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    • CommentRowNumber1.
    • CommentAuthorKarol Szumiło
    • CommentTimeNov 25th 2015

    Are topological vector spaces (over \mathbb{R} with standard topology) monadic over topological spaces? Does the forgetful functor at least have a left adjoint?

    • CommentRowNumber2.
    • CommentAuthorZhen Lin
    • CommentTimeNov 25th 2015

    The solution set condition is satisfied, and the forgetful functor preserves limits of small diagrams, so we may apply the general adjoint functor theorem. (It’s not clear to me what the left adjoint actually is. I imagine the underlying ordinary vector space is the obvious thing, but what is the topology going to be?)

    • CommentRowNumber3.
    • CommentAuthorKarol Szumiło
    • CommentTimeNov 25th 2015

    That’s a good observation, so the free functor exists. I’m curious whether we can describe it explicitly. The best I can say at the moment is that if XX is discrete, then the free TVS on XX is the free VS on XX equipped with the finest topology that makes it a TVS. This can be described quite explicitly as the colimit of A\mathbb{R}^As for all finite AXA \subseteq X with their standard topologies. Perhaps there is some variation of this colimit that takes the topology of XX into account.

    • CommentRowNumber4.
    • CommentAuthorDavidRoberts
    • CommentTimeNov 25th 2015

    Note that there are such things as free topological groups on Tychonoff spaces, so I guess this might be similar.

    • CommentRowNumber5.
    • CommentAuthorTodd_Trimble
    • CommentTimeNov 25th 2015

    It should be noted that in the literature on “free topological groups”, there is often and I think usually imposed extra conditions that are not mandated by the category theory per se, such as the condition that the unit be a topological embedding. There is actually a lot of sweat poured into this by the workers of yesteryear like Markov.

    • CommentRowNumber6.
    • CommentAuthorKarol Szumiło
    • CommentTimeNov 25th 2015

    David, free topological groups exist for all spaces by GAFT. I guess the Tychonoff property is sometimes assumed because of the conditions mentioned by Todd.

    In any case, this reminded me that I already know the construction of free topological abelian groups. I have learned it from Stefan Schwede and it is easily adapted to TVSs.

    For a topological space XX, let X\mathbb{R} X be the free VS on XX. For each mm consider the evaluation map ϕ m: m×X mX\phi_m \colon \mathbb{R}^m \times X^m \to \mathbb{R} X defined as ϕ m(a,x)=a 1x 1++a mx m\phi_m(a, x) = a_1 x_1 + \ldots + a_m x_m. Then the finest topology on X\mathbb{R} X making all ϕ m\phi_ms continuous makes X\mathbb{R} X into the free TVS on XX.

    • CommentRowNumber7.
    • CommentAuthorZhen Lin
    • CommentTimeNov 25th 2015

    That makes perfect sense. There’s an obvious generalisation to arbitrary algebraic theories. Funnily enough, I just remembered that I posted something related about enriched Lawvere theories a few weeks ago…

    • CommentRowNumber8.
    • CommentAuthorKarol Szumiło
    • CommentTimeNov 25th 2015

    You are right, at first I thought that this uses commutativity somehow, but clearly it doesn’t.

    And indeed, my question is a special case of the question you linked and my answer is a special case of yours. Of course, the generality of “topological Lawvere theories” is important since m\mathbb{R}^ms have non-trivial topologies to start with.

    • CommentRowNumber9.
    • CommentAuthorKarol Szumiło
    • CommentTimeDec 7th 2015

    I need to make a correction. The constructions I gave in #3 and #6 work in the category of compactly generated spaces but not in the category of all topological spaces.

    With the topology I described, X\mathbb{R} X may not even be a TVS. The addition X×XX\mathbb{R} X \times \mathbb{R} X \to \mathbb{R} X may fail to be continuous. There is an explicit counterexample for a discrete XX of cardinality 𝔠\mathfrak{c} in Dugundji’s Topology (A.4.3 in Appendix I). The problem is that the product topology on X×X\mathbb{R} X \times \mathbb{R} X may be to coarse. All of this goes away if we work in the category of compactly generated spaces and is of course consistent with Zhen Lin’s math.SE answer linked in #7 where he assumed that products preserve colimits.

    The free TVSs exist in the category of all topological spaces by GAFT but their topologies remain mysterious to me.

    • CommentRowNumber10.
    • CommentAuthorZhen Lin
    • CommentTimeDec 8th 2015

    Ah, I think I misread your post. I thought you meant to give X\mathbb{R} X the finest topology making the maps

    XXX \to \mathbb{R} X n×(X) nX\mathbb{R}^n \times (\mathbb{R} X)^n \to \mathbb{R} X

    continuous, but I suppose that might fail to be a complete lattice.

    • CommentRowNumber11.
    • CommentAuthorKarol Szumiło
    • CommentTimeDec 8th 2015
    • (edited Dec 8th 2015)

    That’s what we would like to say, but it is not clear that there is the finest topology making these maps continuous since X\mathbb{R} X appears in their domains. (And if we know that it exists, it’s still unclear how to construct it from the topology of XX.) I assume that this is what you meant by the complete lattice remark, I haven’t exactly understood that.

    • CommentRowNumber12.
    • CommentAuthorZhen Lin
    • CommentTimeDec 8th 2015
    • (edited Dec 8th 2015)

    Sorry, I missed half a sentence there somehow. What I meant to say: there is a poset of topologies on X\mathbb{R} X making the indicated maps continuous (i.e. the subposet of TVS topologies on X\mathbb{R} X that also makes XXX \to \mathbb{R} X continuous), but I suppose that might fail to be a complete lattice.

    In fact, it must be a complete lattice – this what the GAFT argument boils down to – notice that there is a maximal element of this poset (namely, the indiscrete topology on X\mathbb{R} X), and since the forgetful functor (XTVS)Top(X \downarrow \mathbf{TVS}) \to \mathbf{Top} creates all limits, it follows that the intersection of any set of TVS topologies on X\mathbb{R} X making XXX \to \mathbb{R} X continuous must also be a TVS topology making XXX \to \mathbb{R} X continuous. In particular, there is a minimal TVS topology on X\mathbb{R} X making XXX \to \mathbb{R} X continuous. Not very constructive, but neither is the GAFT…

    • CommentRowNumber13.
    • CommentAuthorKarol Szumiło
    • CommentTimeDec 8th 2015

    Right, I suspected that the existence argument should be similar to GAFT. In any case, what I wrote in #6 is a rather explicit construction of the topology which does work in the category of compactly generated spaces and this should be good enough for most purposes.