TITLE: $f(z)$ is holomorphic on $|z| < 1$, with $|f(z)| ≤ 3$ and $f(\frac{1}{2}) = 2$. Show that $f(z) \neq 0$ when $|z| < \frac{1}{8}.$
QUESTION [6 upvotes]: This is an old Schwarz lemma problem from the August 2020 UMD qualifying exam for analysis, which is posted here. The precise wording from the test is:
Suppose $f(z)$ is a holomorphic function on the unit disk with $|f(z)| \le 3$ for all $|z| < 1$, and
$f(1/2) = 2$. Show that $f(z)$ has no zeros in the disk $|z| < 1/8$. (Hint: first show $f(0) \neq 0$).
There is another kind of standard problem that assumes $|f(0)| \ge r$ and asks to prove $|f(z)| \ge \frac{r - |z|}{1 - r|z|}$ on the disk $|z| < r$; see problem 5 from chapter IX.1 of Gamelin. This problem instead gives you a value for $f(a)$, $a \neq 0$.

REPLY [3 votes]: Another, perhaps more direct approach, but which also doesn't really use the hint: Let $|a|<\frac{1}{8}$. Consider $g_a = f\circ \phi_a$ where $\phi_a(z)=\frac{z-a}{1-\overline{a}z}$ as in your answer. If $f(-a)=0$, i.e. $g(0)=0$, the Schwarz lemma would imply that $|g(z_a)|=|f(\frac{1}{2})|\leq 3|z_a|$ where $z_a = \phi_a^{-1}(\frac{1}{2})=\frac{1+2a}{2+\overline{a}}$. But we have $f(\frac{1}{2})=2$ and $3|z_a|=3\frac{|1+2a|}{|2+\overline{a}|}<3\cdot\frac{1+\frac{2}{8}}{2-\frac{1}{8}}=2$, which would contradict the above inequality.


TITLE: Normalization transformation of a probability density
QUESTION [0 upvotes]: Let $(X,Y,Z)$ be distributed according to the pdf $p(x,y,z)$. What would be the pdf $q(x,y,z)$ of the multivariate random variable $(X',Y',Z') = (X,Y,Z)/\sqrt{X^2+Y^2+Z^2}$? The method that I have studied in school with the Jacobian and the inverse of the transformation doesn't seem to apply to this case (the function doesn't have a well-defined inverse since we lose the length information). Is there any generic method to deal with cases like this?
I would also appreciate any recommendation for a good reference (preferably) for engineers that deals with problems like this too, and not only with the standard problems studied in an engineering degree not focused on mathematics.

REPLY [0 votes]: I found a reference which solves this issue: https://en.wikibooks.org/wiki/Probability/Transformation_of_Probability_Densities


text:
\begin{definition}[Definition:Set of Finite Strings]
Let $\Sigma$ be an [[Definition:Alphabet|alphabet]].
The [[Definition:Set|set]] of all [[Definition:Finite String|finite strings]] from $\Sigma$ is denoted $\Sigma^*$.
\end{definition}

config: wiki


TITLE: First order approximation of the current in ASEP
QUESTION [1 upvotes]: I am searching for an elementary proof of the first order approximation of the current of particles in ASEP (asymmetric simple exclusion process) and TASEP (totally asymetric). To avoid technical details unuseful for the intuition, I will focus on the current at the site $0$.
A simple case : step initial condition
It has been known for a long time (see for exemple theorem 5.12 of the section devoted to exclusion processes in "Interacting particle systems", Liggett) that the current of particles at site $0$ in the ASEP model converges almost surely to $t\gamma/4$ when $t$ tends to infinity, where $\gamma=q-p$, when the initial condition is the so called step initial condition. (i.e. $\eta_0(x)=\mathbf{1}_{x>0}$).
One can explain this result using the density (renormalized) profile of particles in the window $[-\gamma t, \gamma t]$ (or $[-\gamma , \gamma ]$ after renormalization) which is actually a stronger result, but more natural. The fact is that this density profile is linear, hence a number of particles on the left of $0$ like $t\gamma/4$.
Step Bernoulli initial condition
When the initial condition is step Bernoulli, i.e., $\eta_0(x)=\mathbf{1}_{x>0}Ber_x(\rho)$, where $Ber_x(\rho)$ is a Bernoulli random variable with parameter $\rho$, the approximation for the current stays the same for $\rho\geq 1/2$, and the fluctuations follow tracy-widom law on a $t^{1/3}$ scale. For $\rho< 1/2$, the current in $0$ is like $t\gamma\rho(1-\rho)$ (see for example Tracy-Widom, "On ASEP with Step Bernoulli Initial Condition"), and the fluctuations are Gaussian.
In the proof of the fluctuations, the phase transition appears clearly in the calculations (when approximating the fredholm determinant for $t\rightarrow +\infty$), but I would like to find a simple physical interpretation ( I mean a simple and/or direct proof) explaining why the approximation does not depend on $\rho$ as soon as $\rho >1/2$.

REPLY [1 votes]: For me the "physical intuition" is the fact that exclusion processes tend to clog up for large densities and hence you cannot expect a faster current than in the $\rho = \frac{1}{2}$ case, even when you add more particles.
PS: to answer your question in the comments, the phase transition should occur at $\rho = \frac{1}{2}$ by symmetry with respect to interchanging the notion of "particles" and "holes". Indeed (heuristically speaking) if the process has $\rho = \frac{1}{2}$, there are just enough holes that each particle basically behaves like a free particle. If you now increase the density, the front of particles still can only move as fast as the free particle and all the other ones block each other behind it until the first particle in front has moved away a bit. Only then the next one can follow and in this case again it moves like a free particle. This is (my) intuition why the particle current cannot be higher than in the "free" case.


TITLE: Why is there this exact sequence
QUESTION [1 upvotes]: On page 49 in Hartshornes "Deformation Theory", in the proof of theorem 6.4 he states that the following exact sequence $0\to J\otimes O_{X}\to O_{X_{1}} \to O_{X}\to 0$ gives rise to an exact sequence of abelian groups $0\to J\otimes O_{X}\to O^*_{X_{1}} \to O^*_{X}\to 0$, so that the left group is just the additive group and the groups in the middle and on the right are the corresponding groups of units. I would like to know why is there this exact sequence, as well as when in general you can do something like this.
Setting: $X$ is a scheme flat over a local artin ring $C$. $C'$ is another local artin ring that surjects to $C$ with kernel $J$ (that has square 0). $X_{1}$ is an extension of $X$ over $C'$, that is: there is a closed imbedding $X\to X_{1}$ where $X_{1}$ is flat over $C'$ and it holds that $X=X'\times_{C'} C$.

REPLY [1 votes]: Let me do this for rings, other case being similar. Let $A$ be any ring (commutative with identity) and $J\subset A$ an ideal with $J^2=0$. Then, I claim that there is an exact sequence $0\to J\to A^*\to (A/J)^*\to 0$ of abelian groups. The only observation is that if $j\in J$, $1+j$ is a unit in $A$, since $(1+j)(1-j)=1$.
The right side map is obvious and the left side map is $j\mapsto 1+j$. Since $1+(j+k)=(1+j)(1+k)$, we see that this indeed a map of abelian groups. Surjectivity on the right is clear, since any lift of a unit in $A/J$ is a unit in $A$. Exactness is clear too, since if $a\mapsto 1\in (A/J)^*$, this means $a=1+j$ with $j\in J$.


TITLE: The set X is the complex numbers in cofinite topology
QUESTION [0 upvotes]: I have to check if $X\backslash\{0\}$ is homeomorphic to $X \backslash \{n|n\in\mathbb{N}\}$. Here I am considering $X\backslash\{0\}$ and $X \backslash \{n|n\in\mathbb{N}\}$ in subspace topology.

REPLY [2 votes]: Thinking of $X$ as the complex numbers makes this seem more difficult than it really is. Suppose $X$ and $Y$ are two sets with the same cardinality, and let $f: X \to Y$ be a bijection. Then $f$ is a homeomorphism in the cofinite topology, because it preserves cardinality and the topology is defined in terms of cardinality. Said another way, the image of a cofinite set is cofinite, as its preimage, since the cardinality of the complement is preserved by the bijection.


TITLE: Proving the minimality of an element order
QUESTION [1 upvotes]: Assume that I have a finite group G of order n with a generator g, and also assume that I want to prove that $\frac{n}{gcd(n,m)} $ is the order of an element $x = g^m \in G$.
First , I showed that $x^\frac{n}{gcd(n,m)} = e$
let $d = gcd(n,m)$ then $d | n$ and $d|m$ and so $dt=n,ds=m$ for some integers $t,s$ now $t = \frac{n}{gcd(n,m)} = \frac{n}{d}$ then $x^\frac{n}{d} = (g^m)^\frac{n}{d} = g^\frac{mn}{d} = (g^n)^\frac{m}{d} = e$
 then now I want to show that $t$ is minimal , meaning that it's the least positive integer such that $x^t = e$. Here is my attempt
Suppose that $x^k = e$ for some positive integer k, now by the division algorithim, we have $k = tq + r$ where $ 0 \leq r < t$ , so $x^k = x^{tq+r} = x^{tq}x^r = (x^t)^qx^r = e^qx^r=ex^r=x^r$ , now I want to show that $r = 0$ to conclude that $k=tq$ and hence $ k \geq t$ and so $t$ is the order.
But how do I show that $r$ is zero , any suggestions ?

REPLY [0 votes]: Since $g$ has order $n$, $x^k=g^{km}=e\iff n\mid km$. Let $n=n'\gcd(m,n)$ and $m=m'\gcd(m,n)$, so $\gcd(m',n')=1$. Then, $n\mid km\iff n'\gcd(m,n)\mid km'\gcd(m,n)\iff n'\mid km'\iff n'\mid k$. So the smallest $k$, which is the order of $x$, is $n'=\frac{n}{\gcd(m,n)}$.


TITLE: Mapping Points in a Plane to Arc Length of a Spiral
QUESTION [1 upvotes]: Can a spiral, $r=f(n)\theta$, be defined so that all points in a plane are mapped bijectively to an angle?
Related:
Consider $r=(1/n)\theta$.  Is it possible to define the ratio of points in a plane to points along this spiral as $n\to \infty$.
Motivation/Background:
I'm a high school math teacher - confident with multivariable calculus but no experience with real analysis.  Question was a passing thought on a slow day.

REPLY [0 votes]: The number of points in the plane and the number of points in the spiral path both have the same uncountably infinite cardinality. It is possible to put them in one-to-one correspondence,
but the particular attempt at a one-to-one correspondence you have in mind
is another matter altogether.
Consider the points that lie along a particular ray from the origin.
For example, consider the positive $x$-axis. For any particular value
of $n$, if $f(n) > 0$ the intersection of this set
of points with the spiral occurs at angles of the form $\theta=2k\pi$
or $\theta = -\pi-2k\pi$ for any non-negative integer $k$, that is,
$\theta = \ldots, -5\pi,-3\pi, -\pi, 0, 2\pi, 4\pi,\ldots.$
(The negative values come from the fact that the polar coordinates
$(-f(n)\pi, -\pi)$ and $(f(n)\pi, 0)$ identify the same point.)
For any real value of $n$, there are only countably many of those points,
but there are uncountably many points on the $x$-axis.
Since this is true for any real $n$, it's true in the limit as $n \to \infty$.
On the other hand, if $f(n)$ is a continuous function, non-zero for some
positive value of $n$, such that $\lim_{n\to\infty} f(n) = 0$
(for example, $f(n) = 1/n$),
any particular point in the plane (other than the origin) will be on the
spiral $r = f(n)\theta$ for (countably) infinitely many values of $n$.
But for every such value of $n$, there is a larger value of $n$ for which
that same point is not on the spiral $r = f(n)\theta$.
So I would not say that that the spiral $r = f(n)\theta$ maps an angle to
this point "in the limit" as $n\to \infty$,
or even that this point is on the spiral in the limit.
The same reasoning applies to every point in the plane, so no point
other than the origin is on the spiral in the limit.
As I already mentioned,
it is possible to explicitly construct a bijective mapping from the real line
$\mathbb R$ to the plane $\mathbb R^2$.
This answer describes a bijective mapping
between the interval $(0,1]$ and the region $(0,1]\times(0,1]$,
and the "appendix" of that answer indicates how to make a bijection
between $(0,1]$ and $\mathbb R$, which also gives you the tools to make
a bijection between $(0,1]\times(0,1]$ and $\mathbb R^2$.
But the resulting bijection is not a continuous function,
and moreover there cannot be a continuous bijection between the two sets.
Space-filling curves are continuous mappings from a line segment to
a region of the plane, but they are not bijective.
Note that for any $n$, the area of the spiral is zero, whereas the area
of the unit disk around the origin is $\pi$.


TITLE: Correlation and what it tells me
QUESTION [2 upvotes]: OK, I need a little help here.  I have attached two pictures; Data and Chart in which the data shows a correlation coefficient of 0.283168 which was calculated by Excel.  Can someone please tell me what this calculation actually tells me?  Really curious as I am trying to learn this.
Let me add a little more information to this.  The column showing RATES is the Federal Fund Rates and each rate corresponds with the number of sales of homes while at that rate.  Normally, when the Federal rate is higher, sales generally are lower however this is not the case in this example.  With this new information in mind, what does the data show in relation to the correlation coefficient.  Thanks in advance, this is all new to me.

REPLY [0 votes]: The correlation is a measure of the strength of a linear relationship between two variables. Around .28 is a weak relationship, and thanks to the graph we can see why - there are two sales values that ruin what would otherwise be a strong relationship (if the two strange ones weren't there, then the correlation would be .9949, a VERY strong linear relationship).
The other poster's explaination regarding what is explained by the other variable is seemingly mistaking correlation for R-Squared. The R-Squared value for this (as we only have two variables here) is simply the square of the correlation, or 0.0802. That being said, it is actually 8.02% of the variation in sales that can be explained by rates (Again, this would be much higher if it weren't for those two out of line observations).


TITLE: Give an infinite collection of intervals such that $A_{n+1} \subset A_n$ and $\cap^{\infty}_{n=1} A_n = \varnothing$
QUESTION [4 upvotes]: Give an example of an infinite collection $A_n $ of intervals such that $A_{n+1} \subset A_n$ and $\bigcap^{\infty}_{n=1} A_n = \varnothing$.

I have come to the following collection of intervals, but dont know if it is correct nor how to prove it. $A_n = \left(0, \frac{1}{n^2}\right)$. How do I prove my example works?
EDIT: The intervals are on the real line!

REPLY [1 votes]: if $x \leq 0$, $x$ belongs to none of the sets.  If $x>0$, take $n \in \mathbb{N}^+$ with $x > 1/n^2$.  Then $x$ does not belong to $\mathcal{A}_n$, so $x$ does not belong to the intersection.


TITLE: Are the following functions (f and g) examples which are discontinuous at 1, though f x g is continuous at 1?
QUESTION [0 upvotes]: I'm working though (on my own; and unfortunately I don't live near a university) a text on Real Analysis: Bartle and Sherbert, 4th Ed.  Chapter 5 is "Continuous Functions". In Section 5.2 ("Combinations of Continuous Functions") one of the exercises (#3, pp. 133) asks for functions f and g which are discontinuous at some point c in their shared domain, but when the functions are multiplied that function is continuous at the point c.
One of the difficulties in studying on my own is that -- for open-ended questions like this -- it is difficult to know if my own ideas are correct.  I understand the answer suggested in the book, and I think I understand the question.
However, I have no clear way of determining if MY idea is (also) correct.  I came up with the two functions:
f: = (e^(1/(1-x)))/(1-x)
g: = the reciprocal of   f
Thus, when multiplied together, their product-function is the constant 1, and continuous at 1 (and elsewhere).  But each is discontinuous at the point x = 1.
Thus, I believe this answer is also correct?  Is it?
Thanks for taking the time to help me!  (I'm actively looking for a local tutor, as we're a large research-base oriented community and there are at least some mathematicians working here; but no luck so far...)

REPLY [3 votes]: Actually, your $f$ is undefined at $1$, so it's neither continuous nor discontinuous there.
You can take$$f(x)=\begin{cases}1&\text{ if }x\in\mathbb Q\\2&\text{ otherwise}\end{cases}$$and$$g(x)=\begin{cases}2&\text{ if }x\in\mathbb Q\\1&\text{ otherwise.}\end{cases}$$Then $f$ and $g$ are discontinuous everywhere, but $f\times g=2$.


TITLE: Area between a curve and a horizontal line
QUESTION [0 upvotes]: I had 2 make 2 exercises which I found very similar but they have a crucial difference which puzzles me:
1 Calculate the area of the plane $V$, between the graph of the function $f(x) = \dfrac{8x}{x^2+4}$ and the line $y= 1\dfrac{3}{5}$.
2  Calculate the area of the plane $V$, between the graph of the function $f(x) = \dfrac{1}{x^2-4x+5}$, the y-axis and the line $y=\dfrac{1}{2}$.
1  $\displaystyle\int_0^4 (f(x)-1\dfrac{3}{5}) \, \mathrm{dx}$
2 $\displaystyle\int_0^1 (\dfrac{1}{2} - f(x)) \, \mathrm{dx}$
What is the difference between these cases which causes the different calculation? I am puzzled.

REPLY [1 votes]: Your standard formula for calculating areas between curves has the form:
$$
\int_a^b (f(x)-g(x)) dx
$$
where $f(x)$ is the upper curve and $g(x)$ is the lower curve. This makes the integral positive, since we want area. So, draw a picture. For the first question, we see that for $x\in [1,4]$, we have $\dfrac{8x}{x^2+4} \ge 1\dfrac{3}{5}$. (For example, pick a sample point and note that $f(2)=\dfrac{8(2)}{2^2+4}=2 \ge 8/5$). Thus, $f(x)$ is the upper curve, which yields the integral:
$$
\int_1^4 \left(\dfrac{8x}{x^2+4}-1\dfrac{3}{5}\right) dx
$$

On the other hand, for the second question, we see that for $x\in [0,1]$, we have $\dfrac{1}{x^2-4x+5} \le \dfrac{1}{2}$. (For example, pick a sample point and note that $f(0.5)=\dfrac{1}{0.5^2-4(0.5)+5}=4/13 \le 1/2$). Thus, $f(x)$ is the lower curve, which yields the integral:
$$
\int_0^1 \left(\dfrac{1}{2} - \dfrac{1}{x^2-4x+5}\right) dx
$$


TITLE: Expected Value $AB*CD$
QUESTION [0 upvotes]: The digits 1, 2, 3, and 4 are randomly arranged to form two two-digit numbers, AB and CD.
For example, we could have $AB = 42$ and $CD = 13$.
What is the expected value of $AB * CD$?

On Brilliant, there is an explanation of this problem, but I don't get it... FIrst of all, shouldn't $E[AC]$ require ${}_4 \mathrm{ P }_2 = 12$ permutations instead of 6?
https://brilliant.org/wiki/linearity-of-expectation/

REPLY [0 votes]: Using decimal expansion $AB=A\cdot10+B$, $CD=C\cdot10+D$.
$$AB\cdot CD=A\cdot C\cdot 100 + (A\cdot D+C\cdot B)10+B\cdot D$$
HEnce
$$\mathbb{E}[AB\cdot CD]=100 \mathbb{E}[A\cdot C] + 10\mathbb{E}[A\cdot D+C\cdot B] + \mathbb{E}[B\cdot D]$$
$$\begin{align}
\mathbb{E}[A\cdot C]&=\frac{1}{4\cdot 3}(1\cdot 2+1\cdot 3+1\cdot 4+2\cdot 1+2\cdot 3+2\cdot 4+3\cdot 1+3\cdot 2+3\cdot 4 +\\
&\qquad 4\cdot 1+4\cdot 2 + 4\cdot 3)\\
&=\frac{1}{6}(2+3+4+6+8+12)=\frac{35}{6}
\end{align}$$
$$\begin{align}
\mathbb{E}[A\cdot D+C\cdot B]&=\mathbb{E}[A\cdot D]+\mathbb{E}[C\dot B]=2\frac{35}{6}
\end{align}$$
since $A\cdot D$ and $C\cdot B$ have the same distribution as $A\cdot C$.
Similarly, $B\cdot D$ has the same distribution as $A\cdot C$ and so,
$\mathbb{E}[\mathbb{E}[AB\cdot CD]=(100+20+1)\frac{35}{6}$


TITLE: Two cards are drawn without replacement, the find the probabilility
QUESTION [1 upvotes]: Two cards are drawn without replacement, the find the probabilility that the first card is a non face and the second is a black:
Non face would be 40/52 but how would I get the second part?

REPLY [1 votes]: The probability the first pick is a red non-face is $\frac{20}{52}$. That leaves $26$ black out of $51$. So the probability of a red non-face followed by a black card is
$\frac{20}{52}\cdot \frac{26}{51}$.
The probability the first pick is a black non-face is $\frac{20}{52}$. In that case the probability the next pick is black is $\frac{25}{51}$. So the probability of a black non-face followed by a black is $\frac{20}{52}\cdot \frac{25}{51}$.
Add.  There is much simplification.
Another way: By symmetry, the probability of non-face followed by black is the same as the probability of non-face followed by red.
The probability of non-face is $\frac{40}{52}$. The event "first is non-face" is the disjoint union of the events "first is non-face, second is black" and "first is non-face, second is red." Each has equal probability, so each has probability $\frac{20}{52}$.
We can also express this in the language of conditional probability: the probability the second is black given first is non-face is $\frac{1}{2}$. The events "first is a non-face card" and "second is black" are independent. So to find the proability of first non-face and second black, we multiply $\frac{40}{52}$ by $\frac{1}{2}$.


TITLE: Convergence of the series $ \sum\frac{1}{a_na_{n+1}} $ if the sequence $(a_n)$ is arithmetic
QUESTION [4 upvotes]: We know that $(a_n)$ is arithmetic progression and I need to decide if
$$ \sum_{n=1}^{\infty} \frac{1}{a_n \cdot a_{n+1}} $$
convergences.
Firstly I check necessary condition:
$ \frac{1}{a_n \cdot a_{n+1}} \rightarrow 0 $
It is true because $$ a_n \rightarrow +/- \infty $$  because $ a_n = a + (n-1)r $
Ok, now I should check convergence:
Ratio test:
$$ \frac{u_{n+1}}{u_n} = \frac{a_n(a_n+r)}{(a_n+r)(a_n+2r)} \rightarrow 1 $$
So it gives me nothing.
Root tests:
$$ \sqrt[n]{\frac{1}{a_n \cdot a_{n+1}}} \rightarrow 1 $$
so it fails too...
Have somebody any idea how can I check this?

REPLY [2 votes]: First note that  $$\frac{1}{a_na_{n+1}}=\frac{1}{a_{n+1}-a_n}\cdot \left(\frac{1}{a_{n}}-\frac{1}{a_{n+1}}\right)$$
Then since $a_n$ is arrithematic, $a_{n+1}-a_n=a_2-a_1=d.$ So you get:
$$\sum_{n=1}^{N}\frac{1}{a_na_{n+1}}=\frac{1}{d}\left(\frac{1}{a_1}-\frac{1}{a_{N+1}}\right)$$
Which means the series converges to...
We have to assume the $a_i\neq 0$ and that $d\neq 0.$ If $d=0$ then all the $a_i$ are equal, and the sum diverges.

REPLY [1 votes]: If $a_n$ is an arithmetic progression, then $a_n$ grows with order $n$; therefore, $a_n \cdot a_{n + 1}$ grows at least as quickly as $n^2$, and convergence follows from a comparison test. Note that this convergence is too slow to be detected by either ratio test or root test.

To make this a bit more precise, if $a_n$ is a term of an arithmetic progression we can write $a_n =cn + b$ for constants $c$ and $b$. Then
$$\frac{1}{a_n \cdot a_{n + 1}} = \frac{1}{c^2 n^2 + \text{ lower order}}$$
and the comparison is quick from here.


text:
\begin{document}
\maketitle

\section{Introduction and statement of main results}
The notion of exceptional singularity
was introduced by Shokurov \cite{Sh1}. A singularity $(X,P)$ is called
{\it exceptional}, if for any log canonical boundary, there is at most
one exceptional divisor of discrepancy $-1$ over $P$
(see Definition \ref{def-exc}). The reason for
distinguishing these singularities is that
 they have
more complicated multiple anticanonical
systems $|-nK_X|$, than
the nonexceptional ones. However, Shokurov
suggests, and proves in dimension $3$, that
the exceptional singularities are in a sense bounded (loc. cit., Corollary 7.3).
The search of ``good" divisors in $|-nK_X|$,
or so called $n$-{\it complements}, is an essential ingredient of
Shokurov's project of the inductive study of log flips, log contractions
and of the classification of log canonical singularities \cite[Sect.~7]{Sh1}.

According to \cite[5.2.3, 5.6]{Sh}, \cite[1.5]{Sh1}, the exceptional
log terminal singularities in
dimension 2 are exactly the singularities of types
$E_6,E_7,E_8$ in the sense of Brieskorn \cite{Br}. Shokurov's approach
gives a description of the dual graphs of their resolutions (cf \cite{I},
\cite[Chapter 3]{Ut}). According to \cite{Br}, the exceptional graphs correspond
to finite subgroups of $GL_2(\CC )$ of tetrahedral ($E_6$),
octahedral ($E_7$) or icosahedral ($E_8$) types.
We found a classification-free approach to the
proof which works also in dimension 3. The exceptional
groups of types $E_6$, $E_7$, $E_8$ are exactly those finite subgroups of
$GL_2(\CC )$ which have no semiinvariants of degree $\leq 2$. Hence the
following proposition implies Shokurov's statement.

\begin{proposition}
\label{Sh_2}
A two-dimensional quotient singularity $X=\CC^2/G$
by a finite group $G$ without reflections is
exceptional if and only if $G$ has no semiinvariants of degree $\le 2$.
\end{proposition}

In dimension 3, we started the study of exceptional quotient singularities
in our previous publication \cite{MP}, where we showed that the quotients
of $\CC^3$ by Klein's simple group of order 168 and by its unique
central extension contained in $\SL$ (of order 504) are exceptional,
in using the configuration of the action of Klein's group on $\PP^2$
\cite{Kl}, \cite{W}.
The following Theorem is the main result of the present paper.

\begin{theorem}
\label{main}
A three-dimensional quotient singularity $X=\CC^3/G$
by a finite group $G$ without reflections is
exceptional if and only if $G$ has no semiinvariants of degree $\le 3$.
\end{theorem}

Using Miller--Blichfeldt--Dickson classification \cite{MBD}
of finite subgroups of $\GL$, we obtain a complete list of
such subgroups yielding exceptional singularities (Theorem \ref{list-exc}).

Section \ref{s-1} contains basic definitions and preliminary results.
In particular, we
show that the exceptional divisor with discrepancy $-1$ for an exceptional
singularity is birationally unique and that its image is a point, independently
of the choice of the boundary and in any dimension. Section \ref{s-2} contains
the proofs of Proposition \ref{Sh_2}, Theorem \ref{main}, and the list of
finite subgroups of $\GL$ with exceptional quotients.
\par\noindent
{\sc Acknowledgement.}
The second author is grateful to V.~V.~Shokurov
for  discussions on  complements and the exceptionality.

\section{Basics on exceptional singularities}
\label{s-1}

For reader's convenience, we reproduce here some basic facts
about log canonical singularities.
We are using the terminology and notation of \cite{MP}, \cite{Ut},
\cite{Sh} and \cite{Sh1} (see also \cite{KoP} for a nice introduction
to the subject).

\begin{definition}
Let $(X\ni P)$ be a normal singularity (not necessarily isolated) and
let $D=\sum d_iD_i$ be a divisor on $X$ with real coefficients.
$D$ is called a boundary if $0\leq d_i\leq 1$ for all $i$.
It is called a subboundary, if it is majorated by a boundary.
A proper birational morphism $f \colon Y\lra X$ is called a log resolution of
$(X,D)$ at $P$, if $Y$ is nonsingular near $f^{-1}(P)$ and
$\Supp (D)\cup E$ is a normal crossing divisor
on $Y$ near $f^{-1}(P)$, where $D$ is used
to denote both the subboundary on $X$ and its proper transform on $Y$,
and $E=\cup E_i$ is the exceptional divisor of $f$.
The pair $(X,D)$ or, by abuse of language, the divisor $K_X+D$
is called {\it terminal, canonical, Kawamata log terminal,
 purely log terminal,} and, respectively,
{\it log canonical} near $P$,
if the following
conditions are verified:

(i) $K_X+D$ is $\RR$-Cartier.

(ii) Let us write for any proper birational morphism $f \colon Y\lra X$
$$
K_Y\equiv f^\ast (K_X+D)+\sum a(E,X,D)E ,
$$
where $E$ runs over prime divisors on $Y$, $a(E,X,D)\in\RR$,
and $a(D_i,X,D)=-d_i$ for each component $D_i$ of $D$. Then, for some
log resolution of $(X,D)$ at $P$ and for all prime
divisors $E$ on $Y$ near $P$, we have:
$a(E,X,D)>0$ (for terminal), $a(E,X,D)\geq 0$ (for canonical),
$a(E,X,D)> -1$ and no $d_i=1$ (for Kawamata log terminal),
$a(E,X,D)> -1$ (for purely log terminal, without any
restriction on the subboundary $D$), and,
respectively, $a(E,X,D)\geq -1$ (for log canonical).
\end{definition}

The coefficients $a(E,X,D)$ are called {\it discrepancies} of $f$, or of
$(X,D)$; they depend on the discrete valuations of the
function field of $X$ associated to the prime divisors $E$ and on $D$,
but not on the choice of $f$. We will
identify prime divisors with corresponding discrete
valuations, when speaking about `divisors $E$ over $X$'
without indicating, on which birational model $E$ is realized. The conditions
given by inequalities in part (ii) of the above definition do not depend
on the choice of a log resolution.

\begin{definition}
Let $V$ be a normal variety and let $D=\sum d_iD_i$ be
a $\QQ$-divisor on $V$ such that $K_V+D$ is $\QQ$-Cartier.
A subvariety $W\subset V$ is said to be a {\it center of
log canonical singularities} for $(V,D)$ if there exists
a birational morphism from a normal variety
$g\colon \tilde{V}\to V$ and a prime divisor $E$ on
 $\tilde{V}$ with discrepancy $a(E,V,D)\le -1$ such that $g(E)=W$.
(The case $d_i=1$, $W=D_i$ is also possible.)
The union of all centers of log canonical singularities is called the
{\it locus of log canonical singularities} and is denoted by  $LCS(V,D)$
\cite[3.14]{Sh}.
\end{definition}

The following statement is a weak form of \cite[17.4]{Ut} (in dimension $2$
it was proved earlier by Shokurov \cite{Sh}).

\begin{theorem}
\label{conn-1}
Let $X$ be a normal projective variety and let $D=\sum d_iD_i$ be
a $\QQ$-divisor on $X$ such that $K_X+D$ is $\QQ$-Cartier.
If $-(K_X+D)$ is nef and big, then the locus of
log canonical singularities is connected.
\end{theorem}

\begin{theorem}[{\cite[6.9]{Sh}}]
\label{conn-2}
Let $X$ be a normal projective surface and let $D=\sum d_iD_i$ be
a boundary on $X$ such that $K_X+D$ is $\QQ$-Cartier.
If $K_X+D\equiv 0$, then the locus of
log canonical singularities has one or two connected components.
\end{theorem}

Shokurov informed us that the previous theorem is valid in any dimension
modulo Log Minimal Model Program.

\begin{definition}[{\cite[1.5]{Sh1}}]
\label{def-exc}
Let $(X\ni P)$ be a normal singularity and
let $D=\sum d_iD_i$ be a
boundary on $X$ such that $K_X+D$ is log canonical. The pair $(X,D)$ is
said to be {\it exceptional} if there exists at most one exceptional divisor
$E$ over $X$ with discrepancy $a(E,X,D)=-1$. The singularity $(X,P)$ is said to
be {\it exceptional} if $(X,D)$ is exceptional for any $D$ whenever $K_X+D$
is log canonical.
\end{definition}

\begin{lemma}
\label{red}
Assume that there exists a reduced divisor $S=\sum S_i$ passing through $P$
such that $K_X+S$ is log canonical. Then $(X\ni P)$ is nonexceptional.
\end{lemma}

For the proof, see Lemma~1.7 in \cite{MP}. This implies, in particular, that
any three-dimensional cDV-singularity \cite{YPG},
and any three-dimensional terminal singularity
is nonexceptional (see \cite[1.9]{MP}).

\begin{prop}
\label{convex}
Let $(X,P)$ be an exceptional singularity. Then the divisor $E$ from
Definition \ref{def-exc} with discrepancy $a(E,X,D)=-1$ is birationally
unique. This means, that there exists a unique discrete valuation $\nu$
of the field of rational functions $k(X)$, such that for any pair
$(X,D)$, exceptional at $P$, and for any log resolution $f:\tilde{X}
\lra X$, if there is an exceptional divisor $E$ in $\tilde{X}$ with
$a(E,X,D)=-1$, then the corresponding discrete valuation $\nu_E=\nu$.
\end{prop}

\begin{proof}
Let $\{ D_1,\ldots ,D_r\}$ be a finite set of irreducible divisors on $X$,
and $f:\tilde{X}\lra X$ a log resolution of $(X,\sum_{i=1}^r D_i)$ at $P$.
We can represent the boundaries $D=\sum_i d_iD_i$ with components from
$\{ D_1,\ldots ,D_r\}$ as the unit cube $I^r\subset \RR^r$ of
vectors $(d_i)$. Then the subset $\Lambda\subset I^r$ corresponding
to log canonical pairs $(X,D)$ is given by a finite number of linear
inequalities in $d_i$'s
$$
a(E,X,D)=a(E,X,0)-\sum_id_i\mult_E(D_i)\geq -1 ,
$$
where $E$ runs over the exceptional divisors such that $f(E)\ni P$.
Let $\partial_+\Lambda$ be the closure of $\{ (d_i)\in\partial\Lambda\mid
0<d_i<1\;\forall \; i=1,\ldots ,r\}$.
Then the exceptionality of $(X,P)$ implies that there are
no points in $\partial_+\Lambda$ satisfying the equality
$a(E,X,0)-\sum_i\mult_E(D_i)= -1$ for two different $E$'s. So,
$\partial_+\Lambda$ (if nonempty) is an open convex polyhedron lying in exactly
one hyperplane $\Pi_E$ with equation $a(E,X,0)-\sum_id_i\mult_E(D_i)= -1$.

Now, let $(X,P)$ be an exceptional singularity,
$(X,D^{(1)}),(X,D^{(2)})$ two exceptional
pairs, $f_i:X_i\lra X$ two log resolutions with
divisors $E_i$ such that $a(E_i,X,D^{(i)})=-1$. Then we can apply the previous
argument, taking the union of all the components of
$D^{(1)},D^{(2)}$ as $\{ D_1,\ldots ,D_r\}$, and some log resolution
dominating $f_1,f_2$ as $f$.
Then we will have : firstly, by Lemma \ref{red},
$D^{(1)},D^{(2)}$ are represented by two different points of
$\partial_+\Lambda$,
and secondly, $\Pi_{E_1}=\Pi_{E_2}$. But then,
both of the $E_i$ are of discrepancy
$-1$ with respect to any $D$ represented by a point of
$\partial_+\Lambda$. By exceptionality, $E_1=E_2$. This proves the
assertion of the proposition.
\end{proof}

\begin{prop}
\label{LCS-point}
Let $(X,P)$ be an exceptional singularity. Then for any pair
$(X,D)$ exceptional at $P$ and such that $LCS(X,D)\neq\varnothing$,
we have $LCS(X,D)=\{ P\}$.
\end{prop}

\begin{proof}
By Proposition \ref{convex}, the exceptional divisor $E$ with discrepancy
$-1$ is biratonally unique. Suppose, $LCS(X,D)\supsetneqq\{ P\}$.
Then for a generic hyperplane section $H$ of $(X,P)$, we have
$\mult_E(H)=0$. This implies that $a(E,X,D+\epsilon H)=a(E,X,D)=-1$,
and, by exceptionality, $a(E',X,D+\epsilon H)>-1$ for any sufficiently small
positive $\epsi$ and any exceptional divisor $E'\neq E$.
Let $\epsi_\ast =
\sup\;\{\epsi\;\mid\; (X,D+\epsilon H)\ \text{is log canonical}\ \}$.
Then $(X,D+\epsi_\ast H)$
has two different divisors with discrepancy $-1$: $E$ and
either the one of the
$E'\neq E$ with minimal discrepancy, or $H$ in the case when $\epsi_\ast =1$.
This contradicts to the exceptionality of $(X,P)$.
\end{proof}

\begin{prop}
\label{cartesian}
Let $(X_i,P_i)$, $i=1,2,$
be two germs of normal varieties of dimension $>0$. Then
$(X_1\times X_2,(P_1,P_2))$
is nonexceptional.
\end{prop}

\begin{proof}
If at least one of $X_i$ is not log canonical with zero boundary,
we are done, for $X_1\times X_2$ is not log canonical
with any boundary. If $(X_i,0)$ are log canonical,
then, starting from any nonzero boundaries $D^{(i)}$ at $P_i$,
we can find $\epsi_i\geq 0$ such that both pairs $(X_i,\epsi_iD^{(i)})$
are log canonical and possess
at least one discrete valuation of discrepancy $-1$ for $i=1,2$.
Let $W_i$ be their centers. Then
$LCS(X_1\times X_2,\epsi_1 D^{(1)}\times X_2)$
contains $W_1\times X_2$, and
$LCS(X_1\times X_2, X_1\times \epsi_2 D^{(2)})$
contains $X_1\times W_2$. Hence
$(X_1\times X_2,(P_1,P_2))$
is nonexceptional by Proposition \ref{LCS-point}.
\end{proof}

\section{Quotients $\CC^m/G$}
\label{s-2}
In the first part of this section we work
in arbitrary dimension. The following Lemma was proved in \cite[2.1]{MP} in the
$3$-dimensional case; now we generalize it to arbitrary dimension.

\begin{lemma}\label{quot-lem1}
Let $\pi\colon V\to X$ be a finite morphism of normal varieties,
\'etale in codimension $1$. Let $D$ be a boundary on $X$, and $D'=\pi^\ast D$.
Assume that $K_X+D$ is log canonical (and then automatically is $K_V+D'$).
Then $(X,D)$ is exceptional if and only if $(V,D')$ is.
\end{lemma}

\begin{proof}
For every proper birational morphism
$f\colon Y\to X$ consider the following commutative diagram
\begin{equation}
\label{1}
\begin{CD}
W@>{\varphi}>>Y\\
@V{g}VV@VfVV\\
V@>{\pi}>>X.\\
\end{CD}
\end{equation}
where $W$ is the normalization of the dominant component
of $V\times_{X}Y$.
By the ramification formula (cf. \cite[proof of 3.16]{KoP})
for every exceptional divisor $F$ on $Y$ and every
exceptional divisor $E$ on $W$ which dominates $F$ we have
\begin{equation}\label{1.5}
a(E,V,D')+1=r(a(F,X,D)+1),
\end{equation}
where $r$ is the ramification index at the generic point of $E$.
Therefore
\begin{equation}
\label{iff}
a(E,V,D')=-1\qquad\text{ if and only if}\qquad a(F,X,D)=-1.
\end{equation}
Thus if $(X,D)$ is nonexceptional, then $(V,D')$ is so.

Conversely, assume that $(V,D')$ is nonexceptional.
Then there are infinitely many exceptional divisors over
$V$ with discrepancy $a(\phantom{E},V,D')=-1$.
As in \cite[proof of 3.16]{KoP} we can see that
all these divisors appear in some commutative square (\ref{1})
(for suitable $f$), whence $(X,D)$ is also nonexceptional
by (\ref{iff}).
\end{proof}

\subsubsection*{Warning}
The Lemma does not imply that the singularity of $X$ is exceptional if and
only if $V$ is. The assertion concerns {\em pairs} $(V,D')$ with
$D'$ a pullback of a boundary from $X$.

\subsection{}\label{V,G}
Let $\pi\colon V\to X$ be the quotient morphism, where $V=\CC^m$.
Let $D$ be a boundary on $X$ and let $D':=\pi^*D$. By \cite[2.2]{Sh},
(see also \cite[20.3]{Ut}, \cite[3.16]{KoP}) $K_X+D$ is log canonical
(resp., purely log terminal, Kawamata log terminal) if and only if so
is $K_{V}+D'$.

\begin{question}
\label{conj}
In the notation of \ref{V,G}, assume that $G$
has a semiinvariant of degree
$\le m$. Does this imply that $(X\ni P)$ is nonexceptional?
\end{question}

Let $\psi$ be such a homogeneous semiinvariant of minimal
degree $d\le m$ and let
$D'$ be its zero locus.
By Lemma \ref{quot-lem1} it is sufficient, for the positive
answer to 3.3, to prove that
$K_V+D'$ is log canonical. It is clear that $D'$ is a cone.
The following Proposition is an easy particular case of
this question.

\begin{proposition} \label{impr}
Let $G$ be maximally imprimitive, that is,
let $G$ contain a normal abelian subgroup $A$ whose character subspaces
$V_\chi$ are $1$-dimensional and form one orbit under the action of $G$.
Then $V/G$ is nonexceptional.
\end{proposition}

\begin{proof}
$G/A$ acts by permutations of the $V_\chi$, so
$G$ has a semiinvariant of the form $x_1\cdots x_m$, where
$x_i$ are coordinate linear forms. Since
$D'=\{ x_1\cdots x_m=0\}$ is a simple normal
crossing divisor, $(V,D')$ is log canonical. Then $(X,D)$ is also
log canonical, and the assertion follows
from Lemma \ref{red}.
\end{proof}

\subsection{}\label{Vi}
We cannot treat in general the case of {\it imprimitive} groups,
that is, groups $G$ which permute transitively factors
of some direct sum decomposition $V=\oplus V_i$. {\it Maximally
imprimitive} are those for which $\dim V_i =1$. But if the
action of $G$ is not transitive for some direct sum decomposition,
the singularity is also nonexceptional. This follows from
the next proposition.

A group $G$ is called {\it reducible}, if
$V=V_1\oplus V_2$ with $V_1,V_2$ invariant under $G$, $\dim V_i>0\; (i=1,2)$.

\begin{prop}\label{reducible}
If $G$ is reducible, then $V/G$
is nonexceptional.
\end{prop}

\begin{proof}
The proof is similar to that of Proposition \ref{cartesian},
though the result is not a corollary of \ref{cartesian}.
Let $G_i$ denote the image of $G$ in $GL(V_i)$.
Let $D_i\; (i=1,2)$ be any nonzero boundaries for $(V_i,0)$,
given by $G_i$-semiinvariant polynomials from $k[V_i]$.
We can find $\epsi_i>0$ such that both pairs $(V_i,\epsi_iD_i)$ possess
at least one discrete valuation of discrepancy $-1$ for $i=1, 2$.
Let $W_i$ be their centers.
Define $D^{(1)}=D_1\times V_2,D^{(2)}=V_1\times D_2$ in $V$.
Then $LCS((V,\epsi_iD^{(i)})$ contains $W_1\times V_2$ or $V_1\times W_2$.
Using a commutative square of the same type as (\ref{1}), we conclude
from (\ref{1.5}) that the images of $W_1\times V_2$ and $V_1\times W_2$
in $V/G$ are also centers of log canonical singularities.
Hence $V/G$ is nonexceptional by Proposition~\ref{LCS-point}.
\end{proof}

\begin{lemma}\label{quot-lem2}
If $\dim X=m\leq 3$, the answer to \ref{conj} is affirmative.
\end{lemma}
\begin{proof} In \cite[Lemma 2.2]{MP}, we proved the assertion
in dimension $3$. The same and even easier argument
 works in dimension $2$.
\end{proof}

This Lemma shows the ``if" part of Proposition \ref{Sh_2}
and Theorem \ref{main}.

\subsection{}
\label{logic}
Now we will explain the logic of our proof of the exceptionality
of singularities in Proposition \ref{Sh_2} and Theorem \ref{main}.
Assume that $(X\ni P)$ is nonexceptional. Then
there exists a nonexceptional log canonical $K_X+D$.
Further, we will use notations of Lemma~\ref{quot-lem1}.
Take the smallest $n\in\NN$ such that $F:=nD'$ is an
integer divisor. Then $F$ locally near $0$ can be defined by a
semiinvariant function, say $\psi$.
Denote $d:=\mt{mult}_0(\psi)$.
\label{begin}
Let $\sigma\colon W\to V=\CC^m$ be the blow-up of the origin
and let $S\simeq\PP^{m-1}$ be the exceptional divisor.
Then $K_W=\sigma^*K_V+(m-1)S$ and
$\sigma^*F=R+dS$, where $R$ is the proper transform of $F$.

Further
\begin{equation}
K_W+S+\frac{m}{d}R=\sigma^*(K_V+\frac{m}{d}F).
\end{equation}
By \cite[Lemma 3.10]{KoP} $K_V+\frac{m}{d}F$ is log canonical if and only if so
is
$K_W+S+\frac{m}{d}R$. In this case the pair $(V,\alpha F)$ is
exceptional for all $0\leq\alpha\leq \frac{m}{d}$ if and only if
$K_W+S+\frac{m}{d}R$ is purely log terminal.
By the inversion of adjunction (see \cite[5.13]{Sh}, \cite[17.6]{Ut})
the purely log terminal condition for $K_W+S+\frac{m}{d}R$ is equivalent to
that $K_S+\frac{m}{d}C$ is Kawamata log terminal, where $C=R\cap S$.
It is clear that $C$ is given by the equation
$\psi_{{\min}}=0$, where $\psi_{{\min}}$ is
the homogeneous component of $\psi$ of minimal degree $d$.
Therefore we have

\begin{proposition}
\label{check}
In the above notations, if $K_S+\frac{m}{d}C$
is Kawamata log terminal, then $(V,\alpha F)$ is exceptional for any
$0\leq\alpha \leq\frac{m}{d}$.
\end{proposition}\bigskip

\begin{remark}\label{remark}
In the above notations, if $G$ has no
semiinvariants of degree $\le m$, then $\down{\frac{m}{d}C}=0$.
\end{remark}\bigskip

\subsection{Two-dimensional case}
\subsubsection*{Proof of Proposition \ref{Sh_2}}
We obtain Proposition \ref{Sh_2} as an easy corollary of \ref{check}.
Assume that $G$ has no invariants of degree $\le 2$.
Recall that $S\simeq\PP^1$ in our case, so $C$ is a finite set.
By \ref{check} it is sufficient to prove that
$K_S+\frac{2}{d}C$ is Kawamata log terminal, and this is equivalent to that
$\down{\frac{2}{d}C}=0$. The last assertion follows by \ref{remark}.

\hspace*{\fill}\square
\medskip

\subsection{Three-dimensional case}
\subsubsection*{Proof of Theorem \ref{main}}
Assume that $G$ has no invariants of degree $\le 3$.
We shall prove that $\CC^3/G$ is exceptional.
By Proposition \ref{check} it is sufficient to prove that $K_S+\frac{3}{d}C$
is Kawamata log terminal.
Take $c$ to be the log canonical threshold of $(S,C)$, that is,
the maximal $\alp$ such that $K_S+\alp C$ is log canonical.
If $K_S+\frac{3}{d}C$
is not Kawamata log terminal, then $c\le 3/d$.
First we consider the case when $c<3/d$.
Then $-(K_S+cC)$ is ample. By connectedness theorem \ref{conn-1},
the locus of log canonical singularities is connected.
By \ref{remark} $\down{\frac{3}{d}C}=0$, so any divisor of discrepancy
$-1$ should be exceptional.
Therefore the locus of log canonical singularities on $S$ is
a unique point, which must be invariant under the action of $G$.
The dual line gives us a semiinvariant of degree $1$, a contradiction with
our assumptions.
In the case when $c=3/d$, we can use
Theorem \ref{conn-2}. Similarly to the above, we see that
the locus of log canonical singularities is one or two points.
In both cases, there is an invariant line. This ends the
proof of Theorem \ref{main}.

\hspace*{\fill}\square
\medskip

The finite subgroups of $G\subset\GL$ were classified by
Miller--Blichfeldt--Dickson \cite{MBD} modulo central extensions
(compare with \cite{P}). There are 10 types
of such groups, denoted by A,B, \ldots ,J in \cite{MBD}.
A stands for abelian, B for reducible, and C, D are imprimitive.
The groups of type C are called {\it tetrahedral}; their image
in the symmetric group ${\frak S}_3$ permuting the
$V_i,\; i=1,2,3$ (notations as in
\ref{Vi}) is cyclic of order 3.
The groups of type D are called {\it general monomial}; their map
to ${\frak S}_3$ is surjective. The primitive subgroups of $\GL$
belong to the $6$ types E, F, G, H, I, J.
The orders of the associated collineation groups
$PG=G/(G\cap\CC^\ast)\subset\PGL$
are $36$, $72$, $216$, $60$, $360$, $168$;
the first three are solvable,
and the last three are simple. The collineation
groups from G to J have their names:
the Hessian group, the icosahedral one, the alternating group ${\frak A}_6$
of degree $6$, and, finally,
Klein's simple group. We have also $P$E$\triangleleft P$F$\triangleleft P$G.
\medskip

The following assertion is a consequence of Theorem \ref{main}.

\begin{theorem}\label{list-exc}
Let $G$ be a finite subgroup of $\GL$. Then the quotient $\CC^3/G$
is exceptional if and only if $G$ belongs to one of the $4$ types
{\em F, G, I, J}.
\end{theorem}

\begin{proof}
We can eliminate the types A--D by Propositions \ref{impr}, \ref{reducible}.
Further, the icosahedral group in its $3$-dimensional
representation has an invariant of degree two, which is a semiinvariant
of any group of type H. This follows, for example, from the fact that
the $3$-dimensional representation is the
complexification of the standard real one,
which has an invariant scalar product.

\begin{lemma}\label{groupE}
Any group of type {\em E} has two nonproportional
semiinvariants of degree $3$.
Groups of types {\em F, G} have no semiinvariants of degree $\leq 3$.
\end{lemma}

\begin{proof}
According to \cite[Sect. 115]{MBD}, a group $G$
of one of the types E, F, G is an extension
of a group $H$ of type D which leaves invariant the set of
four triangles $t_1,t_2,
t_3,t_4$ defined in appropriate coordinates by the equations
$$
x_1x_2x_2=0,\; (x_1+x_2+x_3)(x_1+\om x_2+\om^{2+i}x_3)
(x_1+\om^{2}x_2+\om^{1+i}x_3)=0 \; $$
$$\phantom{x_1x_2x_2=0,\; (x_1+x_2+x_3)(x_1+\om x_2}
(\om =\exp \frac{\scriptstyle 2\pi\sqrt{-1}}
{\scriptstyle 3},
i=0,1,2),
$$
and the associated collineation group $PG$ is completely characterized
by its image in the group ${\frak S}_4$ of permutations on the set $\{ t_1,t_2,
t_3,t_4\}$. It is the subgroup of order $2$, conjugate to
$\{ 1, (t_1t_2)(t_3t_4)\}$ for the type E, $\{ 1, (t_1t_2)(t_3t_4),
(t_1t_3)(t_2t_4),(t_1t_4)(t_2t_3)\}$ for the type F, and the full
alternating group on four letters for the type G. Moreover, one can easily
verify that the $t_i$ belong to one pencil $\Phi$ of
plane cubics (see \cite[Lemma 4.7.6]{Sp}). So, a group of type E acts on
the pencil $\Phi$ with image $\ZZ /2\ZZ$ in $\mt{Aut}(\PP^1)$. Any involution
on $\PP^1$ has two fixed points, which implies the result for type E.

Since the image of $G$ in $\mt{Aut}(\PP^1)$ has no fixed
points for the groups of types F, G, they
do not have semiinvariants in the pencil $\Phi$. However, any semiinvariant
of $G$ should be also that of $H$. According to \cite[Sect. 113]{MBD},
$PH$ is generated by the cycle $c=(x_1x_2x_3)$, transposition $\tau =(x_2x_3)$
and dilatation $\kappa =\:\mt{diag}(1,\om ,\om^2)$.
Any semiinvariant of ${\frak S}_3=<c,\tau>$ is a
polynomial in elementary symmetric
functions $\sigma_i\; (i=1,2,3)$ in $x_1,x_2,x_3$ and
in $\Delta =(x_1-x_2)(x_2-x_3)(x_1-x_3)$. A direct verification
shows that in degrees $\leq 3$, only the linear combinations of
$x_1^3+x_2^3+x_3^3=\sigma_1^3-3\sigma_1\sigma_2+3\sigma_3$
and of $\sigma_3$ are (skew-)symmetric semiinvariants
under $\kappa$. This yields exactly the pencil $\Phi$.

\end{proof}

Thus, by Lemma \ref{quot-lem2}, the quotients of types E, H are not
exceptional. By Theorem \ref{main}, the quotients of types F, G
are exceptional. It remains to verify that the groups of types
I, J have no semiinvariants of degree $\leq 3$.

For the groups of type I, we can take a representative $I_0\subset\SL$ of
order $1080$. The homogeneous semiinvariants of any group $G$ of type I will
coincide with those of $I_0$, because they are central extensions
of the same group $PI_0\simeq {\frak A}_6$. But all the semiinvariants
of the group $I_0$ are indeed invariants. This follows from the fact that it
has no normal subgroups with abelian quotient: the only nontrivial normal
subgroup of $I_0$ is its center, isomorphic to $\ZZ /3\ZZ$, and its quotient
is the simple group ${\frak A}_6$. The algebra of invariants of $I_0$
was determined by Wiman \cite{Wi} (see also a modern account of invariants
of finite subgroups of $\SL$ in \cite{YY}, where I of \cite{MBD}
is denoted by L); it is generated by basic invariants of
degrees $6$, $12$, $30$ and $45$ with one relation of weighted
degree $90$ between them.
Thus, there are no semiinvariants of degree $\leq 3$, and we are done.

Klein's simple group has a representation $J_{168}\subset\SL$. As above,
the only its semiinvariants are invariants, and they were determined
by Klein \cite{Kl}, see also \cite{W} or \cite{YY}.
The degrees of basic invariants here are $4$, $6$, $14$, $21$,
and there is one
relation of weihgted degree $42$ between them. Again there are no semiinvariants
of degree $\leq 3$, and we are done. This ends the proof of Theorem
\ref{list-exc}.
\end{proof}

Using the classification of finite subgroups in $\SL$ (\cite{P}, \cite{YY}),
one can get the following assertion.

\begin{corollary}
Let $G$ be a finite subgroup of $\GL$ such that
 $\CC^3/G$
is an exceptional canonical singularity.
Then $G$ is, up to conjugation, one of the following subgroups of $\SL$:
\begin{enumerate}
\renewcommand\labelenumi{{\rm (\roman{enumi})}}
\item
Klein's simple group $J_{168}\subset\SL$,
\item
the unique central extension $J'_{504}$ of $J_{168}$ contained in
$\SL$,
\item
the Hessian group $G_{648}\subset\SL$,
\item
the normal subgroup $F_{216}$ of $G_{648}$,
\item
a central extension $I_{1080}$ of $\mathfrak{A}_6$.
\end{enumerate}
\end{corollary}

\begin{proof}
It suffices to show that $G\subset\SL$. Let $r$ be the order of
the center $Z$ of $G$. Then $\CC^3/Z$ is resolved by a single
blow up, giving an exceptional divisor with discrepancy
$-1+\frac{3}{r}$. Construct, as in the proof of Lemma \ref{quot-lem1},
a diagram (\ref{1}) with $\CC^3/Z,\: \CC^3/G$ in place of $V,X$ respectively.
Then the formula (\ref{1.5}) implies that the minimal discrepancy
of exceptional divisors $E$ over $X\;$ $\;\; a_{\min}\leq -1+\frac{3}{r}$.
For canonical singularities, $a_{\min}\geq 0$, hence $r\leq 3$. If
$r=1$ or $3$, $\; G\subset\SL$ and we are done.

Assume that $r=2$. Then $Z=\{ \pm 1\}$. Let $G_0=G\cap\SL$.
The orders of the subgroups of $\SL$ of types
F, G, I, J are even, hence $G_0$ contains an element $g$ of
order 2. Then either $g$ or $-g$ is a reflection.
This contradicts to our hypotheses. Hence $r=2$ is impossible.
\end{proof}

Recall that a variety $X$ is said to have
$\epsilon$-log terminal singularities if
$a(E,X,0)>-1+\epsilon$ for any exceptional divisor $E$ over $X$.
Adapting the proof of the previous corollary to the case
$a_{\min} > -1+\epsi$ instead of $a_{\min}\geq 0$, we obtain the
following result:

\begin{corollary}
Fix $\epsilon>0$. Then the set of subgroups $G\subset\GL$
without reflections such that $\CC^3/G$ is an exceptional
$\epsilon$-log terminal singularity is finite up to
conjugation.
\end{corollary}

A similar finiteness result was obtained by \cite{Bor} for abelian quotients
of any dimension (which are never exceptional by Proposition \ref{reducible}).

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