Quick analysis question that is just killing me, if we know that $\lbrace f_n \rbrace$ converges to $f$ with respect to the $L^p(E)$ norm, does this mean that $f$ is in $L^p(E)$ or that the sequence itself is in $L^p(E)$?
@Ted: So the curvature form on $\Bbb P^n$ is the same as the Kahler form. Is there some more general situation that recovers this, or is this just part of the magic of projective space?
@Ted: Then the appropriate question is "When does there exist a line bundle whose curvature form is the Kahler form?" And the answer is of course always.
@Balarka: I didn't read your email too carefully, but of course you haven't got to the interesting part of that problem yet. Also, look at the theoretical ("challenge") problems.
@Michael: I'm not so fond of this stuff, but you might look at Halmos's Naive Set Theory. He writes well.
@TedShifrin Right. I will dig more after 9th, when my exams are over. Your problem on configuration space is nice. I have no idea what the dimension should be, at first sight.
Ok, what I meant was, I have no idea how the manifold should look like. Not that I have counted degrees of freedom, but that's approximately how I thought about doing it.
@Huy I think you have a wrong homotopy up there. $p$ and $\gamma_t$ do not have the same endpoints. You want $p(ts) \cdot \gamma_t(s) \cdot \bar{p}(ts)$ not?
Hatcher, on his introduction, just talks about simplicial cycle and give some vague ideas about basepoints. I want to know how a singular cycle looks like and make the vagueness precise and make you prove the Hurewicz's theorem.
@Michael: Codomain is where all the values live. The word "range" is used to have two different meanings. Some people mean codomain; other people mean what I call "image," the set of all the actual values of the function.
@Michael: For example, if $f(x)=x^2$, mapping real numbers to real numbers, the codomain (or range) is the set of real numbers, but the set of values is only ...
@Balarka: Well, if it's compact, are you sure it's actually a manifold, then?
@L33ter I don't know how you can know what singular homology is without knowing what singular cycle is. Tell me the definition of singular homology in brief.
I found it a good read. I still remember the meeting I had with the faculty member running the course where I was baffled about the idea of "induction over the length of a string".
One of my homework questions in my multivariable course — the configuration space of 4 rods of fixed lengths all attached to form a (topological) circle.
Speaking of configuration spaces, given a space $X$, one can consider $SX = (X \times X \setminus \Delta)/(\Bbb Z/2)$. This is called the deleted square (or maybe deleted square configuration space, I forget).
@TedShifrin let me know your travel plans in LA. Hopefully, we can get together. It is still windy in LA and it is supposed to get to 79° today, and 86° later this week.
Some people are thinking about this, in particular for $X=L(p,q)$. It's known that the homotopy type of $SX$ tells apart eg $L(7,1)$ and $L(7,2)$, which are homotopy equivalent but not homeomorphic.
I'm not sure which I'd prefer: That this recovers the homeomorphism type of $L(p,q)$ or that it doesn't.
@robjohn I got such an awesome result today! I'm turning now into a proposed problem. The amazing thing (and this is amazing in every way) is that one can finish the proof on half a page. It's beyond mind blowing.
injective means that the function takes on each value at most one time; surjective means that the function takes on each value at least one time (i.e., every value is hit by the function).
Watched a bit of the first lecture of Ted on YouTube now as it was mentioned. He used the "best linear approximation"-explanation for derivatives, so I'm a fan.
@Ted: I'm not getting a lot out of the G&H proof of Kodaira vanishing (or anything they do that invokes formulas with the Lefschetz operator or its adjoint...) Are there alternate proofs?
@AndrewT: Well, everyone knows the derivative of a real-valued function f in a domain D is the Lagrangian section of the cotangent bundle T*(D) that gives the connection form for the unique flat connection on the trivial R-bundle DxR for which the graph of f is parallel.
How do we use the fact that a function is injective to prove that $f(C \cap D) = f(C) \cap f(D)$ where $C$ and $D$ belong to $A$ which is the domain of the function.
since it's only injective I can't use the inverse of the function right?
The high school curriculum for math is actually terrible. I am slowly realizing it. We spent all this time learning practical application problems like $5x+3=8$ instead of doing problems that really taught us how to think
There's a fine line. Its worrisome that many students graduate high school with good grades, but would likely be stumped if asked to define what a function is.
yes but how do I go about it? I mean if for example while trying to prove that $f(C \cap D)$ is contained in $f(C) \cap f(D)$, if I say that $x \belongs to C \cap D$ then is it sensible to assume that $f(x) \belongs to C \cap D$? @TedShifrin
