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• CommentRowNumber1.
• CommentAuthorAndrew Stacey
• CommentTimeOct 2nd 2012

I’ve been pondering the question “What is a tangent vector?” and for inspiration took a look at tangent bundle where the geometric definition started with:

One can also define vectors at $a$ to be curves $c$ such that $c(0) = a$, modulo the equivalence relation that $\dot{c}(0) = \dot{d}(0)$ if $c = d$ on some neighbourhood of $0$.

This seems to me to be a very weak relation. It means that the curve $t \mapsto (t,t^2)$ is not equivalent to the curve $t \mapsto (t,0)$ in $\mathbb{R}^2$.

I’m curious as to whether or not there is a way to say which curves should be equivalent to which without appealing to smooth functions on the space. That is to say, defining $\alpha \simeq \beta$ if $\alpha(0) = \beta(0)$ and for every smooth function $f$ then $(f \circ \alpha)'(0) = (f \circ \beta)'(0)$. (This is the kinematic tangent space). I don’t know of any way to define tangent vectors without appealing to functions (without working in synthetic differential geometry which side-steps the issue).

• CommentRowNumber2.
• CommentAuthorUrs
• CommentTimeOct 2nd 2012

This seems to me to be a very weak relation.

And its wrong (assuming we all mean to speak of the standard definition). I have fixed it.

• CommentRowNumber3.
• CommentAuthorAndrew Stacey
• CommentTimeOct 2nd 2012

Except that now it is circular! What does it mean to say that the first derivatives of two curves agree? In which space will you compare them? The tangent bundle is the space in which one can compare first derivatives of curves, so you can’t start comparing these derivatives until you have constructed the tangent bundle.

• CommentRowNumber4.
• CommentAuthorTodd_Trimble
• CommentTimeOct 2nd 2012
• (edited Oct 2nd 2012)

Of course, one definition of tangent vector at a point $p$ (in the $C^\infty$ setting) is: a derivation on the local ring $\mathcal{O}_p$. (Edit: come to think of it, this works in the $C^n$ setting too.)

In the $C^n$ setting, a standard device is to reduce the problem to ordinary calculus via a coordinate chart. That is, we can easily define what it means to say that two curves $\alpha, \beta: I \to \mathbb{R}^n$ such that $\alpha(0) = \beta(0)$ have the same derivative $\alpha'(0) = \beta'(0)$. Then, say that two curves $\alpha, \beta: I \to M$ are equivalent if…

Is there something about either approach that you don’t like or want?

• CommentRowNumber5.
• CommentAuthorAndrew Stacey
• CommentTimeOct 2nd 2012

Lots! (And this is why I put this in “General discussions” rather than “latest changes”).

Derivations are the Wrong Definition: this gives the bidual for some locally convex topological vector spaces. Essentially, when you define tangent vectors as derivations you are really saying that they are functionals on cotangent vectors (except that you haven’t defined cotangent vectors yet so you get round that by using functions).

The coordinate chart reduction only works if you have charts, and the context that I’m working in is generalised smooth spaces where you don’t have charts. You can reduce it further because when you compose with a chart then you can test the derivative with projections to see if it is zero (or equal to some other curve’s derivative). Following this through means that you can do it with smooth functions on your space instead of charts and this works for generalised smooth spaces. This gives the kinematic tangent space that I referred to above.

What I’m after is some way of defining what a tangent space should look like. Namely, if I wrote down some definition and said “This is the tangent space”, what would you expect it to be like? Here’s my list so far:

1. Ingredients: we have a cartesian closed category $\mathcal{S}$ of generalised smooth spaces into which the category of manifolds embeds fully and faithfully.

2. A tangent structure will be a pair $\mathfrak{T}, \pi$ where $\mathfrak{T} \colon \mathcal{S} \to \mathcal{S}$ is a functor and $\pi \colon \mathfrak{T} \to \mathfrak{Id}$ is a natural transformation from the tangent functor to the identity functor. These satisfy certain conditions.

3. It should extend the “standard” tangent space functor and anchor natural transformation on ordinary smooth manifolds.

4. It should be a strong (enriched) functor in that $\mathcal{S}(X,Y) \to \mathcal{S}(\mathfrak{T}(X),\mathfrak{T}(Y))$ is smooth.

The existence of a zero section follows from these.

But it feels like this is still too loose. I can’t rule out “phantom” tangent vectors, by which I mean ones that exist but vanish whenever you map to a smooth manifold ($\mathbb{R}$ is enough). If you disallow those, then I think that the kinematic tangent space construction is the initial such functor, and the operational tangent space (aka derivations) is the final one.

• CommentRowNumber6.
• CommentAuthorTodd_Trimble
• CommentTimeOct 2nd 2012

So you’re really after a tangent bundle functor on generalized smooth spaces? Is there a preferred notion or axiomatics for generalized smooth spaces that you are working with?

Nitpicky question: what’s up with all the Gothic script? I find it really hard to read. (I confess that I didn’t at first recognize the Gothic script for the identity functor.)

• CommentRowNumber7.
• CommentAuthorAndrew Stacey
• CommentTimeOct 2nd 2012

Well, I did say “the identity functor” so I thought I might get away with that one! Sorry, I’ll tone down the fonts in future. (I now have special “on screen” versions of my latest preprint with different fonts to make it easier to read on screen/iPad. But I do confess a fondness for fancy fonts that goes beyond mere legibility.)

Yes, I’m after a “tangent bundle functor” (though I think that the anchor nat trans should also be part of the data). Or rather, I can construct at least two such so I’m not after a functor as such, more a way of characterising them. To that end, I’m wondering what answers people would give to the question:

Let $X$ and $Y$ be smooth spaces. Suppose I told you that $Y$ was the (total) tangent space of $X$. What would you expect $Y$ to look like/behave like?

(maybe I should make this into a blog post)

I’m trying to be general with my category of smooth spaces. I think it reasonable to assume that it be a $c^5$: complete, cocomplete, cartesian closed category. The category of finite dimensional smooth manifolds should embed fully and faithfully within it. To make my life easier (but for no other reason), I’m assuming that it is concrete and that the concretisation functor has left and right adjoints. I figure if I can get it right for that, then I can drop the concrete stuff later.

• CommentRowNumber8.
• CommentAuthorMike Shulman
• CommentTimeOct 2nd 2012
• (edited Oct 2nd 2012)

Is the notion of ’cartesian differential category’ perhaps relevant, if only for describing what one might expect the end result to look like?

edit: I can’t remember whether that’s the phrase I wanted to say. One of these notions only applies to categories of cartesian spaces — it assumes every object is its own tangent space — but I think there’s some abstract notion of ’category with tangent bundles’ that should apply.

• CommentRowNumber9.
• CommentAuthorUrs
• CommentTimeOct 2nd 2012
• (edited Oct 2nd 2012)

Is the notion of ’cartesian differential category’ perhaps relevant

From a brief look at

• Blute, Cockett, Seely, Cartesian differential categories (pdf)

I am getting away with the impression that the notion of cartesian differential category axiomatizes categories that are like CartSp, but not categories that are like SmthMfd.

I may well be wrong, I haven’t really studied this article. But I am looking at p. 15 and following:

at least there the category CartSp is taken as the motivating example (and it is the only example that I have found so far in the article, somebody please give me a page number to further examples) But morever, if I am reading def. 2.1.1 and the paragraph leading up to it correctly, then the objects of a cartesian differential category are in fact regarded as carrying linear structure.

• CommentRowNumber10.
• CommentAuthorMike Shulman
• CommentTimeOct 3rd 2012

Okay, then, that’s the wrong one. But I thought there was also a right one. I wonder what it’s called, or if my memory is wrong.

• CommentRowNumber11.
• CommentAuthorTodd_Trimble
• CommentTimeOct 3rd 2012

Let X and Y be smooth spaces. Suppose I told you that $Y$ was the (total) tangent space of $X$. What would you expect $Y$ to look like/behave like?

I’m not sure, but here are a few thoughts for what they’re worth:

First I would expect the tangent bundle functor $T$ to be, at the very least, product-preserving. Based on how things work in synthetic differential geometry (SDG), I might expect a whole lot more, such as $T$’s preserving all limits. And maybe all colimits too! For example, $T$ definitely ought to preserve coproducts = disjoint sums.

(In SDG, Lawvere often speaks in slightly awed tones about “the amazing right adjoint” of the tangent bundle functor – in such models where this exists, $T$ would certainly preserve colimits.)

A question that I’d be inclined to ponder in this regard is: to what extent can we probe (or co-probe) general smooth spaces by smaller, say finite-dimensional smooth spaces. For example, we might have a universe where every smooth space is the colimit of the diagram of mappings from small spaces – we’d say the small spaces are dense in the category of all smooth spaces. Then knowledge of how $T$ acts on the small spaces, coupled with $T$’s preserving colimits, would determine how $T$ acts on the whole category.

Density would have to do with probings, but we can also consider co-probings by small spaces, and consider the dual problem of codensity. As you may be aware, there has been a lot of recent discussion at the n-Category Café about codensity and codensity monads and so on, induced from the inclusion of a category of small objects into a category of all objects. I don’t have much feeling for this with regard to smooth spaces, but it might be worth a thought.

A passing thought is that we should have not only a projection $T \to id$, but a zero section functor $id \to T$. With suitable assumptions, this together with $T$’s preserving products would make $T$ a strong functor in the sense we were discussing before.

• CommentRowNumber12.
• CommentAuthorUrs
• CommentTimeOct 3rd 2012

But I thought there was also a right one. I wonder what it’s called

Of course one that comes to mind is: smooth topos.

• CommentRowNumber13.
• CommentAuthorAndrew Stacey
• CommentTimeOct 3rd 2012

Except that I wasn’t asking a categorical question. My original question was very much at the space level. Let me make it even more concrete:

What is $T_0 [0,1]$?

I’ll even make it multiple choice:

1. $0$

2. $[0,\infty)$

3. $\mathbb{R}$

4. Something else

I can give you reasons why each of the first three is “correct”.

You see, the problem I have with “smootheology” is that it takes the “nice category” idea to extremes and I feel as though the smoothness gets lost in the wayside. For example, here’s a perfectly good diffeological space: take $\mathbb{R}$ and declare all continuous maps $U \to \mathbb{R}$ to be smooth. That’s a diffeology on $\mathbb{R}$. But it’s a heck of a weird one! I feel that that space “ought not to be allowed”, to coin a phrase. What would its tangent space be?

So back to tangent spaces. I don’t want to declare that something is the tangent space of $X$, I want to say “to be a tangent space of $X$, here are the properties it must satisfy.”. Then I want to construct a functor so that $T X$ is always a tangent space of $X$ and then examine its properties. Hopefully it will have nice categorical properties, but if not then not. And if not, then interestingly not. Again, back to smootheology: the category of diffeological spaces is cartesian closed for essentially boring reasons. The category of Frolicher spaces is cartesian closed for very interesting ones (to do with the properties of the monoid $C^\infty(\mathbb{R},\mathbb{R})$).

• CommentRowNumber14.
• CommentAuthorUrs
• CommentTimeOct 3rd 2012
• (edited Oct 3rd 2012)

it takes the “nice category” idea to extremes and I feel as though the smoothness gets lost in the wayside

That is the “nice category” idea: allow bad spaces to get a nice category.

The idea is that you are not forced to consider these bad spaces as good spaces. You may ignore them. But the point is that due to their presence in the background, the category is nice and hence all kinds of questions about the good spaces now have good answers.

• CommentRowNumber15.
• CommentAuthorAndrew Stacey
• CommentTimeOct 3rd 2012

I’m happy ignoring them, sure enough. But the problem with that slightly blasé attitude is that there are several of these “nice categories” and the choice sometimes matters. After all, Todd has just told me that it would be nice for the tangent functor to preserve limits and colimits. But these depend on the choice of category. So without a choice of category, I’m unable to determine whether a particular tangent functor satisfies his criteria.

Here’s an example of the sort of thing I’m talking about. In reading around about this, I was reminded of Hadamard/Fermat’s result recalled at Fermat theory. At Fermat theory, this gets turned into an “operation on operations” (is there a name for that, by the way?) which means that it is basically only defined for $C^\infty(\mathbb{R}^n,\mathbb{R})$, or rather for the smooth spaces $\mathbb{R}^n$. Looking at the construction, we can rearrange it to the result that:

$f(y) - f(x) = (y - x)\int_0^1 f'(x + t(y - x)) d t$

For simplicity, let’s make our background space a vector space (so a module over $\mathbb{R}$ in our category of smooth spaces), say $E$. Let’s also apply Hadamard’s result to the case of a curve $\alpha \colon \mathbb{R} \to E$. Then we want to be able to say that:

$\alpha(y) - \alpha(x) = (y - x) \int_0^1 \alpha'(x + t(y - x)) d t$

Now the presence of $\alpha'$ says that the right-hand side ought to be interpreted in a tangent space. Obviously, the tangent space of $E$ at $\alpha(x + t(y - x))$. But as we’re in a vector space, we can translate tangent spaces around. Since we’re going to integrate (and we can think of this as Riemann sums if we want), we’d better put them all in the same tangent space. Setting $t = 0$, we get the tangent space at $\alpha(x)$ which is as good as any. So the RHS we interpret as being in $T_{\alpha(x)} E$.

Now the LHS can be interpreted as a vector in $\alpha(x) + E$ and there is a natural map $\alpha(x) + E \to T_{\alpha(x)} E$. So we can interpret the LHS as being in $T_{\alpha(x)} E$ as well.

Then what does this say? The LHS is in the image of the map $\alpha(x) + E \to T_{\alpha(x)} E$ and if $y \ne x$ then we can recover the integral from the LHS. So the statement says that for $y \ne x$ then the RHS is also in this image. So what it says is that $\int_0^1 \alpha'(x) d t$ is the limit of stuff in the image of $\alpha(x) + E \to T_{\alpha(x)} E$. Thus $\alpha(x) + E$ is dense in $T_{\alpha(x)} E$.

Now this is the sort of thing that I’m trying to get at. Not just that derivatives exist, but also that they have various properties that make them act like we expect derivatives to act. And this will - in all probability - depend hugely on the choice of category and choice of tangent functor.

• CommentRowNumber16.
• CommentAuthorTodd_Trimble
• CommentTimeOct 3rd 2012

Well, I have a naive question at this point. Is there a context you can think of where the tangent bundle of a vector space $E$ is something other than $E \times E$? Or in other words, that the tangent space of an affine space $E$ at a point $p$ is something other than $E$ seen as a vector space with chosen origin $p$?

• CommentRowNumber17.
• CommentAuthorAndrew Stacey
• CommentTimeOct 3rd 2012

Yes! If $E$ is not complete. Take $\ell^0$ (oops: $l^0$ for fancy font phobics) with the $l^2$-topology. Then declare a curve in $l^0$ to be smooth if it is smooth into $l^2$. Then I can find a curve in $l^0$ whose derivative at, say, $0$ is any arbitrary vector in $l^2$. Whence $T l^0 = l^0 \times l^2$.

• CommentRowNumber18.
• CommentAuthorTodd_Trimble
• CommentTimeOct 3rd 2012

@Andrew #17: This looks interesting, but to understand this better I need more information. What’s the category here? I’m guessing it’s defined operationally in terms of certain co-probings.

• CommentRowNumber19.
• CommentAuthorAndrew Stacey
• CommentTimeOct 3rd 2012

What category would you like? I can define $l^0$ as the subobject of $l^2$ in any category. But to make sense of the rest of what I say then I need some construction of smooth curves and their derivatives - which gets us back where we began! Nonetheless, let’s say that $T l^2 = l^2 \times l^2$. Using the denseness of $l^0$ in $l^2$ then I can approximate curves in $l^2$ by curves in $l^0$. In fact, I claim that I can do this in the $C^1$ sense, whereupon $T l^0 = l^0 \times l^2$.

But secretly, the category I’m working in here is that of Kriegl and Michor (whence Frölicher, or at least the linear part thereof). Their starting point is that for a TVS $E$ then the statement $c \colon \mathbb{R} \to E$ is smooth” is unambiguous because you just iterate Leibniz quotients.

There are some spaces where curves fail to be smooth for silly reasons: the limit doesn’t exist but it ought to. This leads to a notion of completeness, which is equivalent to local completeness. For normed vector spaces it is equivalent to ordinary completeness, whence my example of $l^0$.

So for $l^0$ with the topology from $l^2$ we have curves that would be smooth in $l^2$ but aren’t in $l^0$ simply because their derivatives are missing. But in a category of smooth spaces, this doesn’t matter! We can still consider the smooth space $l^0 \subset l^2$ - it just won’t be one of the ones constructible by Kriegl and Michor’s work.

Back to Kriegl and Michor’s convenient vector spaces. In the end, these are defined via co-probings because for those spaces a curve $c \colon \mathbb{R} \to E$ is smooth if and only if $f \circ c$ is smooth into $\mathbb{R}$ for all continuous linear functionals on $E$. But for them, this characterises the spaces they want to study, it isn’t how to take an arbitrary space and put a smooth structure on it: it could equally be defined by taking the smooth curves in the traditional sense.

Incidentally, this is the sort of statement I was looking for: that you expected to have $T E \cong E \times E$ for any (smooth) vector space $E$. Hopefully now you see that that isn’t reasonable. But what I was saying about about Hadamard’s lemma is that I would expect $E$ to be dense in $T_0 E$, in the appropriate sense.

• CommentRowNumber20.
• CommentAuthorTodd_Trimble
• CommentTimeOct 3rd 2012

Andrew, I’m sorry, but I’m still struggling to understand the ground rules you want to play by. My question “which category?” was partly informed by your own desiderata, that you want to be working in a $c^5$ (complete cocomplete cartesian closed category). It’s really not a question of “which category would I like?”, because I want to know what sort of constraints you are interested in. I have some sense of that.

I expected to have $T E \cong E \times E$ because I think that is reasonable at least for certain models of SDG. At the same time, I was prepared to accept that it wouldn’t hold for some context you had in mind, which is why I asked. Pointing to Kriegl and Michor’s convenient vector spaces is a good starting point for me – thanks.

• CommentRowNumber21.
• CommentAuthorUrs
• CommentTimeOct 3rd 2012

Just a side remark, promted by the keywords that appear more by the actual point that Andrew is after (sorry, am absorbed otherwise for the moment):

convenient vector spaces embed faithfully into the smooth topos=model for SDG called the Cahiers topos (discussed there).

• CommentRowNumber22.
• CommentAuthorAndrew Stacey
• CommentTimeOct 3rd 2012

That’s okay, Todd - I’m not explaining myself very well. I’m really attacking this problem from both ends. I want to be able to define what a “tangent functor on a category of generalised smooth spaces” should mean. Namely, if I give an explicit construction of a functor and claim “This is the tangent functor” then I want to know what I need to show to justify that claim. As there is no definition (that I’m aware of) of exactly what a tangent space should be, I’m trying to figure it out. So on the one end, I’m working in a very imprecise environment as I want the eventual conditions to be valid in any (reasonable, let’s start with a $c^5$) category of generalised smooth spaces.

But to figure out what those conditions should be, I want to look at specific examples and say (or ask, rather) “What should the tangent space be here?” Hence my question about $[0,1]$.

So for the very specific case of $l^0$, let me take Frölicher spaces and define $l^0$ to be the sub-smooth space of $l^2$. Then a curve in $l^0$ is smooth if and only if it is smooth after inclusion into $l^2$. Then I need to say what the smooth structure of $l^2$ is, and for that I say that a curve $c \colon \mathbb{R} \to l^2$ is smooth if and only if $\langle v, c \rangle$ is a smooth curve in $\mathbb{R}$ for any $v \in l^2$.

In that setting, then $T l^2 \cong l^2 \times l^2$ (this follows from completeness, you can see the idea from the fact that as $\langle v, - \rangle$ is linear then $\langle v, c' \rangle = \langle v, c\rangle '$). Then from the density of $l^0$ in $l^2$ I can (I claim) get any tangent vector in $l^2$ from a smooth curve in $l^0$, whence $T l^0 \cong l^0 \times l^2$. Here, I guess I’m using the idea of tangent vectors as being derivatives of smooth curves - or at least, that what ever tangent vectors are then derivatives of smooth curves should give me examples of them.

• CommentRowNumber23.
• CommentAuthorEric
• CommentTimeOct 3rd 2012

Hi,

As I read this (with interest!), I can’t help making a comment that is likely irrelevant.

Andrew, I wonder if it may be more natural to look for some kind of cotangent functor?

What is the opposite of SmoothSpace? Could the duality between spaces and algebras be useful here?

• CommentRowNumber24.
• CommentAuthorAndrew Stacey
• CommentTimeOct 3rd 2012

Hi Eric. It certainly would be natural, but I draw the line at more natural.

Certainly, in diffeological spaces there is a natural definition of a cotangent space as “that which pulls back to cotangent spaces” on plots. But then that needs a sure knowledge of what the cotangent space should be on the plot spaces, and thus of tangent vectors. If you look higher up, I claimed that this isn’t obvious for the simple case of $[0,1]$.

• CommentRowNumber25.
• CommentAuthorUrs
• CommentTimeOct 3rd 2012
• (edited Oct 3rd 2012)

But then that needs a sure knowledge of what the cotangent space should be on the plot spaces, and thus of tangent vectors.

I object here. Despite their historical name, you don’t need to define covectors as duals to vectors.

Covectors live happily in great generality on typical model spaces as fundamental entities in themselves, notably by the construction of Kähler differential forms.

And there are arguments that this is more natural. Over at geometry of physics the third section, currently developing, is on differential forms. The tangent bundle, instead, will appear only much later — or rather does not really apear at all yet as far as the section outline goes there (all that really appears is the $GL(n)$-principal bundle of a manifold).

Being, as it is, rooted in a context of examples, this also shows that one does not actually need the tangent bundle as much as it might seem.

(But later developments of the entry will probably see synthetic differential smooth spaces and their functorial tangent bundles, too.)

• CommentRowNumber26.
• CommentAuthorAndrew Stacey
• CommentTimeOct 3rd 2012

I object here. Despite their historical name, you don’t need to define covectors as duals to vectors.

Feel free to object!

However, I return to my motivation which is to understand what pictures enter your head when you hear the words “tangent space”. I could, it is true, replace that by “cotangent space”. And it may well be that the pictures do not involve a duality (though if they do not involve a pairing then I’d be worried). So whether or not you prefer to think that the chicken came before the egg, still I want to know how to tell a chicken from a duck.

• CommentRowNumber27.
• CommentAuthorMike Shulman
• CommentTimeOct 3rd 2012

Ah, what I was thinking of is tangent structure.

• CommentRowNumber28.
• CommentAuthorAndrew Stacey
• CommentTimeOct 3rd 2012

Thanks, Mike. That’s very interesting. I need to follow back to Rosický’s paper, obviously. My initial reaction is that there is some stuff there that I can use (the flip on the double tangent space, for example) but I have to reject some other. I see no reason why a tangent space should be additive.

Here’s an example: take two copies of $\mathbb{R}$ and glue them together at $0$. Essentially, this is the union of the $x$ and $y$ axes. What’s the tangent space at the gluing point? I have tangential directions in each of the two $\mathbb{R}$-directions, but I cannot add them together to get a new direction.

Of course, I could always throw in the sums to force a vector space, but that seems to be taking me one step away from tangents.

• CommentRowNumber29.
• CommentAuthorTodd_Trimble
• CommentTimeOct 3rd 2012

@Andrew: from the looks of things, it appears you favor a notion of tangent space based on equivalence classes of curves, or that you’ve been tending in that direction.

For what it’s worth, my instinctual answer to the $[0, 1]$ question is $\mathbb{R}$. And my instinctual answer to the union of the $x$ and $y$ axes is: $\mathbb{R}^2$. This is very much based on algebraic definitions which focus on things like derivations (or $Ext_{\mathcal{O}_p}^{1}(k, k)$, or maps $\mathcal{m}_p/\mathcal{m}_{p}^{2} \to k$), which you Declared earlier was The Wrong Definition. Hmm…

• CommentRowNumber30.
• CommentAuthorUrs
• CommentTimeOct 3rd 2012
• (edited Oct 3rd 2012)

@Mike:

thanks, I have recorded that reference at tangent bundle. In the intro of that paper they also nicely clarify that “cartesian differential categories” sit inside “tangent structure” (only) as the linear case.

@Todd:

but that’s not the answer that the standard models of SDG give for $T_0([0,1])$. Or am I mixed up?

• CommentRowNumber31.
• CommentAuthorAndrew Stacey
• CommentTimeOct 3rd 2012

Todd, yes I do favour a notion based on equivalence classes of curves (the so-called kinematic tangent space). However, I’m trying to put personal preferences aside and see what a general definition would be. If you would like a target to aim at:

Conjecture The functor assigning the kinematic tangent space to a smooth space is the initial tangent functor on a category of generalised smooth spaces.

So whilst I might prefer it, I need to understand what a “tangent functor” is in order to prove that this one is the initial one.

My reason for preferring it is that if a space is “modelled on its tangent spaces” then (I claim, slightly conjecturally) the only tangent space for which that will work is the kinematic one. Since manifolds ought to be modelled on their tangent spaces, this is the one that makes most sense in that setting.

(If you are really bored, you could read http://loopspace.mathforge.org/discussion/12/yet-more-smooth-mapping-spaces-and-their-smoothly-local-properties to see where all this is coming from.)

Thanks for answering about $[0,1]$. Your reasonings are incompatible with your earlier preference for $T E \cong E \times E$ for vector spaces since for derivations you get $T l^1 = l^1 \times (l^\infty)^*$.

• CommentRowNumber32.
• CommentAuthorAndrew Stacey
• CommentTimeOct 3rd 2012

(Although if you throw out the axiom of choice then that could be $l^1 \times l^1$ I suppose.)

• CommentRowNumber33.
• CommentAuthorUrs
• CommentTimeOct 3rd 2012
• (edited Oct 3rd 2012)

my motivation which is to understand what pictures enter your head when you hear the words “tangent space”. I could, it is true, replace that by “cotangent space”.

I’ll tell you my picture for cotangent space:

If I am supplied with the structure of a “geometry”, then the cotangents to any object are the (homotopy) Yoneda extension of the Kähler differentials to that object: the cotangent complex.

If I am supplied with the structure of cohesion then I have a more fundamental picture:

for any abelian geometric group $\mathbb{A}$ (think: $\mathbb{R}$) and every $n \in \mathbb{N}$ there is then canonically an object $\flat_{dR}\mathbf{B}^n \mathbb{A}$, the de Rham coefficient object for $\mathbb{A}$ in degree $n$. Then cotangents are the things modulated by a 0-truncated object $\Omega^n(-,\mathbb{A})$ such that $\Omega^n(-,\mathbb{A}) \to \flat_{dR} \mathbf{B}^n \mathbb{A}$ is an atlas relative to the given notion of manifold.

With that notion of cotangents one can naturally axiomatize a fragment of differential geometry and physics which is at least “large”.

• CommentRowNumber34.
• CommentAuthorTodd_Trimble
• CommentTimeOct 3rd 2012

Or am I mixed up?

I don’t know; maybe I’m mixed up. Andrew is trying to tell me I’m mixed up about something too.

When I have a good opportunity, I’ll try to work through this more carefully. If you have an argument for what you think the answer is, I’d like to hear it.

• CommentRowNumber35.
• CommentAuthorUrs
• CommentTimeOct 3rd 2012
• (edited Oct 3rd 2012)

If you have an argument for what you think the answer is, I’d like to hear it.

So I am thinking of a smooth topos such as the Cahiers topos or any of the other well-adapted Models for Smooth Infinitesimal Analysis. But I guess all I am using is the validity of the Kock-Lawvere axiom for duals of Weil algebras.

Then $D = Spec(\mathbb{R} \oplus \epsilon \mathbb{R})$, is the space dual to the ring of dual numbers. A morphism $D \to X$ into a manifold is then exactly the standard version of a tangent vector: an equivalence class of smooth curves in $X$ that have the same first derivative at the origin of $D \hookrightarrow \mathbb{R}$.

If $X$ is a manifold with boundary, and the origin of $D$ lands on that boundary, then this cannot have a component normal to the boundary.

It seems to me.

• CommentRowNumber36.
• CommentAuthorMike Shulman
• CommentTimeOct 3rd 2012

I have always thought of a tangent space as a “linear approximation” to a smooth space, which in particular entails additivity. It sounds like maybe you (Andrew) are thinking of something that I might be more inclined to call the “germ” of a space at a point.

• CommentRowNumber37.
• CommentAuthorAndrew Stacey
• CommentTimeOct 3rd 2012

Okay then, so what is your (Mike) answer to the specific examples: $T_0 [0,1]$ and $T_0 \mathbb{R} \vee \mathbb{R}$? Do you “complete” to the nearest vector space?

• CommentRowNumber38.
• CommentAuthorMike Shulman
• CommentTimeOct 4th 2012

I guess. A priori, it seems like the notion of linear approximation is only suited to spaces that are locally linear, but I guess you could consider the ’nearest vector space’ to be the ’best’ linear approximation, even if it’s not as good an approximation as in the locally linear case.

• CommentRowNumber39.
• CommentAuthorAndrew Stacey
• CommentTimeOct 4th 2012

Here’s a possibly interesting result for you, then, Mike. With lots of provisos and “under certain conditions”, suppose I take “tangent space” to not necessarily mean additive meaning that the sum of two tangent vectors need not exist. But I retain the $\mathbb{R}$-action (this is easier to justify than addition). Suppose that the tangent space is an approximation to the space, then it is linear.

So there are conditions under which one can show that the tangent space is additive without assuming that it is so a priori, and one of those is that the space is locally modelled on its tangent spaces.

• CommentRowNumber40.
• CommentAuthorTodd_Trimble
• CommentTimeOct 5th 2012
• (edited Oct 6th 2012)

@Urs #35: I hope you have a moment to go over this with me. If I’m reading you correctly, you think the tangent space at $0$ of $[0, 1]$ should be $0$-dimensional, and you give a reason based on SDG. I’m not sure I follow.

Let $D = Spec(\mathbb{R}[x]/(x^2))$. Smooth maps $D \to [0, 1]$ correspond to $C^\infty$-algebra maps $C^\infty[0, 1] \to \mathbb{R}[x]/(x^2)$. Here we follow standard practice, defining $C^\infty[0, 1] \coloneqq C^\infty(\mathbb{R})/\mathcal{m}_{[0, 1]}$ where $\mathcal{m}_{[0, 1]}$ is the $C^\infty$-ideal of smooth functions $\mathbb{R} \to \mathbb{R}$ which vanish identically on $[0, 1]$. (Moerdijk-Reyes, page 20. We can consider $[0, 1]$ as a closed subspace in other ambient spaces $\mathbb{R}^n$ as well; we get the same algebra regardless.)

That definition is crucial. So for the tangent space at $0$, we need to consider $C^\infty$-algebra maps $\xi$ which fit into a diagram in $C^\infty$-$Alg$:

$\array{ & & \mathbb{R}[x]/(x^2) \\ & ^\mathllap{\xi} \nearrow & \downarrow ^\mathrlap{\pi} \\ C^\infty(\mathbb{R}) & \underset{ev_0}{\to} & \mathbb{R} }$

and such that $\xi$ vanishes on $\mathcal{m}_{[0, 1]}$. At this point it should be clear that we get the same tangent space at $0 \in [0, 1]$ as for $0 \in \mathbb{R}$ (any such $\xi$ will automatically vanish on functions $f$ which are identically zero on $[0, 1]$).

If you see anything wrong with this reasoning, please let me know.

• CommentRowNumber41.
• CommentAuthorUrs
• CommentTimeOct 5th 2012
• (edited Oct 5th 2012)

Thanks, Todd. Yes, I agree with this. Good that you say this.

I was thinking of a different smooth structure on $[0,1]$! Maybe I have been working with the Cahiers topos too much. There $C^\infty(\mathbb{R})/\mathcal{m}_{[0,1]}$ is not (dually) a representable.

So I was thinking of $[0,1]$ as $[0,1] \in Sh(CartSp)$ first by $[0,1] = C^\infty(-,[0,1])$ and then sending that further to the smooth topos along the Yoneda extension of the faithful embedding of smooth manifolds into the smooth topos.

Do you see what I mean?

• CommentRowNumber42.
• CommentAuthorTodd_Trimble
• CommentTimeOct 5th 2012

Re #41, I sort of see what you mean, but let me think on it some more.

• CommentRowNumber43.
• CommentAuthorMirco Richter
• CommentTimeOct 5th 2012
• (edited Oct 5th 2012)

In the category $\mathbf{M}$ of finite dim. smooth manifolds, a global (and I guess standard) definition of the tangent functor $T$ is given by $Hom_{Alg}(C^\infty(.),\mathbb{D})$ i.e the tangent bundle is the set of (commutatice, associative unital) algebra morphisms from the algebra of smooth functions on a manifold into the algebra $\mathbb{D}$ of dual numbers.

Now as far as I understand, any ’smooth context’ should define something like $\mathbb{D}$ or the infinitesimaly thickened point. So maybe it would be reasonable to think about, whether this definition can be generalized to any smooth context?

Or is it already known that it can’t?

• CommentRowNumber44.
• CommentAuthorTodd_Trimble
• CommentTimeOct 5th 2012

Mirco, if by smooth context $C$ you are including things like diffeological spaces, one difficulty is that in those sorts of models, morphisms are distinguished by what they do on points, i.e., the functor $\Gamma = \hom(1, -): C \to Set$ is faithful. This means that for the 1-pointed $D = Spec(\mathbb{R} \oplus \mathbb{R}\epsilon)$, two maps $f, g: D \to M$ which agree on that point would have to be equal, which of course we don’t want! Thus, concrete categories tend not to be too well-adapted for this sort of thing.

• CommentRowNumber45.
• CommentAuthorTodd_Trimble
• CommentTimeOct 5th 2012

So yeah, Urs, I think I see what you mean. Let $Th$ be the Lawvere theory for $C^\infty$-algebras; put $I = [0, 1]$. We have a functor $C^\infty(-, I): Th^{op} \to Set$, and the thing you are considering is the value of $C^\infty(-, I)$ under the left Kan extension $Set^{Th^{op}} \to Cah$ of the canonical functor $Th \to Cah$ sending each arity $n$ to the space $\mathbb{R}^n$ in $Cah$. So you are considering a coend in $Cah$:

$\int^{n \in Th} C^\infty(\mathbb{R}^n, I) \cdot \mathbb{R}^n.$

The basic issue you were bringing up back in #35 can be illustrated by saying that if you have $f \in C^\infty(\mathbb{R}, I)$ and $f(x) = 0$, then in fact $f'(x) = 0$. That fact effectively trivializes the tangent space. Right?

However, for the other toposes you mentioned in #35, I think the situation is how I described it in #40. For example, $I$ viewed as the locus of a germ-determined $C^\infty$-algebra $C^\infty(I)$ (it is already point-determined) belongs to this category $\mathbb{G}$. As a site used to construct $\mathcal{G}$, this $\mathbb{G}$ is subcanonical, and so we calculate the tangent space of $I$ in $\mathcal{G}$ just as we would in the site $\mathbb{G}$, which brings us back to the calculation of #40. Similarly for the well-adapted model $\mathcal{F}$. Or so it seems to me.

• CommentRowNumber46.
• CommentAuthorUrs
• CommentTimeOct 5th 2012
• (edited Oct 5th 2012)

Todd, yes! Good that we agree.

However, for the other toposes you mentioned in #35, I think the situation is how I described it in #40.

Well, you are certainly right about what you say there, but the fact remains that there is a choice involved in what we mean when we say “$[0,1]$”. Because, also for the other toposes, that other meaning of $[0,1]$ exists. The difference is only that for the Cahiers topos only that meaning exists.

One way to think of the situation is as a nice way to answer Andrew’s general point here. Andrew points out that for manifolds with boundary there are different sensible notions of tangent space on the boundary. But from the point of view of synthetic differential geometry the perspective changes: there is one single notion of tangent bundle for synthetic differential spaces, but there are different sensible ways to equip manifolds with boundary with synthetic differential structure. And they differ in what the single canonical tangent bundle construction does to them.

• CommentRowNumber47.
• CommentAuthorTodd_Trimble
• CommentTimeOct 5th 2012

Thanks! This is clarifying (although I still want to ponder on it more).

• CommentRowNumber48.
• CommentAuthorMike Shulman
• CommentTimeOct 5th 2012

Andrew #39, that is a nice result which clarifies the importance of local linearity, but I’m not sure it argues for either choice of meaning for the phrase “tangent space”.

• CommentRowNumber49.
• CommentAuthorzskoda
• CommentTimeNov 12th 2012

I made just a small stub tangent cone with basic references (the best: Shafarevich) to add this discussion. It is now linked at tangent bundle.

• CommentRowNumber50.
• CommentAuthorTodd_Trimble
• CommentTimeNov 13th 2012

Here is another sort of interesting way of considering what the tangent space of $[0, 1]$ might be. Any closed subset of $\mathbb{R}$ is the zero set of some smooth function $f$. For example, we may take $f$ to be identically zero on $[0, 1]$ and $f(x) = \exp(\frac1{x - x^2})$ for $x \notin [0, 1]$, where the zero set of $f$ is $[0, 1]$.

Thus, it is not unreasonable to identify $[0, 1]$ (in the category of smooth spaces one is interested in, for example a model of SDG) with the equalizer of the two maps $f, 0: \mathbb{R} \stackrel{\to}{\to} \mathbb{R}$; it is certainly a correct identification at the underlying set or topological space level.

Now, if the tangent bundle $T$ preserves finite limits (as it indeed will in SDG, since $T$ is a right adjoint $(-)^D$), it will in particular preserve this equalizer, and we may calculate it easily enough as the equalizer of the pair of maps $\mathbb{R} \times \mathbb{R} \stackrel{\to}{\to} \mathbb{R} \times \mathbb{R}$, the first being the constant map $0 \times 0$, and the second being $(x, y) \mapsto (f(x), f'(x)y)$. It’s easy to see from this that the tangent bundle of $[0, 1]$ would be $[0, 1] \times \mathbb{R}$.

As I say, this calculation should hold for any model of SDG, including the Cahier topos (and thus there would appear to be more than one meaning of $I$ here as well). Of course this seems, superficially at least, to conflict with what Urs said in #46 that there was just one interpretation of $I = [0, 1]$ in the Cahier topos (one that led to a different solution to Andrew’s puzzle) – so I ought to leave open the possibility that I am making a silly mistake somewhere.

• CommentRowNumber51.
• CommentAuthorUrs
• CommentTimeNov 13th 2012

conflict with what Urs said

Above you were talking about $C^\infty$-algebras and all I meant was that $C^\infty([0,1])$ is not in the site of definition of the Cahiers topos.

• CommentRowNumber52.
• CommentAuthorTodd_Trimble
• CommentTimeNov 13th 2012

Okay. But this calculation addresses your #30 as well. In other words, I’m now pointing out that even for the Cahier topos there is an interpretation of $[0, 1]$ leading to $T([0, 1]) \cong [0, 1] \times \mathbb{R}$, and I don’t think this was explicitly brought out in the discussion above.