## Monday, April 19, 2010

### Why are vector bundles natural in Physics?

Aren't usual vector fields enough?

When we work with a space $M$ (a differentiable manifold in fact), we may need to consider fields on that space. The fields can be scalar, vector, tensor, spinor fields, depending on the possible values they can take - scalars, vectors, tensors, spinors. But all these can be considered vectors in some spaces, so in general the fields will be considered to be vector fields.

We can think that, considering vector fields on a space, it is as simple as considering functions on that space $M$, valued in a vector space $V$. Unfortunately, this is not the case, and there is a very good reason for this. I will explain it here.

When working with a function $f:M\to V$, we can represent it by its graph, which is in fact a subset of the cartesian product, $\{(x,f(x))|x\in M\}\subset M\times V$. Therefore, we may hope that all the vector fields on $M$ valued in $V$ are subsets of $M\times V$. If the base manifold is the sphere $S^2\subset \mathbb{R}^3$, its tangent vector fields cannot be, in general, represented as subsets of the cartesian product $S^2\times\mathbb{R}^2$ (we say that $S^2$ is not parallelizable). This and other simple examples force us to consider a more general definition of vector fields.

On the other hand, there are spaces with which this representation works always. For example, we can take $M=\mathbb{R}^n$ or a simply connected open subset of it, $M\subset\mathbb{R}^n$. All possible vector fields of such an $M$ can be represented as subsets of $M\times V$.

The idea behind the vector bundles was to consider the base space $M$ as being covered by open sets like above. The restriction of a vector fields to such an open set $U\subset M$ can be represented as a subset of $U\times V$. But the way they are glued together can vary very much, because when they are glued together, the vector space $V$ can be transformed relatively to $V$ on another open set. Take for example a circle as the basis manifold, and consider as a vector space the Euclidean one-dimensional space. We can glue it to each point o the circle in two ways: as a cylinder, and as the Mobius strip. The idea is that we can cover the base manifold with opens such that the way we associate vector spaces to its points is trivial on each open set from the covering.

The basic point is that, in order to have fields of any kind on a manifold, you need bundles. The fields are "sections" in the bundles. Now, these fields can be combined as we do with the vectors. In fact, what we can do with the vector spaces, we can do with vector bundles as well. We can construct direct sums, duals, tensor products.

Vector bundles and quantum entanglement

There is an important difference between two types of tensor products. The fields which are sections of a given vector bundle $E\to M$ form themselves a vector space $\Gamma(E\to M)$. Two such vector spaces of sections can as well be tensored. The tensor product $\Gamma(E_1)\otimes \Gamma(E_2)$ of two vector spaces of sections of two bundles over the same base manifold $M$ is larger than the vector space defined by the sections of the tensor product of the two bundles, $\Gamma(E_1\otimes E_2)$. The first contains nonlocal fields of the form $\phi(x,y)$, while the second contains only local fields, of the form $\varphi(x)=\phi(x,x)$. $\Gamma(E_1)\otimes \Gamma(E_2)$ are no longer sections of a vector bundle. The entangled states in Quantum Mechanics are represented by such nonlocal fields.

One of the most important applications of vector bundles in Physics is related to the Gauge Theory. We will discuss more about this other time.

#### 1 comment:

kiki said...

Thanks for sharing this post with us.