TITLE: Can the anti-torque of a motor be used to spin a space vehicle about it's axis?
QUESTION [1 upvotes]: Imagine an electric motor together with its power supply in space. What exactly will happen to it if it is suddenly powered? Will it spin about its axis, or it will experience no motion relative to an observer on Earth?

REPLY [3 votes]: It will spin about its axis.
In general terms, if you attach a flywheel to a big motor in space and turn it on, the flywheel will spin in one direction with some angular velocity, and the motor and whatever is attached to the motor will spin in the opposite direction with a different angular velocity. This is the basic principle behind a reaction wheel.
Reaction wheels employ the law of conservation of angular momentum, which means that if you start with a net angular momentum of $0kg \centerdot m^2/s$ (nothing is turning) and you add $5kg \centerdot m^2/s$ of angular momentum to the flywheel, the satellite must turn in the opposite direction with an angular momentum of $5kg \centerdot m^2/s$ to maintain a net angular momentum of $0kg \centerdot m^2/s$.
In other words, $L_{wheel}=L_{satellite}$ at all times. If you assume one direction to be positive and the other to be negative, you can restate this relationship as $L_{wheel}+L_{satellite}=0$.
This is sort of like the rotational equivalent of "every action has an equal and opposite reaction".
Reaction wheels usually work very well, but they can become "saturated", which can significantly reduce their effectiveness. In theory, a reaction wheel can function as long as you can add angular momentum to the system. However, in practice, you run into material constraints and equipment limits. For example, your motor may have an upper limit on the number of RpMs it can produce; when getting get close to this limit, the reaction wheel loses accuracy and becomes saturated. To correct this issue, engineers add systems to satellites to bleed momentum from the wheels, slowing them.
Two popular methods of bleeding momentum are:

Magnetorquers. Devices that apply torque to Earth's magnetic field to keep the satellite steady as the reaction wheels slow. This is a very common way to bleed extra momentum, and is even used on the Hubble Space Telescope.
A small engine of some sort, to counteract the torque applied when winding down the reaction wheels. This method is often used on satellites where mass is kept to a minimum, and they don't want to carry a magnetorquer; or on satellites where there is not a magnetic field to torque against.

There is at least one more device worth mentioning. The CMG (Control Moment Gyroscope) is used where efficiency is required, or where a reaction wheel cannot provide the required torque. For example, the ISS has a CMG for attitude changes.
A reaction wheel differs from a CMG in one important respect. A reaction wheel changes its angular momentum to effect a rotation about an axis, but a CMG employs the gyroscopic effect to make attitude changes instead. Greatly simplified, a gyroscope resists changes to the axis of rotation. A CMG applies a torque to a large gyroscope, which then resists the torque. When the gyroscope resists the torque, it rotates the station around the gyroscope.
Related or interesting:
Georgia State University's database for physics (conservation of angular momentum.)
Wolfram Research on reaction wheels
Wolfram Research on conservation of angular momentum (they explain it with diagrams and whatnot. It's pretty good.)
Reaction wheels desaturation using magnetorquers and static input allocation.
Reaction wheels (Google Scholar search)
Three-axis attitude control with two reaction wheels and magnetic torquer bars (About regaining the attitude control systems aboard the FUSE satellite in 1999 after the failure of two of its four onboard reaction wheels.)


TITLE: To show that an entire function is a constant
QUESTION [1 upvotes]: For this problem, can I just simply apply Liouville's theorem?

REPLY [2 votes]: $({1+|z|})^{1/2} $ is not bounded nor smooth at zero. So you can't apply liouvlles thrm.


TITLE: Why the expectation of distance to center of disk is $r/3$ and not $r/2$?
QUESTION [1 upvotes]: A person throw an arrow on a target of radius $r$. The position of the arrow on the target is uniformly distributed. Let $X$ the distance between the arrow and the center of the circle. The score obtained by a person is $r-X$. What is the average score ? The answer is $\frac{r}{3}$, whereas I found $\frac{r}{2}$ as follow
We have that $X$ is uniform on $[0,r]$. If $Y=r-X$, then $$\mathbb E[Y]=\int_0^r (r-x)f_X(x)dx=\frac{1}{r}\int_0^r (r-x)dx=\frac{r}{2}.$$
Maybe there is a subtlety than I don't see ?

REPLY [1 votes]: You may first construct the probability density as follows:

At distance $x$ from the center of the circle a corresponding annulus of "thickness" $dx$ has a probability weight of
$$\frac{1}{\pi r^2}\cdot 2\pi \cdot x \cdot dx$$

So, you get
$$E(Y) = \frac{1}{\pi r^2} \int_0^r (r-x)2\pi \cdot x\; dx = \cdots = \frac{r}{3}$$


TITLE: Positive roots and elements in a Coxeter group.
QUESTION [1 upvotes]: In the paper, a set $L$ associated to an element $w$ in a Coxeter group $W$ is defined as follows. Let $w=s_{i_1} \cdots s_{i_m}$ be a reduced expression. Define $L=\{\beta_1, \ldots, \beta_m\}$, where
\begin{align}
& \beta_1 = \alpha_{i_1}, \\
& \beta_k = s_{i_1} \cdots s_{i_{k-1}}(\alpha_{i_k}), k=2,3,\ldots,m.
\end{align}
The set $L$ is denoted by $R_w$. If $w$ is reduced, $\beta_1, \ldots, \beta_m$ are different from each other. I think that the following is true. If $w=s_{i_1} \cdots s_{i_m}$ is not reduced, then there are two elements $\alpha, -\alpha$ in $\beta_1, \ldots, \beta_m$ whose sum is $0$. Is this true? Are there some references about this?
In the case of type $A_2$, for $w= s_1 s_2 s_1 s_2$ which is not reduced, we have
\begin{align}
\beta_1 = \alpha_1, \beta_2 = \alpha_1+\alpha_2, \beta_3 = \alpha_2, \beta_4=-\alpha_1.
\end{align}
We have $\alpha_1+(-\alpha_1)=0$. Thank you very much.

REPLY [5 votes]: This is a standard observation in Lie theory: if you let the Weyl group act on the root system on the left, $\beta_k$ is the unique root such that $s_{i_k}\cdots s_{i_1}\beta_k$ is negative and $s_{i_{k-1}}\cdots s_{i_1}\beta_k$ is positive.  That is, for a positive root $\gamma$, the number of times it appears as $\pm \beta_k$ is the number of times that $s_{i_k}\cdots s_{i_1}\gamma$ changes sign as $k$ goes from 0 to $m$, and it's $\beta_k$ if the change is positive to negative, and $-\beta_k$ if the change is negative to positive.
An expression is unreduced if and only if there is a positive root that changes sign twice, so the places where this sign change happens gives the cancelling $\beta$'s.


TITLE: Does this sequence diverge $\frac{6^{3n}}{5^{4n}}$
QUESTION [1 upvotes]: I used the ratio test to get the problem in the following form:
$\left|\frac{6^{3n+1}}{5^{4n+1}}\right| \cdot \left|\frac{5^{4n}}{6^{3n}}\right| = \frac{6}{5} \gt 1 $ so the sequence diverges?

REPLY [2 votes]: $\frac{6^{3n}}{5^{4n}} = \left(\frac{6^3}{5^4}\right)^n = \left(\frac{216}{625}\right)^n \rightarrow_n 0.$

REPLY [1 votes]: No!
We have: $\lim_{n \to \infty} a^n = \infty$ for a>1, $\lim_{n \to \infty} a^n = 1$ for a=1 and $\lim_{n \to \infty} a^n =0$ for $a \in (-1,1)$. (for a<-1 the limit doesn't exists).
In your case: $a=\frac{6^3}{5^4}=\frac{216}{625} \in (-1,1)$, so, the limit is $0$.


TITLE: If $r^{n-r}=k^{n-k}$, when is this true other than $r=k$?
QUESTION [4 upvotes]: Let $n,r,k$ be non negative integers such that $r,k\leq n$.
If $r^{n-r}=k^{n-k}$, when is this true other than $r=k$ ?
For example it is holds for $n=6,r=2,k=4$.

REPLY [1 votes]: We have that $n,r,k$ be non negative integers such that $r,k\leq n$.
First of all $(r,k,n)=(0,0,n)$, where $n \neq 0$ is a trivial solution for $r^{n-r} = k^{n-k}$.
For further solutions, we have
$$\begin{align}
r^{n-r} & = k^{n-k} \\
(n-r) \cdot \ln r & = (n-k) \cdot \ln k \\
n & = {\frac{r\ln r - k\ln k}{\ln  r - \ln k}}.
\end{align}$$
Now to provide, that $n$ is an integer let $a$ be a nonnegative integer, and take $k=r^a$. If $a=1$, then $r=k$ and then if $r=k=0$, then $n\neq r$, but else $n$ could be anything. If $a \neq 1$, then
$$n = \frac{r-ar^a}{1-a},$$
which is an integer for appropriate $(r,a)$ pairs. I don't know a condition for this.
So I think there are infinitly many solutions in the form
$$(r,k,n) = \left(r,r^a,\frac{r-ar^a}{1-a} \right).$$
If you can find such condition for $n$, that when is it integer, then you could find all solutions.
Examples: $(2,4,6), (2,8,11), (3,9,15), (3,27,39), (3,81,107), (3,243,303) (4,16,28), (4,64,94), (4,256,340), (4,1024,1279), (5,25,45), \ \dots$


TITLE: Does there exist a set X such that for any set Y, there exists a surjective function f : X → Y?
QUESTION [3 upvotes]: I was thinking about this question and I have my own thoughts on this, please do let me know if my reasoning is correct.
The question asks if I can pick an $X$ such that this condition is fulfilled for all $Y$.
If the function $f$ is to be surjective, it must mean that every element must be an output of $f$. So if $X$ was countably infinite, this would not be possible, since I can pick $Y$ an uncountably infinite set. So $X$ would have to be uncountably infinite set, but then this set $X$ would not work if $Y$ was the null set, because a function only exists from $f:X \rightarrow \emptyset $ only if $X$ is the null set.
So this statement is false. Is my reasoning correct?

REPLY [8 votes]: There's a straight answer to this question :
For any set $X$, there is no surjection from $X$ onto the power set $\mathcal{P}(X)$.
Proof: Assume for contradiction that $f:X \mapsto \mathcal{P}(X)$ is a surjection.
Consider the subset $Y = \{x \in X \big| x \notin f(x) \}$. Since $f$ is surjective, there is some $x_0 \in X$ such that $f(x_0) = Y$.
But one has $x_0 \in Y \Longleftrightarrow x_0 \notin Y$ by definition of $Y$, a contradiction!


TITLE: How to find $z$ with $|\sin z | \le 1$?
QUESTION [1 upvotes]: I am trying to find all $z \in \mathbb C$ such that $|\sin z |\le 1$.
What I did so far:
Clearly, for all real $z$ this is satisfied.
Next I tried to rewrite it like this:
$$ |\sin z |^2 =  |\sin x + iy |^2= \sin^2 x \cosh^2 y + \cos^2 x \sinh^2 y $$
but this doesn't seem to help. I can bound it by $\cosh^2 y + \sinh^2 y $ but that is not useful.
Now my question is:

How can I find all $z$ such that $|\sin z | \le 1$?

REPLY [3 votes]: Use that $\cos^2 x = 1-\sin^2 x$ and $\cosh^2 y-\sinh^2 y=1$, you get:
$$|\sin z|^2 = \sin^2 x\cosh^2 y +(1-\sin^2 x)\sinh^2 y = \sin^2 x + \sinh^2 y$$
So you need:
$$\sinh^2 y \leq 1-\sin^2 x =\cos^2 x$$


TITLE: Lower hemicontinuity of intersection of relations
QUESTION [1 upvotes]: This question is about the proof of a Lemma 7.1 of the paper
E. Michael, Continuous selections I.,Annals of Mathematics
Second Series, Vol. 63, No. 2 (Mar., 1956), pp. 361-382 (22 pages)
Published By: Mathematics Department, Princeton University

Statement of the problem:
Suppose $\phi$ is a lower hemicontinuous  relation from $X$ to $Y$ such that $\overline{\phi(x)}=\phi(x)\neq\emptyset$ for each $x\in X$. For each $x\in X$ let
$$ m(x)=\inf\{\|y\|:y\in\phi(x)\}$$
Assume there is a lower semicontinuous function $\rho:X\rightarrow\mathbb{R}_+$ such that $\rho(x)>m(x)$ whenever $m(x)>0$. Define the relation $\psi$ from $X$ to $Y$ as
$$
\psi(x)=\left\{\begin{matrix}
\phi(x)\cap \{y\in Y:\|y\|<\rho(x)\} &\text{if}&\rho(x)>0\\
\{0\}&\text{if}&\rho(x)=0
\end{matrix}\right.
$$
Then $\psi$ is lower hemicontinuous.

Definition of hemicontinuity: A relation $\phi:X\rightarrow 2^Y$ is lower hemicontinuous
if $\{x\in X: \phi(x)\cap V\neq\emptyset\}$ is open whenever $V\subset Y$ is open.

Some rather obvious observations:

$\phi(x)\neq\emptyset$ for all $x\in X$.
One also can show directly from definition of lower hemicontinuity that for any open set $U\subset Y$ such that $\phi(x)\cap U\neq\emptyset$, the relation $\gamma(x)=\phi(x)\cap U$ is lower semicontinuous.

Is the last relation that the author mentions leads to the validity of the statement. It is not clear to me how to exploit that fact, and may attempts have not been successful thus far. At issue is that the open set $\{y\in Y:\|y\|<p(x)\}$ when $p(x)>0$ depends on $x$, and I have not been able to get some uniform control.

Any hints or a sketch of a proof would be welcome.

REPLY [1 votes]: Here is a sketch of a proof for the statement in the OP.
Notation. I will use l.h.c to denote lower hemicontinuous (of correspondences), and l.s.c to denote lower semicontinuous (of functions). I also use $B(y;r)$ to denote the open ball centered at $y$ with radius $r>0$.
It suffices to show that $\psi$ is lower hemicontinuous at every $x\in X$.
Let $x\in X$.
First we make the observation that for any $x\in X$,  from the definition of $m$, $p(x)>0$ implies that
$$
\begin{align}
\phi(x)\cap B(\mathbf{0};p(x))\neq\emptyset\tag{1}\label{one}
\end{align}
$$
for in this case, $0\leq m(x)<p(x)$.

Case  $p(x)=0$. In this case $m(x)=0$, and since $\phi$ is closed valued, $\mathbf{0}\in\phi(x)$. Also $\psi(x)=\{\mathbf{0}\}$. Suppose $V\subset Y$ is open and $\mathbf{0}\in V$. Let $r>0$ such that $B(\mathbf{0};r)\subset V$. Since $\phi$ is l.h.c., there exists an open $W_x$ neighborhood of $x$ such that
$$
\emptyset\neq \phi(z)\cap B(\mathbf{0};r)\qquad z\in W_x
$$
For $z\in W_x$,

If $p(z)=0$, then $\psi(z)\cap V=\{\mathbf{0}\}\neq\emptyset$.
If $0<p(z)<r$, then by $\eqref{one}$ there exists $y\in \phi(z)\cap B(\mathbf{0};p(z))=\phi(z)\cap B(\mathbf{0};p(z))\cap V$.
If $r\leq p(z)$, $\emptyset\neq \phi(z)\cap B(\mathbf{0};r)\subset\phi(z)\cap B(\mathbf{0};p(x))\cap V$.

Case $p(x)>0$. Suppose $V\subset Y$ is open and that $y_0\in\phi(x)\cap B(\mathbf{0};p(x))\cap V$. Then, there is $\delta>0$ such that $B(y_0;\delta)\subset V$. It follows that $\|y_0\|_Y+\delta<p(x)$, for $y_0+ru\in B(y_0;r)$ fir any unitary vector $u\in Y$. Since $p$ is l.s.c., there exists an open neighborhood $W_x$ of $x$ such that
$$p(z)>\|y_0\|_Y+r,\qquad z\in W_x.$$
Since $\phi$ is l.h.c. there is an open neighborhood $V_x\subset W_x$ of $x$ such that
$$
\emptyset\neq\phi(z)]\cap B(y_0;r),\qquad z\in V_x
$$
If $z\in V_x$, $B(y_0;r)\subset B(\mathbf{0};p(z))\cap V$ and so, $\emptyset\neq \phi(z)\cap B(\mathbf{0};p(z))\cap V$.

Putting things together, we have shown that if $\psi(x)\cap V\neq\emptyset$, then there exists an one neighborhood $W_x$ of $x$ such that
$$ \psi(z)\cap V\neq\emptyset\qquad\text{whenever}\quad z\in W_x$$

Comments:
I took a look at E. Michael's paper and it seems that one can also prove the statement as a consequence of his Proposition 2.5 which states that if $V$ is an open entourage for some uniform structure on $Y$,  $\phi$ and $\psi$ are l.h.c. correspondences from $X$ to $Y$, and $\theta(x)=\phi(x)\cap V(\phi(x))\neq\emptyset$ for all $x\in X$, then $\theta$ is a l.h.c. correspondence from $X$ to $Y$. The proof does not seem complicated, but it does require some knowledge of uniformities which are not in much used nowadays.


TITLE: Is the set of certain polynomials finite or infinite?
QUESTION [5 upvotes]: Let us consider the set of all polynomials with the following properties:
i) all coefficients are integer;
ii) the leading coefficient equals one;
iii) all zeros are real and simple and belonging to $[-1.99,1.99] $.
Is this set finite or infinite?

REPLY [13 votes]: It is finite. This follows by combining two separate results:

If $p \in \mathbb{Z}[x]$ is monic integer having all its complex roots lying in $[-2,2]$, then all these roots are of the form $2\cos(2\pi q)$ with $q \in \mathbb{Q}$.

For the proof of this, write $p = \prod_{i=1}^d (x - 2\cos(2\pi \, t_i))$ and observe that the sequence of degree $d$ monic polynomials $p_n := \prod_{i=1}^d(x - 2\cos(2\pi n t_i))$ is also in $\mathbb{Z}[x]$ and that, of course, it has bounded coefficients (in terms of $d$ alone, independently of $n$). Thus there must be two $p_n = p_{n'}$ with $n \neq n'$; considering their roots, we get the claim with $q(n-n') \in \mathbb{Z}$.

For a given $d \in \mathbb{N}$, the monic polynomial $P_d$ with simple roots at the set $\{2\cos(2\pi m/d) \mid (m,d) = 1\}$ is in $\mathbb{Z}[x]$ and irreducible.

This follows from the corresponding irreducibility property of the cyclotomic polynomial $\Phi_d(z)$, which can be written as $z^{\deg{P_d}} P_d(z+1/z)$.
Combining these two statements, you can see that your set consists precisely of the products of pairwise different minimal polynomials of those $2\cos(2\pi / d)$ that do not exceed $1.99$. There are only finitely many such $d$.


TITLE: If $X$ is Poisson, find the expectation of $\frac{1}{a+X}$
QUESTION [4 upvotes]: If $X$ is a Poisson random variable with $\Pr(X=k)=e^{-\lambda}\frac{\lambda^k}{k!}$ and $a>0$ then find the expectation of $\frac{1}{a+X}$

If I make use of $\displaystyle\frac1{a+x}=\int_0^{\infty}e^{-(a+x)t}dt$
then I obtain;
$\displaystyle\mathbf E\bigg(\frac1{a+X}\bigg)=\sum\limits_{x=0}^{\infty}\bigg(\int_0^{\infty}e^{-(a+x)t}dt\bigg)e^{-\lambda}\frac{\lambda^x}{x!}$
now I can interchange sum and integral, but I don't see how to proceed.

REPLY [2 votes]: The result that I am obtaining using Maple is
$$\sum _{k=0}^{\infty }{\frac {{{\rm e}^{-\lambda}}{\lambda}^{k}}{
 \left( a+k \right) k!}}={\frac {\Gamma  \left( a+1 \right) -a\Gamma
 \left( a,-\lambda \right) }{{{\rm e}^{\lambda}}a \left( -\lambda
 \right) ^{a}}}
$$
or
$$\sum _{k=0}^{\infty }{\frac {{{\rm e}^{-\lambda}}{\lambda}^{k}}{
 \left( a+k \right) k!}}={\frac {{_{1}F_1(a;\,a+1;\,\lambda)}}{{{\rm e}^{
\lambda}}{a}}}
$$
or
$$\sum _{k=0}^{\infty }{\frac {{{\rm e}^{-\lambda}}{\lambda}^{k}}{
 \left( a+k \right) k!}}={\frac {
{{\rm M}\left(a,\,a+1,\,\lambda\right)}}{{{\rm e}^{\lambda}}{a}}}
$$
Other related results are:
$$\sum _{k=0}^{\infty }{\frac {{{\rm e}^{-\lambda}}{\lambda}^{k}}{
 \left( a+k \right) ^{2}k!}}={\frac {{_2F_2(a,a;\,a+1,a+1;\,\lambda)}}{
{{\rm e}^{\lambda}}{a}^{2}}}
$$
$$\sum _{k=0}^{\infty }{\frac {{{\rm e}^{-\lambda}}{\lambda}^{k}}{
 \left( a+k \right) ^{3}k!}}={\frac {
{_3F_3(a,a,a;\,a+1,a+1,a+1;\,\lambda)}}{{{\rm e}^{\lambda}}{a}^{3}}}
$$
$$\sum _{k=0}^{\infty }{\frac {{{\rm e}^{-\lambda}}{\lambda}^{k}}{
 \left( a+k \right) ^{n}k!}}={\frac {
{_nF_n(a,..,a;\,a+1,..,a+1;\,\lambda)}}{{{\rm e}^{\lambda}}{a}^{n
}}}
$$


TITLE: How to Find formula from values x and Y
QUESTION [0 upvotes]: How to find formula, based on this data:
x=1, y=9
x=3, y=6
x=5, y=3
x=7, y=0
If I put this data into the cartesian coordinate, the this graph is produced:

How to calculate the formula for this graph?
Please explain to me in easy way. Thank you.

REPLY [0 votes]: hint
A line has a cartesian equation of the form
$$y-y_0=\frac {\Delta y}{\Delta x}(x-x_0) $$
with $x_0=7\;,\;y_0=0,$$
$$\Delta x=3-2,\Delta y=6-9$, we find

$$y=-\frac {3}{2}(x-7) $$


TITLE: The direction of forces on a mass suspended between two vertical springs
QUESTION [0 upvotes]: I am trying to construct the free body diagram of a mass suspended by two vertical springs such

I concluded that the weight of the mass exerts a force on the upper spring causing a tension force to arise in the upper spring, and the weight of the mass exerts a force on the lower spring causing it to compress producing a thrust force.
Hence my diagram is as shown

However the textbook I am using instead shows the forces exerted by the springs acting in opposite directions,

I do not understand, why would the lower spring exert a tension force instead of a thrust force?

REPLY [3 votes]: It actually doesn’t matter. You can make either assumption and simply do the math to find the value of the force. If you get a negative number then you know your original guess was wrong, and if you get a positive number then you know your original guess was right. Either way works. Just guess one, and be consistent from that point on.


TITLE: For two problems A and B, if A is in P, then A is reducible to B?
QUESTION [3 upvotes]: For any two problems A and B, if A is in P(complexity) then A is reducible to B.
Could anyone explain why this is true?

REPLY [3 votes]: I'll assume you're using this definition of reduction:
$A$ is reducible to $B$ just if there exists a Turing machine running in time polynomial in the length of input that takes an instance $x$ of $A$ and halts with an instance $x'$ of $B$ such that $x\in L(A)$ if and only if $x' \in L(B)$, where $L(A), L(B)$ are the languages for which $A$ and $B$ are decision problems. This is the most standard defintion, but for precision should really be referred to as polynomial-time reducibility, which would probably have helped you out with the problem.
So, the problem statement being true depends on there being strings $y\in B$ and $y' \notin L(B)$ as long as there are strings both in and not in $L(A)$. If so, all we need to do is run a polynomial-time machine deciding whether $x \in L(A)$, then write $y$ at the end if $x \in L(A)$ and $y'$ if not. This last step only adds constant time, so it's certainly a polynomial reduction.


TITLE: Extension of 2D rotation matrix into 3D
QUESTION [0 upvotes]: In 2D euclidean space, rotation matrix is a matrix that tilts every single vector in the 2D space, without changing the scale. For example following matrix rotates the space by 90 degrees counter-clockwise.
$$R_2 = \begin{bmatrix}
0 & -1\\
1 & 0\\
\end{bmatrix}$$
Also, the tip of each unit vector is exactly the same distance from its original position, for example above, the tip of each unit vector moved by $\sqrt{2}$. This is parameterizable and we have a family of 2D rotation matrix given by
$$R_2(\theta) = \begin{bmatrix}\cos \theta &-\sin \theta \\\sin \theta &\cos \theta \\\end{bmatrix}$$
We can say that a rotation matrix is a linear transformation that moves all points on the surface of n-dimensional ball by the same amount.
This definition works for 1D, with the matrix $R_1 = \begin{bmatrix}-1\end{bmatrix}$ or $R_1(\theta) = \begin{bmatrix}(-1)^{\theta}\end{bmatrix}$ in complex plane.
Is there a matrix $R_3$ in 3D that moves each point on the surface of the unit sphere by the same amount? Is there $R_n$ ?

REPLY [2 votes]: In $\Bbb R^n$ with $n$ odd, each rotation (element of $SO(n)$) has $1$ as
an eigenvalue, so has a fixed point on the unit sphere.
If $n$ is even, consider the block matrix
$$\pmatrix{R&&&&\\&R&&&\\&&\ddots&\\&&&R}$$
where $R$ is a $2\times2$ rotation matrix. This matrix shifts each point on the
unit sphere by the same distance.


TITLE: Do planes in space have centers like surfaces and circles do?
QUESTION [2 upvotes]: A circular has a center, so can a plane have a center? A plane extends infinitely in all directions parallel to itself, so it doesn't make sense for a plane to have a center me thinks. If a plane doesn't have a center, does it not make sense to shift a plane right or left to the effect that the equation of the plane z=a is unchanged for some a$\in$ $\mathbb{R}$?.  I know it doesn't make sense for a plane to be centered at some point in space, but I think it makes sense for a surface to be centered at some point in space. If we can change where a  torus is centered in space by editing its equation, can we change where a plane is centered in space or does it not make sense?

REPLY [0 votes]: From a purely geometric point of view, I think you're right: there is no such thing as the center of a plane. But in a way that means that you can treat any point of the plane as the center (this idea can be useful in certain Physics problems that have a certain symmetry; e.g. computing the electric field across a parallel plane capacitor), and in a way we can select a point to be the center by imposing a coordinate system on the space.
For example, let's consider 2D space for a minute and consider the line defined by $y = 2x$. As a geometric object, this line also has no center. But with our usual coordinate system, we are tempted to view it as "centered" about the origin. We could imagine "shifting" the line along itself two units up and one unit to the right, and the "center" would now be located at the point $(1,2)$.
Geometrically, of course, we haven't changed anything. That's the symmetry of a line. But I think it's enlightening to see what happens to the equation of the line when we think about performing this "shift" operation. To shift the graph of some equation involving $x$ and $y$, we subtract the horizontal and vertical shift amounts from the corresponding variables. In this case, $x \to x-1$ and $y \to y-2$. Thus our line equation becomes
$$ y-2 = 2(x-1) $$
which, of course, simplifies into $y = 2x$ upon algebraic manipulation, which corresponds to the geometric symmetry that shifting a line along itself doesn't really change it. Also note that the equation above is the point-slope form of the equation of a line! And for that matter, when we construct an equation for a line using point-slope form, the point we pick could be thought of as the "center".
We can even play this game in 3D with your flat plane example of $z = a$, except, let's rewrite this equation in a way that reflects a coordinate system. I can rewrite it as
$$ z-a = 0(x-h) + 0(y-k) $$
which is something like the point-slope form for a line, but "centered" at the point $(h,k,a)$ in space. And you see that, no matter what $h$ and $k$ we choose, the equation trivially simplifies into $z=a$, which corresponds to the plane's symmetry.
So if you want to think of it this way, to "shift" a plane (or a line, or any other such symmetrical object) doesn't change the plane itself, but it does change the equation of the plane--although it remains algebraically equivalent to the original equation just as, geometrically, a "shifted" plane, though it is shifted, looks identical to the original.


TITLE: When uniformly sampling points from a high dimensional ball, intuitively, why are most of the points at the crust?
QUESTION [1 upvotes]: This is related to the curse of dimensionality. Many proofs can be found showing that the volume is mostly at the crust, but I don't quite understand the intuition.
Consider a $d$ dimensional unit ball - centered at the $d$ dimensional origin and having radius $1$. We can draw the points uniformly by drawing from $x_i \sim N(0, 1)$ for all $d$ components in the vector, and then normalize said vector. (Ref: https://stats.stackexchange.com/questions/7977/how-to-generate-uniformly-distributed-points-on-the-surface-of-the-3-d-unit-sphe)
This is telling me that, on average, each component of the vector will be close to $0$ - so wouldn't the points mostly be concentrated at the center of the sphere, not the crust?

REPLY [0 votes]: Imagine the $d$-ball to be made up of infinitely many concentric $(d-1)$-spheres. The surface of each of these is propositional to $r^{d-1}$. So if you take an infinitesimal or actual range of radii, the volume in the shell corresponding to that range will be proportional to that radius power, too.
For the $1$-ball (line segment) you get uniform weight: equal density at all radii. For the $2$-ball you already have density increasing linearly with radius. The higher the dimension, the higher the exponent and the stronger the concentration of density at the high end of the radius.

Consider a $d$ dimensional unit ball - centered at the $d$ dimensional origin and having radius 1.
We can draw the points uniformly by drawing from $x_i\sim N(0,1)$ for all $d$ components in the vector, and then normalize said vector.

Those instructions are for uniformly sampling from the sphere, not the ball. With the normalization you have to end up on the sphere, and nothing says that without it you would get uniformity in the interior of the ball. You don't.


TITLE: If the torsion subset of a group has "finite index", is the group torsion?
QUESTION [4 upvotes]: Let $T=\mathrm{Tor}(G)$ be the torsion subset of $G$, i.e. $$T=\mathrm{Tor}(G):=\{g\in G: g^n=1 \text{ for some } n\geq 1\}.$$
In general $T$ is not a subgroup of $G$, so it doesn't make sense to talk about the "index" of $T$ in $G$. But we can still ask if finitely many translates can cover $G$.
A set $S\subseteq G$ is called syndetic if finitely many translates of $S$ can cover $G$, i.e. $$G=g_1S\cup\cdots \cup g_n S$$ for some $g_1,\ldots,g_n\in G$. If $S$ is a subgroup then this is equivalent to $[G:S]<\infty$.
Of course if $G$ is a torsion group then $G=T$ so of course $T$ is syndetic. Are there are any other ways that $G$ can be a finite union of translates of $T$?

Problem. Let $T:=\mathrm{Tor}(G)$ be the torsion subset of $G$. Suppose that $T$ is syndetic in $G$. Prove that $G$ is a torsion group.

This checks out when $G$ is finitely-generated abelian, because then $T$ is a direct summand of $G$ with $G=\mathbb{Z}^r\oplus T$. So the only way $[G:T]<\infty$ is if $r=0$ and $G=T$.
What about for nonabelian groups? If $G$ is finitely-generated and the commutator subgroup $G'$ has infinite index, we can bootstrap the f.g. abelian case to solve the problem.

Attempt
Maybe we should try to show that $T$ is a subgroup of $G$. Then $[G:T]<\infty$ which immediately implies that $G$ is a torsion group. It is also clear that the subgroup $\langle T\rangle$ generated by $T$ must have (literal) finite index.
An easy application of the Pigeonhole Principle shows that if $T$ has finite index in $G$, then: for every $g\in G$, there is some $k\in \mathbb{Z}$, $|k|\leq n$ such that $g^k=st$ is a product of two torsion elements $s,t\in T$.

REPLY [7 votes]: Take the infinite dihedral group $D_{\infty} = \{r,s\mid s^2=1, sr=r^{-1}s\}$. The elements of $D_{\infty}$ are precisely the elements of the form $r^is^j$, with $i$ arbitrary and $j=0$ or $1$. The elements of finite order are precisely the identity, and those of the form $r^is$, which have order $2$. Thus, $T=\{r^is\mid i\in\mathbb{Z}\} \cup \{e\}$.
Since $r^i = s(r^{-i}s)\in sT$, then we have that $D_{\infty}=T\cup sT$. But $D_{\infty}$ is not a torsion group, since $r\in D_{\infty}$ has infinite order.
$D_{\infty}$ can also be realized as the semidirect product $\mathbb{Z}\rtimes\mathbb{Z}_2$, where the action is by inversion.
Note that this is an easy example that shows that the set of torsion elements need not be a subgroup in a nonabelian group (though it generates a characteristic subgroup since as a set it is invariant under automorphisms).


TITLE: equivalent property of open/closed immersion
QUESTION [0 upvotes]: I'm studying morphisms of ringed topological spaces. The definition of open/closed immersion in Qing Liu's "Algebraic Geometry and Arithmetic Curves" is as follows:

Definition 2.22. We say that a morphism $(f,f^\#):(X,\mathcal O_X)\to(Y,\mathcal O_Y)$ is an open immersion (resp. closed immersion) if $f$ is a topological open immersion (resp. closed immersion) and if $f^\#_x$ is an isomorphism (resp. if $f^\#_x$ is surjective) for every $x\in X$.

But in many occasions, I think the author is using the following equivalent (not sure) property of open/closed immersion:
My conjecture. A morphism $(f,f^\#):(X,\mathcal O_X)\to(Y,\mathcal O_Y)$ is an open immersion (resp. closed immersion) if and only if $f$ is a topological open immersion (resp. closed immersion) and $f^\#_y:\mathcal O_{Y,y}\to(f_*\mathcal O_X)_y$ is an isomorphism (resp. $f^\#_y$ is surjective) for every $y\in Y$.
Is it really true? I know that $f^\#_x = g\circ f^\#_{f(x)}$, where $g$ is the canonical map $g:(f_*\mathcal O_X)_{f(x)}\to\mathcal O_{X,x}, \ s_{f(x)}\mapsto s_x$, for each $x\in X$ (see the diagram below).
$$\mathcal O_{Y,f(x)}\to(f_*\mathcal O_X)_{f(x)}\to\mathcal O_{X,x}$$
But I can't go any further.

REPLY [1 votes]: No, this is not correct - if $y$ is not in the image of $f$, then $(f_*\mathcal{O}_X)_y$ is zero. This isn't a problem for the case of a closed immersion, but it is for an open immersion.


TITLE: Do we need $\mu, \nu$ to be $\sigma$-finite to show $\int fg \ d(\mu \otimes \nu) = \int f \ d\mu \int g \ d\nu$?
QUESTION [2 upvotes]: The problem statement:

Let $(X, \mathcal F, \mu), (Y, \mathcal G, \nu)$ be $\sigma$-finite and $f \in \mathcal L^1 (\mu), g \in \mathcal L^1 (\nu)$. Show that $fg \in \mathcal L^1 (\mu \otimes \nu)$ and $$\int fg \ d(\mu \otimes \nu) = \int f \ d\mu \int g \ d\nu.$$

I did the usual routine: let first $f,g$ be indicator functions, then simple functions, then non-negative measurable, then integrable.
My question is: where does one (implicitly) use the $\sigma$-finiteness of $\mu, \nu$? I went throw the argument and couldn't spot the place. Since this was a stand-alone exercise, I though it's rather unlikely for the setup to be redundant.

REPLY [1 votes]: $\mu \otimes \nu$ is only well-defined when $\mu$ and $\nu$ are $\sigma$-finite. –  Stefan Hansen


TITLE: What is the minimum value of $x^²+12x$?
QUESTION [0 upvotes]: What is the minimum value of $x^2+12x$?
I do not know what is meant by the minimum value.

REPLY [2 votes]: I realize this has been answered but here is my two bits worth.
Look at $x^2+12 x$. It is an expression. Now pick a value for $x$, say $x=1$. Then the value of the expression is $1 + 12 = 13$. Pick another value, say $x=-10$. Then the value is $100-120 = -20$. Notice that $-20$ is less than $13$. You keep picking different values of $x$. The question asks what is the smallest value can you get?
To answer this problem, you need one fact: Smallest value a square can ever be is zero.
Now to use this we can write
$$
x^2 + 12 x = x^2 + 12 x +36 - 36 = (x+6)^2 - 36
$$
I have not changed anything. If I put $x=-10$, I get $$
(-4)^2 - 36 = -20$$
same as before. But if I look at $(x+6)^2 -36$, I can't change the $-36$ but to make $(x+6)^2$ as small as possible, I have to set $x=-6$. So the complete answer is
The minimum value of $x^2 + 12 x$ is $-36$ and this happens when $x = -6$.
The trick of writing the expression as a square plus a constant is called completion of squares and you may need it a lot. Here is the formula for this:
If you have $a x^2 +  b x$ then add (and subtract) $b^2/(4 a)$ to complete the square.

REPLY [0 votes]: In general, to find the minimum of a quadratic function $ y = ax^2 + bx + c $, we can use a few different methods. Note that this function has a minimum iff $ a \ge 0 $. If $ a \le 0 $, it has a maximum, not a minimum.
Complete the square
Here, we try to write the function in the form $ y = a \cdot (x - h)^2 + k $. Once we have this, we know that $ a \cdot \left( x - h \right)^2 \ge 0 $, so $ y \ge k $. We can see this my graphing it as well. The vertex would be at $ (h, k) $.
An example of this would be the function $ y = x^2 + 12x = \left( x + 6 \right)^2 - 36 $, so the minimum value is $-36$, which occurs at $x=-6$.
Formula
It is a general rule, which can be proved using the complete-the-square method that the vertex of a parabola with equation $ y = ax^2 + bx + c $ occurs at $ x  = - \dfrac {b}{2a} $. Plugging in the $x$-value from here, we can find the $y$-coordinate of the vertex. Another proof of this uses Calculus.
Let $ f(x) = ax^2 + bx + c $. Then, differentiating, we have $ f'(x) = 2ax + b $. Setting this to $0$, we have $$ 2ax + b = 0 \implies x = - \dfrac {b}{2a}. $$


TITLE: Can the average temperature increase in a closed adiabatic (isolated) system?
QUESTION [0 upvotes]: Assume a closed and adiabatic (isolated) system (no energy or mass transfer at the boundaries) then is it possible for the average temperature to change (increase or decrease)?
Corollary:
1. If yes how? Give example of a process.
2. If not. Assume a system that encloses a system that converts mass into energy (Ex. nuclear reaction). The result of the nuclear reaction is producing energy inside the closed system and decreasing the mass inside the closed system. I expect that the resulting increase of energy represents a increase in the internal energy and thus an average temperature increase. Then what is the process that offsets that to maintain the average temperature constant?
I will add an example which we were talking about.
Assume a thermodynamic boundary - system - which does not exchange energy or mass with the surrounding (isolated). That closure includes a nuclear plant and the necessary energy sources to prime and sustain the nuclear reaction. All this system is isolated and has an average temperature T before initiating the reaction. The reaction is initiated - without any input of energy from outside - and a mass to energy conversion takes place. At this point the average temperature inside this system is going up, or stays the same?
The argument for the temperature to go up would be that the conversion of mass to energy produces some sensitive heat which increased the average temperature of the system. But is that violating the first law - as at the boundary no energy exchange took place?
If the average temperature stays the same then this heat production should be offset by something to keep the average the same. What is that?

REPLY [0 votes]: First of all, let's go with the assumption that you are defining the temperature as the average kinetic energy per particle. This is the case for ideal gases where the constituent particles have no internal structure and don't interact with each other, and in this case, the temperature is a direct measure of the average energy in the system.
If the system is closed and isolated, then the total energy of the system is conserved. In the case where we allow nuclear decay, some mass is converted into a different form of energy, and in the end this increases the kinetic energy of the "ideal" gas in the system, which means that the temperature increases, even though the total mass-energy is the same.
The reason there's a disconnect here is that on one hand, we want the temperature to be a measure of the average energy of the system, but on the other hand, temperature-as-average-kinetic energy only works as this measure if we have an ideal gas where the constituents don't have any internal structure.
The upshot is that if you really want to do thermodynamics where nuclear decay goes on in the system, you have two options:

Redefine your temperature so that it takes into account the nuclear degrees of freedom.
Use the ideal gas temperature, and model the nuclear decay process as heating. That is, model your system as not isolated, and model the nuclear decay process as heating from external sources. The problem here is that you might also need to introduce chemical potentials if the nuclear decay is one in which there is transmutation, i.e. where you are changing the numbers of constituent particles.


text:
\begin{document}

\title[3-dimensional gradient Yamabe solitons]
{3-dimensional complete gradient Yamabe solitons with divergence-free Cotton tensor}

\author{Shun Maeta}
\address{Department of Mathematics,
 Shimane University, Nishikawatsu 1060 Matsue, 690-8504, Japan.}
\curraddr{}
\email{shun.maeta@gmail.com~{\em or}~maeta@riko.shimane-u.ac.jp}
\thanks{The author is partially supported by the Grant-in-Aid for Young Scientists,
 No.15K17542 and No.19K14534 Japan Society for the Promotion of Science, and JSPS Overseas Research Fellowships 2017-2019 No.70.}

\subjclass[2010]{53C21, 53C25, 53C20}

\date{}

\dedicatory{}

\keywords{Yamabe solitons; Cotton tensor; Bach tensor; Scalar curvature}

\commby{}

\begin{abstract}
In this paper, we classify 3-dimensional complete gradient Yamabe solitons with divergence-free Cotton tensor. We also give some classifications of complete gradient Yamabe solitons with nonpositively curved Ricci curvature in the direction of the gradient of the potential function.
\end{abstract}

\maketitle

\bibliographystyle{amsplain}

\section{Introduction}\label{intro}
 A Riemannian manifold $(M^n,g)$ is called a gradient Yamabe soliton if there exist a smooth function $F$ on $M$ and a constant $\rho\in \mathbb{R}$, such that
\begin{equation}\label{YS}
(R-\rho)g=\nabla\nabla F,
\end{equation}
where $R$ is the scalar curvature on $M$ and $\n\n F$  is the Hessian of $F$.
If $\rho>0$, $\rho=0$, or $\rho<0$, then the Yamabe soliton is called shrinking, steady, or expanding.

If the potential function $F$ is constant, then the Yamabe soliton is called trivial. It is known that any compact Yamabe soliton is trivial (see for example \cite{CMM12},~\cite{Hsu12}). Yamabe solitons are special solutions of the Yamabe flow which was introduced by R. Hamilton \cite{Hamilton89}.
The Yamabe soliton equation $(\ref{YS})$ is similar to the equation of Ricci solitons.

Ricci solitons are special solutions of the Ricci flow

which was also introduced by R. Hamilton \cite{Hamilton82}.
As is well known, by using the Ricci flow, G. Perelman \cite{Perelman1},~\cite{Perelman2},~\cite{Perelman3} proved Thurston's geometrization conjecture \cite{Thurston} and Poincar\'e conjecture.
In the first paper of Perelman, he mentioned that ``any 3-dimensional complete noncompact $\kappa$-noncollapsed gradient steady Ricci soliton with positive curvature  is rotationally symmetric, namely Bryant soliton".
In \cite{CCCMM14}, H.-D. Cao, G. Catino, Q. Chen, C. Mantegazza and L. Mazzieri gave a partial answer to the conjecture.
Finally, S. Brendle proved the conjecture \cite{Brendle}.
In this paper, we consider the similar problem.
More precisely, we consider the following problem.
\begin{problem}\label{PCY}
Classify nontrivial non-flat complete 3-dimensional gradient Yamabe solitons.
\end{problem}
P. Daskalopoulos and N. Sesum \cite{DS13} showed that ``all locally conformally flat complete gradient Yamabe solitons with positive sectional curvature have to be rotationally symmetric". The proof was inspired by H.-D. Cao and Q. Chen's paper \cite{CC11}.
Furthermore, they constructed some examples of rotationally symmetric gradient Yamabe solitons on $\mathbb{R}^n$ with positive sectional curvature.
Recently,  H.-D. Cao, X. Sun and Y. Zhang relaxed the assumption, and showed that any nontrivial non-flat complete and locally conformally flat gradient Yamabe soliton with nonnegative scalar curvature is rotationally symmetric.
G. Catino, C. Mantegazza and L. Mazzieri's work \cite{CMM12} is also important.
They classified complete conformal gradient solitons with nonnegative Ricci tensor. As a corollary, they classified nontrivial complete gradient Yamabe solitons with nonnegative Ricci tensor. Finally, it is shown that complete gradient Yamabe solitons are rotationally symmetric under (1) nonnegative Ricci tensor is positive definite at some point, by Catino, Mantegazza and Mazzieri \cite{CMM12}, or (2) positive Ricci curvature, by Cao, Sun and Zhang \cite{CSZ12}.
Therefore, in this paper, we consider $3$-dimensional Yamabe solitons without any assumptions for non-negativity of curvatures.
Our main theorem gives an affirmative partial answer to Problem~\ref{PCY}:

\begin{theorem}\label{main}
Let $(M^3,g,F)$ be a nontrivial non-flat $3$-dimensional complete gradient Yamabe soliton with divergence-free Cotton tensor $($i.e., Bach flat$)$.

${\rm I}.$ If $M$ is steady, then $M$ is rotationally symmetric and equal to the warped product
$$([0,\infty),dr^2)\times_{|\n F|}(\mathbb{S}^{2},{\bar g}_{S}),$$
where $\bar g_{S}$ is the round metric on $\mathbb{S}^{2}.$

${\rm II}$. If $M$ is shrinking, then either

$(1)$ $M$ is rotationally symmetric and equal to the warped product
$$([0,\infty),dr^2)\times_{|\n F|}(\mathbb{S}^{2},{\bar g}_{S}),$$
where $\bar g_{S}$ is the round metric on $\mathbb{S}^{2},$ or

$(2)$ $|\n F|$ is constant and $M$ is isometric to the Riemannian product
$$(\mathbb{R},dr^2)\times \left(\mathbb{S}^2\left(\frac{1}{2}\rho|\n F|^2\right),\bar g\right),$$
where $\mathbb{S}^2(\frac{1}{2}\rho|\n F|^2)$ is the sphere of constant Gaussian curvature $\frac{1}{2}\rho|\n F|^2$.

${\rm III}$. If $M$ is expanding, then either

$(1)$ $M$ is rotationally symmetric and equal to the warped product
$$([0,\infty),dr^2)\times_{|\n F|}(\mathbb{S}^{2},{\bar g}_{S}),$$
where $\bar g_{S}$ is the round metric on $\mathbb{S}^{2},$ or

$(2)$ $|\n F|$ is constant and $M$ is isometric to the Riemannian product
$$(\mathbb{R},dr^2)\times \left(\mathbb{H}^2\left(\frac{1}{2}\rho|\n F|^2\right),\bar g\right),$$
where $\mathbb{H}^2(\frac{1}{2}\rho|\n F|^2)$ is the hyperbolic space of constant Gaussian curvature $\frac{1}{2}\rho|\n F|^2$.

\end{theorem}

\begin{remark}
For dimension $n\geq4$, Bach \cite{Bach} introduced the Bach tensor in $1920$'s.
\begin{align}
B_{ij}
=&\frac{1}{n-3}\n^k\n^lW_{ikjl}+\frac{1}{n-2}R_{kl}W_i{}^{k}{}_j{}^l\\
=&\frac{1}{n-2}(\n_kC_{kij}+R_{kl}W_i{}^{k}{}_j{}^l),\notag
\end{align}
where $\nabla$ is the Levi-Civita connection, $W$ is the Weyl tensor, $R_{ij}$ is the Ricci tensor, and $C$ is the Cotton tensor.
 In \cite{CCCMM14}, the Bach tensor for $3$-dimensional manifolds was introduced as follows:
$$B_{ij}=\n_kC_{kij}.$$

\end{remark}

The remaining sections are organized as follows. Section~$\ref{Pre}$ contains some necessary definitions and preliminary geometric results.

Section~$\ref{Proof of main}$ is devoted to the proof of Theorem~$\ref{main}$.
In section~\ref{NRIC}, we consider complete gradient Yamabe solitons with nonpositively curved Ricci curvature in the direction of the gradient of the potential function.

\section{Preliminary}\label{Pre}
The Riemannian curvature tensor is defined by
$$R(X,Y)Z=-\n_X\n_YZ+\n_Y\n_XZ+\n_{[X,Y]}Z.$$

The Ricci tensor $R_{ij}$ (also denoted by $\Ric$) is defined by
$R_{ij}=R_{ipjp}.$
The Weyl tensor $W$ and  the Cotton tensor C are defined by
\begin{align*}
R_{ijkl}
=&W_{ijkl}
+\frac{R}{(n-1)(n-2)}(g_{il}g_{jk}-g_{ik}g_{jl})\\
&-\frac{1}{n-2}(R_{il}g_{jk}+R_{jk}g_{il}-R_{ik}g_{jl}-R_{jl}g_{ik})\\
=&W_{ijkl}+S_{ik}g_{jl}+S_{jl}g_{ik}-S_{il}g_{jk}-S_{jk}g_{il},
\end{align*}
\begin{align*}
C_{ijk}
=&\nabla_iR_{jk}-\nabla_jR_{ik}-\frac{1}{2(n-1)}(g_{jk}\nabla_iR-g_{ik}\nabla_jR)\\
=&\n_iS_{jk}-\n_jS_{ik},\notag
\end{align*}
where $S=\Ric-\frac{1}{2(n-1)}Rg$ is the Schouten tensor.
 The Cotton tensor is skew-symmetric in the first two indices and totally trace free, that is,
$$C_{ijk}=-C_{jik} \quad \text{and} \quad g^{ij}C_{ijk}=g^{ik}C_{ijk}=0.$$

As is well known, a Riemannian manifold $(M^n,g)$ is locally conformally flat if and only if
(1) for $n\geq4$, the Weyl tensor vanishes; (2) for $n=3$, the Cotton tensor vanishes.
Moreover, for $n\geq4$, if the Weyl tensor vanishes, then the Cotton tensor vanishes. We also see that for $n=3$, the Weyl tensor always vanishes, but the Cotton tensor does not vanish in general.

We prove some formulas needed later.
Taking trace of the Yamabe soliton equation \eqref{YS},
\begin{equation}\label{TYS}
n(R-\rho)=\Delta F,
\end{equation}
where $\Delta$ is the Laplacian on $M$.
In general, we have
\begin{equation}\label{p.1}
\Delta {\nabla}_iF={\nabla}_i\Delta F+R_{ij}{\nabla}_jF.
\end{equation}
Substituting
\begin{equation*}
\Delta {\nabla}_iF={\nabla}_k{\nabla}_k{\nabla}_iF={\nabla}_k((R-\rho)g_{ki})={\nabla}_iR,
\end{equation*}
and
\begin{equation*}
{\nabla}_i\Delta F={\nabla}_i(n(R-\rho))=n{\nabla}_iR,
\end{equation*}
into $(\ref{p.1})$, we have
\begin{equation}\label{p.2}
(n-1)\nabla_iR+R_{il}\nabla_lF=0.
\end{equation}
Thus, we have
\begin{equation}\label{p.3}
(n-1)g(\nabla R,\nabla F)=-\Ric(\nabla F,\nabla F).
\end{equation}
On the other hand, by $(\ref{p.2})$ and the contracted second Bianchi identity,
\begin{equation}\label{p.4}
(n-1)\Delta R+\frac{1}{2}g(\nabla R, \nabla F)+R(R-\rho)=0.
\end{equation}
Combining $(\ref{p.3})$ with $(\ref{p.4})$, we obtain
\begin{equation}\label{p.5}
\Delta R=\frac{1}{2(n-1)^2}\Ric(\nabla F,\nabla F)-\frac{1}{n-1}R(R-\rho).
\end{equation}
\quad\\

\section{Proof of Theorem $\ref{main}$}\label{Proof of main}
In this section, we prove Theorem $\ref{main}$.

To prove Theorem \ref{main}, we use the following useful theorem by H.-D. Cao, X. Sun and Y. Zhang:

\begin{theorem}[\cite{CSZ12}]\label{Thm of CSZ12}
Let $(M^n,g,F)$ be a nontrivial complete gradient Yamabe soliton. Then, $|\n F|^2$ is constant on regular level surfaces of $F$, and either

$(1)$ $F$ has a unique critical point at some point $p_0\in M$, and $M$ is rotationally symmetric and equal to the warped product
$$([0,\infty),dr^2)\times_{|\n F|}(\mathbb{S}^{n-1},{\bar g}_{S}),$$
where $\bar g_{S}$ is the round metric on $\mathbb{S}^{n-1},$ or

$(2)$ $F$ has no critical point and $M$ is the warped product
$$(\mathbb{R},dr^2)\times_{|\n F|}(N^{n-1},\bar g),$$
where N is a Riemannian manifold of constant scalar curvature.
Furthermore, if the Ricci curvature of $N$ is nonnegative, then $M$ is isometric to the Riemannian product
$(\mathbb{R},dr^2)\times(N^{n-1},\bar g)$; if $R\geq0$, then either $R>0$, or $R=\overline R=0$ and $(M,g)$ is isometric to the Riemannian product $(\mathbb{R},dr^2)\times(N^{n-1},\bar g)$.
\end{theorem}

\begin{proof}[Proof of Theorem~$\ref{main}$]

We only have to consider the case (2) of Theorem $\ref{Thm of CSZ12}$.
Since
\begin{align*}
\n_iB_{ij}
=&\n_i\n_kC_{kij}\\
=&\n_i\n_k(\n_kS_{ij}-\n_iS_{kj})\\
=&\n_i\n_k\n_kS_{ij}-\n_k\n_i\n_kS_{ij}\\
=&R_{ikkp}\n_pS_{ij}+R_{ikip}\n_kS_{pj}+R_{ikjp}\n_kS_{ip}\\
=&-R_{ip}\n_pS_{ij}+R_{kp}\n_kS_{pj}+(S_{ij}g_{kp}+S_{kp}g_{ij}-S_{ip}g_{kj}-S_{kj}g_{ip})\n_kS_{ip}\\
=&S_{ij}C_{kik}+S_{ip}C_{ijp}\\
=&-C_{jip}R_{ip},
\end{align*}
\begin{align}\label{m1}
\n_i\n_jB_{ji}=-\n_iC_{ijk}R_{jk}-C_{ijk}\n_iR_{jk}.
\end{align}
By the definition and a property of the Cotton tensor,
\begin{align*}
C_{ijk}\n_iR_{jk}
=&C_{ijk}(C_{ijk}+\n_jR_{ik}+\frac{1}{4}(g_{jk}\n_iR-g_{ik}\n_jR))\\
=&|C_{ijk}|^2-C_{jik}\n_jR_{ik}.
\end{align*}
Thus, we have
\begin{equation}\label{m2}
C_{ijk}\n_iR_{jk}=\frac{1}{2}|C_{ijk}|^2.
\end{equation}
Substituting (\ref{m2}) into (\ref{m1}), we have
$$\n_i\n_jB_{ji}=-B_{jk}R_{jk}-\frac{1}{2}|C_{ijk}|^2.$$
By the assumption, the Cotton tensor vanishes.

As in the proof of  Theorem \ref{Thm of CSZ12}, it is shown that

in any open neighborhood $U$ of $N^{2}$ in which $F$ has no critical points,
$$g=dr^2+(F'(r))^2{\bar g}=dr^2+\frac{(F'(r))^2}{(F'(r_0))^2}g_{ab}(r_0,x)dx^adx^b,$$ where $(x^2, x^3)$ is any local coordinates system on $N^{2}$ and $\bar g=(F'(r_0))^{-2}\bar g_{r_0}$, where $\bar g_{r_0}$ is the induced metric on $N^{2}$.

By a direct calculation, we can get formulas of the warped product manifold of the warping function $|\n F|=F'(r).$
 For $a,b,c,d=2,3,$
\begin{align}\label{RT1}
R_{1a1b}&=-F'F'''{\bar g}_{ab},\quad R_{1abc}=0,\\
R_{abcd}&=(F')^2{\bar R}_{abcd}+(F'F'')^2(\bar g_{ad}\bar g_{bc}-\bar g_{ac}\bar g_{bd}),\notag
\end{align}
\begin{align}\label{RT2}
R_{11}=&-2\frac{F'''}{F'},\quad
R_{1a}=0,\\
R_{ab}=&\bar R_{ab}-((F'')^2+F'F''')\bar g_{ab},\notag
\end{align}
\begin{align}\label{RT3}
R=(F')^{-2}\bar R-2\Big(\frac{F''}{F'}\Big)^2-4\frac{F'''}{F'},
\end{align}
where the curvature tensors with bar are the curvature tensors of $(N,\bar g)$.
By (\ref{YS}),
\begin{align}\label{R-rho}
R-\rho=F''.
\end{align}
Since $(N^2,{\overline g})$ is a 2-dimensional manifold,
$${\overline R}_{abcd}=-\frac{\bar R}{2}(\bar g_{ad} \bar g_{bc}-\bar g_{ac}\bar g_{bd}),$$
$$\bar R_{ad}=\frac{\bar R}{2}\bar g_{ad}.$$
Substituting these into (\ref{RT1}) and (\ref{RT2}),  we have
\begin{align}\label{RT11}
R_{1a1b}&=-F'F'''{\bar g}_{ab},\quad R_{1abc}=0,\\
R_{abcd}&=-(F')^3\Big(\frac{1}{2}F'R+2F'''\Big)(\bar g_{ad}\bar g_{bc}-\bar g_{ac}\bar g_{bd}),\notag
\end{align}
\begin{align}\label{RT22}
R_{11}=&-2\frac{F'''}{F'},\quad
R_{1a}=0,\\
R_{ab}=&\Big(\frac{R}{2}(F')^2+F'F'''\Big)\bar g_{ab}.\notag
\end{align}

Hence, the Cotton tensor $C_{ijk}$ is
\begin{equation*}
C_{ijk}=\left\{
\begin{aligned}
&\nabla_1(R_{ab}-\frac{1}{4}Rg_{ab})
&\quad (a,b=2,3),\\
&0\qquad (other).
\end{aligned}
\right.
\end{equation*}
Thus, we have
\begin{equation}\label{key2}
\frac{R}{4}(F')^2+F'F'''=c~(\text{constant}).
\end{equation}
Combining \eqref{key2} with $(\ref{RT3})$, we have
$$(F'')^2=\frac{1}{2}\bar R -2c.$$
Since $\bar R$ is constant, $F''$ is constant.
Thus, $R$ is constant by the Yamabe soliton equation.
Therefore the equation $(\ref{key2})$ is as follows.
\begin{equation}\label{key3}
\frac{1}{4}R(F')^2=c.
\end{equation}

If $c=0,$ then $R=0$. From this and $(\ref{RT11})$, $M$ is flat.

If $c\not=0$, then we have $R\not=0$ and $F'$ is constant. Thus, $R-\rho=F''=0.$

Case I. $M$ is steady: We have $R=\rho=0$, which is a contradiction.

Case II. $M$ is shrinking: Since $R=\rho>0$ and $(\ref{RT3})$,
$\bar R=R(F')^2=\rho|\n F|^2>0$.

Case III. $M$ is expanding: Since $R=\rho<0$ and $(\ref{RT3})$,
$\bar R=R(F')^2=\rho|\n F|^2<0$.

\end{proof}

\section{Complete gradient Yamabe solitons with $\Ric(\nabla F,\nabla F)\leq0$}\label{NRIC}

As mentioned before, H.-D. Cao, X. Sun and Y. Zhang showed that any nontrivial non-flat complete and locally conformally flat gradient Yamabe soliton with $R\geq0$ is rotationally symmetric. Therefore, in this section, we consider Yamabe solitons with $\Ric(\n F,\n F)\leq0$ instead of ``locally conformally flat".

\begin{proposition}\label{prop1}
Let $(M^n,g,F)$ be an $n$-dimensional complete gradient Yamabe soliton with $\Ric(\n F, \n F)\leq0$.
Suppose that $F$ has no critical point. Then, the following holds.

$(1)$ $M$ is shrinking or steady: If $R\geq\rho$, then $R=\rho.$

$(2)$ There exists no expanding soliton with $R\geq0.$
\end{proposition}

As a corollary, by the similar argument as in the proof of Theorem $\ref{main}$, we can classify nontrivial non-flat complete 3-dimensional gradient Yamabe solitons:

If $M$ is shrinking with $R\geq\rho$ and $\Ric(\n F,\n F)\leq0$, then either

(1) $M$ is rotationally symmetric and equal to the warped product
$$([0,\infty),dr^2)\times_{|\n F|}(\mathbb{S}^{2},{\bar g}_{S}),$$
where $\bar g_{S}$ is the round metric on $\mathbb{S}^{2}$, or

(2) $|\n F|$ is constant and $M$ is isometric to the Riemannian product
$$(\mathbb{R},dr^2)\times \left(\mathbb{S}^2\left(\frac{1}{2}\rho|\n F|^2\right),\bar g\right),$$
where $\mathbb{S}^2(\frac{1}{2}\rho|\n F|^2)$ is the sphere of constant Gaussian curvature $\frac{1}{2}\rho|\n F|^2$.

If $M$ is steady or expanding with $R\geq0$ and $\Ric(\n F,\n F)\leq0$, then
$M$ is rotationally symmetric and equal to the warped product
$$([0,\infty),dr^2)\times_{|\n F|}(\mathbb{S}^{2},{\bar g}_{S}).$$
\\

\begin{proof}[Proof of Proposition $\ref{prop1}$]
As in the proof of Theorem $\ref{Thm of CSZ12}$, it is shown that
 in any open neighborhood $U$ of $N^{n-1}$ in which $F$ has no critical points,
$$g=dr^2+(F'(r))^2{\bar g}=dr^2+\frac{(F'(r))^2}{(F'(r_0))^2}g_{ab}(r_0,x)dx^adx^b,$$ where $(x^2,\cdots, x^n)$ is any local coordinates system on $N^{n-1}$ and $\bar g=(F'(r_0))^{-2}\bar g_{r_0}$, where $\bar g_{r_0}$ is the induced metric on $N^{n-1}$.

By a direct calculation, we can get formulas of the warped product manifold of the warping function $|\n F|=F'(r).$
 For $a,b,c,d=2,\cdots,n,$
\begin{align}\label{RT1-2}
R_{1a1b}&=-F'F'''{\bar g}_{ab},\quad R_{1abc}=0,\\
R_{abcd}&=(F')^2{\bar R}_{abcd}+(F'F'')^2(\bar g_{ad}\bar g_{bc}-\bar g_{ac}\bar g_{bd}),\notag
\end{align}
\begin{align}\label{RT2-2}
R_{11}=&-(n-1)\frac{F'''}{F'},\quad
R_{1a}=0,\\
R_{ab}=&\bar R_{ab}-((n-2)(F'')^2+F'F''')\bar g_{ab},\notag
\end{align}
\begin{align}\label{RT3-2}
R=(F')^{-2}\bar R-(n-1)(n-2)\Big(\frac{F''}{F'}\Big)^2-2(n-1)\frac{F'''}{F'}.
\end{align}
By (\ref{YS}),
\begin{align}\label{R-rho-2}
R-\rho=F''.
\end{align}
Since $\n F=F'\frac{\partial}{\partial r}$,
\begin{equation}\label{Ricnf}
\Ric(\n F,\n F)=(F')^2R_{11}=-(n-1)F'F'''.
\end{equation}
By the assumption, $R'=F'''\geq0$.
By the definition of the Laplacian,
\begin{align*}
\Delta R
=&g^{ij}(\partial_i\partial_jR-\Gamma_{ij}^k\partial_kR)\\
=&R''-g^{ij}\Gamma_{ij}^1R',
\end{align*}
where $\partial_1=\frac{\partial}{\partial r}$ and $\partial _i=\frac{\partial}{\partial x_i},~(i=2,\cdots,n).$
The Christoffel symbol is given by
\begin{align*}
\Gamma_{ij}^1
=&\frac{1}{2}g^{1k}(\partial_ig_{jk}+\partial_{j}g_{ik}-\partial_kg_{ij})\\
=&\frac{1}{2}(\partial_ig_{j1}+\partial_{j}g_{i1}-\partial_1g_{ij}).
\end{align*}
Here,
$$\partial_1g_{11}=0, \quad \partial_ag_{11}=0\quad\text{and}\quad \partial_1g_{ab}=2F'F''\bar g_{ab}.$$
Thus, we have
$$\Gamma_{11}^1=0,\quad \Gamma_{1a}^1=0\quad\text{and}\quad\Gamma_{ab}^1=-F'F''\bar g_{ab}.$$
Hence,
\begin{equation}\label{DR}
\Delta R=R''+(n-1)\frac{F''}{F'}R'.
\end{equation}
Combining $(\ref{DR})$ with $(\ref{p.5})$,
\begin{equation}\label{R''}
R''=-(n-1)\frac{F''}{F'}R'-\frac{1}{2(n-1)}F'R'-\frac{1}{n-1}R(R-\rho).
\end{equation}

Case $(1)$, $M$ is shrinking or steady:
 By the assumption, $R\geq\rho(\geq0)$, that is, $F''\geq0$.
From this and $R'\geq0$,
$$F^{(4)}=R''\leq0.$$
Thus, $F''$ is a non-negative weakly concave function, which means that $F''$ must be constant.
Hence, $R$ is constant. By $(\ref{R''})$, $R=\rho$.

Case $(2)$, $M$ is expanding: Assume that there exists an expanding Yamabe soliton with $R\geq0.$ The same argument shows $R=0,$ that is, $F''=-\rho<0$.
From this, $F'(>0)$ is a non-constant linear function, which cannot happen.

\end{proof}

\begin{remark}
If we assume that $\Ric (\n F,\n F)\geq0$ instead of $\Ric(\n F,\n F)\leq0$ on Proposition $\ref{prop1}$, then we immediately obtain $R=\rho$ without the assumption $R\geq\rho$ $($or $R\geq0$$)$.
In fact, by $(\ref{Ricnf})$, $F'''\leq0.$ Thus, $F'$ is a positive weakly concave function, which means that $F'$ must be constant.
Therefore, $R=\rho$.

As a result, we can get the same classification as in Theorem $\ref{main}$ for $n=3$, under $\Ric(\n F, \n F)\geq0$ instead of flatness of the Bach tensor.
\end{remark}

By the similar argument as in the proof of Theorem $\ref{main}$,
we can get the following classification of complete gradient Yamabe solitons with $\Ric(\nabla F, \nabla F)\leq0$.

\begin{lemma}\label{lem1}
Let $(M^n,g,F)$ be an $n$-dimensional complete gradient Yamabe soliton with Ricci curvature bounded from below and $\Ric(\n F, \n F)\leq0$. Suppose that $F$ has no critical point.
Then, the following holds.

$(1)$ There exists no shrinking soliton with $R\leq0$.

$(2)$ $M$ is steady or expanding: If $R\leq \rho$, then $R=\rho.$

\end{lemma}

\begin{proposition}
Let $(M^3,g,F)$ be a nontrivial non-flat $3$-dimensional complete gradient Yamabe soliton with Ricci curvature bounded from below and $\Ric(\n F, \n F)\leq0$.

${\rm I}.$ $M$ is shrinking or steady: If $R\leq0$, then $M$ is rotationally symmetric and equal to the warped product
$$([0,\infty),dr^2)\times_{|\n F|}(\mathbb{S}^{2},{\bar g}_{S}),$$
where $\bar g_{S}$ is the round metric on $\mathbb{S}^{2}.$

${\rm II}$. $M$ is expanding: If $R\leq\rho$, then either

$(1)$ $M$ is rotationally symmetric and equal to the warped product
$$([0,\infty),dr^2)\times_{|\n F|}(\mathbb{S}^{2},{\bar g}_{S}),~\text{or}$$

$(2)$ $|\n F|$ is constant and $M$ is isometric to the Riemannian product
$$(\mathbb{R},dr^2)\times \left(\mathbb{H}^2\left(\frac{1}{2}\rho|\n F|^2\right),\bar g\right),$$
where $\mathbb{H}^2(\frac{1}{2}\rho|\n F|^2)$ is the hyperbolic space of constant Gaussian curvature $\frac{1}{2}\rho|\n F|^2$.
\end{proposition}

\begin{proof}[Proof of Lemma $\ref{lem1}$]
$(1)$ $M$ is shrinking: Assume that there exists a Yamabe soliton with $R\leq0$. Set $L=-R\geq0$.
By $(\ref{p.5})$,
\begin{align*}
\Delta L
=&-\frac{1}{2(n-1)^2}\Ric (\n F,\n F)+\frac{1}{n-1}L(L+\rho)\\
\geq&-\frac{1}{2(n-1)^2}\Ric (\n F,\n F)+\frac{1}{n-1}L^2.
\end{align*}
By the assumption $\Ric(\n F,\n F)\leq0$, we obtain
$$\Delta L\geq\frac{1}{n-1}L^2.$$
Since $L$ is nonnegative, by Omori-Yau's maximum principle, $L=0.$ Thus, $F''=-\rho$ (as in the proof of Proposition~\ref{prop1}). Hence, $F'(>0)$ is a non-constant linear function, which cannot happen.

$(2)$ $M$ is steady or expanding:
 Set $u=\rho-R$. By $(\ref{p.5})$,
\begin{align*}\Delta u
=&-\frac{1}{2(n-1)^2}\Ric (\n F,\n F)+\frac{1}{n-1}R(R-\rho)\\
\geq&-\frac{1}{2(n-1)^2}\Ric (\n F,\n F)+\frac{1}{n-1}u^2.
\end{align*}
By the assumption $\Ric(\n F,\n F)\leq0$, we obtain
$$\Delta u\geq\frac{1}{n-1}u^2.$$
Since $u$ is nonnegative, by Omori-Yau's maximum principle, $u=0,$ that is, $R=\rho.$

\end{proof}

\noindent
{\bf Acknowledgments.}~
The work was done while the author was visiting the Department of Mathematics of Texas A $\&$ M University-Commerce as a Visiting Scholar and he is grateful to the department and the university for the hospitality he had received during the visit.

\bibliographystyle{amsbook}

config: arxiv


text:
\begin{document}

\maketitle

\begin{abstract}
For smooth metric measure spaces $(M, g, e^{-f} dvol)$ we prove a Liuoville-type theorem when the Bakry-Emery Ricci tensor is nonnegative. ÊThis generalizes a result of Yau, which is recovered in the case $f$ is constant. This result follows from a gradient estimate for f-harmonic functions on smooth metric measure spaces with Bakry-Emery Ricci tensor bounded from below.
\end{abstract}

\section{Introduction}

 In this paper we study smooth metric measure spaces.  A smooth metric measure space is a complete Riemannian manifold together with a measure conformal to the Riemannian measure. More precisely a smooth metric measure space is a triple, $(M^n , g , e^{-f} dvol)$, where M is a complete Riemannian manifold of dimension $n$, $f$ is a smooth real-valued function on $M$ and $dvol$ is the Riemannian volume density.

Smooth metric measure spaces carry natural analogs of the Ricci tensor.  In this paper we study the Bakry-Emery Ricci tensor, or $\infty$-Bakry-Emery Ricci tensor, defined by $Ric_f = Ric + \hess (f)$.  This tensor can be thought of as the limit of the N-Bakry-Emery Ricci tensors defined by $Ric^N_f = Ric + \hess(f)- \frac{1}{N} df \otimes df$.  The Bakry-Emery Ricci tensor, as well as its finite dimensional variants, were studied extensively by Bakry and Emery in their investigations of diffusion processes.  It also arises in the theory of the Ricci flow, as $Ric_f = \lambda g$ is precisely the gradient Ricci soliton equation.  Smooth metric measure spaces also carry a natural analog of the Laplace-Beltrami operator, the f-Laplacian, defined for a function $u$ by $\Delta_f (u)= \Delta(u) - \inner{\nabla u}{\nabla f}$.

 A great deal is known about the geometric and topological implications of the Ricci tensor.   The natural question, first studied by Lichnerowicz \cite{MR0268812,MR0300228}, is which of these results hold for the Bakry-Emery Ricci tensor.  This is an active field of current research. See, for example \cite{MR2170766,MR2016700,arXiv:math.DG/0612532,MR2577473, arXiv:1003.0191} or \cite{weisurvey} and the references therein.

In $1976$ Yau  \cite{MR0417452} famously showed that any positive or bounded harmonic function on a complete Riemannian manifold with $Ric \geq 0$ must be constant.  The goal of this paper is to establish a corresponding result for f-harmonic functions on smooth metric measure spaces with $Ric_f \geq 0$.  For positive f-harmonic functions this was shown in \cite{arXiv:math.DG/0612532} for smooth metric measure spaces with uniformly positive Bakry-Emery Ricci tensor.  Such a result, however,  is impossible for positive f-harmonic functions on smooth metric measure spaces with nonnegative Bakry-Emery Ricci tensor, as the following example illustrates:
\begin{example}
\label{example}
Consider $\mathbb{R}^n$ with the usual metric and let $f(x) = x_1$, the first coordinate function.  Clearly $Ric_f \geq 0$ in this case.  Now consider the function $u(x) = e^{x_1}$ and note
\begin{equation}
\Delta_f (u) = \Delta(u) - \inner{\nabla u}{ \nabla f} = u'' - u' =0.
\end{equation}
Thus $u$ is a non-constant positive f-harmonic function on $(\mathbb{R}^n, g , e^{-f} dvol)$.
\end{example}
On the other hand for smooth metric measure spaces with nonnegative Bakry-Emery Ricci tensor we show that bounded f-harmonic functions are constant.
\begin{theorem}
\label{Liouville}
Let $(M,g,{e^{-f}} dvol)$ be a complete smooth metric measure space with ${Ric_f \geq 0}$.  If $u$ is a bounded f-harmonic function defined on $M$ then $u$ is constant.
\end{theorem}
In the case of the N-Bakry-Emery Ricci tensor, for finite N, similar Liouville-type theorems are obtained in \cite{MR2170766, MR2487589, Hua2010, MR2296933}.  The Liouville-type theorem is proved by establishing the following estimate on the gradient of an f-harmonic function.
  \begin {theorem}
  \label{uestimate}
Let $(M,g,{e^{-f}} dvol)$ be a complete smooth metric measure space with ${Ric_f \geq -(n-1)H^2}$ where $H \geq 0$.  If $u$ is a positive f-harmonic function defined on $\overline{B(q;2R)}$ with $R \geq 1$, then for any $q_0 \in \overline{B(q;R)}$,
\begin{equation}
\abs{\nabla u}(q_0) \leq \sqrt{\frac{c_1(n, \alpha)}{R} + c_2(n)H^2} \supover{p}{B(q;2R)}{u},
\end{equation}
\noindent where $\alpha = \maxover{p}{\{p:d(p,q)=1\}}{\Delta_f r}$.
\end{theorem}
\noindent The constants above are given explicitly in \eqref{final}.  In the case $H=0$, Theorem~\ref{Liouville} then immediately follows by taking a limit as $R$ tends to infinity.  The next section is dedicated to the proof of this estimate.  A related estimate is obtained in \cite{arXiv:0902.2226}.

The idea of Yau's original proof was to apply the Bochner formula to the log of a positive harmonic function to obtain an initial estimate, then multiply by a cut-off function and apply the maximum principle and Laplacian comparison.  On smooth metric measure spaces there is a Bochner formula which looks similar to the classical Bochner formula, see \cite{MR2577473} for example.  This formula is given by
\begin{equation}
\frac{1}{2} \Delta_f \abs{\nabla u}^2 = \abs{ \hess \, u}^2 + \inner{\nabla u}{\nabla( \Delta_f u)} + Ric_f (\nabla u , \nabla u).
\end{equation}
The difference between the two Bochner formulae is the Hessian term which is controlled by the Laplacian.  In the context of smooth metric measure spaces the Laplacian is related to the f-Laplacian by $\Delta u = \Delta_f u +\inner{\nabla u}{\nabla f}$.  Thus, when squared, the first and last terms will be positive, so the issue is controlling the cross term.  The cross term is controlled by choosing a suitable function $h$ of the given f-harmonic function and examining two cases depending on $h$ and the relative magnitudes of $\Delta h$ and  $\inner{\nabla h}{\nabla f}$. The cross term is then collected with the larger of the leading and final terms to obtain an initial estimate.  Then we multiply by a cut-off function, use the maximum principle and apply the f-Laplacian estimate of Wei-Wylie \cite[Theorem 2.1]{MR2577473} to control the resulting terms.

In the final section we use the the ideas of the proof of Theorem 2 to obtain an estimate for $\abs{\nabla \log(u)}$ as in Yau \cite{MR0417452}.  However, choosing $h= \log(u)$ a gap develops between the constants required to define the two cases.  Thus the estimate is only established with an additional assumption on the relative sizes of $\nabla \log(u)$ and $\inner{\nabla \log(u)}{\nabla f}$.  Specifically we establish the following:
\begin {theorem}
\label{logestimate}
Let $(M,g,{e^{-f}} dvol)$ be a smooth complete metric measure space with $Ric_f \geq -(n-1)H^2$.  If $u$ is a positive f-harmonic function defined on $\overline{B(q;2R)}$ with $R \geq 1$, such that either $\inner{\nabla f}{\nabla \log(u)} \leq a\abs{\nabla \log (u)}^2$ for some $0\leq a<\frac{1}{2}$ or $\inner{\nabla f}{\nabla \log(u)} \geq 2\abs{\nabla \log (u)}^2$ then for any $q_0 \in \overline{B(q;R)}$,
\begin{equation}
 \abs{\nabla (\log u)}(q_0)  \leq \sqrt{c_1(n, a)H^2 + \frac{c_2(n,a,\alpha)}{R}}.
\end{equation}
\end{theorem}
\begin{remark}
Note that when $f$ is constant this recovers the original estimate of Yau \cite{MR0417452}.
\end{remark}
\begin{remark}
	The assumption on the relative sizes of $\inner{\nabla f}{\nabla \log(u)}$ and $\abs{\nabla \log (u)}^2$ here is required as Example \ref{example} demonstrates.  Note that in Example \ref{example}, $\frac{\inner{\nabla f}{\nabla \log(u)}}{\abs{\nabla \log (u)}} = 1$.
\end{remark}

\section{Proof of Theorems 1 and 2}
The proof of Theorem~\ref{uestimate} will follow from applying the Bochner formula to an appropriate function $h$ of a given positive f-harmonic function $u$.  As noted above, the problem term in the Bochner formula is the Hessian term, which is controlled with the Laplacian.  Indeed, if $\inner{\nabla u}{\nabla f}$ is small in relation to $\Delta u$, then $\Delta u$ can be controlled by $\Delta_f u$ since $\Delta u = \Delta_f u +\inner{\nabla u}{\nabla f}$.   Applying the Bochner formula to $\log(u)$ then yields the desired estimate, just as in the classical case.  On the other hand, if  $\inner{\nabla u}{\nabla f}$ is large in relation to $\Delta u$ this approach fails.  In this case $\Delta u$ is controlled by $\inner{\nabla u}{\nabla f}$.  Here, applying the Bochner formula directly to $u$ will give the estimate.  These two estimates can be combined directly, but a more convenient method suggested by Dr. Zhiqin Lu is to apply the Bochner formula to $u^\epsilon$ for some $\epsilon \in (0,1)$.  This is the proof we follow below.
\begin{proof}[Proof of Theorem~\ref{uestimate}]
\noindent Let $h=u^{\epsilon}$, where $\epsilon \in (0,1)$.  Applying the Bochner formula to $h$, we find:
\begin{equation} \label{bochner}
\frac{1}{2} \Delta_f \abs{\nabla h}^2 = \abs{ \hess \, h}^2 + \inner{\nabla h}{\nabla( \Delta_f h)} + Ric_f (\nabla h , \nabla h).
\end{equation}
Now we compute,
\begin{equation}
\begin{split}
\Delta_f h &= \Delta h - \inner{\nabla h}{\nabla f},\\
                  &= \epsilon (\epsilon -1) u^{\epsilon-2} \abs{\nabla{u}}^2 +\epsilon u^{\epsilon-1}\Delta_f u, \\
                  &=\frac{(\epsilon-1)\abs{\nabla{h}}^2}{\epsilon h}.
\end{split}
\end{equation}
 Where the last equality uses the fact that $u$ is f-harmonic.  Further $Ric_f (\nabla h , \nabla h)  \geq -(n-1)H^2 \abs{\nabla h}^2$ and by the Schwarz inequality we find:
\begin{equation} \label{hessh}
\begin{split}
\abs{\hess \, h}^2 &\geq \frac{\abs{\Delta h}^2}{n},\\
                             &= \frac{1}{n}(\Delta_f h + \inner{\nabla f}{\nabla h})^2,\\
                             &=\frac{1}{n}(\frac{(\epsilon-1)\abs{\nabla{h}}^2}{\epsilon h} + \inner{\nabla f}{\nabla h})^2.\\
 \end{split}
 \end{equation}
 Also,
\begin{equation} \label{mixed}
\begin{split}
\inner{\nabla h}{\nabla (\Delta_f h)}&=\inner{\nabla h}{\nabla (\frac{(\epsilon-1)\abs{\nabla{h}}^2}{\epsilon h})},\\
                                                              &=\frac{\epsilon-1}{\epsilon h}\inner{\nabla h}{\nabla \abs{\nabla h}^2} - \frac{(\epsilon-1)\abs{\nabla h}^4}{\epsilon h^2}.
\end{split}
\end{equation}
Thus substituting \eqref{hessh} and \eqref{mixed} into \eqref{bochner},
\begin{equation}
\begin{split}
\frac{1}{2} \Delta_f \abs{\nabla h}^2 &\geq \frac{ (\epsilon-1)^2\abs{\nabla h}^4}{n\epsilon^2 h^2} + \frac{2(\epsilon-1) \abs{ \nabla h}^2}{n\epsilon h} \inner{\nabla h}{\nabla f} + \frac{1}{n}\inner{\nabla h}{\nabla f}^2\\
                                                               &\qquad+ \frac{\epsilon-1}{\epsilon h}\inner{\nabla h}{\nabla \abs{\nabla h}^2} - \frac{(\epsilon-1)\abs{\nabla h}^4}{\epsilon h^2} -(n-1)H^2 \abs{\nabla h}^2.
\end{split}
\end{equation}

We now examine two cases according to which term in $\Delta h$ dominates.  First, suppose $p \in \overline{B(q;2R)}$ such that $\inner{\nabla h}{\nabla f} \leq a\frac{\abs{\nabla h}^2}{h}$, for some constant $a$ to be determined.  Thus at $p$ we find:

\begin{equation}
\begin{split}
\frac{1}{2} \Delta_f \abs{\nabla h}^2 &\geq \frac{ (\epsilon-1)^2\abs{\nabla h}^4}{n\epsilon^2 h^2} + \frac{2(\epsilon-1) \abs{ \nabla h}^2}{n\epsilon h} (\frac{a\abs{\nabla h}^2}{h}) + \frac{1}{n}\inner{\nabla h}{\nabla f}^2\\
                                                                 & \qquad + \frac{\epsilon-1}{\epsilon h}\inner{\nabla h}{\nabla \abs{\nabla h}^2} - \frac{(\epsilon-1)\abs{\nabla h}^4}{\epsilon h^2} -(n-1)H^2 \abs{\nabla h}^2,\\
&\geq \frac{(\epsilon-1)^2 +2a\epsilon (\epsilon-1) -n\epsilon (\epsilon-1)}{n\epsilon^2}\frac{\abs{\nabla h}^4}{h^2} + \frac{\epsilon-1}{\epsilon h}\inner{\nabla h}{\nabla \abs{\nabla h}^2} -(n-1)H^2 \abs{\nabla h}^2.
\end{split}
\end{equation}
We note that the coefficient of $\frac{\abs{\nabla h}^4}{h^2}$ will be positive whenever
\begin{equation}
a < \frac{n\epsilon - \epsilon +1}{2\epsilon}.
\end{equation}
Now suppose to the contrary that $p \in \overline{B(q;2R)}$ such that $\inner{\nabla h}{\nabla f} \geq a\frac{\abs{\nabla h}^2}{h}$.  Thus at $p$ we find
\begin{equation}
\begin{split}
\frac{1}{2} \Delta_f \abs{\nabla h}^2 &\geq \frac{ (\epsilon-1)^2\abs{\nabla h}^4}{n\epsilon^2 h^2} + \frac{2(\epsilon-1)}{n\epsilon h}(\frac{h}{a}\inner{\nabla h}{\nabla f}) \inner{\nabla h}{\nabla f} + \frac{1}{n}\inner{\nabla h}{\nabla f}^2\\
                                                                & \qquad + \frac{\epsilon-1}{\epsilon h}\inner{\nabla h}{\nabla \abs{\nabla h}^2} - \frac{(\epsilon-1)\abs{\nabla h}^4}{\epsilon h^2} -(n-1)H^2 \abs{\nabla h}^2,\\
                                                                &=\frac{(\epsilon-1)^2 - n(\epsilon-1)}{n\epsilon^2 h^2}\abs{\nabla h}^4 + \frac{2(\epsilon-1) +\epsilon a}{n\epsilon a}\inner{\nabla h}{\nabla f}^2 + \frac{\epsilon-1}{\epsilon h}\inner{\nabla h}{\nabla \abs{\nabla h}^2} -(n-1)H^2 \abs{\nabla h}^2.
\end{split}
\end{equation}
Thus the coefficient of $\inner{\nabla h}{\nabla f}^2 $ will be positive whenever
\begin{equation}
a > \frac{2(1-\epsilon)}{\epsilon}.
\end{equation}
Hence provided $\epsilon \in (\frac{3}{4} ,1)$ there will exist an $a$ satisfying both of these inequalities.  In particular, choosing $\epsilon = \frac{7}{8}$, we may take $a= \frac{1}{2}$ and in either case,
\begin{equation} \label{estimate}
\frac{1}{2} \Delta_f \abs{\nabla h}^2 \geq (\frac{7n-6}{49nh^2}) \abs{\nabla h}^4 - \frac{1}{7h}\inner{\nabla h}{\nabla \abs{\nabla h}^2} -(n-1)H^2 \abs{\nabla h}^2.
\end{equation}
Therefore this estimate holds on all of $\overline{B(q;2R)}$.

Now we wish to multiply $\abs{\nabla h}^2$ by a cut-off function to obtain a function which achieves its maximum inside $B(q;2R)$.  Let $g:[0,2R] \rightarrow [0,1]$ be a $C^3$ function with the following properties: \\
a.)$ g \vert _{[0,R]} =1$,\\
b.) $\supp(g) \subseteq [0,2R)$, \\
c.) $ \frac{-c}{R}\sqrt{g} \leq g\prime \leq 0$,\\
d.) $\abs{g \prime \prime} \leq \frac{c}{R^2}$,\\
for some $c > 0$.  Define $\phi : \overline{B(q;2R)} \rightarrow [0,1]$ by $\phi(x)=g(d(x,q))$ and set $G=\phi \abs{\nabla h}^2$.  We first compute $\nabla G = (\nabla \phi)(\abs{\nabla h}^2) + (\phi)(\nabla \abs{\nabla h}^2)$.  Hence
\begin{equation} \label{gradsquared}
\abs{\nabla h}^2 = \frac{G}{\phi},
\end{equation}
and
\begin{equation} \label{gradgradsquared}
\nabla \abs{\nabla h}^2 = \frac{\nabla G}{\phi} - \frac{\nabla \phi}{\phi ^2}G.
\end{equation}
Further,
\begin{equation}
\begin{split}
\Delta_f G &=(\Delta \phi)(\abs{\nabla h}^2) + 2\inner{\nabla \phi}{ \nabla \abs{\nabla h}^2} + (\phi)(\Delta \abs{\nabla h}^2) - \inner{\nabla f}{ (\nabla \phi)\abs{\nabla h}^2} - \inner{\nabla f}{ (\phi)(\nabla \abs{\nabla h}^2)},\\
&= (\Delta \phi)(\abs{\nabla h}^2) + 2\inner{\nabla \phi}{ \nabla \abs{\nabla h}^2}  - \inner{\nabla f}{ (\nabla \phi)(\abs{\nabla h}^2)} + \phi \Delta_f \abs{\nabla h}^2.
\end{split}
\end{equation}
Thus,
\begin{equation} \label{flaplace}
\begin{split}
\Delta_f \abs{\nabla h}^2 &= \frac{1}{\phi} \Delta_f G - \frac{\abs{\nabla h}^2}{\phi} \Delta_f \phi - 2\inner{\frac{\nabla \phi}{\phi}}{ \nabla\abs{\nabla h}^2},\\
&=\frac{1}{\phi} \Delta_f G- \frac{G}{\phi^2} \Delta_f \phi - 2\inner{\frac{\nabla \phi}{\phi}}{ \frac{\nabla G}{\phi} - \frac{\nabla \phi}{\phi^2}G}.
\end{split}
\end{equation}
Now substituting \eqref{gradsquared}, \eqref{gradgradsquared} and \eqref{flaplace} into \eqref{estimate} and solving for $\frac{1}{\phi} \Delta_f G$ we find that
\begin{equation}
\frac{1}{\phi} \Delta_f G \geq \frac{G}{\phi^2} \Delta_f \phi +2\inner{\frac{\nabla \phi}{\phi}}{ \frac{\nabla G}{\phi}} - 2\inner{\frac{\nabla \phi}{\phi}}{\frac{\nabla \phi}{\phi^2}G}  + (\frac{14n-12}{49nh^2}) \frac{G^2}{\phi^2} - \frac{2}{7h}\inner{\nabla h}{\frac{\nabla G}{\phi} - \frac{\nabla \phi}{\phi ^2}G} -2(n-1)H^2\frac{G}{\phi}.
\end{equation}

Now $G$ must achieve its maximum on the compact set $\overline{B(q;2R)}$ and since $G$ vanishes on the boundary, it must attain its maximum at some point $q_0 \in B(q;2R)$.  We assume $q_0$ is not in the cut locus of $q$, if it is we adjust $\phi$ slightly and interpret the following inequalities in the barrier sense.  Evaluating the above at $q_0$ and solving for the $G$ term,

\begin{equation}\label{almostdone}
(\frac{14n-12}{49nh^2}) G \leq -\Delta_f \phi+\frac{2}{\phi} \abs{\nabla \phi}^2-\frac{2}{7h}\inner{\nabla h}{\nabla \phi} +2(n-1)\phi H^2.
\end{equation}

Now we note that if $q_0 \in B(q;1)$, then since $\phi$ is the identity on $\overline{B(q;R)}$, $q_0$ is the maximum point of $\abs{\nabla h}$ on $\overline{B(q;R)}$.  Thus \eqref{almostdone} yields,

\begin{equation}
\abs{\nabla u} \leq \sqrt{\frac{64n(n-1)}{7n-6}} H \supover{p}{B(q;2R)}{u},
\end{equation}
on ${B(q;R)}$, which is the desired estimate.  Thus we may assume $q_0 \notin B(q;1)$.
Now to control the f-Laplacian term in \eqref{almostdone}, note that $\phi$ is a radial function and hence $\Delta_f \phi = g^\prime \Delta_f r + g^{\prime \prime}$, where $r(x)=d(x,q)$.  Now since ${Ric_f \geq -(n-1)H^2}$ and $d(q,q_0) \geq 1$, the Laplacian estimate of \cite{MR2577473} gives,
\begin{equation}
\Delta_f r(q_0) \leq \alpha + (n-1)H^2 (2R-1).
\end{equation}
Further by the construction of $\phi$, $\frac{-c}{R} \sqrt{g} \leq g^\prime \leq 0$ and $g^{\prime \prime} \geq -\frac{c}{R^2}$.  Hence,
\begin{equation} \label{deltaphibd}
\begin{split}
\Delta_f \phi &\geq \frac{-c}{R}\Delta_f(r) -\frac{c}{R^2},\\
                       & \geq \frac{-c \alpha}{R} - c(n-1)H^2 \frac{2R-1}{R} -\frac{c}{R^2}.
\end{split}
\end{equation}
Finally to control the $\inner{\nabla h}{ \nabla \phi}$ term, note that for any $\delta \geq 0$,
\begin{equation} \label{crossterm}
2\abs{\inner{\nabla h}{ \nabla \phi}} \leq 2\abs{\nabla h} \abs{\nabla \phi}  \leq \frac{\delta \phi}{h} \abs{\nabla h}^2 + \frac{h}{\delta \phi} \abs{ \nabla \phi}^2.
\end{equation}

Thus, substituting \eqref{crossterm} and \eqref{deltaphibd} into \eqref{almostdone} and collecting terms we find:
\begin{equation}
(\frac{14n-12}{49nh^2} - \frac{\delta}{7h^2}) G \leq \frac{c \alpha}{R} +\frac{c}{R^2}+(2+\frac{1}{7\delta})\frac{\abs{\nabla \phi}^2}{\phi} +2(n-1)H^2(\phi + c \frac{R-1}{R}) .
\end{equation}
Now choosing $\delta=\frac{1}{7}$, noting that $\frac{\abs{\nabla \phi}^2}{\phi} \leq \frac{c^2}{R^2}$ and using $R \geq 1$ we find,
\begin{equation}
\frac{13n-12}{49nh^2}G \leq \frac{c \alpha + c+3c^2}{R} +2(n-1)H^2(\phi + c) .
\end{equation}
Finally restricting to $B(q;R)$ and using the definition of $h$ we conclude that
\begin{equation} \label{final}
\abs{\nabla u} \leq \sqrt{ \frac{64n(c \alpha + c+3c^2)}{13n-12} \frac{1}{R} +\frac{128n(n-1)(1+ c)}{13n-12}H^2} \supover{p}{B(q;2R)}{u}.
\end{equation}
\end{proof}

\section{Proof of Theorem 3}

In this section we prove Theorem~\ref{logestimate}.  As noted above the proof begins by applying the Bochner formula to $\log(u)$ and then using the assumption to control the Hessian term with $\Delta_f \log(u)$.

\begin{proof}[Proof of Theorem~\ref{logestimate}]
Define $h=\log(u)$. First we note that,
\begin{equation}
\begin{split}
\Delta_f h &=\Delta h - \inner{\nabla f}{ \nabla h}, \\
                  &=\frac 1{u} \Delta_f u - \abs{ \frac {\nabla u} {u} }^2.
\end{split}
\end{equation}
Thus since $u$ is f-harmonic, $\Delta_f h = - \abs{ \nabla h}^2$.  Now applying the Bochner formula we find:
\begin{equation}
\frac 1{2} \Delta_f \abs{ \nabla h}^2 = \abs{\hess \, h}^2 + \inner{\nabla h}{ \nabla (\Delta_f h)} + Ric_f (\nabla h, \nabla h).
\end{equation}
Further, by the Schwarz inequality $\abs{\hess \, h}^2 \geq \frac {\abs{\Delta h}^2}{n}$. Now,
\begin{equation}
\begin{split}
\abs{ \Delta h}^2 &= \abs{ \Delta_f h + \inner{\nabla h}{\nabla f}}^2,\\
                               &=\abs{ - \abs{ \nabla h}^2 + \inner{\nabla h}{\nabla f}}^2,\\
                               &= \abs{ \nabla h}^4 - 2\abs{\nabla h}^2 \inner{\nabla h}{\nabla f} + \inner{\nabla h}{\nabla f}^2.
 \end{split}
 \end{equation}
Thus if the first assumption holds we find:
\begin{equation}
\begin{split}
\abs{ \Delta h}^2 &\leq \abs{ \nabla h}^4 - 2a\abs{\nabla h}^4,\\
                               &=(1- 2a)\abs{ \nabla h}^4.
 \end{split}
 \end{equation}
If the second assumption holds, then:
 \begin{equation}
\begin{split}
\abs{ \Delta h}^2&\leq \abs{ \nabla h}^4 - \inner{\nabla f}{\nabla h}^2 + \inner{\nabla h}{\nabla f}^2,\\
                              &\leq \abs{ \nabla h}^4.
 \end{split}
 \end{equation}
Also, $Ric_f (\nabla h, \nabla h) \geq -(n-1)H^2 \abs{\nabla h}^2$. Thus substituting the above into the Bochner formula in either case we find that,
 \begin{equation} \label{sboch2}
 \frac{1}{2} \Delta_f \abs{\nabla h}^2 \geq \frac{1-2a}{n} \abs{\nabla h}^4 - \inner{\nabla h}{\nabla(\abs{\nabla h}^2)}-(n-1)H^2 \abs{\nabla h}^2.
 \end{equation}
From here the proof follows the same argument as Theorem~\ref{uestimate}, yielding
\begin{equation}
\abs{\nabla h} \leq \sqrt{\left({\frac{2nc\alpha + 2nc}{3(1-2\alpha)} +\frac{2n(2n+2-4a)c^2}{3(1-2\alpha)^2}}\right) \frac{1}{R} + \frac{4n(c+1)(n-1)}{3(1-2a)}H^2},
\end{equation}
where $a$ is taken to be $0$ in the case the second assumption holds.
\end{proof}

\section{Acknowledgements}
I wish to thank Guofang Wei for her guidance, tutelage and support.  In addition, I'd like to thank Zhiqin Lu and Jeffrey Case for their valuable discussions and advice.

\bibliographystyle{plain}
 \bibliography{citations}
\end{document}


TITLE: Connected Components Graph proof
QUESTION [2 upvotes]: I am trying to do this one problem for a homework set, and am not entirely sure how I would even start this proof. Here is the question
Prove, by induction on k, that a connected component of k nodes has at least k − 1 edges.
Any suggestions? Thanks in advance

REPLY [1 votes]: Outline of Proof: Let $G$ be a connected graph with $k$ vertices. If every vertex of $G$ has at least two edges, then the number of edges must be at least $k$.
If not every vertex of $G$ has two or more edges, let $v$ be a vertex that has only one edge. Remove $v$, and $v$'s single edge. The remaining graph $G'$ is connected, has $k-1$ vertices, and therefore by the induction hypothesis has at least $k-2$ edges. Put our vertex $v$ back, and its edge. That gets us to at least $k-1$ edges.


TITLE: super martingale which dominates
QUESTION [0 upvotes]: Could anyone tell me what it means by ''it dominates $\{E (Z|G_k), G_k\}, E(|Z|)<\infty$'' ? in the following statement? Thanks.
Let $(Y_k, G_k)$ be a super-martingale and it dominates $\{E (Z|G_k), G_k\}, E(|Z|)<\infty$,

REPLY [1 votes]: In general, we say that a random variable $Y$ dominates a random variable $X$ if
\begin{align}
|X(\omega)|\leq Y(\omega) \quad a.s.
\end{align}
This language is used for example in the Dominated Convergence Theorem.
In your case, you have a supermartingale $\{Y_k\}$ and a stochastic process $\{E[Z|G_k]\}$. Thus, that $\{Y_k\}$ dominates $\{E[Z|G_k]\}$ simply means that for all $k$ we have
\begin{align}
\big|E[Z|G_k]\big|\leq Y_k \quad a.s.
\end{align}


TITLE: Probability of Getting Two Specific Cards in 7-Card Poker Hand, Using Ordinary Deck of 52 Cards?
QUESTION [2 upvotes]: If we have an ordinary deck of 52 cards, how many possible 7 card poker hands are there that contain exactly one "2" and 1 "King"?
Originally, I was thinking of it as $$ \frac{\binom{52}{4}*\binom{48}{4}}{\binom{52}{7}}$$.
However, I'm not sure that it takes into account all necessary combinations.
So now I'm thinking that maybe it is just 52*48*(44*43*42*41*40).  Does that seem correct?
Thank you in advance for any help that anyone can provide!

REPLY [1 votes]: Ways. The number of possible hands with exactly one Deuce and exactly one King is
$${4\choose 1}^2 {44 \choose 5},$$
as in the numerator of @Henry's first fraction.
Evaluated in R as:
4^2*choose(44,5)
[1] 17376128

Exact probability. Then the probability of getting that outcome
is
$$\frac{{4\choose 1}^2 {44 \choose 5}}{{52\choose y}}
= 0.1298814,$$
as in his Comment. Evaluated in R as follows:
4^2*choose(44,5)/choose(52,7)
[1] 0.1298814

Simulated probability. This probability can be simulated with reasonable accuracy by looking at results
from 10 million 7-card hands. For simplicity in coding I have
simulated Aces (1) and Deuces (20), instead of Deuces and Kings.
[Code numbers assigned to denominations are arbitrary as long
as there is no ambiguity in the desired total (here 21).]
With ten million iterations, the probability
should be accurate to within $\pm 0.0002$ in 95% of
such simulations.
set.seed(1234)       # for reproducibility
deck = c(1,1,1,1,20,20,20,20, rep(0, 52-8))
tot = replicate(10^7, sum(sample(deck,7)))
mean(tot==21)        # mean of logical vector is its prop of TRUEs
[1] 0.1299681        # aprx P(1 K & 1 D) = 0.1298814
2*sd(tot==21)/sqrt(10^7)
[1] 0.0002126748     # 95% margin of simulation error


TITLE: If $\|Df\|<M$ in a disk, then $M$ is a Lipschitz constant
QUESTION [0 upvotes]: Some notes on multi-variable calculus I was reading, they quote a "standard result":

Suppose $f:\mathbb{R}^n \to \mathbb{R}^n$ is such that $\| f'(x)\|
 \leq M$ for $x\in D$, where $D$ is a closed ball. Then for points $x,
 x+h \in D$, we have $$|f(x+h)-f(x)| \leq M |h|.$$

The way I can think of to prove this is to take a polygonal path from $x$ to $x+h$ whose length is very close to $h$, and where each linear step is parallel to one of the axes. Then we can use the mean value theorem on each little step. Is that the way you would prove it, or am I missing something totally obvious?

REPLY [1 votes]: I would prove it using the convexity of $D$. Define $\gamma: [0,1]\to \mathbb{R}^n$ by $\gamma(t) = x+th$. Note that $\gamma(0) = x$, $\gamma(1) = x+h$, and $\gamma'(t) = h$. The composition $f\circ \gamma$ has derivative:
$$(f\circ \gamma)'(t) = f'(\gamma(t))\,\gamma'(t)$$
(which is shown via a simple application of multivariable chain rule). By the FTC,
$$f(x+h) - f(x) = \int_0^1 f'(\gamma(t))\,\gamma'(t)\,\textrm{d}t.$$
If you take norms and apply a few basic properties of integrals, you should be able to finish the proof from there.


text:
\section{The Real topological Hochschild homology of $H\underline{\mathbb{Z}}$, by Jeremy Hahn and Dylan Wilson}

In this appendix we explain how the results of the main body of the paper allow one to calculate the Real topological Hochschild homology of the Eilenberg--MacLane Mackey functor $\HZ$.  In particular, we deduce the following theorem, which verifies a conjecture of Dotto, Moi, Patchkoria, and Reeh \cite[p. 63]{Real_THH}.

\begin{thm} \label{app:main-thm}
There is an equivalence of $\HZ$-module spectra
$$\THR(\HZ) \simeq \HZ \oplus \bigoplus_{k \ge 2} \Sigma^{k\rho-1} H\underline{\mathbb{Z}/k}.$$
\end{thm}

Dotto, Moi, Patchkoria, and Reeh were able to prove that Theorem \ref{app:main-thm} holds after localization at any odd prime, and so also after localization away from $2$ \cite[Theorem 5.27 \& Corollary 5.28]{Real_THH}.  However, they did not have methods to calculate $\THR(\HZ)_{(2)} \simeq \THR(\HZ_{(2)})$.  On the other hand, the main body of this paper provides methods to calculate the $\THR$ of Thom spectra, and the authors of this appendix previously constructed $\HZ_{(2)}$ as a $C_2$-equivariant Thom spectrum in \cite{HahnWilson}.  These results were combined in Corollary \ref{cor-em}(ii) of the main body to prove that
$$\THR(\HZ_{(2)}) \simeq \HZ_{(2)} \otimes \Sigma^{\infty}_+\Omega^\sigma (S^{\lambda +1}\langle \lambda +1 \rangle).$$
The main contribution of the appendix is to observe that this can be made more explicit:

\begin{lem} \label{app:main-lem}
There is an equivalence of $\HZ_{(2)}$-module spectra
$$\HZ_{(2)} \otimes \Sigma^{\infty}_+ \Omega^\sigma (S^{\lambda +1}\langle \lambda +1 \rangle) \simeq \HZ_{(2)} \oplus \bigoplus_{k \ge 1} \Sigma^{k\rho-1} H\underline{\mathbb{Z}/k}_{(2)}.$$
\end{lem}
We deduce Lemma \ref{app:main-lem} from the non-equivariant calculation of $\mathrm{THH}(H\mathbb{Z})$ together with the following $C_2$-equivariant fact, which we prove before $2$-localization:
\begin{lem} \label{app:red-lem}
There is a cofiber sequence of $\HZ$-module spectra
$$\HZ \otimes \Sigma^{\infty}_+ \Omega^{\sigma} S^{\lambda+1} \langle \lambda+1 \rangle \to \bigoplus_{k \ge 0} \Sigma^{k\rho} \HZ \to \bigoplus_{s \ge 1} \Sigma^{s \rho} \HZ.$$
\end{lem}

\begin{proof}[Proof of Lemma \ref{app:red-lem}]
Applying $\Omega^{\sigma}$ to the definition of $S^{\lambda+1} \langle \lambda+1\rangle$ yields a fiber sequence of $C_2$-equivariant spaces
$$\Omega^{\sigma} S^{\lambda+1} \langle \lambda+1 \rangle \to \Omega^{\sigma} S^{\lambda+1} \to \Omega^{\sigma} \mathrm{K}(\lambda+1,\underline{\mathbb{Z}}),$$
where $\Omega^{\sigma} \mathrm{K}(\lambda+1,\underline{\mathbb{Z}}) = \Omega^{\sigma} \mathrm{K}(2\sigma+1,\underline{\mathbb{Z}}) \simeq \mathrm{K}(\sigma+1,\underline{\mathbb{Z}}) \simeq \mathbb{CP}^{\infty}_{\mathbb{R}}.$

In particular, since $\mathbb{CP}^{\infty}_{\mathbb{R}}$ classifies Real line bundles, it follows that the \emph{cofiber} of the map $\Omega^{\sigma} S^{\lambda+1} \langle \lambda+1\rangle \to \Omega^{\sigma} S^{\lambda+1}$ is the Thom space of a Real line bundle $\mathcal{L}$ over $\Omega^{\sigma} S^{\lambda+1}$.  Using the fact that $\HZ$ is Real oriented, we conclude that there is a cofiber sequence of $\HZ$-modules
$$\HZ \otimes \Sigma^{\infty}_+ \Omega^{\sigma} S^{\lambda+1} \langle \lambda+1 \rangle \to \HZ \otimes \Sigma^{\infty}_+ \Omega^{\sigma} S^{\lambda+1} \to \HZ \otimes \Sigma^{\infty}_+ \left( \Omega^{\sigma} S^{\lambda+1} \right)^{\mathcal{L}} \simeq \Sigma^{\rho} \HZ \otimes \Sigma^{\infty}_+ \Omega^{\sigma} S^{\lambda+1}.$$
By \cite[Theorem 4.3]{HillAlgebras}, there is a James splitting
$$\Sigma^{\infty}_+ \Omega^{\sigma} S^{\lambda+1} \simeq \Sigma^{\infty}_+ \Omega^{\sigma} \Sigma^{\sigma} S^{\rho} \simeq S^0 \oplus S^{\rho} \oplus S^{2\rho} \oplus S^{3\rho} \oplus \cdots.$$
In particular, there is a cofiber sequence of $\HZ$-modules
$$\HZ \otimes \Sigma^{\infty}_+ \Omega^{\sigma} S^{\lambda+1} \langle \lambda+1 \rangle \to \bigoplus_{k \ge 0} \Sigma^{k\rho} \HZ \to \bigoplus_{s \ge 1} \Sigma^{s \rho} \HZ,$$
as desired.
\end{proof}

\begin{proof}[Proof of Lemma \ref{app:main-lem}]
By $2$-localizing the result of Lemma \ref{app:red-lem}, we learn that
$$\THR(\HZ_{(2)}) \simeq \HZ_{(2)} \otimes \Sigma^{\infty}_+ \Omega^{\sigma} S^{\lambda+1} \langle \lambda+1 \rangle$$
may be calculated as the fiber of a map $f$ of $\HZ_{(2)}$-module spectra
$$f:\bigoplus_{k \ge 0} \Sigma^{k\rho} \HZ_{(2)}  \to \bigoplus_{s \ge 1} \Sigma^{s \rho} \HZ_{(2)}.$$
Since the domain of $f$ is a direct sum of free $\HZ_{(2)}$-module spectra, $f$ is determined by a sequence of elements
$f_k \in \pi_{k\rho} \left( \bigoplus_{s \ge 1} \Sigma^{s \rho} \HZ_{(2)} \right).$  The $RO(C_2)$-graded homotopy groups of $\HZ$, as nicely displayed for example in \cite[p.6]{GreenlessApproaches}, show that there are no classes in $\pi_{k\rho} (\Sigma^{s\rho}\HZ_{(2)})$ unless $k=s$.  Furthermore, one sees that any such class is determined by its underlying non-equivariant class in $\pi_{2k}(\Sigma^{2k} H\mathbb{Z}_{(2)}) \cong \Z_{(2)}$, and in particular the map $f$ is entirely determined by its underlying non-equivariant map
$$f_{\text{underlying}}:\bigoplus_{k \ge 0} \Sigma^{2k} H\mathbb{Z}_{(2)}   \to \bigoplus_{s \ge 1} \Sigma^{2s} H\mathbb{Z}_{(2)}.$$
The fiber of $f_{\text{underlying}}$ must agree with the known non-equivariant calculation
$$\mathrm{THH}(H\mathbb{Z}_{(2)}) \simeq H\mathbb{Z}_{(2)} \oplus \bigoplus_{s \ge 1} \Sigma^{2s-1} (H\mathbb{Z}/s)_{(2)},$$
which determines the map $f_{\text{underlying}}$ well enough to determine the fiber of $f$ up to equivalence.
\end{proof}

\begin{proof}[Proof of Theorem \ref{app:main-thm}]
As with any $C_2$-equivariant spectrum, there is a pullback square
$$
\begin{tikzcd}
\THR(\HZ) \arrow{r} \arrow{d} & \THR(\HZ)[\frac{1}{2}]  \arrow{d} \\
\THR(\HZ)_{(2)} \arrow{r} & \THR(\HZ) \otimes H\underline{\mathbb{Q}}.
\end{tikzcd}
$$
The $2$-local Lemma \ref{app:main-lem} allows to calculate the lower left corner of the square, while the result \cite[Corollary 5.28]{Real_THH} of Dotto, Moi, Patchkoria, and Reeh calculates the upper right.  From these results, we learn that the square is a direct sum of squares
$$
\begin{tikzcd}
H\underline{\mathbb{Z}} \arrow{d} \arrow{r} & H\underline{\mathbb{Z}}[1/2] \arrow{d} \\
H\underline{\mathbb{Z}}_{(2)} \arrow{r} & H\underline{\mathbb{Q}},
\end{tikzcd}
$$
and, for all $k \ge 1$,
$$
\begin{tikzcd}
\Sigma^{k \rho-1} H\underline{\mathbb{Z}/k} \arrow{d} \arrow{r} & \Sigma^{k\rho-1} H\underline{\mathbb{Z}/k}[1/2] \arrow{d} \\
\Sigma^{k\rho-1} H\underline{\mathbb{Z}/k}_{(2)} \arrow{r} & 0.
\end{tikzcd}
$$

\end{proof}