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1. • CommentRowNumber1.
   • CommentAuthorKarol Szumiło
   • CommentTimeNov 25th 2015
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   Author: Karol Szumiło
   Format: MarkdownItexAre topological vector spaces (over $\mathbb{R}$ with standard topology) monadic over topological spaces? Does the forgetful functor at least have a left adjoint?

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

2. • CommentRowNumber2.
   • CommentAuthorZhen Lin
   • CommentTimeNov 25th 2015
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   Author: Zhen Lin
   Format: MarkdownItexThe 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?)

   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?)

3. • CommentRowNumber3.
   • CommentAuthorKarol Szumiło
   • CommentTimeNov 25th 2015
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   Author: Karol Szumiło
   Format: MarkdownItexThat'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 $X$ is discrete, then the free TVS on $X$ is the free VS on $X$ equipped with the finest topology that makes it a TVS. This can be described quite explicitly as the colimit of $\mathbb{R}^A$s for all finite $A \subseteq X$ with their standard topologies. Perhaps there is some variation of this colimit that takes the topology of $X$ into account.

   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 $X$ is discrete, then the free TVS on $X$ is the free VS on $X$ equipped with the finest topology that makes it a TVS. This can be described quite explicitly as the colimit of $\mathbb{R}^A$s for all finite $A \subseteq X$ with their standard topologies. Perhaps there is some variation of this colimit that takes the topology of $X$ into account.

4. • CommentRowNumber4.
   • CommentAuthorDavidRoberts
   • CommentTimeNov 25th 2015
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   Author: DavidRoberts
   Format: MarkdownItexNote that there are such things as free topological groups on Tychonoff spaces, so I guess this might be similar.

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

5. • CommentRowNumber5.
   • CommentAuthorTodd_Trimble
   • CommentTimeNov 25th 2015
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   Author: Todd_Trimble
   Format: MarkdownItexIt 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.

   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.

6. • CommentRowNumber6.
   • CommentAuthorKarol Szumiło
   • CommentTimeNov 25th 2015
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   Author: Karol Szumiło
   Format: MarkdownItexDavid, 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 $X$, let $\mathbb{R} X$ be the free VS on $X$. For each $m$ consider the evaluation map $\phi_m \colon \mathbb{R}^m \times X^m \to \mathbb{R} X$ defined as $\phi_m(a, x) = a_1 x_1 + \ldots + a_m x_m$. Then the finest topology on $\mathbb{R} X$ making all $\phi_m$s continuous makes $\mathbb{R} X$ into the free TVS on $X$.

   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 $X$, let $\mathbb{R} X$ be the free VS on $X$. For each $m$ consider the evaluation map $\phi_m \colon \mathbb{R}^m \times X^m \to \mathbb{R} X$ defined as $\phi_m(a, x) = a_1 x_1 + \ldots + a_m x_m$. Then the finest topology on $\mathbb{R} X$ making all $\phi_m$s continuous makes $\mathbb{R} X$ into the free TVS on $X$.

7. • CommentRowNumber7.
   • CommentAuthorZhen Lin
   • CommentTimeNov 25th 2015
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   Author: Zhen Lin
   Format: MarkdownItexThat makes perfect sense. There's an obvious generalisation to arbitrary algebraic theories. Funnily enough, I just remembered that I posted [something related](http://math.stackexchange.com/a/1458112/5191) about enriched Lawvere theories a few weeks ago...

   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…

8. • CommentRowNumber8.
   • CommentAuthorKarol Szumiło
   • CommentTimeNov 25th 2015
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   Author: Karol Szumiło
   Format: MarkdownItexYou 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 $\mathbb{R}^m$s have non-trivial topologies to start with.

   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 $\mathbb{R}^m$s have non-trivial topologies to start with.

9. • CommentRowNumber9.
   • CommentAuthorKarol Szumiło
   • CommentTimeDec 7th 2015
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   Author: Karol Szumiło
   Format: MarkdownItexI 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, $\mathbb{R} X$ may not even be a TVS. The addition $\mathbb{R} X \times \mathbb{R} X \to \mathbb{R} X$ may fail to be continuous. There is an explicit counterexample for a discrete $X$ of cardinality $\mathfrak{c}$ in Dugundji's _Topology_ (A.4.3 in Appendix I). The problem is that the product topology on $\mathbb{R} X \times \mathbb{R} X$ 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.

   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, $\mathbb{R} X$ may not even be a TVS. The addition $\mathbb{R} X \times \mathbb{R} X \to \mathbb{R} X$ may fail to be continuous. There is an explicit counterexample for a discrete $X$ of cardinality $\mathfrak{c}$ in Dugundji’s Topology (A.4.3 in Appendix I). The problem is that the product topology on $\mathbb{R} X \times \mathbb{R} X$ 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.

10. • CommentRowNumber10.
    • CommentAuthorZhen Lin
    • CommentTimeDec 8th 2015
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    Author: Zhen Lin
    Format: MarkdownItexAh, I think I misread your post. I thought you meant to give $\mathbb{R} X$ the finest topology making the maps $$X \to \mathbb{R} X$$ $$\mathbb{R}^n \times (\mathbb{R} X)^n \to \mathbb{R} X$$ continuous, but I suppose that might fail to be a complete lattice.

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

    $$X \to \mathbb{R} X$$ $$\mathbb{R}^n \times (\mathbb{R} X)^n \to \mathbb{R} X$$

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

11. • CommentRowNumber11.
    • CommentAuthorKarol Szumiło
    • CommentTimeDec 8th 2015
    • (edited Dec 8th 2015)
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    Author: Karol Szumiło
    Format: MarkdownItexThat's what we would like to say, but it is not clear that there is the finest topology making these maps continuous since $\mathbb{R} X$ appears in their domains. (And if we know that it exists, it's still unclear how to construct it from the topology of $X$.) I assume that this is what you meant by the complete lattice remark, I haven't exactly understood that.

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

12. • CommentRowNumber12.
    • CommentAuthorZhen Lin
    • CommentTimeDec 8th 2015
    • (edited Dec 8th 2015)
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    Author: Zhen Lin
    Format: MarkdownItexSorry, I missed half a sentence there somehow. What I meant to say: there is a poset of topologies on $\mathbb{R} X$ making the indicated maps continuous (i.e. the subposet of TVS topologies on $\mathbb{R} X$ that also makes $X \to \mathbb{R} X$ 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 $\mathbb{R} X$), and since the forgetful functor $(X \downarrow \mathbf{TVS}) \to \mathbf{Top}$ creates all limits, it follows that the intersection of any set of TVS topologies on $\mathbb{R} X$ making $X \to \mathbb{R} X$ continuous must also be a TVS topology making $X \to \mathbb{R} X$ continuous. In particular, there is a minimal TVS topology on $\mathbb{R} X$ making $X \to \mathbb{R} X$ continuous. Not very constructive, but neither is the GAFT...

    Sorry, I missed half a sentence there somehow. What I meant to say: there is a poset of topologies on $\mathbb{R} X$ making the indicated maps continuous (i.e. the subposet of TVS topologies on $\mathbb{R} X$ that also makes $X \to \mathbb{R} X$ 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 $\mathbb{R} X$), and since the forgetful functor $(X \downarrow \mathbf{TVS}) \to \mathbf{Top}$ creates all limits, it follows that the intersection of any set of TVS topologies on $\mathbb{R} X$ making $X \to \mathbb{R} X$ continuous must also be a TVS topology making $X \to \mathbb{R} X$ continuous. In particular, there is a minimal TVS topology on $\mathbb{R} X$ making $X \to \mathbb{R} X$ continuous. Not very constructive, but neither is the GAFT…

13. • CommentRowNumber13.
    • CommentAuthorKarol Szumiło
    • CommentTimeDec 8th 2015
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    Author: Karol Szumiło
    Format: MarkdownItexRight, 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.

    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.