Quick check of understanding time: in the quotient ring $\mathbb{R}[x] / \langle x^2 \rangle$, for any real coefficients $a,b$, $abx^2 \in \langle x^2 \rangle$ as it is a multiple of $x^2$ by some $r$ in the ring $\mathbb{R}[x]$, correct?
then that point could be said to be equal to the same point on f(x+a)
perhaps you could use better english there
anyway. yes, it is possible for two functions to diverge similarly. in particular, L=lim[f(x)-g(x)] can exist and be 0 at a point, even if limf(x) and limg(x) don't exist. even if L isn't 0, if it exists it can measure the difference between how fast f(x) and g(x) diverge. more generally, asymptotic theory (big O notation, etc.) can classify functions based on growth using equivalence classes, and one can speak of which class f(x)-g(x) is in.
Still writing stuff. Didn't get much done during the conference (which is a good thing because it means I found the talks good enough that I did not do something else instead of listening). Getting close to finishing the paper (I think)
Or at least most of the talks were decent. As always there were a few really awful one (like this one guy who spend about 3 slides defining what a groupoid was)
Also, one guy gave a talk that made no sense at all. Then when I took a look at the paper it was based on it turned out that indeed that paper also made no sense at all (I thought he had chosen to just try to define things more informally in the talk, but it turns out that there were no clear definitions in the paper either).
But this is the sort of thing to be expected when the conference has both invited speakers and talks from people who asked to be able to give one when they signed up (the latter obly getting 30 minutes which is also a bit of a challenge)
Overall, the quality was better than I have experienced before though
Usually, only a few of the short talks are worth anything, because often they are by PhD students talking about some small technical details (this being all they have to talk about yet)
and this was not nearly as much the case this time
@BalarkaSen To give them an opportunity to make people aware of the work they are doing (conferences are supposed to be good networking opportunities after all)
@MikeMiller "$L$ is orientable iff it is trivial" is of course the easy part, because of course it is orientable if it is trivial and if it is orientable, the choice of basis at each point gives a nonzero global section. I am trying to see the "iff $L\setminus X$ is disconnected" part, which I have no idea yet how to approach.
On a related note, if $L$ is nontrivial then any global section has to have zeroes. In pictures, that seems to mean no matter how I perturb $X$ inside $L$, I always get nonzero self intersection of $X$ with the perturbed itself. So I am wondering if something like that can be formulated.
E.g. if $W$ is a $k$-submanifold of $M$ and the cohomology class representing $W$ in $H^k(M)$ has nonzero cup product with itself, shouldn't the tubular neighborhood of $W$ in $M$ be nontrivial?
As in, the circle at infinity in $\Bbb{RP}^2$. There is no trivial tubular neighborhood because otherwise I can take a global section, which a perturbed version of the circle inside the neighborhood, and those two circles have trivial intersection - can't happen in $\Bbb{RP}^2$.
I thought it was a choice of basis at every point in the fiber such that they are equivalent upto the transition functions in the intersection of the charts.
I'm confused.
I am not saying a smooth choice of basis at every point- that'd be a parallelization for a manifold. I am just saying choice of basis at every point such that given charts $U_i$ and $U_j$, the transition functions take one basis to another equivalent basis from $T_p M$ to itself, $p \in U_i \cap U_j$. I.e., transition functions have fiberwise det +1.
Isn't that equivalent to "equivalence class of bases at every point"?
That said, I can choose a unit norm vector in each fiber of the line bundle $L$ such that it's equivalent to the basis coming from orientation at that fiber. That's a smooth section, right?
In fancy language, the structure group of a vector bundle is GL_n (that's where the transition functions live). We're talking about various kinds of "reductions" of the structure group. GL^+ is an orientation; SL is a volume form; O is a Riemannian metric; SO is a Riemannian metric and orientstoon
yes, that's true. note that you wanted to assume you have a Riemannian metric here (a reduction to SO(1)) to get this easily.
ignore the past garbage. yes, once you have a metric, I agree. That's the construction I was looking for.
Also with the above: I wonder if, once you prove that triple wquivalence about line bundles, you can prove that closed hypersurfacees of Euclidean space are orientable
@DanielFischer Hi, If I have a connected metric space $E$ and $A$ a connected subspace of $E$, and for all $x\in E$ there exist a unique $y\in A$ such that $d(x,y)=\min_{z\in P}d(x,z)$, if we look at $y$ as a function of $x:f(x)$ do you think that $f$ is continuous?
I would start trying to prove that the projection is continuous. I'm not entirely sure yet it is, but if it isn't, the difficulties you meet trying to prove continuity will probably help constructing a counterexample.
My goal this Summer is to learn logic, because if you pass the logic qual you get the key to the logic room, and they have an espresso machine in there.
@TedShifrin: I don't have one yet, so I am trying to come up with some smart thing to say and claim it as my approach. :P
Here are some immediate thoughts. $r : M \to M$ is a smooth map with image sitting inside $A$, and leaves $A$ identity. So for a point $a \in A$ in the interior, $Dr(a)$ is identity. I think the same works on the boundary points upto change of charts by continuity of $Dr$.
I want to say if $x \in A$ then there exists chart around $x$ such that $r$ is locally erasing a bunch of coordinates. Intuitively, that seems right, but going to try to prove it.
I mean, there is a chart $U$ around $x$ such that $U \cap A$ is a linear subspace and $r$ is removing coordinates. Is that tautological? Then we'd be done with proving $A$ is a manifold, I mean.
So here mathematik.tu-darmstadt.de/Math-Net/Lehrveranstaltungen/… they claim to do calculus on locally-convex spaces, and there is a book by Colombeau that seems like it should be in Bourbaki's catalogue that does real/complex calculus in locally-convex spaces
Sorry, I got distracted. @MikeMiller Yes, I think that's exactly what I want. If $Dr$ has fixed rank $k$ near $A$, then for every point $p \in A$, then I can choose charts around $p$ such that $r$ is removing $n-k$ coordinates. That does imply $A \cap U$, where $U$ is the chosen chart, is diffeomorphic to $\Bbb R^k$.
I am just wondering why anybody would care to do such a thing, what is the point in going from Banach spaces to Frechet or Locally-convex spaces, I'm guessing the point is for distributions?
Oh, the middle step is not a hard one to make: shouldn't $C^\infty(M,N)$ be a smooth manifold?
There hasn't really been a time when I've wanted to think about manifolds of distributions, and I don't think such a theory would work very well (Frechet manifolds already toe the line). Now it's not hard to justify doing functional analysis on these spaces (though you should talk to Daniel Fischer about that). But differential topology is mostly impossible.
@bolbteppa You certainly need more than just a topological vector space, right? At least a norm, as Akiva mentioned, and completeness. Without completeness a lot of things goes wrong.
E.g., Banach fixed point theorem hence the inverse function theorem.
what map is used in the condition of the elements of the kernel of a group acting on a set $A$? like $ker~ G = \{g \in G ~|~...(g) = I_A\}$? I'm told to show it's the same as the kernel of the corresponding permutation representation $G \to S_A$ I know how to define the kernel of the latter
Wouldn't it be nice if it had a smooth structure so you could do standard calculus to it? So you could say things like "A smooth map $X \to C^\infty(M,N)$ is the same thing as a smooth map $X \times M \to N$"?
So you could have smooth approximation theorems, and much of your intuition about manifolds carries over.
