Part 1:

How big is the category $TVS_{loc.conv.}$ of locally convex topological vector spaces (and continuous maps)?

In other words (and less cheekily), is there a free locally convex TVS having any given set as basis? This would imply the functor $TVS_{loc.conv.} \to Set$ is essentially surjective and has an adjoint.

Part 2:

Consider now the intersection $T$ of $TVS_{loc.conv.}$ (as a subcategory of $Top$) with $CGWH$, the subcategory of $Top$ of compactly generated weak Hausdorff spaces.

How big is $T$? (Or, is $T$ essentially small?)

Note that a Banach space is locally compact iff it is finite dimensional, but I am being stupid and not remembering the relationship between local compactness and compact generation, so I can't immediately use this fact.

  • $\begingroup$ off the top of my head, doesn't any (real or complex) vector space become a LCTVS when it's equipped with the discrete topology? If so, is that relevant to your 1st question? $\endgroup$ – Yemon Choi Jan 21 '11 at 7:08
  • 4
    $\begingroup$ Only if the space is a single point, Yemon. Otherwise scalar multiplication is not continuous. $\endgroup$ – Bill Johnson Jan 21 '11 at 7:30
  • $\begingroup$ Oops. Thanks, Bill, for catching the error. $\endgroup$ – Yemon Choi Jan 21 '11 at 8:18
  • $\begingroup$ David, what is your definition of compactly generated? I see that we are not all using the same definition. $\endgroup$ – Bill Johnson Jan 21 '11 at 9:15
  • $\begingroup$ ncatlab.org/nlab/show/compactly+generated+space $\endgroup$ – David Roberts Jan 21 '11 at 9:25

Part 1: The "cheeky" answer is: huge. There is a left adjoint to the forgetful functor $LCTVS \to Vect$ (in particular there is a left adjoint to the forgetful functor $LCTVS \to Sets$): Equip a vector space $V$ with the locally convex topology induced by all linear functionals on $V$ (or as Pietro Majer put it: the topology given by all semi-norms).

Edit 2:

Every linear map $f: V \to W$ is continuous: every semi-norm $|\,\cdot\,|$ on $W$ gives rise to a semi-norm on $V$ by $v \mapsto |f(v)|$. For every net $v_{i} \to v$ we have $|f(v_{i} - v)| \to 0$, hence $f(v_{i}) \to f(v)$ and thus $f$ is continuous.

Edit: The following summarizes what has transpired from Bill's, Neil's and my answers/comments:

Part 2: If $S$ is any set then the space $\ell^{2}(S) = \left\{\lambda = \sum_{s \in S} \lambda_{s} s \mid \sum_s |\lambda_{s}|^{2} \lt \infty \right\}$ is a Hilbert space with respect to the scalar product $\langle \lambda, \mu \rangle = \sum_{s \in S} \lambda_{s} \overline{\mu}_{s}$ and it contains the free vector space on $S$. Since metrizable spaces are compactly generated and weakly Hausdorff (see N. Strickland's notes, Propositions 1.6 and 1.2), and since the cardinality of $S$ determines the isomorphism type of $\ell^{2}{(S)}$ (see here), the category of compactly generated locally convex topological vector spaces cannot be essentially small.

| cite | improve this answer | |
  • $\begingroup$ So, if I understand you correctly, arbitrary Hilbert spaces are compactly generated and weakly Hausdorff? I'll accept your answer, but as currently written, it doesn't seem to answer the question (one needs to combine your answer, someone else's answer and a comment on a third answer). But I was trying to show it wasn't small, in order to put a lower bound on the size of a class of CGWH spaces containing LCTVS, so I'm happy. $\endgroup$ – David Roberts Jan 21 '11 at 21:52
  • $\begingroup$ @David: You're absolutely right to complain, so I've now tried to synthesize the piecemeal answer into a single one with references. I hope that's fine with you now. Sorry, I just tried to be honest. $\endgroup$ – Theo Buehler Jan 21 '11 at 23:57
  • $\begingroup$ No need to apologise. You came up with the required argument after all. I'm just thinking of posterity (if for some reason the other answers are deleted, then yours would have made no sense). $\endgroup$ – David Roberts Jan 22 '11 at 1:47

Every first countable space is compactly generated (because the topology is determined by the convergent sequences, which are the same thing as continuous maps from the compact space $\mathbb{N}\cup\{\infty\}$). Thus, if the topology on $V$ is determined by countable family of seminorms (or equivalently, it is a Fréchet space) then it is compactly generated.

Someone once told me that it is possible to develop the theory of LCTVS and duality as an application of the theory of CGWH spaces, and that this is very clean and efficient. However, I have never seen an account like this; if anyone can point me to one, I'd be very interested.

| cite | improve this answer | |
  • 1
    $\begingroup$ Re your second paragraph: that would be nice, I would like to see that too. $\endgroup$ – David Roberts Jan 21 '11 at 9:26
  • $\begingroup$ Just because a locally-convex topology is defined by a countable collection of seminorms doesn't mean it need be a Fréchet space. You also need it to be complete and $T_0$. $\endgroup$ – Jonathan Gleason May 19 '17 at 3:52

Since the first question doesn't seem to have been addressed directly in the answers so far, here are some suggestions. Firstly, since we are discussing topological vector spaces, I think the most natural question is to consider the forgetful function onto the category of topological spaces and since functional analysts are interested in function spaces, to completely regular spaces (of course, the case of sets can be incorporated by regarding a set as a discrete topological space). One then has a natural construction of the free locally convex space---one takes the free vector space generated by the topological space $X$ and provides it with the finest locally convex topology which agrees on $X$ with the original one. In our situation, this will be Hausdorff and will contain $X$ as a closed topological subspace. It is simple and natural to carry this one stage further and consider the completion of this space. It will have the corresonding univeral property, now for functions with values in a complete locally convex space. This space has a natural explicit representition, e.g., if we start with $[0,1]$, we get the space of Radon meassures on the interval. One of the nice things about this construction is that it can be varied almost infinitely and provides a unified approach to many spaces whose initial development was slow and painful---some of which are again forgotten lore. As examples, we can consider spaces with the universal property for bounded functions and replace continuity by other smoothness conditions---uniform continuity if $X$ is a uniform space, $C^\infty$ if $X$ is an open subset of some euclidean space, holomorphicity (subsets of the complex plane or its higher dimensional analogues), measurablility if $X$ is a measure space and so on---similarly for functions on suitable manifolds. This provides a unifying approach to such topics as uniform measures, distributions, analytic functionals and so on.

As regards the second question, I have the feeling that functional analysts and topologists use the term compactly generated with different meanings. For the former, a locally convex space (in particular, a Banach space) is compactly generated if it contains a compact subset whose span is dense. For the latter, a topological space is compactly generated if it has the finest topology which agrees with itself on compact sets (otherwise known as a $k$-space or a Kelley space). As remarked above, metric spaces have the latter property and have the former one if they are separable. Further examples of spaces which have the latter property without being metrisable are the so-called Silva spaces, i.e. countable inductive limits of sequences of Banach spaces with compact interconnecting mappings. Many of the important spaces of distributions belong to this class, as do spaces of analytic functionals.

| cite | improve this answer | |

Part 1: If $B$ is a basis for the vector space $X$, put the largest locally convex topology on $X$, sometimes called the direct sum topology. Trivially any mapping from $B$ into any locally convex space extends uniquely to a continuous linear mapping from $X$ into the space.

Part 2: Take a Hilbert space of any dimension but with its weak topology. Its unit ball is weakly compact.

| cite | improve this answer | |
  • $\begingroup$ Ok, then, for part 2 - how big can a Hilbert space be? Or, is there a bound on the cardinality of a (Hamel?) basis of a Hilbert space? $\endgroup$ – David Roberts Jan 21 '11 at 8:45
  • 2
    $\begingroup$ @David: There is no upper bound. If $S$ is any set you can consider the Hilbert space $H(S)$ with Hilbert basis $S$ formed by all sequences $(\lambda_s)_{s\in S}$ such that $\sum_{s\in S}|\lambda_{s}|^{2} < \infty$. This space contains the free vector space on $S$ and the cardinality of $S$ is a complete invariant for $H(S)$ as a topological vector space (or as a Hilbert space, if you prefer). $\endgroup$ – Theo Buehler Jan 21 '11 at 9:10
  • $\begingroup$ Details on part 1: for the largest topology of LCTVS on any real vector space, just topologize it with the family of all seminorms on it. This of course makes continuous any linear operator with values in a LCTVS. $\endgroup$ – Pietro Majer Jan 21 '11 at 9:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.