algebraic surfaces – “Simplification” of the map constructed at the proof of Castelnuovo’s contractibility theorem

I’m reading the proof of the Castelnuovo’s contractibility criterion in Beauville’s book(Theorem II.17), and I guess I could understand all its affirmations. But I still has one question.

For those who don’t remember, Castelnuovo’s contractibility criterion states that “a (smooth, projective, complex) surface $S$ with a curve $E$ such that $E simeq mathbb{P}^1$ and $E^2=-1$ is the blow-up of another surface, and $E$ is the exceptional curve of this blow-up”.

The proof of this is to construct the blow-down map $phi:S rightarrow S’$, and then prove that $S’$ is also smooth.

The map is constructed by linear systems. Beauville’s choose a lot of thins:

  • a hyperplane section $H$
  • an integer $k$
  • a section $sin mathcal{O}_S(E)$ such that $(s)_0 = E$
  • a basis $s_0,ldots,s_n$ of $H^0(S, mathcal{O}_S(H))$
  • a basis $a_{i,0},ldots,a_{i,k-i}$ of $H^0(S, mathcal{O}_S(H+iE))$ for $1le i le k$

With the chooses done correctly, he says that
$${s^ks_0,ldots,s^ks_n,s^{k-1}a_{1,0},ldots,s^{k-1}a_{1,k-1},ldots,sa_{k-1,1},a_{k,0} } $$
is a basis of $H^0(S, mathcal{O}_S(H’) )$, where $H’ = H + kE$. Then, he constructs a rational map $phi: S dashrightarrow mathbb{P}^N$, and want to prove that this map is a blow-down to a smooth surface. That is:

  • $phi$ is a embedding at $S-E$;
  • $phi(E)$ is a point;
  • $phi(S)$ is a smooth surface.

And it’s done at the proof. My doubt arises here: I understand that this map works for the proof, but I think that it can be simplified. I want to consider $psi: S dashrightarrow mathbb{P}^N$ given by the subspace of $|H’|$ with basis ${s^k s_0,ldots,s^ks_0, s a_{k-1,0}, s a_{k-1,1} , a_{k,0} }$. I guess this is enough to prove the three affirmations:

  • $psi$ is an embedding at $S-E$, because $s$ is invertible out of $E$, and as $(s_0: cdots :s_n)$ defines an embedding, then $psi$ also defines at $S-E$
  • $psi(E)$ is a point, because $(s)_0 = E$ and then $phi(E) = (0:cdots:0:1)$ ( $a_{k,0}$ is constructed such that $(a_{k,0})_0 cap E = emptyset$)
  • $psi(S)$ is smooth; first, we just need to prove that $psi(S)$ is smooth at $(0:cdots:0:1)$, because $psi(S-E)$ is already smooth. Beauville do this putting coordinates and do a lot of considerations, but uses just the three last coordinates to do this.

I just want to know if this is right. If it is, I don’t think this is important, because I always like to use the full linear system anyway. But if it’s wrong, I would like to know, because I’m probably missing something.

proof writing – Discuss all rational side right triangle

There are ways to describe all possible rational-sided right triangles. Some basic examples are 3-4-5 triangles or 5-12-13, and those are integer-sided. Something like 3/5-4/5-1 is rational sided, as is 5/13-12/13-1. Given any rational-sided triangle, you can show that it’s similar to a triangle that has side lengths cos(theta), sin(theta), and hypotenuse 1, so now you can reduce the problem to studying right triangles inside the unit disk.. there are clever tricks like this that let you classify all such rational triangles.

why isn’t my proof valid? system of linear equations

The problem was: given the matrix $A in M_{mtimes n}$ and let $b in F^{m}_{col}$.
$y in F^{m}_{col}$ is a solution of the system of equations $AX=b$.
prove: every solution of the system of equations $AX=b$ can be represented as $y+x$, where $x in F^{m}_{col}$ is the solution of the homogeneous system $AX=0$.

so my proof was like that:
because $y in F^{m}_{col}$ is a solution of the system of equations $AX=b$ we can conclude:
$$(1) A cdot y=b$$
and because $x in F^{m}_{col}$ is a solution of the system of equations $AX=0$ we can conclude:
$$(2) A cdot x=0$$
from (1) and (2) we get:
$$ A cdot y+A cdot x=0+b$$
according to the rules of matrix multiplication we get:
$$(3) A cdot ( y+ x)=0+b=b$$
and therefore according to (3) the solution which is represented by $z=x+y$ is also a solution of the system of equations $AX=b$

my instructor gave me only 3 points for that telling that “it isn’t a valid proof”. Why is that? what’s the problem with it? and what should I say in order to appeal his decision.
thank in advance.

Why do some countries require a proof of accommodation in their visa application?

I am applying for a visa to enter a country. The visa application requires a proof of fully-paid accommodation post-quarantine.

I understand the usefulness of a proof of accommodation if:

  1. It was challenging to find accommodation. However, the country in question has a plethora of near-empty hotels desperate to find customers
  2. The visa applicant was impecunious. However, the visa application also requires a proof of funds via bank statements.
  3. The country wants to make sure that the visa applicant has somewhere to stay for the first few nights. However, the country imposes a mandatory 14-day quarantine due to the COVID-19 pandemic in designated quarantine facilities.

In that context, why do some countries require a proof of accommodation in their visa application?

set theory – What’s the proof theoretic ordinal of $KPomega$ with impredicative specification?

Let impredicative specification on a set $A$ relative to formula class $Gamma$ be the principle that
$$exists yforall x(xin y leftrightarrow(xin Aland alpha(x,y))),$$
where $alpha$ is in $Gamma$.

Let $KPomega$ as usual allow precisely $Delta_0$ specification, and let $KPomega IS$ be $KPomega$ plus the the principle of impredicative specification relative to $Delta_0$.

How much stronger is $KPomega IS$ compared with $KPomega$? What about $Sigma_nKPomega IS$ compared with $Sigma_nKPomega IS$?

differential geometry – Proof of smoothness of the Exponential map at a point of a complete regular surface in Euclidean space

Suppose S is a regular, connected surface in Euclidean space and $d_{s}$ is the intrinsic metric on S. When (S,d) is complete, we know that for each geodesic $gamma:Jrightarrow S$ (where J is any interval) there exists a unique geodesic $eta:mathbb{R}rightarrow S$ which is an extension of $gamma$.

If $p$ is any point of S, it is shown in do Carmo (page 284) that if a sufficiently small $win T_{p}S$ is chosen, the geodesic $gamma$ with initial state $gamma(0)=p, gamma'(0)=w$ is well defined at $t=1$. The symbol exp$_{p}(w)$ is made to denote the point $gamma(1)in S$. It is then shown (page 285) that exp$_{p}$ can be made into a smooth map on a sufficiently small neighbourhood of $0in T_{p}S$. This truth depends on the theorem of solutions of systems of ODE’s which says that the solution depends smoothly on initial conditions.

By what I wrote above, when S is complete, $p$ is any point and $w$ any tangent vector to S at $p$, the symbol exp$_{p}(w)$ is well defined, but how can we show smoothness of exp$_{p}$ on all of $T_{p}S$? Am I missing something elementary here? We know it is smooth around $0$, but the proof depended on covering $p$ with a chart and applying the theorem from analysis. It seems we cannot argue this way in the general case.

In do Carmo I did not see this technical question discussed, and neither in Tapp.

functions – Equivalent sets proof

Let $A$ be a set. Show that $P(A)$ – the power set of $A$ – is equivalent (same cardinal) to ${{0, 1}}^A$ – the set of all functions from $A$ to ${0, 1}$. Suppose $A$ was $mathbb{N}$; would this hold? How about if $A$ was $mathbb{N}^mathbb{N}$ (the set of sequences with values in $mathbb{N}$)?

My friend gave me the following proof:

It suffices to construct a bijective function.

Suppose that $f: A rightarrow {0, 1}$, and let $A_f$ be the set of elements such that $ain A_f iff f(a)=1$. Define the map $h(f) =A_f$ then we proceed to show injectivity and surjectivity:

Injectivity:

Suppose that for functions $f, g$, we have $fne g$. Then there is an $xin A$ so $f(x)=1$ and $g(x)=0$ or vice versa. Then $A_fne A_g$. By the contrapositive argument, $h$ is injective.

Surjectivity:

Let $Xin{cal P}(A)$ and then define a function $f$ as follows:

$$f(x)=cases{1, if xin Bcr 0, if x notin B }$$

Then $X = h(f)$ and $h$ is surjective.

We have a bijection $h:{{0, 1}}^A rightarrow {cal P}(A)$ so, we may conclude that the two sets are equivalent.

Is this proof correct? Is there a more formal or more detailed way to express it? In addition, does this proof carry forward to $mathbb{N}$ and $mathbb{N}^mathbb{N}$ defined in the problem above? Any assistance much appreciated.

algorithms – Justifying a claim in the proof of the master theorem

I am trying to understand the proof of the master theorem and I came up with my own proof for why (4.23) is true.

enter image description here

My argument is as follows:

Claim: $g(n)=Oleft(sum_{i=0}^{log_{b}(n)-1}a^i(n/b^i)^{log_b{a}-epsilon}right)$.

$$g(n)=sum_{i=0}^{log_{b}(n)-1}a^if(n/b^i)$$

Now by the definition of big O, we have that $exists cinmathbb{R}, N’in mathbb{R }$ such that $forall n/b^j>N’$, we have that $$f(n/b^j)<c(n/b^j)^{log_b{a}-epsilon}$$

This implies that $forall n>N’b^{log_bn-1}$

$$g(n)=
sum_{i=0}^{log_{b}(n)-1}a^if(n/b^i)
leq sum_{i=0}^{log_{b}(n)-1}a^ic(n/b^i)^{log_b{a}-epsilon}
leq csum_{i=0}^{log_{b}(n)-1}a^i(n/b^i)^{log_b{a}-epsilon}
$$

Which implies that $g(n)=Oleft(sum_{i=0}^{log_{b}(n)-1}a^i(n/b^i)^{log_b{a}-epsilon}right)$ with $M=c$ and $N=N’b^{log_bn-1}$.

Have I found the correct $c$ and $N$ to prove this claim for the upper bound of $g(n)$ and is this proof valid? Thanks!

Does usage of thumbs up/down ratings as social proof actually provide the right feedback?

Let me stay aside from the purpose of this change. There is some business decision behind it, for sure. But regarding the scale itself I am sure it is no good for the accuracy and telling good from better.

Thumb up/down does not allow that because it lacks proper resolution. It does not allow to register “good” and “better” at the very beginning. It is a simple, -1/1 measure. It is, indeed, anchored in the world of quality, but in fact being mostly qualitative.

At the same time, I strongly believe that a 1-5 (or 0-5) scale is just as accurate for that as it should be. It allows me to say a lot about a movie that -/+ does not:

  • “Dude, that was a total massacre, don’t even think about seeing this film!”
  • “It was bad, but not that bad.”
  • “Well, just another movie like a lot of them. A good one to kill time.”
  • “Wow, it was actually quite good!”
  • “That was a masterpiece!”

As a result, thumbs down/up system throws all the masterpieces and quite good movies into the same bin. Moreover, putting more weight on the number of voters, it makes those exquisite niche movies be outscored by that family comedy that just hit the screens. It dangerously pushes the measure from “how good” towards “how many”. And this is bad for both rating and sensing the quality.

EDIT:

So, going further. I think that introducing this kind of measure deteriorates the matching significantly.

  • For a 1-5 rating even one or two votes can shape my opinion somehow, and even if I do not know those users’ preferences, I would give this movie a try.

  • With a thumb up/down mechanism, I need to see e.g. a hundred of them to build up my interest.

In other words, I will get hooked on what is popular, not on what is good. And unless “just an average regular normal guy”, this will not match my preferences.

Is there a way out of this? Well, there is! Should my profile be put against the profiles of other users, so that the preference match is at the user profile level a thumb up/thumb down mechanism could be just enough. In other words, I believe it would be possible to match, to some extent, a movie with my preferences based on a combination of a simple thumb up/thumb down system and a profile matching system, that would recommend me movies “upthumbed” by people with similar taste. Unfortunately, for that, a quite elaborate user preferences profile should be built first. As I am not a Netflix user, I cannot tell if it is somewhere there below.

This is something that one of our local movie services, Filmweb, once introduced. I took a part in a long survey that tested my preferences which should result in presenting better suggestions to me. It is a pity I have never tried to use it after that, though, but I still think it could be a good idea. But even then, it is more about recommendations, not rating.