algebraic topology is nice because we have no conjectures
at least in homotopy theory
somebody conjectures something and it gets proved or disproved like 15 seconds later
and the whole thing is unintellible to the average joe so we don't get any... lay mathematicians throwing fake proofs around
although I guess that's mostly a problem for number theorists and nobody else
oh and all conjectures can pretty much be summarised as "do the homotopy groups of spheres have even more structure than we suspected?" and the answer is always yes
Problems in number theory are very easy to state, and generally very, very difficult to resolve. So early career mathematicians who focus on problems in number theory are highly likely to produce almost nothing publishable. It is career suicide.
If you have an interest in number theory, get tenure first, then start working on it.
(easy to state, hard to resolve, and often hard to predict how hard they will be to resolve)
There are a few programs where one might go to graduate school specifically for number theory, but you have to be pretty hot sh*t to even get into those programs.
@BenSteffan Indeed, though one might expect that some non-trivial theory will have to be created to resolve the conjecture, and that the new theory might be useful.
Like, answering the question "Is it possible to trisect an angle with just a compass and straight-edge?" is not a question which has obvious applications, but to resolve the problem, you have to invent Galois theory, which is useful.
@BenSteffan Oh, I don't think it is very worth studying.
@Jakobian Hard to say? He died pretty young, and I think he was more interested in solving polynomial equations by radicals. But my point is that you can't know if the solution to a problem is going to be interesting or not if you haven't solved the problem.
What I'm saying is, sometimes the theory is invented for the sake of inventing it, and problems are resolved because the theory is powerful, and not because someone studied hard to resolve a problem
@Jakobian I understand the point you are making---it is just completely orthogonal to the point that I was making.
I was not saying "Galois was interested in trisecting angles," rather I am postulating a contemporary of Galois, who (in principle) could have invented Galois theory to resolve the trisection problem.
@Jakobian Okay... but Galois was chasing an open problem ("is it always possible to solve a polynomial equation by radicals?"). But I didn't want to use that example, as being able to factor polynomials has a ton of applications.
It goes back to the Greeks... in "modern" mathematics, a lot of people hit their heads against it, starting with Cardano and Tartaglia.
Descartes was interested in it. The Bernoulli's. Name a mathematician prior to Galois (and, really, prior to Gauss), and they were probably interested in the problem, and had probably made some attempts at it.
Gauss failed. Hilbert, Russell, Whitehead, etc failed. Tao has, thus far, failed. Pomerance has failed, so far. And so on, and so on.
@Jakobian I wonder if one can really say that he "abandoned" functional analysis? I feel like a lot of the work he did was motivated by tough questions in functional analysis, and he just saw a much, much more general approach.
@Jakobian I've wondered that as well. Probably not, because his functional analysis work was not nearly as transformative (which isn't to say that it's not important).
I thought it's generally agreed he abandoned functional analysis?
@BenSteffan yeah, but that's not what I said. I said that it was motivated (initially, at least) by problems in funky anal (or, perhaps, in the physics which underlie funky anal).
Differential geometry and the like is more analysis than it is topology, so it's not something I want to learn Differential topology on the other hand is what I consider learning more in the future
I am trying to prove that each finite set is compact. My attempt: let $A$ be an arbitrary finite set. Hence, there exists $N \in \mathbb{N}$ such that $A=\{a_1,\dots,A_N\}$. Let $\mathcal{F}$ be an arbitrary cover of $A$, so $A \subseteq \bigcup_{F \in \mathcal{F}} F$. By definition of union, this means that $a_i \in F_i$ for each $i \in \{1,\dots,N\}$ and so $A \subseteq \bigcup_{i=1}^N F_i$. Since $\{F_i\}_{i=1}^N$ is a finite subcover of $\mathcal{F}$, hence $A$ is compact. Does this works?
I don't really care about Riemannian manifolds as an end in and of itself, but there are some interesting connections between, say, curvature and topology
and some basic geometric ideas like geodesics do go a long way in making topological arguments
Uhm, indeed I was a little suspicious of that. What I mean is that each $a_i$ is in some $F$ of the family, but in a finite amount because there are at most $N$ elements of $A$.
Bott & Tu is like the go-to source that explains very clearly how Poincaré duality, transversal intersection and Thom spaces are related in the de Rham-setting
I don't know if I can call something that's a tool in doing topology, to be topology itself So I don't think I believe you that de Rham cohomology is topology
I just feel that some non-trivial amount of the questions I read about it the first thing that goes through my mind is "oh, this is easy to proof using singular theory"
There are some cohomology/homology theories that clearly you would call topology, but maybe not all of them. And I think de Rham cohomology is something you wouldn't
Ok, so $A = \{a_1,\dots,a_N\} \subseteq \bigcup_{F \in \mathcal{F}} F$ means that there exist $i_1,\dots,i_N$ such that $a_1 \in F_{i_1}, \dots, a_N \in F_{i_N}$. Hence, $A \subseteq \bigcup_{k=1}^N F_{i_k}$ and $\{F_{i_k}\}_{k=1}^N$ is a finite subcover of $A$.
If I have descriptive set theory, and I am applying topology to set theory, am I actually doing set theory, or am I using topology as a tool to do set theory
Just to be sure: the first writing was sloppy because saying "$a_i \in F_i$ for each $i \in \{1,\dots,N\}$" means that the first element of the set $A$ is in the first element of the family $\mathcal{F}$, and so on, but this is not necessarily the case because they don't need to be indexes-ordered in that way?
you can frame many questions and answers of topology in the language of category theory. the point is that the actual methods used to study the questions of topology are not purely abstractly categorical in nature. in the same way, tho set theory is concerned with discrete spaces, the methods of set theory are not topological.
Its just that I think that being able to apply something to topology isn't enough to call it topology, unless that thing has a topological flavour. And de Rham has analysis flavour
The process of applying this thing is topology, but not the thing I am applying
the technique of integration has a deep relationship with the shape of the domain you are integrating over. this insight is in equal parts topological and analytical.
for a related statement, I would agree that, say, geodesics are not topology, though they are still a useful tool that can be applied to topology
the difference you make between de Rham cohomology and other cohomologies here is not in terms of their actual mathematical content, it's purely if they're defined in terms of things you consider topological to begin with or not
@BenSteffan Even though it's the same, I still wouldn't call it topological if you explicitly use de Rham. This is still applying integration to topology
again, it's very wrong to assume that mathematical fields are disjoint, which is why de Rham cohomology being analytical does not preclude it from also being topological
And the same with topology, I can define topological spaces however, but it doesn't matter as long as its a topological space and I am studying it in that setting
in fact, one could very well make the point that dR-cohomology is just another way to define singular cohomology, so the comment I cited basically implies that it has the same status as singular cohomology
but you seem to agree that singular cohomology is topological, whence dR-cohomology is topological
I have asked you what grounds for a distinction you have left, argued that if you ignore the definition you have none, in fact that what you said above can be used to argue that it is topological
@Jakobian so I have no problem with this per se, but applied to the case at hand I'm asking what other reason for distinction you have
I don't remember if I claimed I always base what belongs where based on definition alone, but if I did, that was definitely not entirely true. I use it as a suggestion, but other things are also important in decisive process.
Plus, things can belong to multiple things at once
@BenSteffan from the standpoint of homology theory
@Jakobian it doesn't matter whether it is based on definition alone. My point is that you have no other characteristics of dR-cohomology other than the definition to make a distinction
@Jakobian this is like bedrock stable homotopy theory
it's stuff you can build directly out of spaces, not that we really do that anymore
oh, so you're saying it is an object of homotopy theory, namely the part that's not algebraic topology?
if that's the case I'm going to leave you with the recommendation to look at, say, Hatcher Ch. 4.3 and ponder whether this is really where you want to draw the line between homotopy-theory-that's-topology and homotopy-theory-that's-not :)
@BenSteffan No I would say this is still algebraic topology. But just like $\ell^2$ with weak topology and questions like "is this countably paracompact" or whatever, is topology, that doesn't mean $\ell^2$ itself is
Okay if $U$ is open in $\ell^2$ then $U = \bigcup_{i=1}^\infty F_i$ where $F_i$ are closed in $\ell^2$ and then I want to claim that if I take closed ball $B$ and consider $F_i\cap n\cdot B$ in weak topology then it's compact
but I don't think that's true
my argument was wrong
it can't be always compact because they $\ell^2$ would have a compact closed ball which is not possible
yeah why would open sets of $\ell^2$ be $\sigma$-compact
they're not sigma-compact
But they still should be countable unions of closed balls
and that should prove that in weak topology, open sets are sigma-compact
So it turns out that my space $X$ will be perfectly normal with less effort
You see, $\ell^2$ has countable network, so it follows that $X$ has countable network, so is hereditarily Lindelof, and since it's also regular it will be perfectly normal
Consider a simple function $\phi$ in its unique standard representation $\sum_1^n a_j\chi_{E_j}$. It's been said that through linearity, you can show that any other representation of the simple function yields the same integral. How would you go about this? Given a simple function in a non-standard representation, the integral is undefined, since it is only defined for standard representations. It seems like you need to work a little harder other than using just linearity.
A simple function in standard representation has finite range with distinct image points, and where the disjoint sets of the indicator functions partition the domain of the function.
psie you say "it's been said," but all of this depends on the definitions of course. who is saying this and what definitions are they actually using? (anyway it sounds like it isn't you)
i would generally agree that it is circular or at least potentially circular to appeal to the 'linearity' of something that maybe hasn't been defined yet to answer a question like this
it might help to reduce whatever this is to a statement that doesn't involve integrals at all and to try proving it e.g. by induction
psie e.g. in a particularly simple example (pun intended) this is likely to specialize to a statement along the lines of, if c chi_E represents the zero function, then c m(E) = 0, where m is your measure
