# Do locally convex topological vector spaces embed into diffeological spaces?

The nLab casually remarks that locally convex tvs embed into diffeological spaces by (discussion around) a corollary in Kriegl and Michor, namely 3.14, but this deals with Boman's theorem and results about smooth maps from a euclidean space to a locally convex spaces.

The definition of smoothness for maps $U \to V$ where $U$ is $c^\infty$-open in a lctvs $W$, and $V$ is a lctvs, is given so that smooth maps are automatically the same as between maps between the associated diffeological spaces.

However, there is a notion of smoothness for maps between lctvs (due to Michal [1,2] and Bastiani [3]) not relying on K&M's definition, but I don't how they relate (Keller [4] treats this, I think, but I don't have access to that book at the moment).

Is it still true that lctvs embed into diffeological spaces using Michal-Bastiani smoothness?

[1] Michal, A. D. Differential calculus in linear topological spaces, Proc. Nat. Acad. Sci. USA 24 (1938), 340-342 (pdf)

[2] Michal, A. D. Differential of functions with arguments and values in topological abelian groups, Proc. Nat. Acad. Sci. USA 26 (1940), 356–359. (pdf)

[3] Bastiani, A., Applications différentiables et variétés différentiables de dimension infinie, J. Anal. Math. 13 (1964), 1–114. (article, paywalled)

[4] Keller, H. H., Differential Calculus in Locally Convex Spaces, Lecture Notes in Mathematics 417, Springer-Verlag, 1974 (Springerlink, paywalled)

The answer is no. In what follows (see OP's comments below) we assume that the arrows of the category of lctvs are Michal-Bastiani smooth maps. Recall that a map $\Phi:E\rightarrow F$ from a lctvs $E$ into another lctvs $F$ is said to be Michal-Bastiani smooth if its directional (Gâteaux) derivatives of order $k$

$$D^k\Phi[x](y_1,\ldots,y_k) = \left.\frac{\partial^k}{\partial\lambda_1\cdots\partial\lambda_k}\right|_{\lambda_1=\cdots=\lambda_k=0}\Phi\left(x+\sum^k_{j=1}\lambda_j y_j\right)$$

exist and the maps $D^k\Phi:E\times E^k\rightarrow F$ are (jointly) continuous for all $k\in\mathbb{N}$. This notion of smoothness is the one used in Milnor's treatment of infinite-dimensional Lie groups (J. Milnor, "Remarks on Infinite-Dimensional Lie Groups". In: B. DeWitt, R. Stora, eds., Les Houches Session XL, Relativity, Groups and Topology II (North-Holland, 1984), pp. 1007-1057) and Hamilton's exposé of the Nash-Moser inverse function theorem (R.S. Hamilton, "The Inverse Function Theorem of Nash and Moser". Bull. Amer. Math. Soc. 7 (1982) 65-222).

Michal-Bastiani smooth maps are also smooth in the sense of Kriegl and Michor (i.e. composing with a smooth curve leads to another smooth curve; some call the latter convenient smoothness), since the chain rule holds for Michal-Bastiani smooth maps. The converse is not necessarily true - both notions of smoothness coincide for Fréchet spaces, but in general they differ: conveniently smooth maps need not even be continuous, whereas Michal-Bastiani smooth maps are always so. A counter-example to continuity for conveniently smooth maps may be found in the paper of H. Glöckner, "Discontinuous Non-Linear Mappings on Locally Convex Direct Products". Publ. Math. Debrecen 68 (2006) 1-13, arXiv:math/0503387. In particular, if neither $E$ nor $F$ are Fréchet spaces, there may be maps between $E$ and $F$ which are smooth in the diffeological sense and need not even be continuous, let alone Michal-Bastiani smooth. Therefore, the natural inclusion functor of lctvs into diffeological spaces is not fully faithful. It is not clear from the above counterexample whether there are convenient diffeomorphisms which fail to be continuous or Michal-Bastiani smooth. (see however TaQ's answer below for such a counterexample)

• "there are maps between E and F which are smooth in the diffeological sense and need not even be continuous, let alone Michal-Bastiani smooth." <-- ah, this is what I wanted to know! Although it wasn't clear from how I asked it, since 'embedding between categories' doesn't have a universal definition. So I guess the only thing to worry about is if there diffeological isomorphisms (not necc. linear) that are not MB-smooth. This seems like a difficult question. – David Roberts Jun 15 '15 at 6:17
• @DavidRoberts so, what is the definition of "embedding of categories" you have in mind in your question? Do you put any restriction on the arrows in the category of lctvs besides being continuous linear maps (e.g. do you require them to be topological isomorphisms, or something)? – Pedro Lauridsen Ribeiro Jun 15 '15 at 14:43
• I wasn't thinking of linear maps, and I was trying to compare to the statement for Fréchet spaces, which by embedding means a full and faithful functor (and taking all smooth maps between Fréchet spaces). My residual query is whether a convenient diffeomorphism is necessarily MB-smooth, which is intermediate between the inclusion being fully faithful and being just faithful. – David Roberts Jun 15 '15 at 22:39
• So, you were considering smooth maps (in either sense) as the arrows of the category of lctvs in your original question, and your residual question is whether the answer changes if you restrict the arrows to be smooth diffeomorphisms. Is this corrrect? – Pedro Lauridsen Ribeiro Jun 15 '15 at 23:04
• I have edited my answer to better suit your original question, based on your comments. I've also added a remark at the end concerning your remaining question, to which unfortunately I don't know an answer. Papers who cite Glöckner's work quoted above, according to MathSciNet, apparently don't even raise that question, let alone answer it. – Pedro Lauridsen Ribeiro Jun 16 '15 at 1:25

$\def\sp{\kern.4mm}\def\bbN{\mathbb N}\def\bbZ{\mathbb Z}\def\bbR{\mathbb R}$Here is the answer to David Roberts' residual query: is a convenient diffeomorphism necessarily MB-smooth? No. A counterexample follows:

Let $E$ be the vector space of all real (two-sided) sequences $x=\langle\sp x_i:i\in\bbZ\sp\rangle$ for which $x_i=0$ for sufficiently small $i$ , topologized so that we get a linear homeomorphism $E\to\bbR\,^{\bbN}\times\bbR\,^{(\bbN)}$ defined by $x\mapsto(u,v)$ where $u_i=x_{i-1}$ and $v_i=x_{-i}$ for $i\in\bbN$ . Then define $f:E\to E$ by $x\mapsto y$ where $y_i=x_i$ for $i\in\bbZ\setminus\{\sp 0\sp\}$ and $y_0=x_0+\sum_{i\in\bbN}(x_i\cdot x_{-i})$ . The inverse is given by $y\mapsto x$ where $y_i=x_i$ for $i\in\bbZ\setminus\{\sp 0\sp\}$ and $x_0=y_0-\sum_{i\in\bbN}(y_i\cdot y_{-i})$ . Since the duality map $\bbR\,^{\bbN}\times\bbR\,^{(\bbN)}\to\bbR$ is a discontinuous but bornological, and hence conveniently smooth bilinear map, the assertion follows.

• To clarify a bit: " $x_i=0$ for sufficiently small $i$ " means more precisely that " there is $i_0\in\mathbb Z$ such that $x_i=0$ for all $i\in\mathbb Z$ satisfying $i\le i_0$ ". This $i_0$ of course depends on $x$ . The discontinuous (and hence not MB−smooth) convenient diffeomorphism is the map $f:E\to E$ . – TaQ Jun 17 '15 at 20:03
• Thanks, TaQ, I've added your example to the nLab page ncatlab.org/nlab/show/Michal-Bastiani+smooth+maps, citing this answer. I may ask another question in future about MB-smooth bijections with only convenient smooth inverse. But that's another question. – David Roberts Jun 17 '15 at 23:46