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I'm trying to generalize a method which can solve a problem involving discrete group to continuous group. And the method involve construct a "free vector space" from discrete space. So I'm trying to find a way to construct some kind of "free Hilbert space" from a topological space. But I have no idea how to achieve this goal. Then I find that one can get a free group or free vector space by considering the left adjoint of forgetful functor, so I wonder if one can go this way to get "free Hilbert space". But I don't know how to construct this adjoint.

Let $\mathbf{Hil}$ be the category of Hilbert spaces, $\mathbf{Top}$ be the category of topological spacces, and consider the forgetful functor $ F : \mathbf{Hil} \to \mathbf{Top}.$ Does this functor $F$ have a left adjoint? If it does, what is this left adjoint?

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    $\begingroup$ Quick beginner guide for asking a well-received question + please avoid "no clue" questions: include some work in your post. $\endgroup$
    – Anne Bauval
    Commented Oct 3, 2023 at 17:40
  • $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$
    – Community Bot
    Commented Oct 3, 2023 at 17:51
  • $\begingroup$ I'm sorry I didn't clear about what my question is, I add the problem I had in the question. $\endgroup$
    – Chen
    Commented Oct 3, 2023 at 17:55
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    $\begingroup$ @Chen Ignore the bot. $\endgroup$
    – FShrike
    Commented Oct 3, 2023 at 17:57
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    $\begingroup$ Your question was already clear enough. What was/is lacking is some thoughts and/or attempts. $\endgroup$
    – Anne Bauval
    Commented Oct 3, 2023 at 17:59

1 Answer 1

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The left adjoint does not exist. Otherwise, also the forgetful functor $\mathbf{Hilb} \to \mathbf{Top} \to \mathbf{Set}$ to the category of sets had a left adjoint (since left adjoints "compose"), but it doesn't:

Is there a concept of a "free Hilbert space on a set"?

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  • $\begingroup$ Thank you for your answer. I think I have to try other way to generalize the method. $\endgroup$
    – Chen
    Commented Oct 3, 2023 at 18:11

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