Is $U(1)\times SU(2) \times SU(3)$ a vector space over a field? I saw an article here that seemed to me that a similar concept to a field extension was being used.
In QFT, is each particle considered to be its own vector space?
And are the individual vector spaces then bridged together by these group extensions?
If I have asked too many questions, feel free to respond to just one.
Answer
No, the group $U(1)\times SU(2)\times SU(3)$ isn't a vector space of any kind because it doesn't have any (commuting) addition operation (curved group manifolds can rarely have such a structure).
The article "group extension" you linked to makes it very clear that the group extension does not have to be a vector space and the group operation does not have to be Abelian.
Field extensions are commutative (and vector spaces) because a field itself is a commutative ring. But the Standard Model group isn't based on any field in this mathematical sense and the group is non-Abelian i.e. non-commuting – that's what physicists describe as "groups in Yang-Mills theory".
No, it is not true that individual particles "are" vector spaces. QFT, like any quantum theory, has an important complex vector space, the Hilbert space. No other spaces appearing in the formalism of QFT are vector spaces in general. This also answers your last question by negating its assumptions.
Just to be sure, one may consider "first-quantized" theories or one-particle sectors of QFTs. They have their own Hilbert space you could interpret as the vector space for "one particle". But you could have meant something completely different; it wasn't clear what role your hypothetical vector space connected with one particle should play.
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