Xerxes Posted January 15, 2016 Posted January 15, 2016 This is just so much fun. We will suppose that [math]U,V,W[/math] are vector spaces, and that the linear operators (aka transformations) [math]f:U \to V,\,g:V \to W[/math]. Then we know that the composition [math]g \cdot f: U \to W[/math] as shown here. (Remember we compose operators or functions reading right to left) Notice the rudimentary (but critical) fact, that this only makes sense because the codomain of [math]f[/math] is the domain of [math]g[/math] Now, it is a classical result from operator theory that the set of all operators [math]U \to V[/math] is a vector space (you can take my word for it, or try to argue it for yourself). Let's call the vector space of all such operators [math]L(U,V)[/math] etc. Then I will have that [math]f \in L(U,V),\, g\in L(V,W),\,g \cdot f \in L(U,W)[/math] are vectors in these spaces. The question naturally arises: what are the linear operators that act on these spaces? Specifically, what is the operator that maps [math]f \in L(U,V)[/math] onto [math]g \cdot f \in L(U,W)[/math]? By noticing that here the "[math]U[/math]" is a fixed domain, and that [math]g: V \to W[/math], we may suggest the notation [math]L(U,g): L(U,V) \to L(U,W)[/math]. But, for reasons which I hope to make clear, I will use a perfectly standard alternative notation [math]L(U,g) \equiv g_*:L(U,V) \to L(U,W),\, g_*(f) = g \cdot f[/math]. Now, looking up at my diagram, I can think of this as "pushing" the tip of the f-arrow along the g-arrow to become the composite arrow. Accordingly, I will call this the push-forward along [math]g[/math], or, by an abuse of language, the push-forward of [math]g[/math] So, no real shocks here, right? Ah, just wait, the fun is yet to begin, but this post is already over-long, so I'll leave you to digest this for a while......... 1
Xerxes Posted January 16, 2016 Author Posted January 16, 2016 Well,you guys are no fun! Am I boring you? Then try this..... Recall that, given [math]g: V \to W[/math] as a linear operator on vector spaces, we found [math]g_*: L(U,V) \to L(U,W),[/math] as the linear operator that maps [math]f \in L(U,V)[/math] onto [math]g \cdot f \in L(U,W)[/math], and called it the push-forward of [math]g[/math]. In fact let's make that a definition: [math] g_*(f) = g \cdot f[/math] defines the push-forward. This construction arose because we were treating the space [math]U[/math] as a fixed domain. We are, of course, free to treat [math]U[/math] as as fixed codomain, like this. This seems to make sense, certainly domains and codomains come into register correctly, and we easily see that [math]h \in L(W,U),\, h \cdot g \in L(V,U)[/math]. Using our earlier result, we might try to write the operator [math]L(g,U): L(W,U) \to L(V,U), h \mapsto h \cdot g[/math], but something looks wrong; [math]g[/math] is going "backwards"! Nothing daunted, let's adopt the convention [math]L(g,U) \equiv g^*[/math]. (We will see this choice is no accident) Looking up at my diagram, I can picture this a pulling the "tail" of the h-arrow back along the g-arrow onto the composite arrow, and accordingly (using the same linguistic laxity as before), call [math]g^*[/math] the pull-back of [math]g[/math], and make the definition: [math]g^*(h) = h \cdot g[/math] defines the pullback (Compare with the pushforward) This looks weird, right? But it all makes beautiful sense when we consider the following special case of the above. where I have assumed that [math]\mathbb{F}[/math] is the base field for the vector spaces[math]V,W[/math]. As before, the composition makes sense, and I now have [math]\phi \in L(W, \mathbb{F}),\, \phi \cdot g \in (V, \mathbb{F})[/math], and the pullback [math]g^*: L(W,\mathbb{F}) \to L(V,\mathbb{F})[/math]. But, hey, lookee here.... [math]L(U, \mathbb{F})[/math] (say) is the vector space of all linear maps [math]U \to \mathbb{F}[/math], which defines the dual vector space, so we quite simply have that [math] L(W,\mathbb{F}) = W^*,\,L(V, \mathbb{F}) = V^*[/math], the dual vector spaces. Putting this all together I find that, for [math]g: V \to W[/math] I will have [math]g^*:W^* \to V^*[/math] as my pullback. I say this is just about as nice as it possibly could be. I have one further trick up my sleeve.........
ajb Posted January 17, 2016 Posted January 17, 2016 This reminds me of the quite standard notions of lifting morphisms to mapping spaces using the push-forward and pull-back. You can consider similar things for (super)manifolds. Anyway, do you have a question or some clear point you want to make?
Xerxes Posted January 17, 2016 Author Posted January 17, 2016 No, I don't have a question, and I thought my "point" was clear This is just so much fun.Assuming fun is allowed here.... Anyway. Recall that for any vector space [math]V[/math] its dual [math]V^*[/math] is also a vector space defined over the same field [math]\mathbb{F}[/math], let us agree to call the ensemble of all such vector spaces as the category F-Vec. Then the mapping [math]^*:\text{F-Vec} \to \text{F-Vec},\,\,V \mapsto V^*[/math] is referred to as a functor. Moreover it is an Endofunctor. Even more over, since we saw the mapping [math]^*:(f:V \to W) \mapsto [f^*:W^* \to V^*][/math] it is called a contravariant functor. So if we adopt the rather outmoded convention that "ordinary" vectors are contravariant and that dual vectors are covariant, the contravariant functor here maps contravariant vector spaces onto covariant vector spaces. Whereas the identity functor [math]Id_{\text{F-Vec}}[/math], which is obviously also an Endofunctor, maps covariant vector spaces onto covariant vector spaces, likewise maps contravariant spaces onto contravriant spaces. Which I think is rather nice Since nobody seems especially interested in this thread, I will leave it there (but there are questions that I haven't addressed)
Keen Posted January 18, 2016 Posted January 18, 2016 We might not have the same notion of fun.... I think it would be much more interesting if you showed some problems that naturally arise from what you just showed or at least some reason why this deserves any attention. Perhaps you could show some concrete vector spaces and give an example of a specific pull-back or push-forward? Otherwise it just seems (at least to me) like some pointless abstract symbol manipulation.
Xerxes Posted January 18, 2016 Author Posted January 18, 2016 Thank you for your candour. .....it just seems (at least to me) like some pointless abstract symbol manipulation.The fact that this seems to me a curious thing to say in a mathematics forum is of course neither here nor there.
wtf Posted January 18, 2016 Posted January 18, 2016 (edited) Xerxes, your subject matter is very interesting to me since it's just beyond what I currently know, hence I would be motivated to learn more. However, your exposition is unclear. Just to take one example, you said that the mapping from a vector space to its dual is called a "functor." Now if one knows what a functor is (from category theory) then your usage is totally unmotivated and far too specific. A mapping is a functor because it has certain properties, which you didn't mention. On the other hand, if one doesn't know what a functor is, they would be baffled. Likewise your discussion of pushforwards. You said it's something in operator theory, as if you either don't know or don't care to explain that a pushforward is a very general construction that abstracts particular ideas in many fields of math. Unfortunately the relevant Wiki pages aren't very good, and I ended up reading the section on tangent bundles in an online pdf of Spivak's big differential geometry book. At that point I understood what you were trying to get at. Remember, exposition is a separate problem from math. It's clear that you know some differential geometry but perhaps haven't given much thought to exposition. As a start, you might consider asking yourself who is your audience and what is their background. If someone has had a year of calculus and a semester of linear algebra, you could build up everything you've said from first principles, using a curve in the plane and the tangent spaces (ie tangent lines) at each point. This would be a beautiful writeup that I'd love to read. If that's your aim, it's something you could challenge yourself to do. But as it is, all you've done is toss out some random factoids devoid of context that are either badly stated (to a knowledgable reader) or baffling and pointless to everyone else. I hope you will take this as constructive criticism. I would LOVE to read a clear explication of your topics. As it stands, you have major expositional problems and people are trying to tell you that. Edited January 18, 2016 by wtf 2
Xerxes Posted January 18, 2016 Author Posted January 18, 2016 I ended up reading the section on tangent bundles in an online pdf of Spivak's big differential geometry book. At that point I understood what you were trying to get at.Yes, but I deny that the constructions I "failed" to explain are restricted to differential geometry. you might consider asking yourself who is your audience and what is their background.As to the first, I had assumed it would be those members interested in pure mathematics. As to the second, how on Earth would I know? If someone has had a year of calculus and a semester of linear algebra, you could build up everything you've said from first principles, using a curve in the plane and the tangent spaces (ie tangent lines) at each point. This would be a beautiful writeup that I'd love to read. If that's your aim, it's something you could challenge yourself to do. You're kidding, right? How does anyone explain (in the way you demand) tangent spaces on differential topological manifolds without first going into point set topology? And a lot more besides I can do it (mod my "expository deficiencies"), but it would take whole book! My final word: If it is the consensus that this thread is rubbish, so be it.There is no point my insisting otherwise. Apologies for wasting forum disc space. Ben
ajb Posted January 18, 2016 Posted January 18, 2016 My final word: If it is the consensus that this thread is rubbish, so be it.There is no point my insisting otherwise. It is just that unless you ask some questions, or make some clear point for other to discuss, this thread is better off as a blog post.
wtf Posted January 19, 2016 Posted January 19, 2016 (edited) My final word: If it is the consensus that this thread is rubbish, so be it.There is no point my insisting otherwise. Not at all. I never studied differential geometry and your posts have gotten me reading up and trying to understand the point about the pushforwards. Apparently this is an abstract way of looking at the differential as a map from one manifold to another. The Wiki article on fiber bundles is very good. It's a hairbrush! The base space is the handle, the fibers are the bristles, there's a natural map from each bristle to its basepoint. A continuous right inverse of this map is a section, aka a vector field. This relates to ajb's recent comment on another thread. I'm finding this all very interesting. Your very lack of clarity is motivating me to try to understand. Don't stop on my account. Edited January 19, 2016 by wtf
Xerxes Posted January 19, 2016 Author Posted January 19, 2016 Apparently this is an abstract way of looking at the differential as a map from one manifold to another.Since, by your own admission, you do not know any differential geometry, you seem to have failed to grasp that the article you read is a special case of the more general construction I gave. At the risk of attracting more criticism from other members here I will try to explain. First suppose that [math]M,\,\,N[/math] be manifolds with [math]p \in M,\,\,\,q \in N[/math] Let us denote the class (it's not a set!) of all Real-valued functions [math]f: M \to \mathbb{R}[/math] at [math]p \in M[/math] by [math]F^{\infty}_p[/math] Then the vector space of all vectors tangent at the point [math]p[/math] denoted by [math]T_pM[/math]is the set of all mappings [math] v:F^{\infty}_p \to \mathbb{R}[/math]. Now if I have a smooth mapping [math]\phi:M \to N[/math] then, exactly as before I will have the pushforwrd [math]\phi_*:T_pM \to T_q N[/math] as a mapping of maps onto maps. Why your Wiki article says the pushfoward is equivalent to the differential of the mapping [math]\phi:M \to N[/math] requires a greater knowledge of differential geometry, which my recent experience here demotivates me from attempting to explain. Your very lack of clarity is motivating me to try to understand. There could have been a more tactful way to phrase that! But whatever, I am now truly done here
studiot Posted January 19, 2016 Posted January 19, 2016 Xerxes Anyway. Recall that for any vector space its dual is also a vector space defined over the same field , let us agree to call the ensemble of all such vector spaces as the category F-Vec. What do you mean by "the ensemble of all such vector spaces" please?
wtf Posted January 19, 2016 Posted January 19, 2016 (edited) There could have been a more tactful way to phrase that! But whatever, I am now truly done here You are taking this way too seriously. Thank you for the explanation. I'll get to work on it. Let us denote the class (it's not a set!) of all Real-valued functions [math]f: M \to \mathbb{R}[/math] ... It's a set, of course. A manifold is a topological space; and a topological space is a set. And the real numbers are a set. Hence the collection of functions from a manifold to the reals forms a set within ZF. What do you mean by "the ensemble of all such vector spaces" please? The context of Xerxes's discussion is Category theory. In Category theory we want to consider all the possible instances of a given type of object. For example, the collection of all the vector spaces over some given field [math]\mathbb{F}[/math]. The problem is that within set theory, there is no "set of all vector spaces" for pretty much the same reason there's no set of all sets. What shall we do? In Category theory we simply assume that we have the entire collection but that it's not necessarily a set. So Xerxes is using the word "ensemble" to avoid having to deal with this point. It's sufficient to realize that the collection of all vector spaces over some field is a collection that's not a set. I would say that "ensemble" is actually NOT a good word to use, since it is the word French mathematicians use to mean set! So an ensemble really is a set, hence using that word in this context is a little inaccurate. Now a question arises: What happens if we want to make sure our categorical reasoning is legitimate? At some point we might want to frame Category theory within set theory, for example so that we can apply set-theoretical manipulations to our categories. There has been a considerable amount of work done in this direction. The usual fix is to place Category theory within a Grothendieck universe, which is an extension of ZFC that includes an extra axiom, one that posits a particular universe that has enough sets to do Category theory. In particular, the set-theoretical universe must contain an inaccessible cardinal, a transfinite number so big it can't be proven to exist in ZFC. Once again we see the principle than math is not based on the axioms; rather, our choice of axioms is based on what math we want to do! But really, that's more than anybody needs to care about. The collection of vector spaces over a given field is a collection that's not a set. That's all anyone cares about in practice. Edited January 19, 2016 by wtf
studiot Posted January 19, 2016 Posted January 19, 2016 wtf Once again we see the principle than math is not based on the axioms; rather, our choice of axioms is based on what math we want to do! Which is exactly the point I was trying to make in another thread. Thank you for phrasing it better than I did.
Keen Posted January 20, 2016 Posted January 20, 2016 Thank you for your candour. The fact that this seems to me a curious thing to say in a mathematics forum is of course neither here nor there. Look. You could call me an advocate of mathematics for the sake of mathematics and I am not criticizing the fact that you are exposing theories that require a higher level of abstraction. What I am criticizing is your approach at exposing them. For me to consider some theory interesting, it must either have some sort of problems or solve other problems from other fields. Also usually I find a much better approach when explaining abstract mathematics: to give a lots of examples to make your audience grasp what kind of objects you are dealing with. Also what does your abstraction generalize? Does this generalization help to tackle some problems? And so on... I once took courses in Arakelov geometry, which were unfortunately pretty useless to me, because our teacher made something similar to what you are doing right now. He started with some very abstract notions like Grothendieck topology or some very general functorial inductive limits (not the usual ones with a directed set), without ever motivating these notions, so most students got lost in these abstractions pretty quickly. It's a pity that you are over dramatizing people's criticism and lack of enthusiasm, because I am sure there are interesting things to be learned from your posts. You are taking this way too seriously. Thank you for the explanation. I'll get to work on it. It's a set, of course. A manifold is a topological space; and a topological space is a set. And the real numbers are a set. Hence the collection of functions from a manifold to the reals forms a set within ZF. The context of Xerxes's discussion is Category theory. In Category theory we want to consider all the possible instances of a given type of object. For example, the collection of all the vector spaces over some given field [math]\mathbb{F}[/math]. The problem is that within set theory, there is no "set of all vector spaces" for pretty much the same reason there's no set of all sets. What shall we do? In Category theory we simply assume that we have the entire collection but that it's not necessarily a set. So Xerxes is using the word "ensemble" to avoid having to deal with this point. It's sufficient to realize that the collection of all vector spaces over some field is a collection that's not a set. I would say that "ensemble" is actually NOT a good word to use, since it is the word French mathematicians use to mean set! So an ensemble really is a set, hence using that word in this context is a little inaccurate. Now a question arises: What happens if we want to make sure our categorical reasoning is legitimate? At some point we might want to frame Category theory within set theory, for example so that we can apply set-theoretical manipulations to our categories. There has been a considerable amount of work done in this direction. The usual fix is to place Category theory within a Grothendieck universe, which is an extension of ZFC that includes an extra axiom, one that posits a particular universe that has enough sets to do Category theory. In particular, the set-theoretical universe must contain an inaccessible cardinal, a transfinite number so big it can't be proven to exist in ZFC. Once again we see the principle than math is not based on the axioms; rather, our choice of axioms is based on what math we want to do! But really, that's more than anybody needs to care about. The collection of vector spaces over a given field is a collection that's not a set. That's all anyone cares about in practice. When it comes to category theory I usually prefer to use the Godel-Neumann class theory. It's most likely just a cosmetic change to talk about the class of all sets instead of Grothendieck universe, but in theory it should be larger since it encompasses all the sets, while Grothendieck universe only encompasses the sets we "care" about. I doubt it changes much in practice, but it's a thing I wanted to point out.
imatfaal Posted January 20, 2016 Posted January 20, 2016 ! Moderator Note conway do not attempt again to shoehorn your locked topic into other discussions. this is hijacking and will result in sanctions. i have hidden your post and the quite correct follow up message. do not respond to this moderation within the thread 1
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