by conformal_gk (3,625 points) [ revision history ]

So according to 40227's answer, connected vector bundles have constant rank but (as your last remark shows), working instead with sheaves allows the rank requirement to be relaxed?

As far as sheaves are concerned they are defined locally and come equipped with some Chern character $(r,c_1,ch_2)$ and for most of the points of $Χ$ they behave as vector bundles. The funny thing with the rank only happens with the so-called point like instantons, that is over the singular points of the moduli space.  In general the idea as 40227 said is that coherent sheaves defined on open sets have constant rank but when considering subsets of lower dimension their rank can change. 

+ 4 like - 0 dislike

The definition of a vector bundle given in the question is quite confusingly written (do we have $k=n$? $F_U$ is probably inside $E$ rather than inside $F$...). Fibers of a (finite rank) vector bundle are (finite dimensional) vector spaces. If the base manifold $X$ is connected, then it follows from the definition that all these vector spaces have the same dimension, which is called the rank of the vector bundle. In other words, the rank is always defined locally and if $X$ is connected, it automatically extends to a global notion (by definition, a vector bundle locally looks like a product so the rank is locally constant). If $X$ is disconnected, you just have various vector bundles of various ranks on the various connected components of $X$.

A generalization of the notion of vector bundle, where the rank can jump discontinuously, is given by the notion of sheaf. Typical examples of sheaves include vector bundles supported on submanifolds (the rank is zero outside the submanifold and possibly non-zero on the submanifold). This is relevant to describe gauge theory on a D-brane of worldvolume this submanifold.

answered Nov 27, 2017 by 40227 (5,140 points) [ no revision ]

Thanks for these nice explanations.

I thought that $k$ is the dimension of the fibres and $n$ is the dimension of the base space, the dimension of the vector bundle $E$ is then $n+k$...

$F_U$ is meant to be the part of the fibre space "above" the local environment $U$ of $x\in X$.

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