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In QFT, there is a vector field of electromagnetism, usually notated by $A$, which transforms as a 1-form under coordinate changes. Since quantum fields are operator-valued, I thought it is a section of a bundle like $T^*M \otimes \mathrm{End}(\mathcal{H})$, or a complexified version of that. Here, $M$ is our spacetime, $T^*M$ is its cotangent bundle, $\mathcal{H}$ is a Hilbert space.

Soon I realized my first guess has a problem. If we consider quantum fields as sections of $(some\,geometric\,bundle)\otimes \mathrm{End}(\mathcal{H})$, it generally lacks a multiplicative structure due to the first part. For example, tensor product of $dx^{\mu}$ and $dx^\nu$ of two distinct points on $M$ would not make any sense, because they live in different cotangent fibers.

In contrast, we can define composition of two operators at two points in QFT. For example, we often compute the propagator $\langle 0| \varphi(x)\varphi(y)|0\rangle$.

Question: Quantum fields are sections of which bundles?

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    $\begingroup$ In QFT, the background spacetime is usually assumed to be flat. It can be formulated for curved space, but things look different (I don't know how different though, I am only starting to learn about this) $\endgroup$
    – paulina
    Commented Jul 6 at 9:50
  • $\begingroup$ @paulina Thank you for your answer. Perhaps things are not very straightforward as I guessed... $\endgroup$
    – Sung Kan
    Commented Jul 6 at 10:11
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    $\begingroup$ Yes, there are a variety of important physical subtleties missing from your first guess. Until someone can write a useful basic answer, let me mention the issue of which bundles classical fields live in is discussed in books like Nakahara's 'Geometry, Topology, and Physics' or Frankel's 'The Geometry of Physics'. $\endgroup$
    – SethK
    Commented Jul 6 at 16:20
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    $\begingroup$ Quantum fields are operator-valued distributions, not operador-valued functions. This is likely an extra complication. Also, it is not very productive to use a Hilbert space when doing QFT in curved spacetimes. You may be interested in this answer. $\endgroup$
    – Níckolas Alves
    Commented Jul 6 at 20:03
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    $\begingroup$ Recall also that gauge invariance means the electromagnetic field is better understood as a connection on a principal $\mathrm{U}(1)$-bundle, rather than merely a one-form. $\endgroup$
    – Níckolas Alves
    Commented Jul 6 at 20:05

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