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• CommentRowNumber1.
• CommentAuthorUrs
• CommentTimeNov 14th 2012
• (edited Nov 14th 2012)

I looked again after a long while at the entry manifold structure of mapping spaces, looking for the statement that for $X$ a compact smooth manifold and $Y$ any smooth manifold, the canonical Frechet structure on $C^\infty(X,Y)$ coincides with the canonical diffeological structure.

So this statement wasn’t there yet, and hence I have tried to add it, now in Properties – Relation between diffeological and Frechet manifold structure.

To make the layout flow sensibly, I have therefore moved the material that was in the entry previously into its own section, now called Construction of smooth manifold structure on mapping space.

While re-reading the text I found I needed to browse around a good bit to see where some definition is and where some conclusion is. So I thought I’d equip the text more with formal Definition- and Proposition environments and cross-links between them. I started doing so, but maybe I got stuck.

Andrew, when you see this here and have a minute to spare: could you maybe check? I am maybe confused about how the $\{P_i\}$ and $\{Q_i\}$ are to be read and what the index set of the charts of $C^\infty(M,N)$ in the end is meant to be. For instance from what you write, what forbids the choice of $\{P_i\}$ and $\{Q_i\}$ being the singleton consisting just of $M$ and $N$ itself, respectively?

• CommentRowNumber2.
• CommentAuthorAndrew Stacey
• CommentTimeNov 14th 2012

1. The whole mapping-space stuff needs overhauling. I’ve never been totally happy with the treatment so far, and I think that I now have figured out the right way to think about all of this. Bits are appearing on the nLab, but more is in my preprint Yet more smooth mapping spaces and their smoothly local properties. I’m polishing that and as I do so, I’ll clean up the nLab as well. Quite a few of these pages (like the one in question) have been in a state of flux while I’ve tried to sort out my own thinking on various issues.

2. I need to look at the Frechet to diffeological bit in more detail. I remember having a few questions about it when skimming it a while back but never got round to looking at them more closely.

• CommentRowNumber3.
• CommentAuthorUrs
• CommentTimeNov 14th 2012

Okay, thanks for the information!

I have added to the entry a pointer to your preprint. I like the Flamingos! (No joke, I really do. :-)

• CommentRowNumber4.
• CommentAuthorAndrew Stacey
• CommentTimeNov 14th 2012

The beamer version has a tortoise as well!

• CommentRowNumber5.
• CommentAuthorDavidRoberts
• CommentTimeNov 14th 2012

And don’t forget the little ducks at the end of the proofs… :-)

• CommentRowNumber6.
• CommentAuthorAndrew Stacey
• CommentTimeNov 15th 2012

Yeah, maybe they were a little bit silly:

\usepackage{duck}
\def\qedsymbol{\raisebox{-2pt}{\resizebox{1em}{!}{\drawduck}}}


Keeps me … well, I was going to say “sane”.

• CommentRowNumber7.
• CommentAuthorDavidRoberts
• CommentTimeNov 15th 2012
• (edited Nov 15th 2012)

A bit like \makeToast, which I think is awesome btw. Now you just need to get to the level of making Talkie Toaster… :)

More on topic, I had some questions too, but I’ve forgotten what they were :-( When I remember I’ll ask here.

• CommentRowNumber8.
• CommentAuthorAndrew Stacey
• CommentTimeNov 15th 2012

Please do ask questions! I’m starting to feel that I’d like to get this one polished and arXived so I’d welcome feedback that would make it better. I’ll post the latest version on loopspace later today.

(And Talkie Toaster in TeX would be extremely scary. I don’t think the world is ready for it. Mind you, if he were programmed in TeX that might explain the obsessive behaviour. It’s not a large leap from being obsessive about line breaks to being obsessive about toast.)

• CommentRowNumber9.
• CommentAuthorjim_stasheff
• CommentTimeNov 15th 2012
Where can I see Talkie Toaster and where di it appear?
• CommentRowNumber10.
• CommentAuthorAndrew Stacey
• CommentTimeNov 15th 2012

Talkie Toaster is a character from the BBC Sci-Fi Comedy series Red Dwarf. He is noted for his obsessiveness towards toast (and anything that can be put in a toaster). His catchphrase is “I toast therefore I am” (or that might be the other way around). What David is referring to is a blog post I wrote recently about playing with a Raspberry Pi (a cheap computer). You can read about that at http://tex.blogoverflow.com/2012/10/i-tex-therefore-i-toast/. There is a slight connection between these two topics.

• CommentRowNumber11.
• CommentAuthorjim_stasheff
• CommentTimeNov 17th 2012
Although I didn't recognize Talkie Toaster, I had caught a few episodes of that zany BBC Sci-Fi Comedy series Red Dwarf and recognized one of the characters.
• CommentRowNumber12.
• CommentAuthorTim_Porter
• CommentTimeNov 18th 2012

If you type talkie toaster into Google you will see what us Brits are capable of! (no comment)

• CommentRowNumber13.
• CommentAuthorTodd_Trimble
• CommentTimeNov 18th 2012

Just watched a Youtube video of Red Dwarf. Sounds like Talkie Toaster and Kryten are American – do they reserve American accents for the stupid characters? :-)

• CommentRowNumber14.
• CommentAuthorTim_Porter
• CommentTimeNov 18th 2012
• (edited Nov 18th 2012)

Kryten is an English gentleman’s gentleman android. Remember Steven Hawkin”s artificial voice is sort of US. Please note that many of the characters who are the ‘badies’ in American films speak with an English accent. :-)

(PS. I should point out that not all gentleman’s gentlemen were gentlemen as well!)

• CommentRowNumber15.
• CommentAuthorAndrew Stacey
• CommentTimeNov 18th 2012

Todd, Not sure what you mean. Lister, Rimmer, and Holly (in both incarnations) have unmistakeably British accents. (Only slightly more seriously, part of the point is that none of them can truly be classed as “not stupid”.)

I think that Talkie Toaster is Canadian, and the actor who voices him played Kryten in the first episode where he appeared but when they decided to make Kryten a member of the crew then he wasn’t available so they got someone else, but kept the accent to keep the connection with his original appearance.

Anyway! This is all getting just a smidgeon off-topic …

• CommentRowNumber16.
• CommentAuthorDavidRoberts
• CommentTimeNov 18th 2012

Andrew, is the new version of the mapping spaces paper up? I can haz link? (trying to put the worms back in the can here :-) …)

• CommentRowNumber17.
• CommentAuthorTodd_Trimble
• CommentTimeNov 19th 2012

Andrew, I didn’t mean anything – I never even watched it before! (Lived for 2 years in Australia and six months in London, but managed never to have seen Red Dwarf!) I was just asking.

• CommentRowNumber18.
• CommentAuthorAndrew Stacey
• CommentTimeNov 22nd 2012
• CommentRowNumber19.
• CommentAuthorDavidRoberts
• CommentTimeNov 23rd 2012

Here are some questions:

1. What is the definition of standard tangent space of an ’ordinary manifold’? (Although the standard version I just downloaded to my computer seems different to the one I downloaded a couple of hours ago to my phone (might be a caching thing), but the iPad and laptop ones are the same, including the coloured background) Maybe an ordinary manifold is a finite dimensional manifold?

2. You don’t seem to use the concept of kinematic extension anywhere after the definition (ah, just before Prop 5.10, but not using the same term. In fact, where did you state that you would only be using smooth spaces which are kinematic extensions of manifolds?

3. If I have some finite number of tangent vectors such that they are pairwise summable, then can I talk about a bracketed sum? e.g. given $v_1,v_2,v_3$ elements of the tangent space at a point, with $v_i$ and $v_j$ summable for $i,j\in\{1,2,3\}$, can I define $(v_1+v_2) + v_3$ (and permutations thereof)? Then the span of the given finite set of tangent vectors forms a vector space. More generally, if this works, can I find infinite sets of tangent vectors which are pairwise summable? Clearly this last part is really only relevant for infinite dimensional manifolds.

4. In the sentence “If it exists, the Fermat map $F$ is unique for $s\neq t$..” you should probably say “Given two Fermat maps $F$ and $F'$ for $f$, they agree whenever $s\neq t$”, or possibly “If it exists, the Fermat map $F$ is uniquely determined away from the diagonal given by $s=t$”. I don’t think it makes sense to say a function is unique except for some subset of its domain (it may be clear what you mean, but it seems like an undergraduate style typing error)

5. Would you like to put in a bibliography at some point? I can guess that almost all references are to Comparative smootheology, or at least one of your papers, but it’s hard to know for sure…

6. (This is more a desirable, and a selfish one at that) Could either theorem 5.1 or corollary 5.2 be extended to Banach manifolds? I guess the first hurdle is knowing that the space of smooth sections of a smooth, locally trivial bundle of Banach spaces over a compact (and finite-dimensional) smooth manifold is Frechet (I’ve quickly checked the path space case and get everything but completeness of the resulting locally compact space, but this is only my weakness is analysis, rather than failure of this fact)

7. (another selfish desirable) Can we also recover 5.1 or 5.2 for Frechet spaces which are mapping spaces? Namely, given a smooth regular map between mapping spaces $f\colon M^K \to N^L$, not necessarily induced by maps between the finite dimensional manifolds $M$, $N$ or compact $K$, $L$, do we know that it is sent to a regular maps by mapping space functors $(-)^J$? What about the case $f\colon M^K \to M^L$ induced by some map $L\to K$?

8. Ditto to the previous two, but now looking at theorem 5.6

• CommentRowNumber20.
• CommentAuthorAndrew Stacey
• CommentTimeNov 23rd 2012

Thanks!

1. I guess I should figure out some good terminology for referring to “standard” stuff from differential topology. It’s a bit of a mouthful to be always saying “finite dimensional smooth manifold” (though maybe that’s not a bad idea - it is being read not spoken). As for a “standard” definition of the tangent space, take any you like! They’re all equivalent. I like equivalence classes of smooth curves but pointwise derivations would do equally well.

2. Just seemed an important concept so worth naming. I refer to it at the start of Section 5, but again I get the name wrong.

3. No. Take the union of the $x{-}y$, $y{-}z$, and $z{-}x$ planes in $\mathbb{R}^3$. The tangent vectors along the axes are pairwise summable but the triple does not have a sum.

4. Thanks.

5. Yes, bibliography is definitely needed! Lots to Comparative Smootheology but I should also reference where the local addition construction stems from (Kriegl and Michor, and further back to Eells and Elworthy - need to track down the history of it).

6. I’m not au fait with the inverse function theorem for Banach spaces. I mean, I know it exists but I don’t know enough about it to be able to say confidently that the result of 5.1 would extend. Certainly worth looking at. Would Hilbert be good enough, or would you want Banach? One useful fact for Hilbert is that the (square of the) norm is smooth so one can test smooth functions using it, whereas not all Banach spaces have smooth norms (or equivalent smooth functions).

7. Certainly if $f \colon M^K \to M^L$ is induced by a map $L \to K$ then we get regularity. This comes from the fact that the exponential map is functorial in the source space so once we’ve chosen a smoothly locally additive structure on $M$ then morphisms $L \to K$ induce morphisms of smoothly locally additive spaces $M^K \to M^L$. But for a general map, there you’d need to do some analysis. The whole point of this is that good local properties are not enough to assume good local properties on the mapping space. In finite dimensions then the inverse function theorem is your rescuer but in infinite dimensions things are going to get very tricky.

This approach isn’t a way to avoid hard analysis. It helps sometimes in that you can do the analysis on the smaller space (the target space).

Thanks for all of your comments! Keep them coming - I’d quite like to get this to the arXivable stage fairly soon.

• CommentRowNumber21.
• CommentAuthorDavidRoberts
• CommentTimeNov 24th 2012

Re 6. That depends on the answer to this MO question: http://mathoverflow.net/questions/114303/what-sort-of-manifold-is-puh. I’d like to apply the results of my paper which relies on the results in your paper to lifting bundle gerbes arising from principal PU(H) bundles.

• CommentRowNumber22.
• CommentAuthorAndrew Stacey
• CommentTimeNov 24th 2012

Depends a bit on your choice of topology, but in the norm topology it would be a Banach manifold. The only other sensible topology is the weak topology whence it would be modelled on an LCTVS with no particularly special properties (that I’m aware of).

I’m not sure that the Hilbert-Schmidt bit that you mention makes much sense. It acts on HS by conjugation, but whilst the action is faithful I don’t know that it produces an obvious embedding $P U(H) \to \mathcal{H}$ because the action isn’t free.

• CommentRowNumber23.
• CommentAuthorDavidRoberts
• CommentTimeNov 25th 2012

What I was thinking of when I mentioned Hilbert-Schmidt was the free action of $PU(H)$ on $U(HS)$, rather than the action on $HS$ itself. I was writing from memory, and should have checked…

OK, so the best I can do is Banach. Hmm, maybe that can wait for future work. Or perhaps we can mention it to Michor, and he might just have something up his sleeve. I mean the generalisation of 5.2 doesn’t require all the smocally machinery if we need to consider the mapping space involving a compact, fin. dim. manifold…

PS might $PU(H)$ be paracompact?

• CommentRowNumber24.
• CommentAuthorAndrew Stacey
• CommentTimeNov 26th 2012

Thought that might be what you meant with HS after reading the link to John’s stuff. No better, $P U(H) \to U(HS)$ is a norm-isometry.

Metrisable implies paracompact so you get that in both cases.

• CommentRowNumber25.
• CommentAuthorDavidRoberts
• CommentTimeNov 26th 2012

Well, paracompact Banach manifolds have to be better than arbitrary Banach manifolds.

• CommentRowNumber26.
• CommentAuthorAndrew Stacey
• CommentTimeJan 24th 2013

Finally got round to putting this on the arXiv, complete with flamingoes and ducks: http://arxiv.org/abs/1301.5493.

• CommentRowNumber27.
• CommentAuthorDavidRoberts
• CommentTimeJan 25th 2013

Hooray!