Apparently the following problem goes back to Banach and is still open: Let $B$ be a separable Banach space with 1-homogeneous unit sphere. Must it be Hilbert?
And the most suprising part (to me, at least) is that this is known to fail for nonseparable spaces
And the counterexample is much less ugly than I expected
Take $\Omega=A\times [0,1]$ where $A$ is some discrete uncountable set and put on it the $\sigma$-algebra $\Sigma$ consisting of sets $E\subseteq\Omega$ such that $E_a$ is Lebesgue measurable in $[0,1]$ for all $a\in A$. Put on $(\Omega,\Sigma)$ the measure $\mu(E)=\sum_{a\in A}\lambda(E_a)$ where $\lambda$ is the Lebesgue measure on $[0,1]$
Then $L^p(\Omega,\Sigma,\mu)$ has a 1-homogeneous sphere for all $p$ but is only Hilbert for $p=2$.
@Ted: You of course are aware of the fact that if $G$ is a connected Lie group, $p : X \to G$ is a covering space, then $X$ has to also be a Lie group in a way that makes $p$ a group homomorphism, yes?
So I am trying to make sense of the intersection pairing of line bundles(in general coherent sheaves). So suppose I have two curves $C, C'$ on a surface $S$ and let $O_S(C)$ and $O_S(C')$ be the invertible sheaves. Then I can consider the sequence $$0 \to O_S(-C-C') \to O_S(C) \oplus O_S(C') \to O_S \to O_{C \cap C'} \to 0$$ I want to show this is exact. Before this, I want to first see if I have got the maps right.
So I have some $f$ in the ideal sheaf of the curve $C \cup C'$ and I restrict this to both $C$ and $C'$ and get something in the direct sum, say $(a,b)$ and which I put inside $O_S$ as $a+b$ which naturally goes into the skyscraper sheaf at the end (This is all happening on some $U$ containing a point $x$ ). This works right?
Yeah the second term is the one that I fail to understand. But you were right, I did not try this on any examples. So let me just do that and I will get back
My guess is you specify two elements $a \in \mathcal O_S(C)$ and $b \in \mathcal O_S(C')$, and the first map is $x\mapsto (ax, bx)$ and the second $(u,v)\mapsto b u-av$
But the thing that I was thinking of was that since we have the sky scraper sheaf it's higher cohomologies vanish and when I take the euler poincare characteristic I will just have $\operatorname{dim}H^0(S, O_{C \cap C'})$ which is precisely the intersection number of $C$ and $C'$ and which will be equal to the sum of euler poincare characteristic of all the other terms in the sequence, giving the actual definition of the intersection pairing of two line bundles
@Astyx Yeah I think what you are saying is correct.
Okay right here is what the map should look like. Set $a \in H^0(S,O_S(C))$ and $b \in H^0(S,O_S(C'))$. You want something that has zeroes in both $C$ and $C'$ and put it in something that has zeroes on $C$ or $C'$, so you have to multiply it with an element in the opposite guys. So $ x \mapsto (bx, ax)$ and for the second map $(\alpha, \beta) \mapsto (a\alpha - b \beta)$ because we want to make it regular
To check exactness all I have to work with is how $C$ and $C'$ look like in the local ring $O_{X,x}$. I can call them $f,g$ for $C,C'$ respectively. Then since they are all on $C$, the fact that the local rings are Cohen-Macaulay will work
I had completely forgotten the powerful intuition that comes with thinking of these are poles and zeroes. Thanks @Ted @Astyx!
Also @Ted, I remember how you were once talking about thinking of performing blowup two times being related to curvature. So a few days back I was working with the Hirzebruch surface $\Sigma_n = \Bbb{P}(\mathcal{O} \oplus \mathcal{O}(-n))$. The first Hizrebruch surface is the blowup of $\Bbb{P}^2$ at a point and you can talk about connections forms on this pretty explicitly thanks to that projective bundle description. I was wondering if, maybe what you said could be made proper this way?
@robjohn Thanks for your reply. What do you mean by $k_\alpha(x)=\frac{[0\le x\le \alpha]}\alpha$? Is $[0\le x\le \alpha]$ an interval or a number somewhere between $0$ and $\alpha$? In case of the latter, should it not be $k_\alpha^2(x)=\frac x\alpha k_\alpha(x)$?
By the way, my question in the chat regarded a question I asked on SE here -->(stats.stackexchange.com/a/530541/263915). As mentioned earlier, it concerns kernel density estimation, but my question is purely math related I would say. In the linked answer, I don't understand the use of $\int |k^2(u)u|du \le \int k(u)|u|du<\infty$ or if these inequalities at all are true. They are supposedly used in the calculation of $\mathrm{Var}[f_n(t)]$.
It is the calculation given in 3. under Edit in the linked answer.
@robjohn But when you said "...you can adjust the ratio between the integrals to be anything you wish...", did you mean one can adjust it to make the inequality true?
I already know that if $X$ is a finite CW complex and if $p:\tilde{X}\to X$ is a $n$-sheeted covering space then the euler characteristic $\chi(\tilde{X}) = n\chi(X)$. Can I also say the statement is true for the case when $p$ is a finite covering space i.e. $|p^{-1}(x)|<\infty$ for each $x\in X$ ?
lately ive been experimenting with a 3adic coordinate system. it plots $0,1,2$ then subdivides by 3 again to $00,10,20,01,11,21,02,12,22$ and so on, it maps all integers to a "number line" between $[0,3)$, i can do this on the x and y axis and plot functions like y=x^2. does anyone know if this is a standard way of visualizing padic functions?
I was wondering if it's true that a function need not be surjective to have an inverse. It obviously needs to be injective but per your previous reply @robjohn (from a while back), the preimage of an element that is not hit is just the empty set. So it doesn't need to be surjective?
