TITLE: What computational error am I making?
QUESTION [3 upvotes]: Okay, I'm reading over the textbook notes and I'm making up my own notes for the section. There is an example problem that I'm trying to solve out; however, I'm having some problem and  I'm not getting the same answer as them. I erased all the work and tried to do it again but I got the same answer. So, can someone check out my work?
$A = \{a_1,a_2,a_3\}$ where $a_1 = [2,1,1]$ and $a_2 = [1,2,1]$ and $a_3 = [1,1,2].$ $a_1,a_2,a_3$ are row matrices.
$$A=\left(\begin{matrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1& 1& 2\end{matrix}\right)$$
They want to find the characteristic polynomial of A and the eigenvalues.
$$\det[A-\lambda I] = 0$$
$A = \{a_1,a_2,a_3\}$ where $a_1 = [2-\lambda , 1, 1]$ and $a_2 = [1 , 2-\lambda, 1]$ and $a_3 = [1,1, 2-\lambda].$
I decided to expand by column one to get the characteristic polynomial:
$$(2-\lambda)[(2-\lambda)^2 - 1] + [(2-\lambda) - 1] + [1- (2-\lambda)] = 0$$
This simplifies to:
$$6-11\lambda+6\lambda^2-\lambda^3 = 0$$
This is not correct. The book's characteristic polynomial is $-\lambda^3 + 6\lambda^2 - 9\lambda + 4 = 0$
What am I doing wrong?

REPLY [1 votes]: When you take the determinant of a matrix by column expansion, you need to alternate the signs.
$$\det\begin{pmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{pmatrix} = a_1 \det\begin{pmatrix} b_2 & c_2 \\ b_3 & c_3 \end{pmatrix} - b_1 \det\begin{pmatrix} a_2 & c_2 \\ b_3 & c_3 \end{pmatrix} + c_1 \det\begin{pmatrix} a_2 & b_2 \\ a_3 & b_3 \end{pmatrix}$$
When you computed your characteristic polynomial, you forgot the negative sign of the second term.


TITLE: Sheaf cohomology of $\mathbb{A}^3$ minus the origin
QUESTION [7 upvotes]: For each $i\geq 0$, describe $H^i(X,\mathcal{O}_X)$ where $X:=\text{Spec}(k[x,y, z])\setminus\{(x,y,z)\}$.

The first definition of cohomology I've learned involves injective resolutions, which I have no idea how to apply here.
I've read some authors who claimed that Cech cohomology is often useful to compute sheaf cohomology in real life, so I decided to take that road.
If $A:=k[x,y,z]$ and $Y:=\text{Spec}(A)$, I've thought about using the affine open cover $U_x:=Y\setminus V(x)$, $U_y:=Y\setminus V(y)$ and $U_z:=Y\setminus V(z)$.
That way, $\mathcal{O}_X(U_x)=A_x$, $\mathcal{O}_X(U_y)=A_y$ and $\mathcal{O}_X(U_z)=A_z$ and consequently:
$$C^0:=C^0(\mathcal{O}_X)=A_x\times A_y\times A_z$$
$$C^1=A_{xy}\times A_{xz}\times A_{yz}$$
$$C^2=A_{xyz}$$
I've checked that $\left(\frac{a}{x^n},\frac{b}{y^m},\frac{c}{z^\ell}\right)\in\ker(d^0)\Leftrightarrow\frac{a}{x^n}=\frac{b}{x^m}=\frac{c}{z^\ell}\in k[x,y,z]$, so $\ker(d^0)\simeq A$ and $H^0(X,\mathcal{O}_X)=A$.
Now, determining $H^1(X,\mathcal{O}_X)=\ker(d^1)/\text{im}(d^0)$ and $H^2(X,\mathcal{O}_X)=C^1/\text{im}(d^1)$ is considerably more complicated and made me wonder whether or not this is the best option.
I've also tried the simpler case $\text{Spec}(k[x,y])\setminus\{(x,y)\}$, and even then I found hard to describe $H^1(X,\mathcal{O}_X)$ explicitly.
This looks like a standard problem, so I can't help but wonder if there isn't a simpler approach which could work even for the general case $\text{Spec}(k[x_1,...,x_n])\setminus\{(x_1,...,x_n)\}$.

REPLY [2 votes]: Here is the general result:
Let $X=\operatorname {Spec}k[X_1,\cdots, X_n]\setminus \{\langle X_1,\cdots,X_n\rangle\}=\mathbb A^n_k\setminus \{O\}$. Then:

For $n=1$:
$ H^0(X,\mathcal O_X)=k[X_1,X_1^{-1}]$
$H^i(X,\mathcal O_X)=0 $ for $i\gt 0$
For $n\gt 1$:
$H^0(X,\mathcal O_X)=k[X_1,\cdots, X_n]$,
$ H^i(X,\mathcal O_X)=0$ for $i\neq 0,n-1$,
$  H^{n-1}(X,\mathcal O_X)=\bigoplus_{r_1\cdot\cdots,r_n\gt 0} k\cdot X_1^{-r_1} X_2^{-r_2} \cdots X_n^{-r_n}.$

The proof is by Čech cohomology. A small part can be found here.
Remarks
a) Remember  that if a topological space has a covering $\mathcal U$  consisting of $n$ open subsets, then for any sheaf $\mathcal F$ of abelian groups, we have $\check H^i(\mathcal U,\mathcal F)=0$ as soon as $i\geq n$ (because Čech cohomology can be computed by alternating cochains).
b) Remember also that for $n=1$ the variety $X$ is affine, so that $\mathcal O_X$ is acyclic (as is any coherent sheaf).


TITLE: De Rham Cohomology Question
QUESTION [3 upvotes]: Let $M$ be a compact connected $n$-manifold without boundary. Let $\mu\in\Omega^{n-1}(M)$, show that there exists a point $p\in M$ such that $d\mu(p)=0$.

REPLY [8 votes]: First, assume that $M$ is oriented.  Then, by Stokes's theorem, we have $\int_M d\mu = \int_{\partial M} \mu = 0$ since $\partial M = \emptyset$.  This can't happen if $d\mu$ is never $0$.
If $M$ is nonorientable, let $N$ by any orientable cover of $M$ (say, the universal cover, or the orientation covering).  Let $\pi:N\rightarrow M$ be the covering map.
Then by the previous case, $d(\pi^*\mu)$ is $0$ on some point $p$ of $N$.  But $0 = d(\pi^*\mu) = \pi^* (d\mu)$.  Since $\pi$ is a local diffeomorphism, $\pi^*$ induces an isomorphism from $\Omega(M)_{\pi(p)}\rightarrow \Omega(N)_{p}$.  Since the image of $d\mu$ under this isomorphism is $0$, we must have $d\mu(\pi(p)) = 0$.


TITLE: Possible combinations of two sets of three into a set of two
QUESTION [3 upvotes]: I'm trying to find the mathematical term for a certain phenomenon and hopefully a more general way to solve such problems. Say there are two sets, each $= \{A, B, C\}.$ I would like to know how many different possible ways I can combine these sets into a set of two $\{...,...\}.$ The combination $\{A, B\}$ is considered the same as $\{B, A\}.$ The only way I can think to do this, is:
$$ 3*3 - \binom{3}{2} = 6 $$
Basically, I am counting up the total number of possible combinations (3*3) and subtracting out the combinations I overcounted. Is there a name for this kind of a calculation? Is there anyway to describe this more mathematically, as opposed to this intuitive approach?
Thanks!

REPLY [2 votes]: You are actually counting the number of subsets of $\{A,B,C\}$ of a fixed size ($2$, in your example), without regarding order, and allowing duplicates. A more general way to look at it, is taking some set $A=\{1,\ldots,n\}$, and trying to assign a number of occurrences to each of $1\ldots n$, that is, "$1$ appears once, $2$ appears five times, etc.", while demanding that the sum of these numbers be some constant $k$ - that is, you're only selecting a set of $k$ elements.
This is equivalent to the following situation:
You have $n$ buckets and $k$ balls. You wish to toss the balls into the buckets, putting as many balls in one bucket as you like. In order to calculate the number of ways to do that, it's more convenient to look at the $n$ buckets as $n-1$ partitions - defining $n$ different spaces, that you can place among the balls, like in the following illustration:
$$\cdot|\cdot\cdot\cdot||\cdot|\cdot\cdot||$$
This has $7$ buckets, with $1,3,0,1,2,0,0$ balls respectively.
Now we can give a general equation.
Sort the partitions and the balls together - there are $(n-1+k)!$ ways to do that.
Disregard the internal ordering of the partitions and the internal ordering of the balls, as you're only interested in the number of balls in each "bucket", and the "buckets" are defined by their position regardless of which partitions surround them.
Therefore, the number of ways to select a set of size $k$, with duplicates allowed and without regarding order, from a set of size $n$ is $\frac{(n+k-1)!}{(n-1)!k!}=\binom{n+k-1}{k}$.
In your particular case, $n=3$ and $k=2$, so this equals $\binom{4}{2}=6$.


TITLE: A weird property of $\sum_{k = 1}^{n} \sin k$
QUESTION [3 upvotes]: I was playing around with the sum $\sum_{k = 1}^{n} \sin k$, and using very loose rigour I arrived at the following:
Proposition. Let $n \equiv n_0 \pmod {44}$ and $n_0 \equiv n_1 \pmod {6}$. Then $$\sum_{k = 1}^{n} \sin k \sim \frac {1} {2} \left ( 3 - |3 - n_1| \right ).$$
Does heuristics support this claim?
Note: I can provide my "proof" on request.

REPLY [0 votes]: I think it appropriate to present my rigour here now that it created sufficient curiosity.
Consider $$f (n) = \sum_{k = 1}^{n} \sin k.$$ Let $a_k = \left [ \frac {2k} {\pi} \right ]$ for $1 \leqslant k \leqslant n$ and $r_1, r_2, \cdots, r_n$ be real numbers in the interval $[0, 1]$. Then, denote $$\sin k = \begin {cases}
r_k, & \text {if } a_k \equiv 0 \pmod {4} \\
1 - r_k, & \text {if } a_k \equiv 1 \pmod {4} \\
-r_k, & \text {if } a_k \equiv 2 \pmod {4} \\
r_k - 1, & \text {if } a_k \equiv 3 \pmod {4} \\
\end {cases}$$ Here $r_{\theta}$ is the side of the right triangle just in front of the angle $\theta$. By Weyl's criterion, we see that $r_k$ is equidistributed modulo $1$, so the average of real numbers $r_k$ is $\frac {1} {2}$. Hence, $$\sin k \approx \begin {cases}
\frac {1} {2}, & \text {if } a_k \equiv 0 \pmod {4} \\
\frac {1} {2}, & \text {if } a_k \equiv 1 \pmod {4} \\
- \frac {1} {2}, & \text {if } a_k \equiv 2 \pmod {4} \\
- \frac {1} {2}, & \text {if } a_k \equiv 3 \pmod {4} \\
\end {cases}$$ Since $\frac {2mk} {\pi}$ is close to an integer when $m = 11 t$ and the period of $\sin k$ with respect to $a_k$ is $4$, it is safe to assume $f (44t + n_0) \approx f (n_0)$ for any integer $t$. Let $n = 44 t + n_1$. Also, since $a_{6j + i} \equiv a_i \pmod {4}$ for any integers $i, j$, we have by setting $i = n_1$ the desired result.
This is a slightly simplified and fast version of what I'd done to arrive at the ridiculous expression that my question was about.


TITLE: show that $\sum_{i=1}^n i^2$ is $O(n^3)$
QUESTION [2 upvotes]: To start, am I on the right track?
$\sum_{i=1}^n i^2$ = $1^2 + 2^2 + 3^2 + ... + n^2 \le n^2 + n^2 + n^2 + ... + n^2$
Where would I go from here?

REPLY [0 votes]: It's equivalent to

$$
\lim_{n \to \infty}{\sum_{k = 1}^{n}k^{2} \over n^{3}} =
\lim_{n \to \infty}
{\left(\,n + 1\,\right)^{2} \over \left(\,n + 1\,\right)^{3} - n^{3}} =
\lim_{n \to \infty}
{\left(\,1 + 1/n\,\right)^{2} \over 3 + 3/n + 1/n^{2}}\ =\ \bbox[10px,#ffe,border:1px dotted navy]{\displaystyle{1 \over 3}}
$$


TITLE: Evaluate the double integral $\int\int_R(x^2-2y) $ $dxdy$
QUESTION [0 upvotes]: Evaluate the double integral $\int\int_R(x^2-2y) $ $dxdy$, where $R$ is a region bounded by $y=kx$, $y=(kx)^3$, $x \geq 0$, and $k$ is a fixed positive constant. (The answer will depend on $k$.)
I have completed this question but it doesn't match up with the solution I have been given.
My solution, where the region is
$R = \{ (x,y) : \frac{y}{k}\leq x \leq \frac{\sqrt[3]{y}}{k}, 0\leq y \leq k \}$
\begin{align}
\int_{0}^{k}\int_{\frac{y}{k}}^{\frac{\sqrt[3]{y}}{k}}(x^2+2y) dxdy &= \int_{0}^{k}\bigg[\frac{x^3}{3}+2yx\bigg]_\frac{y}{k}^\frac{\sqrt[3]{y}}{k} dy \\
&= \int_{0}^{k}\frac{y-y^3}{3k^3}+\frac{2y\sqrt[3]{y}-2y^2}{k}dy \\
&= \bigg[\frac{y^2}{6k^3}-\frac{y^4}{12k^3}+\frac{6y^\frac{7}{3}}{7k}-\frac{y^2}{k}\bigg]_0^k \\
&= \frac{1}{6k}-\frac{k}{12}+\frac{6k^\frac{4}{3}}{7}-k
\end{align}

REPLY [1 votes]: Change the order of integration. I get
$$\int_{x=0}^{1/k}\int_{y=(kx)^3}^{kx}(x^2-2y)\; dy\; dx =
\frac{1}{12k^3}-\frac{4}{21k}$$

Note that the curves intersect when
$$0=(kx)^3-(kx) = (kx)((kx)^2-1)=(kx)(kx+1)(kx-1)$$
so when $x=0$ or $x=\pm1/k$. In the right half-plane, then, the bounded region is between $x=0$ and $x=1/k$.
 None of the answers given is correct. I would say they intend the second answer, since it is nearly correct but with a sign error in the first term.


TITLE: Three Approaches to a Markov Chain Calculation with Three Different Answers
QUESTION [3 upvotes]: I've approached this question three different ways and arrived at three different answers. Something is wrong here...
Consider a Markov chain $X = \{X_n : n \in \mathbb{N}\}$ with probability transition matrix
$$ P = \begin{pmatrix}
1/5 & 3/5 & 1/5 \\
1/2 & 1/2 & 0 \\
0 & 1/3 & 2/3
\end{pmatrix} $$
and state space $\{0,1,2\}$.
Suppose we want to find the conditional probability $\alpha$ that $X_2 = 2$ given that the chain started at time $0$ in state $0$ and has not yet entered state $1$ by time $2$. That is, (I believe) we want to find
$$ \alpha_1 = \mathbb{P}(X_2 = 2 | X_0 = 0, X_1 \ne 1).$$
It seems to me that this should just be
$$ \alpha_1 = \mathbb{P}(X_2 = 2 | X_0 = 0, X_1 = 0) + \mathbb{P}(X_2 = 2 | X_0 = 0, X_1 = 2); $$
and since $X$ is a Markov chain, it follows that
$$ \alpha_1 = \mathbb{P}(X_2 = 2 | X_1 = 0) + \mathbb{P}(X_2 = 2 | X_1 = 2) = P_{02} + P_{22} = \frac{1}{5} + \frac{2}{3} = \frac{13}{15}. $$
However, another approach suggests that we can compute this probability by first modifying $P$ (and calling the modified matrix $Q$) like so:
$$ Q = \begin{pmatrix}
1/5 & 3/5 & 1/5 \\
0 & 1 & 0 \\
0 & 1/3 & 2/3
\end{pmatrix}. $$
We then compute $Q^2$:
$$ Q^2 = \begin{pmatrix}
1/25 & 59/75 & 13/75 \\
0 & 1 & 0 \\
0 & 5/9 & 4/9
\end{pmatrix}. $$
Since the modified process gets absorbed if it hits state $1$, it follows that
$$ \alpha_2 = Q^2_{02} = \frac{13}{75} \ne \alpha_1, $$
which contradicts the calculation in Approach #1.
Yet another approach is as follows. We can condition only on the event that $X_0 = 0$ and compute the probability of an intersection of events; i.e.,
$$ \alpha = \mathbb{P}(X_2 = 2, X_1 \ne 1 | X_0 = 0). $$
Using the "chain rule" for the conditional probabilities yields (I will omit the tedious calculations)
$$ \alpha_3 = P_{01}P_{02} + P_{01}P_{12} = \frac{1}{25} + 0 = \frac{1}{25}. $$
So we get yet another contradiction.
My Question:
Are any of these approaches correct? If so, what is wrong with the other two approaches?
UPDATE: After properly rephrasing the question (properly), my thought is that Approach #1 is correct, but that the computation of $\alpha_1$ is incorrect. Also, although Approaches #2 and #3 are incorrect, my thought is that they should yield the same result; i.e., $\alpha_2$ should be equal to $\alpha_3$ (even though my computations suggest otherwise). I'm still trying to get all of this sorted out...

REPLY [1 votes]: I don’t understand the third approach, where you left out the details, so I can only speak to the first two.
It seems to me that the problem starts with the imprecise formulation “Suppose we want to find the probability of the chain starting at state $0$ and being in state $2$ on the second step, while avoiding state $1$.”
Here you list three events, all in the same grammatical form, even though some of them are given conditions and others are events whose probability is to be calculated.
You don’t really want to find the probability of the chain starting at state $0$; you take it as given that it started at state $0$. We could write that off as a slight inaccuracy, but from your subsequent calculations it seems that this inaccuracy extends beyond the initial state and also affects your handling of $X_1$.
In your first approach, you’re treating $X_1$ as given; that is, you’re calculating the probability that, given that the chain was in state $0$ at time $0$ and not in state $1$ at time $1$, it will be in state $2$ at time $2$. This is wrong under a literal interpretation of “the probability of ... while avoiding state $1$”, but given that you apparently didn’t mean that literally for state $0$, perhaps you didn’t mean it literally for state $1$, either.
In your second approach, you’re treating $X_1$ as the literal interpretation would suggest, as an unknown random state.
So I think if you think about that sentence that specifies what you want to calculate, and precisely formulate it accordingly, you’ll realize that one of the calculations corresponds to that and the other doesn’t.
By the way, your modification of $P$ is unnecessary, since state $1$ anyway doesn’t transition to state $2$; you get the same result simply from $P^2_{02}$.


TITLE: How to find relations between the roots of a fourth degree polynomial which has only complex roots?
QUESTION [0 upvotes]: Let f(x) is a fourth degree polynomial such that $f(x) = x^4+x^3+x^2+x+1$. Let p be a root of f(x). Then which one of the following cannot be root of f(x):

$p^2$
$p^3$
$p^4$
$p^5$

I know f(x) has no real solutions and four complex solutions as given here
http://www.enotes.com/homework-help/x4-x3-x2-x-1-0-4-3-2-gt-exponents-could-anyone-205747
These solutions are very complicated and it is not possible to square or cube etc. each one to find which of the four options is correct. May someone help?

REPLY [1 votes]: Hint.
In this case, as noted in the comments, the solutions of the $4-$degree equation are simple, since: $x^5-1=(x-1)(x^4+x^3+x^2+x+1)$ so :
$
x^4+x^3+x^2+x+1=0
$
for $x=e^{i2k\pi/5}$ with $k=1,2,3,4$, that are the four roots of $x^5-1=0$ different from $x=1$.


text:
\begin{document}

\title[$*$-exponential of slice-regular functions]
{$*$-exponential of slice-regular functions}

\author[A. Altavilla]{A. Altavilla${}^{\dagger,\ddagger}$}\address{Altavilla Amedeo: Dipartimento Di Matematica, Universit\`a di Roma "Tor Vergata", Via Della Ricerca Scientifica 1, 00133, Roma, Italy}\email{altavilla@mat.uniroma2.it}

\author[C. de Fabritiis]{C. de Fabritiis${}^{\dagger}$}\address{Chiara de Fabritiis: Dipartimento di Ingegneria Industriale e Scienze
Matematiche, Universit\`a Politecnica delle Marche, Via Brecce Bianche, 60131,
Ancona, Italy}\email{ fabritiis@dipmat.univpm.it}
\thanks{${}^{\dagger}$GNSAGA of INdAM, ${}^{\ddagger}$FIRB 2012 {\sl Geometria differenziale e teoria geometrica delle funzioni}, SIR grant {\sl ``NEWHOLITE - New methods in holomorphic iteration''} n. RBSI14CFME and SIR grant {\sl AnHyC - Analytic aspects in complex and hypercomplex geometry} n. RBSI14DYEB.}

\date{\today }

\subjclass[2010]{Primary 30G35; secondary 30C15, 32A30, 47A60}
\keywords{Slice-regular functions, quaternionic exponential, $*$-product of slice-regular functions}

\begin{abstract}
According to~\cite{C-S-St-2} we define the $*$-exponential of a slice-regular function, which can be seen as a generalization of the complex exponential to quaternions.
Explicit formulas for $\exp_*(f)$ are provided, also in terms of suitable sine and cosine functions.
We completely classify under which conditions the $*$-exponential of a function is either slice-preserving or $\C_J$-preserving for some $J\in\SF$ and show that $\exp_*(f)$ is never-vanishing.
Sharp necessary and sufficient conditions are given in order that $\exp_*(f+g)=\exp_*(f)*\exp_*(g)$, finding an exceptional and unexpected case in which equality holds even if $f$ and $g$ do not commute.
We also discuss the existence of a square root of a slice-preserving regular function, characterizing slice-preserving functions (defined on the circularization of simply connected domains) which admit square roots.
Square roots of this kind of functions are used to provide a further formula for $\exp_{*}(f)$.
A number of examples is given throughout the paper.
\end{abstract}
\maketitle

\section{Introduction}

In classical complex analysis the exponential map has a tremendous role in the study of growth of holomorphic functions, differential equations and uniformization theorem. In this paper we investigate the behaviour of a quaternionic  analogous of the complex exponential map.
In particular, special features of the exponential in the complex case include the fact that it never vanishes, the expression of the exponential of a pure imaginary in terms of sine and cosine, and the formula which gives the exponential of a sum as the product of the exponentials of the two summands.
These properties will be the object of our study in the setting of slice-regularity.

We briefly introduce quaternions and quaternionic slice-regular functions.
Given an alternating triple $i,j,k$ with $i^2=j^2=k^2=-1$ and $k=ij=-ji$, we denote by $\HH$ the real algebra of quaternions
$$\HH=\{q=q_0+q_1i+q_2j+q_3k\ :\ q_0,q_1,q_2,q_3\in \R\}.$$ The conjugation on $\HH$ is given by $q^c=q_0-(q_1i+q_2j+q_3k)$ and
we sometimes write ${\rm Re} (q)=q_0=\frac12(q+q^c)$ and
$\vec q=q- {\rm Re} (q)$, so that $q=q_0+\vec q$; the quaternion $\vec q$ is often denoted by ${\rm Im}(q)$.

The following subsets of $\HH$ have special interest:
\begin{align*}
\SF&=\{q\in\HH\ : \ q^2=-1\}=\{q_1i+q_2j+q_3k\ : \ q_1^2+q_2^2+q_3^2=1\}\\
{\rm Im}(\HH)&=\{q\in\HH: {\rm Re} (q)=0\}=\bigcup_{I\in \SF} \, \R I;
\end{align*}
the first can be identified with the standard sphere of $\R^{3}\simeq {\rm Span}_{\R}(i,j,k)$ and the second with $\R^3\simeq {\rm Span}_{\R}(i,j,k)$. For any $I\in\SF$ we set  $\C_I=
{\rm Span}_{\R}(1,I)$.

Given any set $D\subset \C=\{\alpha+\imath \beta\,|\,\alpha,\beta\in\R\}$, we define its \textit{circularization} as
$$
\Omega_D=\{\alpha+\beta I\in\HH\, | \, \alpha+\imath \beta\in D, I\in\SF\}\subseteq \HH.
$$
Subsets of $\HH$ of this form will be called \textit{circular sets}. If $D$ is a singleton $\{q\}$ then we also denote the sphere $\Omega_{\{q\}}$ with real center ${\rm Re}(q)$ by $\SF_{q}$. Notice that, if $D$ is open in $\C$, then $\Omega_{D}$ is open in $\HH$. Here we assume that $\HH\simeq \R^{4}$ is equipped with the Euclidean topology. In order to simplify the notation we will often drop the subscript $D$ writing $\Omega$ in place of $\Omega_{D}$.

The following definitions identify the functions we will work with (see~\cite{G-P})

\begin{definiz}
Let $D\subset \C$ be invariant under conjugation.
A  function $F:D\to\HH\otimes_{\R} \C$ is a {\sl stem} function if $F(\overline z)=\overline{F(z)}$ where $\overline{p+\imath q}=p-\imath q$ for any $p+\imath q\in\HH\otimes_{\R} \C$.
A {\sl slice} function $f:\Omega_D\to \HH$ is a function induced by a stem function $F=F_1+\imath F_2:D\to\HH\otimes_{\R} \C$ in the following way:
$f(\alpha+\beta I)=F_1(\alpha+\imath \beta)+I F_2(\alpha+\imath \beta)$. Such a function will also be denoted by $f=\mathcal I(F)$.
\end{definiz}

Notice that a slice function $f$ is induced by a unique stem function $F = F_{1} + \imath F_{2}$, given by
$F_{1}(\alpha+\imath \beta) = \frac{1}{2}(f(\alpha+\beta I)+f(\alpha-\beta I))$ and $F_{2}(\alpha+\imath \beta) = -\frac{1}{2}I(f(\alpha+\beta I)-f(\alpha-\beta I))$ for
any $I \in \SF$.

\begin{definiz}
Let $\Omega\subset \HH$ be a circular domain, that is a circular connected open subset of $\HH$.
A {\sl slice} function $f=\mathcal I(F):\Omega\to \HH$ is slice {\sl regular} if $F$ is holomorphic with respect to the natural complex structures of $\C$ and $\HH\otimes_{\R} \C$. We denote by $\So(\Omega)$ the set of all slice-regular functions on $\Omega$.
\end{definiz}
Notice that $\So(\Omega)$ has a natural structure of right $\HH$-module.\\
In what follows we always assume $\Omega$ is a (non-empty) circular domain of $\HH$.
If $\Omega$ contains real points, a function is slice-regular if and only if it is regular in the sense of Cullen (see~\cite{G-S-St}). A useful result for slice-regular functions is the following

\begin{prop}[Representation formula] Let $f\in\So(\Omega)$ and let $\alpha+\beta J\in\Omega$. For all $I\in\SF$ we have
$$
f(\alpha+\beta J)=\frac{1-JI}2f(\alpha+\beta I)+\frac{1+JI}2f(\alpha-\beta I).
$$
\end{prop}
Therefore if
$f_I:\Omega_I=\Omega\cap \C_I\to \HH$  is a holomorphic function with respect to the left multiplication by $I$, then there exists a unique slice-regular function $g:\Omega\to\HH$ such that $f_I=g|_{\Omega_I}$. Such a function will be called the {\sl regular extension of $f_I$}, (see~\cite{G-S-St}, p 9).

In general the pointwise product of two slice-regular functions is no more slice. Nonetheless, this problem can be overcome by defining the following non-commutative product (see~\cite{C-G-S-St} also).
\begin{definiz}\label{prodotto-stella}
Let $f=\mathcal I(F)$ and $g=\mathcal I(G)$ be two slice functions on $\Omega$. We denote by $f*g$ their {\sl $*$-product} defined by $f*g=\mathcal I(FG)$ where $FG$ is the pointwise product with values in $\HH\otimes_{\R} \C$, {\sl i.e.}
$(p+\imath q)(p'+\imath q')=pp'-qq'+\imath (pq'+qp')$.
\end{definiz}

In $\So(\Omega)$ we can also define the {\sl conjugate function} of $f$ that is
$f^c=\mathcal I(F_1^c+\imath F_2^c)$ if $f=\mathcal I(F_1+\imath F_2)$ and the functions $F_1^c$ and $F_2^c$ are obtained from $F_1$ and $F_2$ by pointwise conjugation in $\HH$.
Given $f\in \So(\Omega)$ its {\sl symmetrized function} $f^s$ is given by $f^s=f^c*f=f*f^c$.

In~\cite{G-P} the $*$-product $f*g$ and the symmetrized function $f^{s}$ of $f$ are called slice product and normal function of $f$, and are denoted by $f\cdot g$ and $N(f)$, respectively.

It is known that if $\alpha+\beta I$ is such that $f(\alpha+\beta I)=0$, then any point in the set $\SF_{\alpha+\beta I}=\{\alpha+\beta J\,|\,  J\in\SF\} $ is a zero for $f^s$. In particular if $f^s$ is never-vanishing then also $f$ is. Moreover, the zero set of a slice-regular function $f\not\equiv0$ is closed with empty interior and, if $f^s\not\equiv0$, it is a union of isolated points and isolated spheres of the form  $\SF_{\alpha+\beta J}$ for suitable $\alpha,\beta\in\R$  (see~\cite{G-P, G-S}).

We now introduce two special classes of slice-regular functions.
A function $f=\mathcal I(F_1+\imath F_2)\in\So(\Omega)$ is {\sl slice-preserving} if both $F_1$ and $F_2$ are real-valued. The class of such functions will be denoted by  $\So_\R(\Omega)$
(they are called $\HH$-intrinsic in \cite{G-M-P} and quaternionic-intrinsic in \cite{C-S-St-2}).
We remark that for any $f\in\So(\Omega)$ its symmetrized function $f^s$ belongs to $\So_\R(\Omega)$.

Given $J\in\SF$, a function $f=\mathcal I(F_1+\imath F_2)\in\So(\Omega)$ is said to be
  {\sl $\C_J$-preserving} if both $F_1$ and $F_2$ are $\C_J$-valued;  the class of such functions will be denoted by $\So_J(\Omega)$ (see~\cite{G-M-P}).

For slice-preserving and $\C_J$-preserving functions, the $*$-product has special features:
\begin{itemize}
\item for any $f\in\So_\R(\Omega)$ and $g\in\So(\Omega)$ we have $f*g=g*f=fg$ (that is, the $*$-product $f*g$ coincides with the pointwise product $fg$);
\item choosen $J\in\SF$, for any $f,g\in\So_J(\Omega)$ we have $f*g=g*f$ (and $f*g$ coincides with the regular extension of the pointwise product of $f_J$ and $g_J$);
\item if $\rho_1,\rho_2\in \So_\R(\Omega)$ and $a_1,a_2\in\HH$ then $(\rho_1a_1)*(\rho_2a_2)=\rho_1\rho_2a_1a_2$.
\end{itemize}

We have now the main tools to define the $*$-exponential of a slice-regular function.
The first definition is the natural extension of the complex exponential to $\HH$ and indeed its restriction to any complex line $\C_I$, $I\in\SF$, coincides with the complex exponential.

The function $\exp:\HH\to\HH$ is given by
$\exp q=e^q=\sum_{n\in\N}\frac{q^n}{n!}$. Trivially $\exp\in\So_\R(\HH)$.

Given $f\in\So(\Omega)$ the composition $\exp\circ f$ is not always slice-regular. In~\cite{C-S-St-2}, Colombo, Sabadini and Struppa gave the following definition which coincides with $\exp\circ f$ if $f\in\So_\R(\Omega)$ (see also~\cite{R-W} and~\cite{V}, where several different regular compositions are introduced).
\begin{definiz} If $f\in\So(\Omega)$ the $*$-exponential of $f$ is defined as
$\exp_*(f)=\sum_{n\in\N}\frac{f^{*n}}{n!}$.
\end{definiz}

Let $f\in\So(\Omega)$ and let $n$ be a positive integer. Notice that, given any (non-empty) circular compact subset $K$ of $\Omega$, $\max_{q\in K} |f^{*n}(q)|\leq\max_{q\in K}|f^{n}(q)|$. This follows easily by induction on $n$ observing that $f^{*n}(q) = 0$ if $f(q) = 0$ and $|f^{*n}(q)| = |f(q)| \cdot |f^{*(n-1)}(f(q)^{-1}qf(q))|$
if $f(q)\neq 0$ (see~\cite[Theorem 3.4]{G-S-St}). Thanks to these estimates, the above series converges uniformly on compact subsets of $\Omega$ and hence $\exp_{*}(f)$ is a well defined function in $\So(\Omega)$.

The function $\exp_{*}(f)$ will be the main object of our study, which will be organized as follows.

Section 2 contains a new interpretation of the $*$-product given by Formula~(\ref{star-formula}). This result allows us to give a necessary and sufficient condition for the commutation of two functions with respect to this product.
We will exploit Formula~(\ref{star-formula}) extensively in Section 4, where it will be a new, useful tool to simplify calculations.

In Section 3 we investigate on the possibility of finding a square root of a slice-preserving regular function in the space $\So_\R(\Omega)$, when each connected component of $\Omega_I=\C_I\cap \Omega$ is simply connected, obtaining a complete answer to this question. Notice that, since $\Omega$ is connected by assumption, the set $\Omega_I$ has at most two connected components.

A part of the necessary and sufficient condition in order that a function in $\So_\R(\Omega)\setminus\{0\}$ has a square root in $\So_\R(\Omega)$ is given in terms of the spherical and isolated multiplicities of its zeroes (see~\cite{G-S-St}, Paragraph 3.6, and Definition 3.37 in particular).
The result is the following.
\begin{prop}
Suppose each connected component of $\Omega_I$ is simply connected for some (and hence any) $I\in\SF$.
Given $h\in \So_\R(\Omega)\setminus\{0\}$ there exists $f\in \So_\R(\Omega)$ such that $f^2=h$ iff
\begin{enumerate}[(i)]
\item the zero set of $h$ consists of real zeroes of even isolated multiplicity and isolated spheres $\SF_{q}$ with spherical multiplicity multiple of $4$;
\item if $\Omega\cap \R\neq\emptyset$ then $h(\Omega\cap \R)\subseteq [0,+\infty)$.
\end{enumerate}
\end{prop}

In particular we apply this result when $h=g^s$ for $g\in\So(\Omega)$.

Last section is devoted to the study of the $*$-exponential of a slice-regular function.
We first give a sufficient condition, namely commutativity of the $*$-product of $f$ and $g$, for the equality
$\exp_*(f+g)=\exp_*(f)*\exp_*(g)$ to hold. Given $f=f_0+f_v$ (see Definition~\ref{vector} and subsequent remark for the notation), we then find an explicit expression for $\exp_*(f)$ in terms of $f_0$ and $f_v$.
In Corollaries~\ref{sine-cosine} and~\ref{true-sine}
this allows us to rewrite $\exp_*(f)$ in terms of suitable sine and cosine functions.
Moreover we completely classify under which conditions the $*$-exponential of a function is either in $\So_\R(\Omega)$ or in $\So_J(\Omega)$ for some $J\in\SF$.

In Proposition~\ref{real-part} we compute the symmetrized function of the $*$-exponential of $f$ in terms of the real part of $f$, proving as a consequence that $\exp_*(f)$ is never-vanishing.
Finally, in Theorem~\ref{super-equality} we give necessary and sufficient conditions on $f$ and $g$ in order that $$\exp_*(f+g)=\exp_*(f)*\exp_*(g).$$
Quite surprisingly we are able to find a large bunch of slice regular functions which do not commute under the $*$-product and for which nonetheless the above equality holds.
This last result is followed by four examples which illustrate the sharpness of the required conditions.

The authors warmly thank the anonymous referee for many suggestions which helped to improved the quality of the paper.

\section{regular product: a further interpretation}

The following result, which is due to Colombo, Gonzales-Cervantes and Sabadini (see~\cite{C-GC-S}, Proposition~3.12), gives a way to decompose a given slice-regular function by means of four slice-preserving regular functions (see also \cite{G-M-P}, Lemma 6.11).

\begin{prop}\label{GMP}
Let $\{1,I,J,K\}$ be a vector basis of $\HH$. Then the map
$$
\left(\So_\R(\Omega)\right)^4\ni (f_0,f_1,f_2,f_3)\mapsto f_0+f_1I+f_2J+f_3K\in \So(\Omega)
$$
is bijective. In particular it follows that given any $f\in \So(\Omega)$ there exist and are unique $f_0,f_1,f_2,f_3\in\So_\R(\Omega)$ such that
$$
f=f_0+f_1I+f_2J+f_3K.
$$
Moreover if $I\in\SF$ then $f\in \So_I(\Omega)$ iff $f_2\equiv f_3\equiv 0$.
\end{prop}

\begin{remark}
If $I,J,K\in\SF$, then $(f_0,f_1,f_2,f_3)$ is the $4$-tuple associated to $f$  if and only if $(f_0,-f_1,-f_2,-f_3)$ is the $4$-tuple associated to $f^c$; in particular $f_0=\frac{f+f^c}2$ does not depend on the chosen
vector basis $\{1,I,J,K\}$ of $\HH$, provided $I,J,K\in\SF$.

\end{remark}

\begin{definiz}\label{vector}
Given $f\in\So(\Omega)$, we define the slice-regular function $f_v$  on $\Omega$ by
$f_v=\frac{f-f^c}2$.
\end{definiz}

\begin{remark}
Trivially $f_0+f_v=f$ and $f_0-f_v=f^c$.
According to the notation of Proposition~\ref{GMP} we have $f_v=f_1I+f_2J+f_3K$.
\end{remark}

The reason we choose the underscore $v$ for the above function is that, in a certain sense, it represents the ``vector'' part of the quaternionic-valued function $f$, according to the splitting $q=q_0+q_1i+q_2j+q_3k=q_0+\vec{q}$.

The above result and definition allow us to describe the $*$-product of two slice-regular functions in terms of intrinsic operators.

Let now $I,J,K$ be an orthonormal basis of ${\rm Im}(\HH)$ with $IJ=K$.

\begin{definiz}
Given $f,g\in\So(\Omega)$, we denote by $f\pv g$
and $\langle f, g\rangle_*$ the slice-regular functions given by
$$
(f\pv  g)(q)=\frac{(f*g)(q)-(g*f)(q)}2,
\qquad
\langle f, g\rangle_*(q)=(f*g^c)_0(q)
$$
for any $q\in\Omega$.
\end{definiz}

\begin{remark}
In terms of the notation of Proposition~\ref{GMP} we can rewrite the above intrinsic expressions in the following form:
\begin{align*}
f\pv  g &=(f_2g_3-f_3g_2)I+(f_3g_1-f_1g_3)J+(f_1g_2-f_2g_1)K \\
\langle f, g\rangle_*&=f_0g_0+f_1g_1+f_2g_2+f_3g_3.
\end{align*}
which in particular show that
\begin{align*}
f\pv  g &=f_v\pv  g_v=-g_v\pv  f_v\\
\langle f, g\rangle_*&=\langle g, f\rangle_*=\langle f^c, g^c\rangle_*= f_0g_0-(f_v*g_v)_0.
\end{align*}
\end{remark}

These operators can be used to write the $*$-product of two slice-functions in a more explicit form which recalls the formula for the product of two quaternions in the scalar-vector form. This different expression of the $*$-product will turn out to be a useful tool in the study of slice regular functions, simplifying and speeding formerly long and complicated computations.

\begin{prop}\label{star-product}
The $*$ product of $f=f_0+f_v$ and $g=g_0+g_v$
is given by
\begin{equation}\label{star-formula}
f*g=f_0g_0-\langle f_v, g_v\rangle_*+f_0g_v+g_0f_v+f_v\pv  g_v
\end{equation}
\end{prop}
\begin{proof}
Choosing the standard vector basis $\{1,i,j,k\}$ of $\HH$ on $\R$ and writing $f=f_0+f_1i+f_2j+f_3k$ and $g=g_0+g_1i+g_2j+g_3k$ we have
\begin{align*}
f*g&=(f_0+f_1i+f_2j+f_3k)*(g_0+g_1i+g_2j+g_3k)\\
&=
f_0g_0-f_1g_1-f_2g_2-f_3g_3+f_0(g_1i+g_2j+g_3k)+g_0(f_1i+f_2j+f_3k)\\
&+(f_2g_3-f_3g_2)i+(f_3g_1-f_1g_3)j+(f_1g_2-f_2g_1)k\\
&=f_0g_0-\langle f_v, g_v\rangle_*+f_0g_v+g_0f_v+f_v\pv  g_v,
\end{align*}
where the second equality is a consequence of the definition of the $*$-product and of the fact that $f_0,\ldots,f_3,g_0,\ldots,g_3$ belong to $\So_\R(\Omega)$.
\end{proof}

\begin{remark}\label{symmetrized}
Given $f\in\So(\Omega)$, we can recognize in $\langle f, f\rangle_*
$ its symmetrized function $f^s$.
Written $f$ as $f_0+f_1i+f_2j+f_3k$,
since $f^c=f_0-f_v$
we have
$$\langle f, f\rangle_*=f_0^2+f_1^2+f_2^2+f_3^2=f_0g_0-\langle f_v, -f_v\rangle_*+f_0(-f_v)+f_0f_v+f_v\pv  (-f_v)=f*f^c.$$
\end{remark}

\begin{remark}\label{null-symmetrized}
Notice that if $\Omega$ contains real points, then
 $f_v^s\equiv0$ if and only if $f_v\equiv 0$. If $\Omega$ does not contain any real point, there exist examples in which $f_v^s\equiv 0$ and $f_v\not\equiv 0$.
Indeed if $\Omega$ contains real points and
 $f_v^s\equiv0$ then $f_1^2+f_2^2+f_3^2\equiv 0$. On real points  $f_1$, $f_2$ and $f_3$ take real values and therefore
 $f_1^2+f_2^2+f_3^2\equiv 0$ implies $f_1=f_2=f_3=0$ on the intersection between $\Omega$ and $\R$, which entails $f_1\equiv f_2\equiv f_3\equiv0$ by the identity principle (see~\cite{G-S-St}, Theorem 1.12). If the domain $\Omega$ does not contain any real point there exist $f\in\So(\Omega)$ such that $f_v\not\equiv0$ and $f_v^s\equiv0$ (see~\cite{AA}, Example 2).
\end{remark}

In a certain sense, Formula~\ref{star-formula} allows us to untangle the $*$-product on $\So(\Omega)$, confining the skewness to $\HH$ and thus simplifying its computaion with respect to Definition~\ref{prodotto-stella} or Theorem~3.4 in \cite{G-S-St}.
Next result characterizes slice-regular functions whose $\pv$-product vanishes identically on $\Omega$ and shows the effectiveness of Proposition~\ref{star-product}. Notice that by definition the vanishing of the $\pv$-product is equivalent to the fact that $f$ and $g$ commute with respect to the $*$-product.

\begin{prop}\label{wedge-product}
Given $f,g\in\So(\Omega)$, the function $f\pv  g$ vanishes identically if and only if $f_v$ and $g_v$ are linearly dependent over $\So_\R(\Omega)$, namely there exist $\alpha,\beta\in\So_\R(\Omega)$ such that $\alpha f+\beta g\equiv 0$ and either $\alpha\not\equiv0$ or $\beta\not\equiv0$.
\end{prop}
\begin{proof}
First suppose $f\pv  g\equiv0$.
If $f_v\equiv 0$ the assertion is trivial by setting $1\cdot f_v+0\cdot g_v\equiv 0$. Otherwise,
without loss of generality, we can suppose $f_1\not\equiv 0$.
The identity
$f\pv  g\equiv0$ is equivalent to
$(f_2g_3-f_3g_2)i+(f_3g_1-f_1g_3)j+(f_1g_2-f_2g_1)k\equiv 0$.
Thanks to Proposition~\ref{GMP}, this gives
$$
\begin{cases}
f_2g_3-f_3g_2\equiv 0, \\ f_3g_1-f_1g_3\equiv 0, \\ f_1g_2-f_2g_1\equiv 0.
\end{cases}
$$
Last two equations of the previous system give $f_1g_v-g_1 f_v\equiv 0$; as $f_1\not\equiv 0$, the functions $f_v$ and $g_v$ are  linearly dependent over $\So_\R(\Omega)$.

Vice versa, if $f_v$ and $g_v$ are  linearly dependent over $\So_\R(\Omega)$, we can
suppose, up to a rearrangment, that there exist $\alpha,\beta\in\So_{\R}(\Omega)$,
with $\alpha\not\equiv 0$, such that $\alpha f_{v}+\beta g_{v}\equiv 0$.
By taking the $\pv$-product with $g_{v}$ we obtain
$$
0\equiv \alpha f_{v}\pv g_{v}+\beta g_{v}\pv g_{v}= \alpha f_{v}\pv g_{v}.
$$
Since $\alpha\in\So_{\R}(\Omega)$ is not identically zero, then its
zero set is a disjoint union of isolated real zeros and isolated spheres with
real center. Therefore the equality $\alpha f_{v}\pv g_{v}\equiv 0$ implies that
$f_{v}\pv g_{v}\equiv 0$ since the product of $\alpha$ and $f_v\pv g_v$ is the pointwise product.
\end{proof}

\section{Square roots of slice-preserving regular functions}

Since the symmetrized of any slice-regular function always belongs to $\So_\R(\Omega)$ and the symmetrized of a
slice-preserving function coincides with its square, it is natural to ask when an element in $\So_\R(\Omega)$ has a square root in $\So_\R(\Omega)$ and, in particular, when the symmetrized function of a slice-regular function has a square root in $\So_\R(\Omega)$.  The following results completely solve the question, giving necessary and sufficient conditions on a non-zero slice-preserving function in order to be the square of a slice-preserving function; in particular they apply when we look for the square root of the symmetrized of a given function.

As it will be seen in Corollary~\ref{true-sine}, the results contained in this section can be applied to obtain a more explicit form for the $*$-exponential of a slice-regular function.

Throught the whole section we assume that  each connected component of
$\Omega_I=\Omega\cap \C_I$ is simply connected for some $I\in\SF$ (and so for all $I\in\SF$, being $\Omega$ a circular domain).

\begin{prop}\label{real-square-root}
Given $h\in \So_\R(\Omega)\setminus\{0\}$ there exists $f\in \So_\R(\Omega)$ such that $f^2=h$ if and only if
\begin{enumerate}[(i)]
\item the zero set of $h$ consists of  real zeroes of even isolated  multiplicity and isolated spheres $\SF_{q}$ with spherical multiplicity multiple of $4$;
\item if $\Omega\cap \R\neq\emptyset$ then $h(\Omega\cap \R)\subseteq [0,+\infty)$.
\end{enumerate}
\end{prop}

\begin{proof}
The necessity of the first condition is due to the fact that functions belonging to $\So_\R(\Omega)\setminus\{0\}$ only have real  isolated zeroes and isolated spherical zeroes with real center (see \cite{G-P}). Squaring a slice-preserving function therefore produces real  isolated zeroes with even multiplicity and  isolated spherical zeroes with real center with multiplicity multiple of $4$, as the spherical multiplicity is always an even number~(see~\cite{G-S-St}, Definition 3.37).
The necessity of the second condition is straightforward.

We first perform the proof of the sufficiency of conditions (i) and (ii) in the case when $\Omega$ intersects the real axis.
As $h\not\equiv 0$, we can choose a point $q_0\in\Omega   \cap \R$ with $h(q_0)\neq 0$. Now
fix $I\in \SF$ and choose a family $\{\Omega_n\}$ of nested domains with compact closure $\overline{\Omega_{n}}$ in $\Omega_I=\Omega\cap \C_I$ such that each $\Omega_n$ is simply connected, symmetric with respect to the real axis and contains $q_0$ (such a family certainly exists thanks to Riemann mapping theorem for domains symmetric with respect to the real axis, see~\cite{Gal-S} Theorem 3.3).

The restriction $h_{I}$ of $h$ to $\Omega_I$ is a holomorphic function from $\Omega_I$ to $\C_I$ which does not vanish at $q_0$ and, by condition (i), has zeroes of even multiplicity both at real points and at conjugate points outside $\R$. Now denote by $Z_n$ the intersection of the zero set  of $h_I$ with $\Omega_n$: since $h_I(q_0)\neq0$ then $Z_n$ is finite and we can find a polynomial $P_n(z)$ with real coefficients and a holomorphic function $g_n:\Omega_{I}\to \C_I$ such that
$g_n$ does not vanish on $\overline{\Omega_n}$ and $h_I(z)= P_n^2(z)g_n(z)$ for any $z\in\Omega_I$.
As $g_n$ does not vanish on $\Omega_n$, we can find a holomorphic square root $\gamma_n$ of $g_n$ on $\Omega_n$ such that $P_n(q_0)\gamma_n(q_0)$ is real and positive. Since $P_n(z)$ has real coefficients, condition (ii) implies that $\gamma_n(\Omega_n\cap \R)\subset \R$.
Then the function $\varphi_n=P_n\cdot\gamma_n$ defined on $\Omega_n$ has the following properties:
\begin{itemize}
\item it is holomorphic on $\Omega_n$,
\item it is a square root of $h_I$ on $\Omega_n$, that is $\varphi_n^2(z)=h_I(z)$ for any $z\in\Omega_n$,
\item it maps $\Omega_n\cap \R$  into $\R$,
\item it takes positive value at $q_0$.
\end{itemize}
It is easily seen that the above properties entail the uniqueness of $\varphi_n$ and therefore $\varphi_n\equiv \varphi_{n+1}$ on $\Omega_n$; hence setting $\varphi(z)=\varphi_n(z)$ for any $z\in\Omega_n$ defines a holomorphic function on $\Omega_I$ which is a square root of $h_I$. Now denote by $f$ the regular extension of $\varphi$ to $\Omega$; it is easily seen that such a function is a square root of $h$. Indeed $f\in \So_\R(\Omega)$ because $f$ coincides with $\varphi$ on $\Omega\cap \R$, moreover $f^2=h$ on $\Omega\cap \R$ which ensures $f^2\equiv h$ on $\Omega$.

Now suppose that $\Omega   \cap \R=\emptyset$.
Fix $I\in \SF$ and consider $\Omega_I^+=\Omega\cap \{z\in\C_I\,|\, {\rm Im}(z)>0\}$. As $\Omega_I^{+}$ is simply connected and the set of zeroes of $h_I$ in $\Omega_I$ consists of conjugate points of even multiplicity,

by reasoning as above, we can find a square root $\varphi^+$ of $h_I$ on $\Omega_I^+$. As the function $h \in\So_\R(\Omega)$, then $h_I$ coincides with its Schwarz reflection on $\Omega_I^-=\Omega_I\setminus\Omega_I^+$, so we can define a holomorphic map $\varphi$ on $\Omega_I$ by extending $\varphi^+$ by Schwarz reflection on $\Omega_I^-$. In this way we obtain a holomorphic square root $\varphi$ of $h_I$ on $\Omega_I$; now denote by $f$ the regular extension of $\varphi$ to $\Omega$; as above it is easily seen that $f\in\So_\R(\Omega)$ is a square root of $h$.
\end{proof}

The above result can be applied to infer the existence of the square root of the symmetrized of a regular function with suitable zeroes. As previously stated, we assume that each connected component of $\Omega_I$ is simply connected.

\begin{cor}\label{square-root}
Given $g\in \So(\Omega)$ there exists $f\in \So_\R(\Omega)$ such that $f^2=g^s$ if and only if
the zero set of $g$ does not contain non real zeroes of odd isolated multiplicity.
\end{cor}

\begin{proof}
If $g^s\equiv 0$, we can take $f\equiv 0$. Then we are left to study the case when $g^s\not\equiv 0$.
In \cite{G-P} the zero set of such $g$'s is explicitely described as the union of isolated points and isolated spheres $\SF_q$.
As $g^s$ maps $\Omega\cap \R$ in $ [0,+\infty)$, thanks to Proposition~\ref{real-square-root} it is enough to check that the zeroes of the symmetrized function $g^s$ are only real zeroes of even isolated multiplicity and spherical zeroes of multiplicity multiple of $4$. This request exactly means that the zero set of $g$ cannot contain non real zeroes of odd isolated multiplicity.
\end{proof}

Next example gives an explicit application of the above corollary.

\begin{example}
Set $g(q)=q+i$. Then $g^s(q)=q^2+1$ has a spherical zero given by $\SF$ with multiplicity $2$ and so it is not the square of any slice-preserving  regular function.
\end{example}

\section{The $*$-exponential of a slice-regular function}

We now enter in the discussion of the $*$-exponential. In~\cite{C-S-St-2}, Colombo, Sabadini e Struppa introduced the following definitions, motivated by the natural challenge of functional calculus in the non-commutative setting, see also \cite{C-S-St-1, G-R}.

\begin{definiz}
Given $f\in \So(\Omega)$ we set
$$
\exp_*(f)=\sum_{n\in\N}\frac{f^{*n}}{n!}; \qquad
\cos_*(f)=\sum_{n\in\N}\frac{(-1)^nf^{*(2n)}}{(2n)!}; \qquad
\sin_*(f)=\sum_{n\in\N}\frac{(-1)^nf^{*(2n+1)}}{(2n+1)!}.
$$
\end{definiz}
As we have just seen, natural estimates on circular compact subsets
of $\Omega$ show that the series converge uniformly and therefore $\exp_*(f)$, $\cos_*(f)$, $\sin_*(f)$ are well defined and belong to $\So(\Omega)$.

In some special cases, $\exp_*$, $\cos_*$, $\sin_*$ take the usual form of the complex case and we sometimes denote them also dropping the underscore $*$.

\begin{remark}
If $f\in \So_\R(\Omega)$ then $\exp_*(f)$ can be written as $\sum_{n\in\N}\frac{f^{n}}{n!}$ and belongs to $ \So_\R(\Omega)$; the same holds for $\cos_*(f)$ and $\sin_*(f)$. If there exists $J\in\SF$ such that $f\in \So_J(\Omega)$ then $\exp_*(f)$ can be obtained as the regular extension to $\Omega$ of the complex function $e^{f_J}$, where $f_J$ denotes the restriction of $f$ to $\Omega_J=\Omega\cap \C_J$; in this case $\exp_*(f)$ belongs to $ \So_J(\Omega)$. Again, the same holds for $\cos_*(f)$ and $\sin_*(f)$, too.
\end{remark}

In the complex case, one of the most peculiar features of the exponental is its behaviour with respect to the sum; in the quaternionic case this happens under special conditions. The most natural hypothesis is the commutation of $f$ and $g$ which, due to Proposition~\ref{wedge-product}, is equivalent to the fact that $f_v$ and $g_v$ are linearly dependent over $\So_\R(\Omega)$. At the end of the paper we will prove a more refined result, see Theorem~\ref{super-equality}, containing an unexpected couple of conditions on the functions $f$ and $g$ which are equivalent to the equality $\exp_*(f+g)=\exp_*(f)*\exp_*(g)$.

\begin{prop}\label{lin-dep}
If $f_v$ and $g_v$ are linearly dependent on
 $\So_\R(\Omega)$, then
$$
\exp_*(f+g)=\exp_*(f)*\exp_*(g).
$$
In particular the above equality holds if either $f$ or $g$ belong to $\So_\R(\Omega)$.
 \end{prop}
\begin{proof}
It holds:
\begin{align*}
\exp_*(f+g)&\!=\!\sum_{n\in\N}\frac{(f+g)^{*n}}{n!}\!=\!
\sum_{n\in\N}\frac{1}{n!}\sum_{m\leq n} \!\binom{n}{m} f^{*m}* g^{*(n-m)}\!=\!
\sum_{n\in\N}\sum_{m\leq n}\!\frac{1}{m!(n-m)!} f^{*m} *g^{*(n-m)}\\
&=\sum_{m\in\N}\sum_{n\geq m} \frac{f^{*m}}{m!}*\frac{g^{*(n-m)}}{(n-m)!}
=\sum_{m\in\N} \frac{f^{*m}}{m!}*\sum_{n\geq m}\frac{g^{*(n-m)}}{(n-m)!}=
\left(\sum_{m\in\N} \frac{f^{*m}}{m!}\right)*\left(\sum_{\nu\in\N}\frac{g^{*\nu}}{\nu!}\right)\\
&=\exp_*(f)*\exp_*(g),
\end{align*}
where in the second equality we took into account the fact that $f$ and $g$ commute because of Proposition~\ref{wedge-product}.
\end{proof}

As an immediate consequence we obtain the following
\begin{cor}\label{exp-reale}
Let $f=f_0+f_v\in\So(\Omega)$, then
$$
\exp_*(f)=\exp_*(f_0)*\exp_*(f_v)=\exp(f_0)\exp_*(f_v).
$$
In particular if $f_v(q_0)=0$ then $\exp_*(f)(q_0)=\exp(f_0)(q_0)$.
\end{cor}

The above corollary allows us to interpretate the $*$-exponential in a cosine-sine form.

\begin{prop}\label{star-exponential}
Let $f=f_0+f_v\in\So(\Omega)$, then
\begin{equation}\label{cosine-sine}
\exp_*(f)=\exp_*(f_0)\left(\sum_{m\in\N} \frac{(-1)^m (f_v^s)^m}{(2m)!}+\sum_{m\in\N} \frac{(-1)^m (f_v^s)^m}{(2m+1)!} f_v\right).
\end{equation}
\end{prop}
\begin{proof}
Thanks to Corollary~\ref{exp-reale} we can perform the computation in the case when $f_0\equiv 0$.

\noindent
Notice that Proposition~\ref{star-product} and Remark~\ref{symmetrized} imply
$(f_v)^{*2}=f_v*f_v=-\langle f_v,f_v\rangle_*=-f_v^s$.
Now
\begin{align*}
\exp_*(f_v)&=\sum_{n\in\N}\frac{f_v^{*n}}{n!}=
\sum_{n\ \rm even}\frac{f_v^{*n}}{n!}+\sum_{n\ \rm odd}\frac{f_v^{*n}}{n!}=
\sum_{m\in\N}\frac{f_v^{*(2m)}}{(2m)!}+\sum_{m\in\N}\frac{f_v^{*(2m+1)}}{(2m+1)!}\\
&=\sum_{m\in\N}\frac{(f_v^{*2})^{*m}}{(2m)!}+\sum_{m\in\N}\frac{(f_v^{*2})^{*m}}{(2m+1)!}*f_v=
\sum_{m\in\N}\frac{(-f_v^{s})^{*m}}{(2m)!}+\sum_{m\in\N}\frac{(-f_v^{s})^{*m}}{(2m+1)!}*f_v\\
&=\sum_{m\in\N}\frac{(-1)^m(f_v^{s})^{m}}{(2m)!}+\sum_{m\in\N}\frac{(-1)^m(f_v^{s})^{m}}{(2m+1)!}f_v
\end{align*}
because $f_v^{s}$ is slice-preserving and the $*$-product becomes the pointwise product.
\end{proof}

Formula~\ref{star-exponential} allows us to write $\exp_*(f)$ as a local cosine-sine analogous of the complex case.

\begin{cor}\label{sine-cosine}
Let $f=f_0+f_v\in\So(\Omega)$. If
\begin{enumerate}
\item $f_v^s(q_0)=0$, then
 $$\exp_*(f)(q_0)=\exp_*(f_0)(q_0)\left(1+ f_v(q_0)\right); $$
\item $f_v^s(q_0)$ is a positive real, denote by $x_0$ one of the square roots of $f_v^s(q_0)$, then
 $$\exp_*(f)(q_0)=\exp_*(f_0)(q_0)\left(\cos x_0+\frac{\sin x_0}{x_0} f_v(q_0)\right); $$
\item $f_v^s(q_0)$ is a negative real, denote by $x_0$ one of the square roots of $-f_v^s(q_0)$, then
 $$\exp_*(f)(q_0)=\exp_*(f_0)(q_0)\left(\cosh x_0+\frac{\sinh x_0}{x_0} f_v(q_0)\right);$$
\item $f_v^s(q_0)=\alpha_0+\beta_0 J$  with $\beta_0\neq 0$ denote by $a_0+b_0 J$ one of the square roots of  $f_v^s(q_0)$, then
$$\exp_*(f)(q_0)=\exp_*(f_0)(q_0)\left(\cos (a_0+b_0 J)+\frac{\sin (a_0+b_0 J)}{a_0+b_0 J} f_v(q_0)\right) .$$
\end{enumerate}
\end{cor}
\begin{proof}
The proof is a trivial application of Formula~\eqref{cosine-sine} since in (1) the two power series sum both up to $1$; in
$(2)$ we have $f_v^s(q_0)=x_0^2$; in $(3)$ we have $f_v^s(q_0)=-x_0^2$ and in $(4)$ we have $f_v^s(q_0)=(a_0+b_0 J)^2$.
Notice that the expression $\frac{\sin(a_0+b_0 J)}{a_0+b_0 J}f_v(q_0)$  in $(4)$ is well defined because all factors lie in the same $\mathbb{C}_J$.
\end{proof}

If the zero set of $f_v$ does not contain non real isolated zeroes of odd multiplicity, the above result can be made more precise.  Indeed in this case Corollary~\ref{square-root} ensures we can find a regular function in $\So_\R(\Omega)$ whose square is $f_v^{s}$ and which therefore gives a global determination of the square root of $f_v^{s}$; in order to stress the analogy with the complex case we denote it by $\sqrt{f_v^s}$.

\begin{cor}\label{true-sine}
Let $f=f_0+f_v\in\So(\Omega)$. If  each connected component of $\Omega_I$ is simply connected,
$f_v^s$ is not identically zero and the zero set of $f_v$ does not contain non real zeroes of odd isolated multiplicity then
\begin{equation}\label{true-sine-eq}
\exp_*(f)=\exp_*(f_0)
\left(\cos_*\left(\sqrt{f_v^s}\right)+\frac{\sin_*\left(\sqrt{f_v^s}\right)}{\sqrt{f_v^s}} f_v\right),
\end{equation}
where $f_{v}^{s}$ does not vanish.
\end{cor}
\begin{proof}
As
$$\sum_{m\in\N}\frac{(-1)^m(f_v^{s})^{m}}{(2m)!}=\sum_{m\in\N}\frac{(-1)^m(\sqrt{f_v^s})^{2m}}{(2m)!}=
\cos_*\left(\sqrt{f_v^s}\right)$$ and $$\sum_{m\in\N}\frac{(-1)^m(f_v^{s})^{m}}{(2m+1)!}=
\sum_{m\in\N}\frac{(-1)^m(\sqrt{f_v^s})^{2m}}{(2m+1)!}=\frac1{\sqrt{f_v^s}}\sum_{m\in\N}\frac{(-1)^m(\sqrt{f_v^s})^{2m+1}}{(2m+1)!}=\frac{\sin_*\left(\sqrt{f_v^s}\right)}{\sqrt{f_v^s}},$$
the assertion follows from~\eqref{cosine-sine}.
\end{proof}

\begin{remark}
In the hypothesis of Corollary~\ref{true-sine}, at points where $f_v^s$ does not vanish, we can write Formula~\eqref{true-sine-eq} in the form
\begin{equation*}
\exp_*(f)=\exp_*(f_0)
\left(\cos_*\left(\sqrt{f_v^s}\right)+\sin_*\left(\sqrt{f_v^s}\right)\frac{f_v}{\sqrt{f_v^s}}\right).
\end{equation*}
We underline that $\left(\frac{f_v}{\sqrt{f_v^s}}\right)^s\equiv 1$ outside the zero set of $f_v^s$ and therefore the function $\frac{f_v}{\sqrt{f_v^s}}$ imitates
the behaviour of the imaginary unit in $\C$.
\end{remark}

As a further consequence of Proposition~\ref{star-exponential} we can describe all functions whose $*$-exponential is either slice-preserving or $\C_J$-preserving for a suitable $J\in\SF$.
To simplify computations we set
$$
\mu(f)=\sum_{m\in\N} \frac{(-1)^m (f_v^s)^m}{(2m)!},\qquad
\nu(f)=\sum_{m\in\N} \frac{(-1)^m (f_v^s)^m}{(2m+1)!}
$$
which are slice-preserving regular functions. Then~\eqref{cosine-sine}
 can be reformulated as
\begin{equation}\label{mu-nu}
\exp_*(f)=\exp_*(f_0)\left(\mu(f)+\nu(f) f_v\right).
\end{equation}

Notice that, thanks to Corollary~\ref{sine-cosine}, the equality $\nu(f)(q_0)=0$ holds if and only if $f_v^s(q_0)$ belongs to $\mathcal{Z}=\{n^{2}\pi^{2}\,|\,n\in\Z\setminus\{0\}\}$, that is there exists $n\in\Z\setminus\{0\}$ such that $f_v^s(q_0)=n^2\pi^2$. Notice that at such points
$\mu(f)(q_0)$ is equal to $(-1)^n$.

Formula~\eqref{mu-nu} enables us to characterize when the $*$-exponential of a regular function preserves one or all slices. Next example shows that things are not as simple as one could imagine.

\begin{example}
If $f(q)=\pi\cos(q)i+\pi\sin(q)j$, then $f_1(q)=\pi\cos(q)$,  $f_2(q)=\pi\sin(q)$ and therefore $f_v^s(q)=f_1^2(q)+f_2^2(q)=\pi^2\cos^2(q)+\pi^2\sin^2(q)=\pi^2$ for all $q\in\HH$. By Corollary~\ref{true-sine}
this implies that $\exp_*(f)=\cos(\pi)=-1$. It is easy to check that the function $f$ does not preserve any slice but its $*$-exponential trivially preserves all slices.
\end{example}

\begin{prop}
Let $f\in\So(\Omega)$. Then
\begin{enumerate}[(i)]
\item $\exp_*(f)\in\So_\R(\Omega)$ if and only if either $f_v\equiv 0$ or
$f_v^s$ is a constant which belongs to $\mathcal{Z}$,
\item  there exists  $J\in\SF$ such that $\exp_*(f)\in\So_J(\Omega)\setminus \So_\R(\Omega)$  if and only if $f_v\in \So_J(\Omega)$ and it is not a constant which belongs to $\Z\pi J$.
\end{enumerate}
\end{prop}
\begin{proof} $(i)$
If $f_v\equiv 0$ then $f\in \So_\R(\Omega)$ and therefore $\exp_*(f) \in \So_\R(\Omega)$.
If $f_v\not\equiv 0$ and $f_v^s$ is a constant which belongs to $\mathcal{Z}$, let us denote by $n\neq 0$ an integer such that $f_v^s\equiv n^2\pi^2$;
then Corollary~\ref{true-sine}
gives $\exp_*(f)=\exp_*(f_0)
\left(\cos_*\left(n \pi\right)+\frac{\sin_*\left(n\pi\right)}{n\pi} f_v\right)
=\exp_*(f_0)\cos(n\pi)=(-1)^n\exp_*(f_0)\in\So_\R(\Omega)$.

Now if $\exp_*(f)\in\So_\R(\Omega)$, then the function $\exp_*(f_0)\left(\mu(f)+\nu(f) f_v\right)$ does; since $\exp_*(f_0)$  is never-vanishing, this entails that $\mu(f)+\nu(f) f_v$ belongs to $\So_\R(\Omega)$.
This fact implies that $\nu(f) f_v$ is identically zero. If $\nu(f)\equiv 0$, then
$\sum_{m\in\N} \frac{(-1)^m (f_v^s)^m}{(2m+1)!}\equiv 0$ and therefore $f_v^s$ belongs to one of the zeroes of the power series which are given by $\mathcal{Z}=\{n^2\pi^2\, |\, n\in\Z\setminus\{0\}\}$ and hence $f_v^s$ is a constant which belongs to $\mathcal{Z}$. If $\nu(f)\not\equiv0$, we have that
$f_v=0$  outside the zero set of $\nu(f)$ which is closed and has empty interior, therefore by continuity $f_v\equiv 0$, that is $f\in\So_\R(\Omega)$.

$(ii)$ If $f_v\in \So_J(\Omega)$ and it is not a constant which belongs to $\Z\pi J$, then we can write $f=f_0+f_1J$ where $f_1\not\equiv n\pi$ for some $n\in\Z$. As $f_v^s=f_1^2$, by Formula~\eqref{cosine-sine} we obtain that
$\exp_*(f)=\exp_*(f_0)\left(\cos_*(f_1)+\sin_*(f_1)J\right)$. As  $\exp_*(f_0)$ is never-vanishing and $f_1$ is not a constant which belongs to $\Z\pi$, then $\exp_*(f)$ belongs to $\So_J(\Omega)\setminus \So_\R(\Omega)$.

Now, if $\exp_*(f)\in\So_J(\Omega)\setminus \So_\R(\Omega)$, then the function $\exp_*(f_0)\left(\mu(f)+\nu(f) f_v\right)$ does; since $\exp_*(f_0)$  is never-vanishing, this entails that $\mu(f)+\nu(f) f_v$ belongs to $\So_J(\Omega)\setminus \So_\R(\Omega)$.
This fact implies that $\nu(f) f_v$ is equal to $g_1J$ for some $g_1\in \So_\R(\Omega)\setminus\{0\}$. Since $\nu(f)
\in \So_\R(\Omega)\setminus\{0\}$ we obtain that $f_v\in \So_J(\Omega)\setminus\{0\}$. Would $f_v\equiv n\pi J$ for some $n\in\Z\setminus\{0\}$, then $f_v^s\equiv n^2\pi^2$ and therefore $\nu(f)\equiv 0$, which is a contradiction.
\end{proof}

Formula~\eqref{mu-nu} can also be used to obtain an explicit expression for
$(\exp_*(f))*(\exp_*(g))$.

\begin{prop}\label{star-product-exponential}
Let $f,g\in \So(\Omega)$. Then $\exp_*(f)*\exp_*(g)$ is equal to
$$\exp_*(f_0)\exp_*(g_0)\left(\mu(f)\mu(g)-\nu(f)\nu(g)\langle f_v,g_v\rangle_*+\nu(f)\nu(g) f_v\pv g_v
+\mu(f)\nu(g) g_v+\mu(g)\nu(f)f_v\right).
$$
\end{prop}

\begin{proof}
The assumption is a direct offspring of Formulas~\eqref{mu-nu} and~\eqref{star-formula}.
\end{proof}

Since $f^c=f_0-f_v$, Formula~\eqref{mu-nu} also gives $\exp_*(f^c)$ in terms of $\exp_*(f)$. Indeed we have

\begin{remark}
Let $f\in\So(\Omega)$. Then
$\exp_*(f^c)=\left(\exp_*(f)\right)^c$.
\end{remark}

When applied to $f$ and $f^c$ the above Proposition shows that, as it is natural to expect, the $*$-exponential of a regular function never vanishes. Indeed we have the following

\begin{prop}\label{real-part}
Let $f\in\So(\Omega)$. Then
\begin{align}
(\exp_*(f))^s&=\exp_*(2f_0) \label{eq1},\\
\frac{\exp_{*}(f)+\exp_{*}(f^{c})}{2}&=\exp_{*}(f_0)\mu(f),\label{eq2}\\
\frac{\exp_{*}(f)-\exp_{*}(f^{c})}{2}&=\exp_{*}(f_0)\nu(f)f_{v},\label{eq3}\\
\exp_{*}(f)*\exp_{*}(-f)&\equiv 1.\label{eq4}
\end{align}
In particular $\exp_*(f)$ is a never-vanishing function.
\end{prop}

\begin{proof}
Indeed we have
\begin{align*}
(\exp_*(f))^s&=(\exp_*(f))*(\exp_*(f))^c=(\exp_*(f))*(\exp_*(f^c))\\
&=\exp_*(f_0)\exp_*(f_0)\left(\mu(f)\mu(f)+\nu(f)\nu(f)\langle f_v,f_v\rangle_*-\mu(f)\nu(f) f_v+\mu(f)\nu(f)f_v\right)\\&=
\exp_*(2f_0)\left(\mu(f)\mu(f)+\nu(f)\nu(f) f_v^s\right)=
\exp_*(2f_0)\left(\mu(f)^2+\nu(f)^2 f_v^s\right)
\end{align*}
Now using the same notation as in Corollary~\ref{sine-cosine} we have that $\mu(f)^2+\nu(f)^2 f_v^s\equiv1$ and therefore we obtain Equality~\eqref{eq1}.
As $\exp_*(2f_0)=\exp(2f_0)$ is never-vanishing, then also $\exp_*(f)$ has no zeroes on $\Omega$.
Equalities~\eqref{eq2} and~\eqref{eq3} follow immediately from~\eqref{mu-nu}, while~\eqref{eq4} can be obtained from Proposition~\ref{star-product-exponential} following
the same kind of computations as above.
\end{proof}

Proposition~\ref{star-product-exponential} enables us to deepen our understanding of the comparison between the $*$-product of $\exp_*(f)$ and $\exp_*(g)$ and $\exp_*(f+g)$, thus giving, when $\Omega$ contains real points, a necessary and sufficient condition for them to be equal. We underline that, while Condition (i) appears quite natural, Condition (ii) is in some sense surprising: in particular it holds for a class of functions larger than one could initially presume (see  Example~\ref{non-constant}).

\begin{teo}\label{super-equality}
 Let $\Omega$ contain real points. Take $f,g\in\So(\Omega)$.
If
\begin{equation}\label{equality}
\exp_*(f+g)=
\exp_*(f)*\exp_*(g)
\end{equation}
then either
\begin{enumerate}[(i)]
\item $f_v$ and $g_v$ are linearly dependent over $\So_\R(\Omega)$ or

\item there exist $n,m,p\in\mathbb Z\setminus\{0\}$ such that $f_v^s\equiv n^2\pi^2$,  $g_v^s\equiv m^2\pi^2$, $2\langle f_v, g_v \rangle_*=(p^2 -n^2 -m^2)\pi^2$ and $n+m\equiv p$ mod 2.

\end{enumerate}
Vice versa, if either $(i)$ or $(ii)$ are satisfied,
then~\eqref{equality} holds.
\end{teo}

\begin{proof}
By Formula~\eqref{mu-nu}  and Proposition~\ref{star-product-exponential}
we can suppose that $f_0$ and $g_0$ vanish identically.
Then we have
\begin{align*}
\exp_*(f+g)&=\mu(f+g)+\nu(f+g)(f+g)_v\\
\exp_*(f)*\exp_*(g)&=\mu(f)\mu(g)-\nu(f)\nu(g)\langle f_v,g_v\rangle_*\\
&\qquad\qquad\quad+\nu(f)\nu(g) f_v\pv g_v
+\mu(f)\nu(g) g_v+\mu(g)\nu(f)f_v.
\end{align*}

First we study the case when neither $\nu(f)$ nor $\nu(g)$ are identically zero.
Since $(f+g)_v=f_v+g_v$ belongs to ${\rm Span}_{\So_\R(\Omega)}(f_v,g_v)$,  if   $\exp_*(f+g)=\exp_*(f)*\exp_*(g)$ then
$\nu(f)\nu(g) f_v\pv g_v$ belongs to ${\rm Span}_{\So_\R(\Omega)}(f_v,g_v)$ too.
Then there exist $\alpha,\beta\in\So_\R(\Omega)$ such that
$\nu(f)\nu(g)f_v\pv g_v=\alpha f_v+\beta g_v$. If $\alpha$ and $\beta$ are both identically zero,
then $\nu(f)\nu(g)f_v\pv g_v\equiv 0$.
Thus $f_v\pv g_v\equiv 0$ and Proposition~\ref{wedge-product} ends the proof. If $\alpha$ and $\beta$
are not both identically zero, up to rearrangement, we can suppose $\alpha\not\equiv 0$.
Now $0\equiv\langle\nu(f)\nu(g)f_v\pv g_v ,f_v\rangle_*=\alpha f_v^s+\beta \langle f_v,g_v\rangle_*$
and $0\equiv\langle \nu(f)\nu(g)f_v\pv g_v,g_v\rangle_*=\alpha \langle f_v,g_v\rangle_*+\beta g_v^s$ that is
$$
\begin{cases}
\alpha f_v^s+\beta \langle f_v,g_v\rangle_*\equiv0,\\
\alpha \langle f_v,g_v\rangle_*+\beta g_v^s\equiv0.
\end{cases}
$$
Taking a suitable combination of these equations we obtain $\alpha\left(f_v^sg_v^s-\langle f_v, g_v\rangle_*^2\right)\equiv 0$.
As $\alpha\in\So_{\R}(\Omega)\setminus\{0\}$, we obtain that $f_v^sg_v^s-\langle f_v, g_v\rangle_*^2=(f_v\pv g_v)^s$ is identically zero. Since $\Omega$ contains real points, then $\So(\Omega)$ is an integral domain, and therefore $f_v\pv g_v$ is identically zero. Again, Proposition~\ref{wedge-product} entails that $f_v$ and $g_v$ are linearly dependent over $\So_\R(\Omega)$.

If $\nu(f)\equiv0$ (the case $\nu(g)\equiv 0$ is treated analogously), then $f_v^s$ is identically equal to $n^2\pi^2$ for a suitable $n\in\Z\setminus\{0\}$.
In this case $\exp_*(f+g)=
\exp_*(f)*\exp_*(g)
$ gives
$$\mu(f+g)+\nu(f+g)(f+g)_v=\mu(f)\mu(g)+\mu(f)\nu(g) g_v$$
which, taking the equality of the vector parts, implies
$$\nu(f+g)(f_v+g_v)=\mu(f)\nu(g) g_v,$$
that can be written as
$$\left(\mu(f)\nu(g)-\nu(f+g)\right)g_v=\nu(f+g)f_v.$$
This gives the linear dependency of $f_v$ and $g_v$ over $\So_\R(\Omega)$
 unless  $\mu(f)\nu(g)-\nu(f+g)\equiv\nu(f+g)\equiv0$. In this last case, we obtain
$\mu(f)\nu(g)\equiv\nu(f+g)\equiv0$. A simple computation gives $\mu(f)=(-1)^n$ and therefore $\nu(g)\equiv\nu(f+g)\equiv0$.

Then there exist $m,p\in\mathbb{Z}\setminus\{0\}$ such that $g_v^s\equiv m^2\pi^2$ and $(f_v+g_v)^s= p^2\pi^2$ and therefore $\mu(f+g)=(-1)^p$.

Since $(f_v+g_v)^s=-(f_v+g_v)*(f_v+g_v)=n^2\pi^2+m^2\pi^2+2\langle f_v,g_v\rangle_*$ we obtain $2\langle f_v, g_v \rangle_*=(p^2 -m^2 -n^2)\pi^2$.

Under these assumptions we have
$$
\exp_*(f+g)=\mu(f+g)\equiv(-1)^p\quad\mbox{and}\quad
\exp_*(f)*\exp_*(g)=\mu(f)\mu(g)\equiv(-1)^n(-1)^m=(-1)^{n+m}
$$
which holds if and only if $n+m$ and $p$ have the same parity.

If $(i)$ is satisfied, then~\eqref{equality} holds because of Proposition~\ref{lin-dep}. If $(ii)$ is satisfied, then $\nu(f)=\nu(g)=\nu(f+g)\equiv0$ and $\mu(f)=(-1)^n$, $\mu(g)=(-1)^m$, $\mu(f+g)=(-1)^p$ so that
$\exp_*(f+g)=\mu(f+g)=(-1)^p$ and  $\exp_*(f)*\exp_*(g)=(-1)^n(-1)^m=(-1)^{n+m}=(-1)^{p}$ because $n+m$ and $p$ have the same parity.
\end{proof}

The following two examples show cases where~\eqref{equality} holds while $f_v$ and $g_v$ are not linearly dependent over $\So_\R(\HH)$.  In particular the second one gives a large class of non constant functions which do not commute but satisfy~\eqref{equality}.

\begin{example}\label{constant_orthogonal}
If $I,J\in\SF$ are orthogonal, $f_v\equiv 3\pi I$ and $g_v\equiv 4\pi J$ then
$$\exp_*(f+g)=-\exp_*(f_0+g_0)=\exp_*(f)*\exp_*(g).$$
\end{example}

\begin{example}\label{non-constant}
Let $n,p,m\in\N$ such that $m^2=n^2+p^2$.
Now choose $\alpha,\beta\in\So_\R(\Omega)$ and set
\begin{align*}
f(q)&=-n\pi\left(\cos(\alpha(q))\cos(\beta(q)) i+\cos(\alpha(q))\sin(\beta(q)) j+\sin(\alpha(q))k\right),\\
g(q)&=\pi((n\cos(\alpha(q))\cos(\beta(q))+p\sin(\alpha(q))\cos(\beta(q))) i+\\
&+(n\cos(\alpha(q))\sin(\beta(q))+p\sin(\alpha(q))\sin(\beta(q))) j\\
&+(n\sin(\alpha(q))-p\cos(\alpha(q)))k).
\end{align*}
As $\alpha,\beta$ are slice-preserving functions then $f,g\in\So(\Omega)$, moreover
$f_v^s=n^2\pi^2$, $g_v^s=m^2\pi^2$ and $\langle f_v, g_v\rangle_*=-n^2\pi^2$. Since $n,p,m$ is a Pythagorean triple then $2\langle f_v, g_v\rangle_*=-2n^2\pi^2=\pi^2(-n^2-p^2-n^2+p^2)=\pi^2(p^{2}-n^{2}-m^{2})$ and hence the given functions satisfy
the requirements of the statement because the parity condition is satisfied for any Pythagorean triple.

\end{example}

Next  example gives an explicit couple of functions $f$ and $g$ where the parity relation between $n,m$ and $p$ does not hold.

\begin{example}\label{constant_non_orthogonal}
If $I,J\in\SF$ satisfy $IJ+JI=-\frac{13}{20}$, $f_v\equiv 2\pi I$ and $g_v\equiv 5\pi J$ then

$n=2$, $m=5$ and $p=4$ since

$((f+g)_v)^s=(2^2+5^2+2\cdot 2\cdot 5(IJ+JI))\pi^2=4^2\pi^2$. Then we have $\mu(f)=\mu(f+g)\equiv1$, $\mu(g)\equiv-1$ and $\nu(f)=\nu(g)=\nu(f+g)\equiv0$ which entail
$\exp_*(f+g)=\exp_*(f_0+g_0)$ while $\exp_*(f)*\exp_*(g)= -\exp_*(f_0+g_0)$.
\end{example}

The example below illustrates a case where $\Omega$ does not contain real points.

\begin{example}\label{constant_non_real}
Denote by $\tau:\HH\setminus \R\to\SF$ be given by $\tau(\alpha+\beta I)=\begin{cases} I &\text{if } \beta>0,\\
-I &\text{if } \beta<0.
\end{cases}$

Consider $f,g:\HH\setminus \R\to\HH$ defined as
\begin{align*}
f(\alpha+\beta I)&=2\pi i-2\pi(1+\tau(\alpha+\beta I)i)j=2\pi(i-j-\tau(\alpha+\beta I)k) \\
g(\alpha+\beta I)&=\tau(\alpha+\beta I)i+\pi j+k.
\end{align*}
Now we have $(f+g)(\alpha+\beta I)=(2\pi+\tau(\alpha+\beta I))i-\pi j+(-2\pi\tau(\alpha+\beta I)+1)k$, $f_v^s=4\pi^2$, $g_v^s=\pi^2$ and $((f+g)_v)^s=(2\pi+\tau(\alpha+\beta I))^2+\pi ^2+(-2\pi\tau(\alpha+\beta I)+1)^2=\pi^2$.

This entails $\mu(f)\equiv1$, $\mu(g)\equiv-1$, $\mu(f+g)\equiv-1$, $\nu(f)\equiv\nu(g)\equiv \nu(f+g)\equiv0$
and therefore
$\exp_*(f+g)\equiv-1\equiv\exp_*(f)*\exp_*(g)$ holds.
\end{example}

\bibliographystyle{amsplain}

config: arxiv


TITLE: Action of $\operatorname{Gal}(K(X))$ on the normalization of $X$ in $K(X)^\text{sep}$
QUESTION [0 upvotes]: I'm reading this part of The Stacks Project regarding ramification theory.
Right before defining decomposition and inertia groups for schemes there is the following passage:

Let $X$ be a normal integral scheme with function field $K$. Let $K^\text{sep}$ be a separable algebraic closure of $K$. Let $X^\text{sep}\rightarrow X$ be the normalization of X in $K^\text{sep}$. Since $G=\operatorname{Gal}(K^\text{sep}/K)$ acts on $K^\text{sep}$ we obtain a right action of G on $X^\text{sep}$.

It is not clear to me why and how $G$ acts on $X^\text{sep}$. Little hint?

REPLY [2 votes]: Saying that $X^{sep}$ is the normalization means that affine-locally on $U=\operatorname{Spec} A\subset X$ the map $X^{sep}\to X$ corresponds to the natural inclusion from $A$ to it's integral closure $\overline{A}\subset K^{sep}$. As $Gal(K^{sep}/K)$ acts on $\overline{A}$ preserving the inclusion $A\to \overline{A}$, we see that $Gal(K^{sep}/K)$ acts by automorphisms on the map $U^{sep}\to U$. As normalization commutes with localization, we may patch these actions on open affines together to see that $Gal(K^{sep}/K)$ acts via automorphisms on $X^{sep}\to X$.


TITLE: Proof that if group $N$ and $G/N$ are p-groups, than $G$ is p-group
QUESTION [2 upvotes]: I have another question about correctness of my proof. We know, that $N$ and $G/N$ are p-groups, than $G$ is p-group.
Proof: $|N|=p^m$ and thus $|G|=p^m q$. Suppose that $(p^m,q)=1$, than in $G$ exists element of order $k$, such that $k | q$ and $(k,p^m)=1$ - suppose it's $b \in G$. Element $bN \in G/N$ is also of order $k$, and thus $k | \left| G/N \right|$ which contradicts with prerequisite, that $G/N$ is p-group.
It's correct? It's here simply way to prove that? Thanks.

REPLY [11 votes]: There is a slight mistake in your argument: a priori, you don't know if the order $bN$ in $G/N$ is exactly $k$, but you do know that it is of order dividing $k$, hence relatively prime to $p$. In order to show that the order cannot be $1$, you need to use the fact that $N$ itself is a $p$-group (did you notice that you never used it?) so that $g\notin N$. Once you know that, then your contradiction would follow.
As to a simpler way, it depends on your definition of $p$-group! If to you a $p$-group is a group whose order is a power of $p$, then the very simplest way is simply to remember that for any group $G$ and any subgroups $H\subseteq K\subseteq G$, you have $[G:H]=[G:K][K:H]$ (cardinal multiplication in the case of infinite indices if necessary). So $|G|=[G:1]=[G:N][N:1] = |G/N||N|$ is a power of $p$, hence $G$ is a $p$-group.
However, there are meanings of $p$-group: some authors define $G$ to be a $p$-group if for every $g\in G$ there exists $k\gt 0$ such that $g^{p^k}=1$ (that is, the order of every element is a power of $p$). Lagrange's and Cauchy's Theorems tell you that for finite groups, the two definitions coincide. But  there are infinite groups that satisfy the second meaning (but obviously not the first); the Prüfer group for $p$ for example.
How do we prove the statement under this definition, for possibly infinite groups (it's also true)? Suppose $N$ is a $p$-group and normal in $G$, and $G/N$ is a $p$-group. Let $g\in G$. We want to show that the order of $g$ is some power of $p$. Look at $gN$ in $G/N$: since $G/N$ is a $p$-group by hypothesis, then there exists $k\gt 0$ such that $(gN)^{p^k} = eN$; that is, $g^{p^k}N = eN$. That means that $g^{p^k}\in N$. Now use the fact that $N$ is itself a $p$-group (under this definition) to deduce that the order of $g$ must be a power of $p$.


TITLE: blow-ups and fields of definition
QUESTION [3 upvotes]: Let $X$ be a smooth variety over some field $k$, not necessarily algebraically closed. Let $Z$ be a closed subvariety of $X$ which is only defined over an extension of $k$ (e.g. a closed point which is not $k$-rational). Consider the blow-up $\mathrm{Bl}_Z X$ of $X$ along $Z$.
Is $\mathrm{Bl}_Z X$ defined over $k$? I think this is the case but I would appreciate some help to understand what is going on.

REPLY [1 votes]: Yes - blowups are defined intrinsically, without reference to a base field. (See Hartshorne for a full treatment). You can blow up any closed subscheme of a scheme (i.e. you can consider non-reduced structures), and the operation is defined intrinsically, by the relative Proj construction on a graded algebra determined by the corresponding quasi-coherent sheaf of ideals. However, if a subvariety is not rational over $k$, I don't think the blowup will neceasarily have the structure of a variety over $k$.
Edit: this answers an original form of the question which asked if the blowup is defined at all. See the comments for my thoughts on whether it is defined over $k$: as I'm uncertain, I'll leave this in the comments.


TITLE: Does every dimension $n\geq4$ admit a manifold with an exotic smooth structure?
QUESTION [12 upvotes]: It is known that $\mathbb{R}^4$ has exotic smooth structures, and there are many such examples in higher dimensions, such as the famous 7-sphere. My (probably very naive) question is, for every $n\geq4$, does there exist an $n$-manifold with exotic smooth structures?
In other words, for every $n\geq4$, does there exist topological $n$-manifolds which admit more than one diffeomorphism class of smooth structures?

REPLY [17 votes]: Yes. For every $n\ge 5$ there are exotic tori.
In fact, the PL-structures on $T^n$ are in one-to-one correspondence with $H^3(T^n;\mathbb{Z}/2)$, and every one of these is smoothable (Reference: "Surgery on Compact Manifolds" by C. T. C. Wall, Chapter 15A). Since any smooth manifold admits a unique PL-structure up to PL-isomorphism, it follows that there are many manifolds homeomorphic but not diffeomorphic to the standard torus.


TITLE: Find local and global minimizer(s)?
QUESTION [2 upvotes]: Consider the problem:
minimize $f(x)=x_1$ subject to
$(x_1-1)^2+x_2^2=1$, $(x_1+1)^2+x_2^2=1.$
Are there any local minimizers? Are there any global minimizers?
Is $(-2,0)$ both the local and global minimizer?

REPLY [1 votes]: You constrain your function to the intersection of the circles of radius 1 centred at $(-1,0)$ and $(1,0)$, i.e. you minimize the function f over the set $\{(0,0)\}$. Consequently, $(0,0)$ is the only local and global optimiser.


TITLE: In how many solutions of equation : $x_1+x_2+...+x_n=m$
QUESTION [0 upvotes]: In how many solutions of equation: $x_1+x_2+...+x_n=m$ satisfied: $x_i\in \mathbb{N}(i=\overline{1,n}),1\le x_i\le 26,n\le m\le 26n,m\in \mathbb{N}$

This is my try:
  Let $t_i=x_i-1\implies \sum\limits_{i=1}^{n}t_i=\sum\limits_{i=1}^{n}x_i-n=m-n$, where $0\le t_i\le 25$.
  And I don't know how to solve it when $0\le t_i\le 25$.

REPLY [2 votes]: The question is asking how many integer solutions are there for $\displaystyle\sum_{i=1}^nx_i$ = m. We can use combinations with repetition. To find the number of such solutions, we can think of $m+n-1$ slots and we pick $n-1$ "dividers". The number of slots between the dividers d$_i$ and d$_{i+1}$ = x$_{i+1}$. We designate the slots before the first divider as x$_1$ and after the last divider as x$_n$. Thus every arrangements of lots and dividers is a different solution. So the number of solutions = $\binom{n+m-1}{n-1}$ = $\binom{n+m-1}{m}$.


text:
\begin{document}

\maketitle

\begin{abstract}
Problem 10 of the Landelijke Interuniversitaire Mathematische Olympiade 2010 asks for a proof that all matrices in a certain family are nilpotent. Both model solutions prove this using the Cayley-Hamilton theorem. I give a purely combinatorial proof.
\end{abstract}

The LIMO (Landelijke Interuniversitaire Mathematische Olympiade) is a mathematics competition that is organized annually in the Netherlands \cite{limosite}. Mathematics students from the Netherlands and from Flanders participate in teams. The 2010 competition was held on May 28 in Utrecht. Problem number 10 was proposed by Jaap Top from the University of Groningen and reads as follows:

\begin{quotation}
Let $m$ be a positive integer and $n = 2^m - 1$. The $n \times n$ matrix $A = (a_{i,j})$ over $\Z / 2\Z$ is given by $a_{i,j} = 1$ if $|i - j| = 1$ and $a_{i,j} = 0$ otherwise. Show that $A$ is nilpotent.

\textbf{Remark:} \emph{A matrix $A$ is nilpotent if some power of $A$ equals the zero matrix.}
\end{quotation}

Two model solutions were given \cite{uitwerkingen}, proving in particular that \mbox{$A^n = \mathbf{0}$}. Both model solutions use the Cayley-Hamilton theorem: every square matrix over a commutative ring is a \emph{root} (in the obvious sense) of its own characteristic polynomial. The rest of the solution is to ingeniously calculate that characteristic polynomial, which turns out to be equal to $\lambda^n$.

However, it is striking that $A$ is the adjacency matrix of the path graph $P_n$. And it is well known that the entries of the $k$'th power of an adjacency matrix correspond to the number of walks of length $k$ between two particular vertices of the graph. This suggests a combinatorial approach to the problem.

\section{Background from graph theory}

A \textbf{graph} $G$ consists of a finite set $V$ of \textbf{vertices} and a set $E$ of \textbf{edges}. Every edge is a (unordered) pair of vertices. If $\{x,y\}$ is an edge of $G$, then we say that the vertices $x$ and $y$ are \textbf{connected} by this edge. (Note that a vertex is \emph{not} connected to itself.) A \textbf{walk} in $G$ is a finite sequence $x_0, x_1, \ldots, x_k$ of vertices, such that every vertex in the sequence is connected to the next one. Vertex $x_0$ is the \textbf{start}, vertex $x_n$ is the \textbf{end} and $k$ is the \textbf{length} of the walk.

Let $n$ be a positive integer. The \textbf{path graph} $P_n$ is the graph with vertex set $\{1,2,\cdots, n\}$ and in which two integers are connected if and only if they are consecutive. The path graph $P_7$ for example, can be depicted as follows:

\pstrick{p7}{The path graph $P_7$.}{p7}

Let $G$ be a graph with $n$ vertices. We assume from now on that the vertices are always numbered from $1$ to $n$. The \textbf{adjacency matrix} of $G$ is the $n \times n$ matrix A with
\[
a_{x,y} =
	\begin{cases}
		1 & \mbox{if the vertices $x$ are $y$ connected,} \\
		0 & \mbox{otherwise.}
	\end{cases}
\]

For example, the path graph $P_7$ has adjacency matrix
\[
\begin{pmatrix}
	0 & 1 & 0 & 0 & 0 & 0 & 0 \\
	1 & 0 & 1 & 0 & 0 & 0 & 0 \\
	0 & 1 & 0 & 1 & 0 & 0 & 0 \\
	0 & 0 & 1 & 0 & 1 & 0 & 0 \\
	0 & 0 & 0 & 1 & 0 & 1 & 0 \\
	0 & 0 & 0 & 0 & 1 & 0 & 1 \\
	0 & 0 & 0 & 0 & 0 & 1 & 0
\end{pmatrix}.
\]

\begin{lemma}
\label{lemma}
Let $G$ be a graph with $n$ vertices and adjacency matrix $A$. Let $k$ be a positive integer and let $B = A^k$. Then $b_{x,y}$ equals the number of walks of length $k$ that start in vertex $x$ and end in vertex $y$.
\end{lemma}

\begin{proof}
We prove be induction on $k$. The case $k=1$ is satisfied by definition of adjacency matrix. Indeed, the unique walk of length $1$ from vertex $x$ to vertex $y$ exists only if $x$ and $y$ are connected.

Suppose now that the lemma is correct for $k=l$. Let $C = A^l$ and consider $B = A^{l+1} = C \cdot A$. By definition of matrix multiplication we have
\[ b_{x,y} = \sum_{z=1}^n c_{x,z} \cdot a_{z,y}. \]
Here $c_{x,z}$ is the number of walks of length $l$ from $x$ to $z$ by induction hypothesis. Moreover $a_{z,y}$ equals $1$ if $z$ are $y$ connected, and $0$ otherwise, by definition of adjacency matrix. So $b_{x,y}$ is the sum of the number of walks of length $l$ from $x$ to $z$ over all vertices $z$ that are connected to $y$. This is the number of walks of length $l+1$ from $x$ to $y$, summed over the penultimate vertex of the walk. Hence the lemma is also satisfied for $k=l+1$.

By induction, the lemma is proven for all positive integers $k$.
\end{proof}

\section{A combinatorial solution}

\begin{theorem}
Let $m$ be a positive integer, $n = 2^m - 1$ and $k$ an integer with $k \geq n$. Take any two vertices $x$ and $y$ of the path graph $P_n$. The number of walks in $P_n$ of length $k$ from $x$ to $y$ is even.
\end{theorem}

\begin{proof}
We give a proof by induction on $m$.

The case $m=1$ (and hence $n=1$) is trivial, as there are no walks of positive length in $P_1$.

So suppose the theorem is true for $m=l$ and consider $n=2^{l+1}-1$ and $k \geq 2^{l+1}-1$. Let $x$ and $y$ be two vertices of $P_n$. We partition the walks of length $k$ from $x$ to $y$ in three classes:
\begin{description}
	\item[\textbf{Class 1:}] the walks that never visit $2^l$.
	\item[\textbf{Class 2:}] the walks that visit $2^l$ exactly once.
	\item[\textbf{Class 3:}] the walks that visit $2^l$ at least twice.
\end{description}
We will prove that each of these classes contains an even number of walks, which immediately implies that the theorem is also true for $m=l+1$.

Essential in applying the induction hypothesis is the fact that the two subgraphs of $P_{2^{l+1}-1}$ which are to the left and to the right of vertex $2^l$, are isomorphic to $P_{2^l-1}$. Hence, the walks in $P_{2^{l+1}-1}$ that are completely on the left (or on the right) of $2^l$ correspond to the walks in $P_{2^{l}-1}$.

The three cases are summarized in Figure \ref{drieklassen}.

\pstrick{drieklassenenglish}{An illustration of how the parity of the number of walks in each class is counted. Here we have $m=3$, $n=7$, $k=7$, $x=3$ and $y=2$.}{drieklassen}

\begin{description}
	\item[\textbf{Class 1:}] In case $x$ and $y$ are on different sides of $2^l$, there are no walks in this class. If $x$ and $y$ are on the same side of $2^l$, then the walks in this class are exactly the walks of length $k$ from $x$ to $y$ in the subgraph of $P_{2^{l+1}-1}$ which is on that side of $2^l$. This subgraph is isomorphic to $P_{2^{l}-1}$, so by induction hypothesis the number of these walks is even.
	\item[\textbf{Class 2:}] We claim that for all $i \in \{0,1,2,\cdots,k\}$, there's an even number of walks in this class that visit $2^l$ after exactly $i$ steps. A such walk
\[ x, a_1, \cdots, a_{i-1}, 2^l, a_{i+1}, \cdots, a_{k-1},y \]
is determined by two subwalks, each completely at one side of $2^l$: one walk of length $i-1$ from $x$ to $a_{i_1}$ and one walk of length $k-i-1$ from $a_{i+1}$ to $y$. (If $i=0$ or $i=k$, the walk is determined by \emph{one} such subwalk of length $k-1$.) Since \[k \geq 2^{l+1} - 1\] we have either \[ i-1 \geq 2^l - 1 \] or \[ k-i-1 \geq 2^l - 1.\] By induction hypothesis, we can choose one of these subwalks in an even number of ways. Therefore the total number of such walks is also even.
	\item[\textbf{Class 3:}] In the walks in this class, we can \emph{reflect} the subwalk between the first to occurrences of $2^l$. To reflect here means: each step to the \emph{right} (i.e. from $i$ to $i+1$) is replaced by a step to the \emph{left} (i.e. from $i$ to $i-1$) and the other way around. Like this we obtain a different walk in this class. Furthermore, on applying this operation twice, we return to the original walk. Hence we can partition the walks in pairs (a walk together with its \emph{mirror image}), so the number of walks in this class is even.
\end{description}
By induction the theorem is valid for all positive integers $m$, as required.
\end{proof}

Note that the bound $k \geq n$ is optimal. Indeed, there is only one walk in $P_n$ of length $n-1$ from vertex $1$ to vertex $n$.

Also note that it is not possible to simply consider the first vertex $x$ with maximal $2$-exponent (i.e. the exponent of $2$ in the prime factorization) which is visited at least twice in the walk, and to reflect the subwalk between the first two occurrences of $x$. Like that, $x$ remains the first vertex with maximal $2$-exponent that is visited twice in the new walk, so that reflecting twice takes us back to the original walk. But the reflection is not always well-defined, as is illustrated in Figure \ref{tegenvb}. The above proof avoids this problem by splitting the walk into two subwalks where the walk visits $4$, whereby the subwalk that Figuur \ref{tegenvb} tried to reflect is broken in two (see the middle picture in Figure \ref{drieklassen}).

\pstrick{tegenvb}{How a naive method goes awry. The middle picture in Figuur \ref{drieklassen} illustrate how this example can be treated correctly.}{tegenvb}

A solution to the LIMO question is now an immediate corollary.

\begin{corollary}
Let $m$ be a positive integer and $n = 2^m - 1$. The $n \times n$ matrix $A = (a_{i,j})$ over $\Z / 2\Z$ is given by $a_{i,j} = 1$ if $|i - j| = 1$ and $a_{i,j} = 0$ otherwise. Then $A^n = \mathbf{0}$.
\end{corollary}

\begin{proof}
The matrix $A$ is the adjacency matrix of $P_n$. Write $B = A^n$. By Lemma \ref{lemma}, and because we are working over $\Z / 2\Z$, $b_{x,y}$ is the parity of the number of walks of length $n$ from $x$ to $y$ in $P_n$. This number is even by the above theorem, so $b_{x,y} = 0$ for all $x,y$. Consequently $A^n = B = \mathbf{0}$, as required.
\end{proof}

\bibliographystyle{plain}
\bibliography{limo10english}

\end{document}


TITLE: Multiply by $t$ to account for $e^t$ in the homogeneous solution
QUESTION [1 upvotes]: $$y'' - y' = e^t \sin{t}$$
So far I have $\lambda(\lambda - 1) = 0 \quad\implies\quad \lambda_1 = 0 \quad\land\quad \lambda_2 = 1 \qquad\implies y_h(t) = c_1+c_2e^t $
This line brings me my question: $$y_p(t) = t\left(c_1e^t\sin{t} + c_2e^t\cos{t} \right)$$
Why do I not have to multiply by $t$? I noticed WolframAlpha disagreed.
The general method I was under the impression of, was that if anything in the form of a solution I am currently looking for was in the homogeneous solution, I had to expand by $t$. Clearly that is not correct(?).. Would a kind soul be able to shed some light on this matter?
Thank you so much for any help. :-)

REPLY [1 votes]: I'll expand on my comment slightly.
Note that while $e^t$ is a solution to the homogeneous equation, neither $e^t\cos(t)$ nor $e^t\sin(t)$ are solutions to the homogeneous equation, so it suffices to try a particular solution of the form $$Ae^t\sin(t) + Be^t\cos(t)$$
If instead you had an equation such as $y'' - 2y' + 2y = e^t\sin(t)$, then because the complementary solution is $y_c = c_1 e^t\cos(t) + c_2 e^t\sin(t)$, here, you do need to multiply by $t$, so that you would have a trial solution for the particular solution of the form
$$y_p = t(Ae^t\cos(t) + Be^t\sin(t))$$
One thing I might add is (as I said in my comment), this might be easier to see in terms of annihilators.  Looking at your equation, which was (with $D$ the derivative)
$$y'' - y' = (D^2-D)y = e^t\sin(t)$$
we notice that $D^2 - D = D(D-1)$ exactly annihilates all functions of the form $c_1 + c_2e^t$, which is our complementary solution, and then we notice that $(D-1)^2+1$ annihilates $e^t\sin(t)$ (this is an easy thing to check, and is just a computation).  So our equation could be written
$$((D-1)^2+1)D(D-1)y = 0$$
which has solution precisely
$$y = c_1 + c_2 e^t + Ae^t\cos(t) + Be^t\sin(t)$$
and since the first two terms are exactly our complementary solution, the rest of the terms must be the trial solution which will give the particular solution after we use the method of undetermined coefficients.
Also, returning to the example above, in terms of annihilators, if we had the equation
$$y'' - 2y' + 2y = e^t\sin(t)$$
then this can be rewritten (as above) in terms of annihilators as
$$((D-1)^2+1)^2y = 0$$
which has solution of the form
$$y = c_1 e^t \cos(t) + c_2 e^t\sin(t) + t\left( A e^t \cos(t) + B e^t\sin(t)\right)$$
and since the first two terms are exactly the complementary solution, the rest of the terms give the form of the trial solution we use the method of undetermined coefficients on to find the particular solution.


TITLE: Locally connected spaces
QUESTION [3 upvotes]: Let $X$ be a locally connected Tychonoff space without isolated points. If $f:X\rightarrow Y$ is a continuous bijection, then can we prove that $Y$ is locally connected (A space $X$ is locally connected if each $x\in X$ has a neighborhood base of open connected sets)?

REPLY [5 votes]: The answer in general is no. For a simple counterexample, choose any bijection $\varphi: \mathbb{N} \to \mathbb{Q}$. Notice that $\varphi$ would already be a counterexample, except for the fact that $\mathbb{N}$ has isolated points. But we can easily fix that: we just take $X = \mathbb{N} \times \mathbb{R}$ and $Y = \mathbb{Q} \times \mathbb{R}$, and define $f: \mathbb{N}\times\mathbb{R}  \to \mathbb{Q}\times \mathbb{R}$ by $(n, x) \mapsto (\varphi(n), x)$. Now $f$ is a product of two continuous bijections and therefore a continuous bijection. Moreover, $\mathbb{N}\times \mathbb{R}$ has all properties we want, and $\mathbb{Q} \times \mathbb{R}$ is not locally connected.

Since the previous example feels a bit like cheating the 'no isolated points' rule, let's give another counterexample, where this time we take $X$ to be an open half-open interval, let's say $[0, 1)$. For the space $Y$ we take the graph in $\mathbb{R}^2$ of $\sin(1/x)$ for $x \in (0, 1]$ and, we join it back to itself. This space is sometimes known as the Warsaw Circle. See the image below for a better desciption. The space $X$ is shown in blue and $Y$ is shown in red. The arrows are just to indicate the direction. Notice the importance of the fact that $X$ (the blue space) is open on the lower side, otherwise this wouldn't work.

Now, clearly $X$ has all properties we want, but what about $Y$? Consider a point on the left of the red figure, somewhere at the height of the sine curve, and a small open ball around this point (see figure below). Then in this ball, we see the sine curve enter and leave infinitely often, so every small enough open neighbourhood of this point has infinitely many connected components. So this space is not locally connected at this point.


TITLE: Determining (parity of) maximum path length from adjacency matrix. Can I avoid having to square?
QUESTION [1 upvotes]: Setup: I have an undirected (multi)graph $G$ with $n+1$ vertices. Each vertex has degree $\leq 2$, and vertex number $0$ has degree exactly one.
What I'm looking for: I know that any graph with maximum degree $\leq 2$ can be split up into just paths and cycles. Since vertex zero has degree exactly $1$, it must be part of a path. What I'd really like to know is the maximum length (diameter, I guess) of this path.
My thoughts: Since vertex $0$ is part of a path, the part of the graph that it's in is not a multigraph. That means that multiplying the adjacency matrix $A$ by itself ($A^k$) will eventually yield a matrix with all zeros, except for a $1$ in the $0,j$ entry, where $j$ is the end of the path. So one way to get the path length would be to keep squaring $A$ until it looks like this, and then note the parity of $k$.
My question: Is there any way to determine the parity just by looking at the adjacency matrix? I know that maximum path length is in the general case very hard (NP-hard, I think), but given the number of constraints I have on my graph, and that I don't even need path length, just parity, I figured there might be some easier way of doing it.

REPLY [1 votes]: In general, (i.e., general graphs $G$ on $n$ vertices), if you take a high enough power $k$ of adjacency matrix $A$ of a graph $G$ (where the $ij$-th entry of $A$ is 1 iff $i$ and $j$ are adjacent in $G$), then $A^k$ will be a block matrix: You can partition $V(G)$ into sets $V_1,\ldots, V_m$ for some $m$ where
(i) the $ij$-th entry of $A$ is positive iff $i$ and $j$ are both in the same $V_l$, and
(ii) the $V_l$'s are the connected components of $G$; i.e., $i$ and $j$ are in the same connected compoent of $G$ iff $i$ and $j$ are in the same $V_l$.
In fact, taking $k = 2^{\lceil\log n\rceil}$ will do where $n$ is the number of vertices of $G$.
Meanwhile is there a reason why you just cannot follow the path starting from vertex 0. If the graph has maximum degree 2, then storing $G$ as an adjacency matrix is extremely inefficient use of space.


TITLE: Is this problem/counterexample stated correctly/valid?
QUESTION [0 upvotes]: This problem was given by the teacher as a practice exercise. Is it valid?
If $f:M \rightarrow N,$ $g:M \rightarrow P$, and $h: P\rightarrow N$ are maps with $g$ surjective and $h$ injective, show that $\mathrm{Im}(f)=P.$
If valid, how can it be proven?
If invalid, is the following counterexample valid?
Counterexample: Take $M=\mathbb{R}, P=\{0\}, N = \mathbb{R}.$ Define $f:x \mapsto x,$ $g:x \mapsto \{0\}$, and $h:\{0\} \mapsto x^3.$ Then $g$ is surjective, $h$ is injective, and $\mathrm{Im}(f) \neq P.$

REPLY [1 votes]: I think your counterexample is very close, just a bit of notation is off.
In $g$, you probably really mean $g: x \to 0$ not $g: x \to \{0\}$, because the former assigns a real number value of $0$ to all possible inputs whereas the latter assigns a value of the SET $\{0\}$ to all inputs.
Your mapping $h$ is also little funny because $x^3$ is not a member of $\mathbb{R}$, it is an expression involving an unbound variable $x$.  That sounds more technical than it is, I just mean that $x$ has no meaning in that expression!  And the same comment about the $\{0\}$ holds here.. Since $P$ is just $\{0\}$ both $h: x \to x^3$ and $h: 0 \to 0$ would work for you.


TITLE: Sum of part of consecutive terms of expansion of (x+y)^n
QUESTION [1 upvotes]: Let $S(x, y, m_1,m_2, n) = \sum\limits_{i=m_1}^{m_2} \binom{n}{i}x^i y^{n-i}$, where $0 < m_1\leq m_2 < n$. I want to derive the relation between $S(x, y, m_1, m_2, n)$ and $S(x, y, m_1-1, m_2, n-1)$.
Is there any formulas I can use? Similarly to Sum of the first m terms of the expansion $(x+y)^n$.

REPLY [2 votes]: Use the fact that, with
$$
T(x, y, m, n) = \sum\limits_{i=0}^m \binom{n}{i}x^i y^{n-i}
$$
you have
$$
S(x, y, m_1, m_2, n)= T(x, y, m_2, n)-T(x, y, m_1-1, n)
$$
Now using the formula for $T$ in the provided link, you can connect now $S(x,y,m_1, m_2,n)$ to $S(x,y,m_1, m_2,n-1)$. Finally
$$
S(x,y,m_1, m_2,n-1)+\binom{n}{m_1-1}x^{m_1-1} y^{n-m_1}=S(x,y,m_1-1, m_2,n-1)
$$
allows you connect the two you want.


TITLE: Finding a unit in $\mathcal{O}_{\mathbb{Q}(\theta)}$ where $\theta$ is the root of a particular monic polynomial
QUESTION [0 upvotes]: The notes of a module I'm doing has an example I don't quite understand.
If $\theta$ is the root of $t^3-4t+2$, then $2\theta-1$ is a unit of $\mathcal{O}_{\mathbb{Q}(\theta)}$.
First note that $2\theta-1$ is a root of the monic integer polynomial $\frac{1}{8} \left[(t+1)^3-4(4(t+1))+16\right]=t^3+3t^2-13t+1$.
How did they find the polynomial above? I understand why they did it (the $+1$ in the end is useful later one to determine $2\theta -1$ is indeed a unit).
Could you please explain? thanks in advance.

REPLY [1 votes]: Short answer: that comes out from trying to get a $\pm 1$ as the constant term.
The general $y=a\theta^2+b\theta+c\in\mathbb{Z}[\theta]\subset\mathbb{Q}(\theta)$ has minimal polynomial
$$
y^3
- (3 c + 8 a) y^2
+ (16 a^2 + 6 a b + 16 a c - 4 b^2 + 3 c^2) y
- (4 a^3 + 8 a^2 b + 16 a^2 c + 6 a b c + 8 a c^2 - 2 b^3 - 4 b^2 c + c^3)
$$
So you want, for a unit,
$$
-4 a^3 - 8 a^2 b - 16 a^2 c - 6 a b c - 8 a c^2 + 2 b^3 + 4 b^2 c - c^3 = \pm 1
$$
For simplicity try $a=0$ and by switching signs we can assume RHS is $1$:
$$
2 b^3 + 4 b^2 c - c^3 = 1
$$
so $(a,b,c)=(0,2,-1)$ is a choice, as are $(0,-1,1)$, $(0,-1,-1)$ or $(0,-3,5)$ (and of course $(a,b,c)=(0,0,-1)$ which is trivial).  I think they are all the solutions with $a=0$ but haven't checked it.


TITLE: Question about the ring $A_P$ of regular functions on a variety $V$
QUESTION [1 upvotes]: Note: I am doing an introductory course in Commutative Algebra and not Algebraic Geometry; This is just a quick application I am trying to understand. Also, rings are commutative with $1$.
When discussing localizations $S^{-1}A$ for a ring $A$ and $S\subset A$, we made the following example.

Def. Let $k$ be a field, and $P$ be a prime ideal in $A=k[X_1,\dots, X_n]$. Let $V=\{(a_1,\dots,a_n)\in k^n:f(a_1,\dots,a_n)=0, \forall f\in P\}$ be the variety of $P$, and let $S=A\setminus P$.

Then $S^{-1}A$ is denoted $A_P$, and we have $A_P=\{\dfrac{f}{g}:f\in A,g\not\in P\}$. In class, this was called the ring of regular functions on $V$ (EDIT: I learned this is apparently not how the ring of regular functions on V is defined, see the comments). These are apparently well defined rational functions $k^n\to k$ since $g$ does not vanish on $V$, i.e. it seems we have
$g\not\in P\Rightarrow g(a_1,\dots a_n)\not =0$ for all $(a_1,\dots,a_n)\in V$.
But, I think it's also evident that we have
$g\in P\Rightarrow g(a_1,\dots,a_n)=0$ for all $(a_1,\dots,a_n)\in V$
which is logically equivalent to $(g\not\in P) \vee (g(a_1,\dots,a_n)=0$ for all $(a_1,\dots,a_n)\in V)$.
It just feels like I'm missing something obvious, since I can't seem to deduce a contradiction from $g\not\in P$, and $g(a_1,\dots,a_n)=0$ for some point of $V$.
Question: Why couldn't we have $g\not\in P$, but still have $g(a_1,\dots,a_n)=0$, for some $(a_1,\dots,a_n)\in V$?

REPLY [2 votes]: We absolutely can have $g\notin P$ and $g(a_1,\dots,a_n)=0$ for some $(a_1,\dots,a_n)\in V$.
Take $n=2$ and the prime ideal $P=(x)\subset k[x,y]$, so that $V=\{(a,b)\in k^2\mid a=0\}$. The function $g=y$ will vanish at the point $(0,0)\in V$, but it will not vanish at any other point $(0,b)$ with $b\neq0$.


TITLE: Burgers' equation / Perturbation theory
QUESTION [2 upvotes]: Let the Burgers' equation:
$$u_t+u u_x=\epsilon u_{xx} ,\qquad x\in\mathbb{R}, \; t>0.$$
With the initial condition: $u(x,0)=1_{\{x<0\}}(x).$
Studying it in the theory of perturbation:
I did not know how to find the shock wave $s(t)$

REPLY [2 votes]: Let us seek traveling wave solutions of the form $u(x,t) = v(\xi)$ with $\xi = x-ct$. According to the chain rule, we have $u_x = v'$ and $u_t = -cv'$. Thus, the PDE leads to the ODE
$$
-cv' + vv' = \epsilon v'' \, ,
$$
with the boundary conditions $v(-\infty) = 1$ and $v(+\infty) = 0$.
Integrating once w.r.t. $\xi$ gives Riccati's equation
$$
\tfrac12 v^2 - c v + c - \tfrac12 = \epsilon v' \, ,
$$
which bounded solutions are of the form
$$
v(\xi) = \frac{1}{1 + A e^{\xi/(2\epsilon)}} \, ,
$$
where $A>0$ is a constant, and where we have set $c=1/2$ to satisfy the boundary conditions. These traveling wave solutions are displayed below for several values of $\epsilon$:

While the solution converges towards a discontinuity with infinite slope as $\epsilon \to 0$, the speed of the wave $c=1/2$ remains unchanged. Therefore, the shock trajectory in the $x$-$t$ plane is $s(t) = t/2$ for $\epsilon \to 0$. However, note that the traveling wave isn't a Riemann solution as it doesn't satisfy the discontinuous boundary condition. Nevertheless, it does satisfy the original problem in the limit of vanishing viscosity $\epsilon\to 0$.
With classical perturbation theory, we make the power series Ansatz
$$
u = u^0+\epsilon u^1 +\dots
$$
and inject it in the PDE. By separating powers of $\epsilon$, the following system is obtained:
$$
\begin{aligned}
u^0_t + u^0 u^0_x &= 0\\
u^1_t + u^0 u^1_x &= u^0_{xx} - u^0_x u_1 \\
&\dots
\end{aligned}
$$
However, solving at 0th order amounts to solving the inviscid Burgers' equation (i.e, the case
$\epsilon = 0$), which doesn't make much sense in the present strong form... So the traveling wave approach is more appropriate here.


TITLE: Independent and dependent events
QUESTION [1 upvotes]: Suppose that $A,B,C$ are three events such that $A$ and $B$ are independent but $A$ and $C$ are dependent. Can anything be said about the independence of $B$ and $C$? Intuitively, I would think that $B$ and $C$ are dependent, too, but I was unable to prove that.

REPLY [2 votes]: They might be dependent, but they don't have to be.
You're drawing a card out of a deck
A - you draw a black card
B - you draw an ace
C - you draw a spade
Here B and C are independent
A - you draw a black card
B - you draw an ace
C - you draw the ace of spades
Here B and C are dependent


TITLE: What's the solution to this conditional probability problem?
QUESTION [0 upvotes]: Knowing that in the first 3 draws without replacement from a deck of 52 french cards (26 red|26 black) at least one red card has been drawn, what is the probability that the 2nd draw has been a red card?
How would you solve this problem?

REPLY [0 votes]: At the first 3 cards at least one red card has been drawn.
1) 2 black & 1 red cards
2) 1 black & 2 red cards
3) 3 red cards
We don't know it is case 1, 2, or 3.
We have to consider all the possibilities.
Case 1)
First, what is the probability that we are in case 1?  It is important, because case 1 happens much more frequently than the other cases, we have to give a higher weight.
And then calculate the probability the next one is red under the condition of case 1.
Pr(case1) = $ \frac{{26 \choose 2}{26 \choose 1}}{52 \choose 3}$
This is not the probability I want to use. What I want to use is the probability of case 1 under the condition of at least 1 red card drawn.
Therefore Pr(case1|at least 1 red) = Pr(case 1| Case1 or Case 2 or Case3)
$=\frac{\text{Pr(case1)}}{\text{Pr(case 1)+Pr(case 2)+Pr(case 3)}}$
Pr(red|case1) = $\frac{25}{49}  $
where, 25 is the number of red cards remain under the case 1.
And 49 is the number of total cards.
Pr(red & case 1|at least 1 red) = Pr(red|case1)xPr(case1|At least 1 red)
And repeat the calculation for cases 2 and 3.
Pr(red|Case1 or case2 or case3) = Pr(red & case 1|at least 1 red) + Pr(red & case 2|at least 1 red) + Pr(red & case 3| at least 1 red)