TITLE: Fixed point theorem for half open half closed interval
QUESTION [2 upvotes]: I am thinking whether Every continuous function from $[0,1)$ on to $ [0,1)$ has a fixed point. I know about fixed point theorem for compact sets. The proof I came up fory mentioned problem is, if $f(0)=0$ then, we are done. If not, then it has to reach zero Somewhere in between and go to 1. So a compact subset of image is contained in the compact subset of domain or compact subset of domain is contained in the compact subset of image and thus it has a fixed point. Is it right??
Is it true that every interval mapped to itself with a continuous onto function, has a fixed point?

REPLY [0 votes]: There is no general result to this effect.  Consider $f: [0,1) \to [0,1)$ via:
$$
f(x) = \frac{1}{2}x + \frac{1}{2}.
$$
The issue here is that if I'm missing any point in the closure of $(0,1)$, I can shunt the fixed point off to a missing bit of the boundary to come off with a counterexample like the above.


TITLE: Calculate complex integral with pole at zero
QUESTION [0 upvotes]: Calculate for $\alpha >0$ and $n \in {\mathbb Z}$.
$$
\oint_{\left\vert\,z\,\right\vert\ =\ \alpha} z^{n}\,{\rm d}z.
$$

REPLY [0 votes]: Line integrals $\int_\gamma f(z)\>dz$ are defined for  chains $\gamma$ (formal sums of directed arcs), and not for sets $S:=\{z\>|\>|z|=\alpha\}$. But never mind.
We all know that
$$z^n={1\over n+1}\>{d\over dz}\> z^{n+1}\qquad(n\ne-1)\ ,$$
and this implies
$$\int_\gamma z^n\>dz=0$$
for all $n\ne-1$ and all closed curves $\gamma$ not passing through $0$. Therefore it remains to compute
$$\int_\gamma{dz\over z}\ ,$$
where $$\gamma:\quad t\mapsto z(t):=\alpha\>e^{it}\qquad(0\leq t\leq2\pi)$$
is the circle of radius $\alpha>0$ around the origin. One obtains
$$\int_\gamma{dz\over z}=\int_0^{2\pi}{\dot z(t)\over z(t)}\>dt=\int_0^{2\pi} i\>dt=2\pi i\ .$$


TITLE: Can we define Mobius function for any real number and any complex number ?
QUESTION [1 upvotes]: All:
To me, Mobius function is a bit mysterious. I just want to know if we can define
Mobius function for any real number or any complex number ?
Can anyone point out any resource on this ?
Thank you.

REPLY [1 votes]: I don't know what you mean by defining it for any real or complex number.
But (assuming that by Möbius function you mean the standard thing whose value at square-free numbers is $\pm1$ depending on whether the number of prime factors is even or odd, and equal to $0$ elsewhere) there is a thing called an incidence algebra.
Every locally finite partially ordered set has an incidence algebra, in which the identity element for multiplication is called the $\delta$ function, the function everywhere equal to $1$ is the $\zeta$ function, and the inverse of the $\zeta$ function is the $\mu$ function.  In the special case in which the poset is the positive integers ordered by divisibility, the $\mu$ function is the standard $\mu$ function mentioned in the paragraph above.
PS: I just realized that what you meant was: "Can one define $\mu(x)$ for $x\in\mathbb R$ or $x\in\mathbb C$?".  I would guess not in any reasonable way.


TITLE: How to get uniform distribution with two dice rolls?
QUESTION [1 upvotes]: The sum of two dice rolls will not have uniform distribution. Never realized...
Is there an easy way to cheat?
Will this work?
1st die roll, 1-6...
2nd die roll, if 1-3, add 0 to first die, if 4-6, add 6 to first die.
Is this sum uniformly distributed?
ps... inspired by "How to generate a random number between 1 and 10 with a six-sided die?"

REPLY [0 votes]: Yes, this will give you a uniform distribution on the numbers from 1 to 12.


TITLE: Determining whether a set is a group under matrix multiplication
QUESTION [0 upvotes]: I am aware of the process of determining groups (closure, identity, inverse and associativity) and have managed previous questions such as the set of square matrices when $\det A =\pm 1$. However I am struggling with getting my head around this set/group:
Let $S\subset M_n(\mathbb{R})$, with the property that if $A\in S$, then $A^{−1}$ exists and is in $S$. Let $G$ be a set of all possible products of a finite number of elements of $S$, that is $$G=\{A_1A_2\dots A_k:A_i\in S, i=1,\dots,k\text{ for some }k\in\mathbb{N}\}$$ Any help would be appreciated.

REPLY [1 votes]: Let $M, N\in G$. That means that both $M$ and $N$ can be written as a finite product of elements in $S$:
$$
M = S_1S_2\cdots S_m, \quad N = T_1T_2\cdots T_n\\
S_i, T_i\in S
$$
Then clearly $MN$ is the product of finitely many elements of $S$, namely
$$
MN = S_1S_2\cdots S_mT_1T_2\cdots T_n
$$
so $G$ is closed under product.
Showing that the identity matrix is an element of $G$ requires a minor subtlety.
Finally, for any matrix $M\in G$, can you show that $M^{-1}\in G$?


text:
\section{$\B$-modules}

\begin{definition}
The Boolean semi-ring $\B$ is defined as the set $\{0,1\}$ equipped with two operations: a binary $\pplus$ and $\cdot$, with the following properties:
\begin{itemize}
\item $(a\pplus b)\pplus c=a\pplus (b\pplus c)$, $a\pplus b=b\pplus a$;
\item $0\pplus a=a\pplus 0=a$, hence $0$ is an \textbf{neutral element};
\item $a\pplus a=a$;
\item $(a\cdot b)\cdot c=a\cdot (b\cdot c)$, $a\cdot b=b\cdot a$;
\item $1\cdot a=a\cdot 1=a$.
\end{itemize}
\end{definition}
Given this algebraic structure, a $\B$-module is defined as follows.

\begin{definition}
\label{def:bmodule}
A \textbf{$\B$-module} is a structure $(V,0,-,\pplus)$ such that the following identities hold:
\begin{itemize}
\item $(a\pplus b)\pplus c=a\pplus (b\pplus c)$, $a\pplus b=b\pplus a$;
\item $a\pplus a=a$, $a\pplus 0=a$.
\end{itemize}
A notions of \textbf{submodule}, \textbf{congruence}, \textbf{quotient module} and \textbf{module homomorphism}, \textbf{finitely generated module} are defined as usual.
\end{definition}
Note that a $\B$-module is join-semilattice with a minimal element (0), the join of two elements is exactly $\pplus$.

We say that a $\B$-module $P$ is \textbf{projective}, if the usual lifting property holds: given any surjective morphism of $\B$-module $f:M\ra N$ and a $\B$-module morphism $g:P\ra N$, there exists a $\B$-module morphism $h:P\ra M$ so that $g=f\circ h$. It is not hard to show that a finitely generated $\B$-module $P$ is projective if and only if there exists a splittable surjection $F\rightarrow P$ from a free $\B$-module $F$ to $P$ (by splittable surjection, we mean a surjection which has a right inverse). In this way, the projective $\B$-modules can be equipped with a structure of a \textbf{finite distributive lattice} \cite{horn1971}.

\begin{Example}\label{ex:b}
Consider the following finite distributive lattices of size $4n-3$ (see the figure below) denoted by $D_{n}$ (here $n>1$). The elements of the lattice are denoted by $a_{ik}$ representing the position of the element in the lattice. The operations (join and meet) $\vee$ and $\wedge$ are defined as follows:
\[a_{ik}\vee a_{lm}=a_{\max(i,l),\max(k,m)}, \quad a_{ik}\wedge a_{lm}=a_{\min(i,l),\min(k,m)}.\]
\begin{center}
\definecolor{ududff}{rgb}{0,0,0}
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\begin{tikzpicture}[line cap=round,line join=round,>=triangle 45,x=.5cm,y=.5cm]
\clip(5.6721097057209438,4.741855581495774) rectangle (19.18008870615289,28.127014890185116);
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\draw [line width=2.pt] (12.,24.)-- (14.,26.);
\draw [line width=2.pt] (14.,26.)-- (12.,28.);
\draw [line width=2.pt] (12.,20.)-- (10.,18.);
\draw [line width=2.pt] (10.,18.)-- (16.,12.);
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\draw [line width=2.pt] (10.,10.)-- (12.,8.);
\draw [line width=2.pt] (12.,6.)-- (12.,8.);
\begin{scriptsize}
\draw [fill=ududff] (12.,6.) circle (2.5pt);
\draw[color=ududff] (12.41700685660759,5.263624437166554) node {$O$};
\draw [fill=ududff] (12.,8.) circle (2.5pt);
\draw[color=ududff] (12.61700685660759,7.263624437166554) node {$a_{11}$};
\draw [fill=ududff] (14.,10.) circle (2.5pt);
\draw[color=ududff] (14.834755945813466,10.014645006989227) node {$a_{12}$};
\draw [fill=ududff] (12.,12.) circle (2.5pt);
\draw[color=ududff] (13.12560306398616,12.223797888816526) node {$a_{22}$};
\draw [fill=ududff] (10.,10.) circle (2.5pt);
\draw[color=ududff] (9.12430415693911,9.681187968222842) node {$a_{21}$};
\draw [fill=ududff] (14.,14.) circle (2.5pt);
\draw[color=ududff] (14.793073815967668,14.182857991569035) node {$a_{23}$};
\draw [fill=ududff] (16.,12.) circle (2.5pt);
\draw[color=ududff] (16.376994750108,12.765665576811902) node {$a_{13}$};
\draw [fill=ududff] (10.,14.) circle (2.5pt);
\draw[color=ududff] (9.249350546476505,14.683043549718613) node {$a_{32}$};
\draw [fill=ududff] (12.,16.) circle (2.5pt);
\draw[color=ududff] (13.12560306398616,16.266964483858942) node {$a_{33}$};
\draw [fill=ududff] (14.,18.) circle (2.5pt);
\draw[color=ududff] (14.709709556276072,18.51779949553204) node {$a_{34}$};
\draw [fill=ududff] (16.,16.) circle (2.5pt);
\draw[color=ududff] (16.793816048565983,16.308646613704738) node {$a_{24}$};
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\draw[color=ududff] (12.292146025219775,20.768634507205135) node {$a_{44}$};
\draw [fill=ududff] (10.,18.) circle (2.5pt);
\draw[color=ududff] (9.3743969360139,18.72621014476103) node {$a_{43}$};
\draw [fill=ududff] (12.,24.) circle (2.5pt);
\draw[color=ududff] (13.12560306398616,23.444701466565206) node {$a_{n-1,n-1}$};
\draw [fill=ududff] (10.,26.) circle (2.5pt);
\draw[color=ududff] (9.3743969360139,25.47871517978032) node {$a_{n,n-1}$};
\draw [fill=ududff] (12.,28.) circle (2.5pt);
\draw[color=ududff] (13.12560306398616,27.77160343759837) node {$a_{nn}$};
\draw [fill=ududff] (14.,26.) circle (2.5pt);
\draw[color=ududff] (15.109709556276072,26.020582867775694) node {$a_{n-1,n}$};
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\draw[color=black] (14.626345296584475,24.895165361939146) node {$$};
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\draw[color=ududff] (12.29288825781809,22.769376739803445) node {$\dots$};
\end{scriptsize}
\end{tikzpicture}
\end{center}
We remark that the $\B$-module structure is given by the join operation $\vee$.

It is easy to see that there exists a splittable surjection $F\ra D_n$. Indeed, the surjective map $g:\B[A_1,...,A_{2n-1}]\ra D_n$ from the free $\B$-module generated by $A_1$, $A_2$, ..., $A_{2n-1}$ given by
\[g(A_1)=a_{11}, g(A_2)=a_{12}, g(A_3)=a_{21},...,g(A_{2i})=a_{i-1,i+1},g(A_{2i+1})=a_{i+1,i}, ...\]
splits: the injection $h:D_n\rightarrow \B[A_1,...,A_{2n-1}]$ generated by
\[h(a_{11})=A_1, h(a_{12})=A_1+A_2, h(a_{21})=A_1+A_3 ...,\]
\[h(a_{i-1,i+1})=A_1\pplus A_2\pplus ...\pplus A_{2i-2}\pplus A_{2i},h(a_{i+1,i})=A_1\pplus A_2\pplus ...\pplus A_{2i-1}\pplus A_{2i+1}, \]
provides a right inverse.
\end{Example}

We have the following property for these distributive lattices.

\begin{proposition}\label{prop:bnonnot}
Let $D_n$ and $D_m$ be lattice as in Example \ref{ex:b}, with elements $a_{11}$, ..., $a_{nn}$, and $b_{11}$, ..., $b_{mm}$ respectively. Then, any injective $\B$-module morphism $f:D_n\ra D_m$ satisfying
\[f(a_{11})=b_{11}, f(a_{12})= b_{12}, f(a_{21})= b_{21}, f(a_{22})= b_{22}, \mbox{and}\]
\[f(a_{n-1,n-1})=b_{m-1,m-1}, f(a_{n-1,n})= b_{m-1,m}, f(a_{n, n-1})=b_{m, m-1}, f(a_{nn})= b_{mm}\]
is the identity morphism (and hence $n=m$).
\end{proposition}

\begin{proof}
First, we show that $f(a_{13})=b_{13}$. Indeed, $a_{13}$ is the only element (other than $a_{11}$, $a_{12}$, $a_{21}$ and $a_{22}$) in the lattice which is not bigger than $a_{22}$.

As a consequence, since $f$ is a $\B$-module map, we have to have that
\[f(a_{23})=f(a_{22}+a_{13})=b_{22}+b_{13}=b_{23}.\]
Now, we show that $f(a_{32})=b_{32}$. Indeed, $a_{32}$ is the only element (other than the previously discussed elements) which is not bigger than $a_{23}$. Hence, $f(a_{32})=b_{32}$. By continuing this process, we can show step by step that $f(a_{ij})=b_{ij}$ which concludes the proof.
\end{proof}

\begin{Example}\label{ex:bbmod}
Similarly, we consider the finite distributive lattice of 9 elements depicted in the figure below. We will denote this lattice by $\D_0$. Note that this is a sub-lattice of every $D_n$ for $n>3$.

\begin{center}
\definecolor{ududff}{rgb}{0,0,0}
\definecolor{cqcqcq}{rgb}{0,0,0}
\begin{tikzpicture}[line cap=round,line join=round,>=triangle 45,x=.5cm,y=.5cm]
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\draw [line width=2.pt] (12.,16.)-- (12.,12.);
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\draw [line width=2.pt] (12.,12.)-- (14.,10.);
\draw [line width=2.pt] (10.,10.)-- (12.,8.);
\draw [line width=2.pt] (12.,6.)-- (12.,8.);
\begin{scriptsize}
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\draw[color=ududff] (12.61700685660759,7.263624437166554) node {$A_{11}$};
\draw [fill=ududff] (14.,10.) circle (2.5pt);
\draw[color=ududff] (14.834755945813466,10.014645006989227) node {$A_{12}$};
\draw [fill=ududff] (12.,12.) circle (2.5pt);
\draw[color=ududff] (13.12560306398616,12.223797888816526) node {$A_{22}$};
\draw [fill=ududff] (10.,10.) circle (2.5pt);
\draw[color=ududff] (9.12430415693911,9.681187968222842) node {$A_{21}$};

\draw [fill=ududff] (12.,16.) circle (2.5pt);
\draw[color=ududff] (13.12560306398616,16.266964483858942) node {$A_{33}$};
\draw [fill=ududff] (14.,18.) circle (2.5pt);
\draw[color=ududff] (14.709709556276072,18.51779949553204) node {$A_{34}$};

\draw [fill=ududff] (12.,20.) circle (2.5pt);
\draw[color=ududff] (12.292146025219775,20.768634507205135) node {$A_{44}$};
\draw [fill=ududff] (10.,18.) circle (2.5pt);
\draw[color=ududff] (9.3743969360139,18.72621014476103) node {$A_{43}$};
\draw[color=black] (14.626345296584475,27.854596580990812) node {$$};
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\draw[color=black] (12.62560306398616,13.84940095280265) node {$$};
\draw[color=black] (13.625974180285318,10.848287603905188) node {$$};
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\draw[color=ududff] (12.29288825781809,22.769376739803445) node {$\dots$};
\end{scriptsize}
\end{tikzpicture}
\end{center}
\end{Example}

\begin{lemma}\label{lem:splinjb}
The injections $i:\D_0\rightarrow D_n$ for $n>3$ defined as
\[i(A_{11})=a_{11}, i(A_{12})=a_{12}, i(A_{21})=a_{21}, i(A_{22})=a_{22},\]
\[i(A_{33})=a_{n-1,n-1}, i(A_{34})=a_{n-1,n}, i(A_{43})=a_{n,n-1}, i(A_{44})=a_{nn}\]
are splittable injections of $\B$-modules.
\end{lemma}

\begin{proof}
It is easy to see that the map $j:D_n\ra \D_0$ given by
\[j(a_{kl})=\begin{cases}
A_{kl} & k,l\leq 2,\\
A_{33} & \mbox{if }a_{12}<a_{kl}\leq a_{n-1,n-1}\mbox{ and } a_{kl}\ne a_{22}\\
A_{34} & a_{kl}=a_{n-1,n}\mbox{ or }a_{kl}=a_{n-2,n},\\
A_{43} & a_{kl}=a_{n,n-1},\\
A_{44} & a_{kl}=a_{nn}
\end{cases}\]
provides a left inverse.
\end{proof}

config: arxiv


TITLE: K3-surface is not the blow-up of any other smooth complex surface?
QUESTION [3 upvotes]: Good evening,
I'm stuck in the following exercise in Huybrechts, Complex Geometry, chapter 2, page 103.
Let $X$ be a K3 surface, i.e. X is a compact complex surface with $K_X \cong \mathcal{O}_X$ and $h^1(X,\mathcal{O}_X)=0.$ Show that X is not the blow-up of any other smooth surface. Here $K_X$ is the canonical line bundle of $X.$
Does anyone have some ideas to solve this exercise?
Thanks in advance,
Duc Anh

REPLY [4 votes]: By the adjunction formula we have $$2g(C) - 2 = C^2$$ for the self-intersection of any curve C on X. Hence X does not contain any -1 curve, which were necessary for a blow-up, q.e.d.