I'm teaching myself some differential geometry in the hope to understand gauge theory properly. In the definition of the the pullback bundle I came across a strange notation that I've never seen before. The pullback bundle was defined, with $f: M \to N$, as

\begin{equation} f^\ast \mathcal{E} \equiv (f^\ast E, f^\ast \pi, M , F), \end{equation} where the total space is given by \begin{equation} f^\ast E \equiv N \times_M E := \{ (x,z) \in N \times E | f(x) = \pi(z) \}. \end{equation}

What does the subscript on the $\times$ operator mean? Is it something to do with the operation occurring on the manifold $M$?

  • 1
    $\begingroup$ Are you familiar with Pullbacks in category theory? If not i think reading through this will help you understand the notation: en.wikipedia.org/wiki/Pullback_(category_theory) $\endgroup$ – Riquelme Jul 12 at 13:15
  • $\begingroup$ Ah perfect, thanks! $\endgroup$ – SBrents Jul 12 at 14:55
  • $\begingroup$ SBrents Can you please give a reference where you found it? $\endgroup$ – magma Jul 13 at 14:04

This is an operation in that is sometimes referred to as "fiber product", and which has other names and a general formalization in category theory (see the comment of @Riquelme). I have never seen this particular notation for it, but it nonetheless makes a kind of sense.

The idea is that you are given two functions both with the same target $M$, in this case $f : N \to M$ and $\pi : E \to M$. You want to form a kind of restricted product of $N$ and $E$, which could be called the "product of $N$ with $E$ over $M$". This is a subset of the "true" product, and it's defined by the formula that you gave.

You can form a fiber product in many different categories: groups and topological spaces are two places where I see fiber products now and then (probably related to the fact that I am a geometric group theorist).

In your situation the fiber product is being put to work to form pullback bundles in various bundle categories, such as vector bundles or fiber bundles.

  • $\begingroup$ Makes sense actually, thanks for the clear answer. $\endgroup$ – SBrents Jul 12 at 14:57

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.