TITLE: It is true that every group that has a finite number of subgroups is finite?
QUESTION [3 upvotes]: It is true that every group that has a finite number of subgroups is finite?
I think not, but I can not find counterexamples.

REPLY [6 votes]: Yes, true - look at $\langle g\rangle$ for every $g \in G$ and note that an infinite cyclic group has an infinite number of subgroups.

 $\langle g \rangle$ must be finite, and since $G$ has only a finite number of subgroups, we get that $G=\bigcup_{g \in G}\langle g \rangle$ is a finite union. hence $G$ must be finite after all!


TITLE: Sitting around a round table in circular permuations
QUESTION [2 upvotes]: If $3$ people are selected at random from $15$ people sitting at a round table, then what is the probability that no two of them are adjacent to each other?

REPLY [2 votes]: This can be generalized to x people in y seats. As Andre noted, you can use symmetry, but I will assume seat 0 is open.
Let's have "O" refer to an open seat and "F" refer to a filled seat. Then every table combination can be expressed as a combination of (O) and (F) strings, and every legal table combination with seat 0 open can be expressed by a combination of (O) and (OF) strings.
Any seating combination, legal or not, is made up of x (F) and y-x (O) strings. But since it must start with an (O), that leaves y-x-1 (O) and x (F) in the last y-1 slots, so you have ${ y-1 \choose x}$ total combinations.
Any legal seating combination will contain x (OF) and y-2x (O) strings. That means you have ${ y-x \choose x }$ legal combinations.
That makes ${ \frac{y-x \choose x}{y-1 \choose x}}$ our more general answer. I don't think the combinations can be reduced to something simpler, though
${ \frac{(y-2x+1) * ... * (y-x-1)}{(y-x-1) * ... * (y-1)} }$ might help to visualize this. This assumes x > 1, which is okay, since x=0 or 1 is trivial.


TITLE: Graph Theory, the amount of labeled trees
QUESTION [2 upvotes]: Let us consider all labeled trees with nodes $\{1, 2, 3,...,n\}$. The question is how many of them contains the edge $\{1, 2\}$.
I would really appreciate an answer with some explanation.

REPLY [2 votes]: Combinatorially  we can  construct these  trees by  choosing a  set of
trees  to attach  at the  node labeled  one, and  another at  the node
labeled two.  These must together contain a total of $n-2$ nodes. Note
also that we must choose the labels  that go into the first set, where
the rest goes into  the second set. The two labeled  sets of trees are
then re-labeled with the chosen  labels respecting the ordering in the
source, which is the standard  construction. Therefore with $Q(z)$ the
EGF of sets of rooted labeled trees where
$$Q(z) = \sum_{n\ge 0} q_n \frac{z^n}{n!}$$
the desired quantity is given by
$$\sum_{k=0}^{n-2} {n-2\choose k} q_k q_{n-2-k}.$$
We use rooted trees  because we attach the root nodes  of the trees in
the two sets to the corresponding  node (labeled one or two). Here the
parameter $k$ gives  the number of nodes in the  set of trees attached
to node  one, with $n-2-k$  nodes in the  second set attached  to node
two.
Observe that when we multiply  two exponential generating functions of
the sequences $\{a_n\}$ and $\{b_n\}$ we get that
$$ A(z) B(z) = \sum_{n\ge 0} a_n \frac{z^n}{n!}
\sum_{n\ge 0} b_n \frac{z^n}{n!}
= \sum_{n\ge 0}
\sum_{k=0}^n \frac{1}{k!}\frac{1}{(n-k)!} a_k b_{n-k} z^n\\
= \sum_{n\ge 0}
\sum_{k=0}^n \frac{n!}{k!(n-k)!} a_k b_{n-k} \frac{z^n}{n!}
= \sum_{n\ge 0}
\left(\sum_{k=0}^n {n\choose k} a_k b_{n-k}\right)\frac{z^n}{n!}$$
i.e. the  product of the  two generating functions is  the exponential
generating function of $$\sum_{k=0}^n {n\choose k} a_k b_{n-k}.$$
This means that in the present case our answer is
$$(n-2)! [z^{n-2}] Q(z)^2.$$
Recall  the  combinatorial  species   of  labeled  rooted  trees  with
specification
$$\mathcal{T} = \mathcal{Z} \mathfrak{P}(\mathcal{T})$$
which gives for the corresponding EGF the functional equation
$$T(z) = z \exp T(z).$$
We thus have for sets of trees $\mathfrak{P}(\mathcal{T})$ that
$$Q(z) = \exp T(z) = \frac{1}{z} T(z).$$
It follows that
$$(n-2)! [z^{n-2}] Q(z)^2
= \frac{(n-2)!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n-1}} Q(z)^2 \; dz
\\ = \frac{(n-2)!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} T(z)^2 \; dz.$$
Now putting $T(z)=w$ the functional equation yields $w = z \exp(w)$ or
$z = w \exp(-w)$ and $dz =  (\exp(-w) - w \exp(-w)) \; dw.$ Here $z=0$
gives  $w=0$ and  since  $T(z) =  z  + \cdots$  the  image contour  of
$|z|=\epsilon$ in $w$  may be deformed to a circle  being traced once,
giving
$$\frac{(n-2)!}{2\pi i}
\int_{|w|=\gamma} \frac{\exp((n+1)w)}{w^{n+1}}
w^2 \exp(-w) (1-w) \; dw
\\ = \frac{(n-2)!}{2\pi i}
\int_{|w|=\gamma} \frac{\exp(nw)}{w^{n-1}}
(1-w) \; dw.$$
Extracting the residue we find
$$(n-2)! \times
\left( [w^{n-2}] \exp(nw) - [w^{n-3}] \exp(nw)\right)
\\ = (n-2)! \times
\left( \frac{n^{n-2}}{(n-2)!} - \frac{n^{n-3}}{(n-3)!}\right)
\\ = n n^{n-3} - (n-2) n^{n-3}
= 2 n^{n-3}.$$
This matches the result from the probabilistic argument that was first
to appear.


TITLE: Show that the Hessian matrix of f at the origin is not symmetric.
QUESTION [0 upvotes]: Let$f:R^2 \rightarrow R$ be defined by
$f(x,y)=\frac{xy(x^2-y^2)}{x^2+y^2}$ if $(x,y)\neq(0,0)$
and $f(x,y)=0$ if $(x,y)=(0,0)$
I was asked to show that the Hessian matrix of f at the origin is not symmetric, but I don't know where to begin from.
I also don't know how to find the Jacobian matrix of these kind of functions.
Edit: I forgot to add “at the origin” in the problem.

REPLY [0 votes]: I don't think this statement holds correctly. This is because we have most often the following property:
$$
\frac{\partial^2 f}{\partial x\partial y}
=
\frac{\partial^2 f}{\partial y\partial x},
$$
as holds here, implying that the Hessian matrix is symmetric.


TITLE: When is $\sum_{n=1}^\infty (a_n-b_n) = \sum_{n=1}^\infty a_n - \sum_{n=1}^\infty b_n$ true?
QUESTION [0 upvotes]: Given that $a_n, b_n > 0 $ and that $\sum_{n=1}^\infty a_n$ and $\sum_{n=1}^\infty b_n$ both converge, does the property hold? Is there a particular case when it is true like maybe when $\sum_{n=1}^\infty (a_n-b_n)$ is absolutely convergent?

REPLY [1 votes]: If $\sum_{n=1}^\infty a_n$ and $\sum_{n=1}^\infty b_n$ both converge, then yes the equality must hold. This is a consequence of properties of limits. Namely, that if $\lim_{n \to \infty} s_n$ and $\lim_{n\to \infty} t_n$ both exist, then $\lim_{n\to \infty} (s_n-t_n)$ exists and equals the difference of the limits. So,
\begin{align*}
\sum_{n=1}^\infty (a_n-b_n)&=\lim_{N\to \infty} \sum_{n=1}^N (a_n-b_n)\\
&= \lim_{N \to \infty} \sum_{n=1}^N a_n - \lim_{N\to\infty} \sum_{n=1}^N b_n\\
&=\sum_{n=1}^\infty a_n - \sum_{n=1}^\infty b_n.
\end{align*}


TITLE: Existence of first order PDEs?
QUESTION [2 upvotes]: It seems the follow theorem is classical, but I don't know how to proof it:
For $x\in\Omega\subset R^n$, where $\Omega$ is a domain with smooth boundary, consider the system of PDEs:
\begin{equation}\label{eq:1}
\frac{\partial f(x)}{\partial x_i}=f_i(x),\quad i=1,2,\ldots,n, f_i\in C^\infty(\Omega),\tag{1}
\end{equation}
if the compitable condition
$$
\frac{\partial f_i}{\partial x_j}=\frac{\partial f_j}{\partial x_i},\quad\forall i,j=1,2,\ldots,n,\forall x\in\Omega
$$
is satisfied, then there exists $f\in C^\infty(\Omega)$, which solves \eqref{eq:1}.
So, can anyone give a detailed proof? OR, at least, provid some exactly references?

REPLY [2 votes]: Think in $\Bbb R^3$ for a minute.  The hypothesis
$\dfrac{\partial f}{\partial x_i}(x) = f_i(x) \tag{1}$
says that the vector field
$\vec F(x) = (f_1(x), f_2(x), f_3(x))^T \tag{2}$
is the gradient of the function $f(x)$:
$\vec F(x) = \nabla f(x); \tag{3}$
as such,
$\nabla \times \vec F = 0.  \tag{4}$
As is well-known, this situation is reversible:  for $\vec F(x)$ satisfying (4), there is a scalar function $f(x)$ such that (3) applies, at least locally.  And if one carefully examines the relations
$\dfrac{\partial f_i}{\partial x_j} = \dfrac{\partial f_j}{\partial x_i}, \; \; 1 \le i, j \le 3, \tag{5}$
one sees that (4) is exactly what they say; so that resolves the situation for $n = 3$, again, locally at least.  For $n > 3$, it is perhaps easiest to revert to the language of differential forms; setting
$\omega = \sum_1^n f_i dx_i, \tag{6}$
we have
$d\omega = \sum_{1 \le i < j \le n} (\dfrac{\partial f_i}{\partial x_j} - \dfrac{\partial f_j}{\partial x_i}) dx_i \wedge dx_j = 0 \tag{7}$
this last equality holding by our hypothesis on the derivatives of the $f_i(x)$.  In the parlance of differential forms, $\omega$ is said to be closed.  It is substance of the Poincare lemma, which may be found at http://en.m.wikipedia.org/wiki/Closed_and_exact_differential_forms, that such forms, of no matter degree $p$ they may be,  are also exact, that is such $\omega = d\alpha$ for some $p-1$ form $\alpha$, provided the domain $\Omega$ is simple enough; "star-shaped" is the criterion oft' applied, though the simpler notion convex will do to ensure such $\alpha$ exists, locally at least.  In the present case, this means $\alpha$ is a function such that
$\dfrac{\partial \alpha}{\partial x_i} = f_i \tag{8}$
on appropriate $\Omega$.  The linked citing explains the conditions which must be satisfied by $\Omega$ for the Poincare lemma to bind.
The preceeding presents, in broad outline, how the existence of solutions to the given equation may be established.  Such are "funny" problems as far as PDEs go; for example it is not at all clear to me how issues such as boundary/initial data fit into the differential forms-based view of these matters; but it is certain that the domain $\Omega$ must be somewhat carefully specified if a solution is to be had.  But then again, that's cohomology for you!
Hope this helps.  Cheers,
and as ever,
Fiat Lux!!!


TITLE: Prove that $T$ is continuous.
QUESTION [1 upvotes]: I am having a really hard time trying to solve this problem:
Let $X$ and $Y$ be normed linear spaces and $T : X \to Y$ a linear operator with closed graph and finite-dimensional range $R(T)$. Prove that T is continuous.
Obviously the closed graph theorem cannot be applied since our spaces are not neccessarily Banach ones. Nonetheless, at least "Y" is a Banach space because of its finite-dimensionality. I do not know how I can go further from here. Thanks in advance.

REPLY [0 votes]: Assume $T$ is not continuous at zero. Then there exist a sequence $(x_n)\subset X$ with $x_n\to 0$ and $\epsilon>0$ such that $\|Tx_n\|\ge\epsilon$ for $n\in\mathbb N$. Set $u_n := \frac{x_n}{\|Tx_n\|}\in X$. Then $u_n\to 0$ and $\|Tu_n\|=1$ for all $n\in\mathbb N$. Since $R(T)$ is finite-dimensional, there exists a subsequence $(u_n')$ of $(u_n)$ such that $(Tu_n')$ converges. Let $y\in Y$ be its limit. Then from $u_n'\to 0$ and $Tu_n'\to y$ we conclude that $y=0$. But $\|y\| = \lim_n\|Tu_n'\| = 1$. Contradiction!


TITLE: Find the limit of a function using the definition
QUESTION [2 upvotes]: I want to prove the limit of the function sqrt(x) equals 2 using the definition of the limit of a function.
The definition of the limit of a function is: Let I be an open interval that contains the point c and suppose f is a function defined on I, except possibly at the point c. Then f has limit L at c if for each e>0, there exists a number q>0 such that |f(x)-L|

So far, I have the following completed: Let e>0 be given. We must find q>0 such that |sqrt(x)-2|

I have completed these proofs when the function is a polynomial, but I am a little confused of where to go with this problem next. If anyone is willing to assist me, it would be much appreciated. Thank you!

REPLY [0 votes]: Assuming you mean $\lim\limits_{x\to4}\sqrt{x}=2$ then the question would read
for a fixed $\varepsilon$, can we find a $\delta$ such that $|x-4|<\delta$ implies that $|2-\sqrt{x}|<\varepsilon $?
Start by choosing $\varepsilon>0$.
If $0<x<4$ then $|\sqrt{x}-2|=2-\sqrt{x}<\varepsilon \implies (2-\varepsilon)^2 < x$.
If $4<x$ then $|\sqrt{x}-2|=\sqrt{x}-2<\varepsilon \implies (2+\varepsilon)^2 > x$.
I think you can figure out the rest. :)


TITLE: Request for references in computational complex analysis
QUESTION [4 upvotes]: We know complex analysis is one of the most important branches of mathematics connecting myriad areas. It is replete with profound results and theorems and theorems. However, a good number of the theorems both in elementary and advanced complex analysis are of existential sort. As a prime example take the Riemann open mapping theorem. I would be highly obliged if somebody could kindly recommend me some books that takes a computational and algorithmic approach to the subject combined with a geometric approach. Books that illustrate examples with coding examples, if any such  books are there, are also requested. Thanking  you in advance

REPLY [1 votes]: This edited volume is specific to Riemann surfaces but might be of interest:
Computational Approach to Riemann Surfaces, Bobenko & Klein (eds), Springer 2011


TITLE: Example of two integrable functions/random variables whose product is not integrable?
QUESTION [1 upvotes]: If two functions/random variables are integrable and independent, then their product is integrable. What if they are not independent? What is an example?
What I tried:

Let $X, Y \in \mathscr L^{1}(\Omega, \mathscr F, \mathbb P)$.
Consider $X$ and $X - Y$ w/ $X$ having an infinite second moment but finite first moment.
Then
$$E[X(X-Y)] = E[X^2] - E[XY] = \infty$$
assuming $-\infty \le E[XY] < \infty$ and indeterminate otherwise. In either case, $X(X-Y)$ is not integrable.
An example is $X$ having a student-t distribution with two degrees of freedom and $Y$ can be anything integrable, I guess.

If that's right, what are examples are better/simpler (in your opinion) or more well-known (from what you've observed)?
If that's wrong, which part is wrong, why, and please provide examples.

REPLY [4 votes]: Let $U$ be a random variable which is uniform on $(0,1]$. Let $X=Y=U^{-1/2}$. Then $E[X]=E[Y]=2$ but $E[XY]=+\infty$.
You can make lots of examples of functions like this, which are not integrable but their square root is integrable, because it diverges "more slowly".
On an infinite measure space you can also have the opposite phenomenon with tailing: $x^{-2}$ is integrable over $[1,\infty)$ but its square root is not, because it decays more slowly. But this is impossible on a finite measure space.


TITLE: A couple of questions on the ADM formalism in general relativity
QUESTION [3 upvotes]: I've been reading up on the ADM formalism in general relativity and have been stuck on a couple of concepts.

The first is to do with the foliation of spacetime into space-like hypersurfaces.  I understand that the hypersurfaces must be space-like as they are surfaces of constant time and hence any two points on a given surface must necessarily be space-like separated, but I'm slightly unsure as to why the normal to such a surface is necessarily time-like?! Does this follow because the normal is,  by definition, orthogonal to the surface, thus contains no space-like components, and so necessarily must be time-like? Can this be proven mathematically, or is it just a requirement (I understand it intuitively, as each hypersurface describes the 3-dimensional space at a given instant in time and thus consecutive hypersurfaces must be causally related by a time-like path).
The second question I have is what is the so-called lapse function, $N(t) $,  as in what is it physically describing? Is it just quantifying the change in description of time as one moves between two given hypersurfaces, or is there more to it than that?

REPLY [2 votes]: I have tried to understand it too, for me.. I will describe it along with the normal and the shift vector.
For free particles free-falling in the spacetime (following the geodesic) their world-lines define the time flow. We call vector tangents to this flow at each point a $\textit{time-flow vector}\, t^\mu $. In their free-falling  frames at a particular point in the space-time
the metric field to describe the particle motion is the flat Minkowksian metric adapted from metric for curved metric at that point. The observer in such frame may expect that a particle will follow the geodesic defined by the flat Minkowksian metric since she don't feel gravitation, but, indeed the shift in position occurs since a particle really follows the geodesic defined by general space-time metric. $\textit{Such shifts occur when the gravity is not uniform distribution}$, equivalently we can say that such free falling frame $\textit{ has inertia }$. Imagine that we are live inside the falling elevator in an uniform gravitational field and place the ball at some height above the floor of this elevator the deviation of path of such a ball will be not detected, but if so, the non-uniformity of gravitational field was detected because the shifts in position of this ball occurred. Such shifts are described by the shift vector $N^\mu (x)$.  At each infinitesimal region the shift vector is a projection of the time vector as shown in  figure.

Therefore we can associate the  vector perpendicular to the hypersurface  with the time-flow and the shift vectors
\begin{equation}
 Nn^\mu := t^\mu - N^\mu \perp \Sigma_t ,
\end{equation}
where $n^\mu$ is a unit vector orthogonal to the hypersurface($\Sigma_t$)
\begin{eqnarray}
 & g_{\mu\nu} n^\mu n^\nu =-1 , & \\
 & g_{\mu\nu} n^\mu N^\nu =0 , &
\end{eqnarray}
and we call $N(x)$ the lapse function. It is dynamical and capture the deviation of speed of the clock from the local Minkowksian sense caused by non-uniformity of the gravitational field.
The vector $n^\mu$ can be interpreted as a gradient of some scalar function that is constant at each hypersurface. Such time function, $T(x)%^\mu)
$, is usually defined via
\begin{equation}
 n_\alpha = -N\nabla_\alpha T.
\end{equation}


TITLE: Problem with probability
QUESTION [0 upvotes]: If we choose randomly an infinite, countable set of disks in $\mathbb R^2$, what is the probability that intersection of every pair of disks from the set is an empty set?
EDIT: Because the problem in the above stated form is too hard to handle, or undecidable, or it may have many different solutions depending on the way we choose the centers and the radii I will state it with extra conditions.
First variant of the problem: Add to the original problem: "all disks have the same radius $R$".
Second variant of the problem: Add to the original problem "all disks have the same radius $R>1$ and every center has the coordinates $(x,y)$, where $x,y \in\mathbb Z$"
Question for the first and second variant of the problem: Is the problem solvable in at least one of these two variants or we need more extra conditions?

REPLY [1 votes]: Sadly, some natural formulations of this problem are extremely difficult. Take a finite square with side-lengths $n$ and then make it periodic by identifying opposite sides, i.e. making it a torus. Then you can uniformly pick the centers of each circle within the rectangle. This is called the "Pennies on a carpet problem" or "2-D hard-spheres model." The problem is extremely difficult and not too much is known about it. You can find a bit more info here


TITLE: What is the distinctive characteristic/structure of Polish Space?
QUESTION [3 upvotes]: I am trying to understand the geometric structure of Polish space. While reading I came up with the wikipedia link:
http://en.wikipedia.org/wiki/Descriptive_set_theory
and on the second paragraph of the link it says:
A Polish space is a second countable topological space that is metrizable with a complete metric. Equivalently, it is a complete separable metric space whose metric has been "forgotten"
I did not understand the context of the word "forgotten". I think this relates to some concept which I am missing. Any comment would be appreciated.

REPLY [2 votes]: A Polish space in this context doesn't actually have a metric, but it is completely metrizable. I can think of two equivalent ways to view this statement:

Starting from a Polish space you can (non-uniquely) choose a metric which induces the topology, obtaining a complete metric space.
Starting from a complete metric space you can (uniquely) record the topology and then throw out the metric, to leave behind a Polish space.

There is probably a way of viewing the statement as somehow being about equivalence classes, but I would rather not get into it.


TITLE: Find an example of K, A, B, and I to show $\beta_i\neq \beta_i(A)+\beta_i(B)-\beta_i(A\cap B)$
QUESTION [0 upvotes]: I’ve been trying to find a simplicial complex $K$, where $A$ and $B$ are subcomplexes, and $K=A\cup B$ to show that the Betti number
$\beta_i(K)\neq \beta_i(A)+\beta_i(B)-\beta_i(A\cap B)$
but every complex I come up with contradicts this for $i=0$, what would be a good example for this?

REPLY [1 votes]: Here's an example (for $i=0$) where $\beta_0(K)\neq\beta_0(A)+\beta_0(B)-\beta_0(A\cap B)$. Let simplicial complex $K$ be the boundary of a triangle, with three vertices $[0]$, $[1]$, $[2]$, and three edges $[0,1]$, $[1,2]$, $[2,0]$. Let subcomplex $A$ be the union of two edges, containing all three vertices $[0]$, $[1]$, $[2]$, and only two edges $[0,1]$, $[1,2]$. Let subcomplex $B$ be the third remaining edge, containing vertices $[0]$, $[2]$ and only one edge $[2,0]$. We have $K=A\cup B$.
Note that $K$, $A$, and $B$ are each connected, giving $\beta_0(K)=\beta_0(A)=\beta_0(B)=1$. However, $A\cap B$ consists of two connected components (two vertices $[0]$ and $[2]$), giving $\beta_0(A\cap B)=2$. So we have
$$\beta_0(K)=1\neq 0=1+1-2=\beta_0(A)+\beta_0(B)-\beta_0(A\cap B),$$
as desired. Note in this example that $\beta_1(K)=1$ is nonzero, agreeing with the Mayer-Vietoris long exact sequence (https://en.wikipedia.org/wiki/Mayer%E2%80%93Vietoris_sequence).


TITLE: Exercise 3.2.13 Introduction to Real Analysis by Jiri Lebl
QUESTION [1 upvotes]: Let $f:S \to \mathbb{R}$ be a function and $c \in S$, such that for every sequence $\{x_n\}$ in $S$ with $\lim x_n = c$, the sequence $\{f(x_n)\}$ converges. Show that $f$ is continuous at $c$.

Suppose that $f$ is not continuous at $c$. Then, $\exists \epsilon > 0$ s.t. for every $\delta> 0$, $\exists x$ s.t. $|x-c| < \delta$, but $|f(x) - f(c)| \ge \epsilon$. Let $\{x_n\}$ be a sequence s.t.   $|x_n - c| < \frac1n$. Then, $|f(x_n) - f(c) | \ge \epsilon$ when $n$ is large enough. But, I think that $f(x_n)$ can still converges to somewhere else other than $f(c)$. How can I proceed from here?
I appreciate if you give some help.

REPLY [2 votes]: Consider any $(x_{n})$ such that $x_{n}\rightarrow c$, consider also the constant sequence $(c,c,...)$ and consider further that $(y_{n}):=(x_{1},c,x_{2},c,...)$, the later also converges to $c$ and $(f(y_{n}))$ has the constant subsequence $(f(c),f(c),...)$, so $f(y_{n})\rightarrow f(c)$ and also that $f(x_{n})\rightarrow f(c)$ as $(f(x_{1}),f(x_{2}),...)$ is also a subsequence of the convergent sequence $(f(y_{n}))$.
So we have proved that, for any $(x_{n})$ such that $x_{n}\rightarrow c$, then $f(x_{n})\rightarrow f(c)$, this is another characterization of continuity.


text:
\DOC EXPAND_TAC

\TYPE {EXPAND_TAC : string -> tactic}

\SYNOPSIS
Expand an abbreviation in the hypotheses.

\DESCRIBE
The tactic {EXPAND_TAC "x"}, applied to a goal, looks for a hypothesis of the
form {`t = x`} where {x} is a variable with the given name. It then replaces
{x} by {t} throughout the conclusion of the goal.

\FAILURE
Fails if there is no suitable assumption in the goal.

\EXAMPLE
Consider the final goal in the example given for {ABBREV_TAC}:
{
  val it : goalstack = 1 subgoal (1 total)

   0 [`12345 + 12345 = n`]

  `n + f n = f n`
}
\noindent If we expand it, we get:
{
  # e(EXPAND_TAC "n");;
  val it : goalstack = 1 subgoal (1 total)

   0 [`12345 + 12345 = n`]

  `(12345 + 12345) + f (12345 + 12345) = f (12345 + 12345)`
}

\SEEALSO
ABBREV_TAC.

\ENDDOC

subset_name: curated


TITLE: Can someone what this notation means?
QUESTION [1 upvotes]: I don't understand what does $\phi_I$ mean
The proof includes writing $\phi_I$ as a product of $\phi_{i_1}\phi_{i_2},\dots$, but it doesn't explain what the LHS really means

REPLY [0 votes]: $\phi_I$ is a $k$-multilinear form. This means $\phi_I$ takes in a $k$-tuple of vectors and is linear in slot if you fix the others. For example:
$$\phi_I(c{\bf v}+{\bf w},{\bf v}_2,\dots,{\bf v}_k) = c\phi_I({\bf v},{\bf v}_2,\dots,{\bf v}_k)+ \phi_I({\bf w},{\bf v}_2,\dots,{\bf v}_k)$$
where $c$ is a scalar and the v's are vectors. Likewise, $\phi_I$ is linear in the other arguments. By "form" I mean that the output: $\phi_I({\bf v}_1,\dots,{\bf v}_n)$ is a scalar ($\phi_I$ maps $k$-tuples of vectors to scalars).
Now $I$ is a $k$-tuple of integers. Given a list of vectors from the given basis: $a_{j_1},\dots,a_{j_k}$ (repeats allowed), $\phi_I(a_{j_1},\dots,a_{j_k})$ spits out 0 unless $(j_1,\dots,j_k)=I$ (the subscripts match identically).
For example: $\mathbb{R}^2$ has the basis ${\bf i}=(1,0)$ and ${\bf j}=(0,1)$.
Then...
$$\phi_{(1,1)}({\bf i},{\bf i})=1, \quad \phi_{(1,1)}({\bf i},{\bf j})=0, \quad \phi_{(1,1)}({\bf j},{\bf i})=0, \quad \phi_{(1,1)}({\bf j},{\bf j})=0$$
$$\phi_{(1,2)}({\bf i},{\bf i})=0, \quad \phi_{(1,2)}({\bf i},{\bf j})=1, \quad \phi_{(1,2)}({\bf j},{\bf i})=0, \quad \phi_{(1,2)}({\bf j},{\bf j})=0$$
$$\phi_{(2,1)}({\bf i},{\bf i})=0, \quad \phi_{(2,1)}({\bf i},{\bf j})=0, \quad \phi_{(2,1)}({\bf j},{\bf i})=1, \quad \phi_{(2,1)}({\bf j},{\bf j})=0$$
$$\phi_{(2,2)}({\bf i},{\bf i})=0, \quad \phi_{(2,2)}({\bf i},{\bf j})=0, \quad \phi_{(2,2)}({\bf j},{\bf i})=0, \quad \phi_{(2,2)}({\bf j},{\bf j})=1$$
Also, for example,
$$\phi_{(1,2)}((1,2),(3,4)) = \phi_{(1,2)}({\bf i}+2{\bf j},3{\bf i}+4{\bf j})
=\phi_{(1,2)}({\bf i},3{\bf i}+4{\bf j})+2\phi_{(1,2)}({\bf j},3{\bf i}+4{\bf j})$$
$$=3\phi_{(1,2)}({\bf i},{\bf i})+4\phi_{(1,2)}({\bf i},{\bf j})+6\phi_{(1,2)}({\bf j},{\bf i})+8\phi_{(1,2)}({\bf j},{\bf j})=3(0)+4(1)+6(0)+8(0)=4$$


TITLE: How can something happen when time does not exist?
QUESTION [14 upvotes]: I saw this documentary hosted by Stephen Hawkins.
And if I didn't get it wrong, it says that there was no time before the big bang, time was created there.
So how can anything happen when there is no time (eg: the creation of time and the universe)? Is this what he calls a 'singularity'? Which I could think like this: when there is no time nothing can happen, except during this singularity.
Of course I don't know of a law that says 'when there is no time nothing can happen..', I just came up with it, as something intuitive.

REPLY [1 votes]: It amazes me that there's a very simple explanation coming straight from special relativity and that nobody appears to be realizing (I never see it anywhere) : proper time of particles in a collision or creation/annihilation process.
The universal time of cosmology is the "proper time" of the universe, as defined by observers locally at rest with the cosmic fluid (i.e. comoving observers).  Thus, for the following, you can conceptually reduce the whole universe as a simple particle coming out of some collision/creation process.
In special relativity, the proper time $\tau$ of a particle doesn't have any sense before the particle have been created in a given process.  It doesn't mean that there wasn't any time before the particle is created, it just mean that the particle's proper time doesn't exist since the particle isn't there (time is a relative concept in relativity).  Before the creation process, you may have other particles with their own proper time.  In the creation process, some particles annhiled and created a new one.  This is the "start" of your new universe (seen here as a "particle").
So asking what was before the Big Bang event (and what created it) is like asking what proper time was before the particle was created.  There was no proper time at all before its own particle !  Proper time got created at "the same time" as the particle itself !  There may (or may not) have been other universes that "collided" (i.e. collapsed/interacted/annhiled/wathever), but we cannot know them by just following our universal proper time !
This is where General Relativity is breaking appart since it can't retrodict the events that "produced" the Big Bang (see the answer of John Rennie).  For that, we would need a kind of "relativity theory" of General Relativity, i.e. a theory of colliding/interacting/annhilating universes, i.e. a theory of topology changes.  Since General Relativity is based on Differential Geometry, it asks for a manifold with a given topology.  Once you have a topology, you have to stay with it, within General Relativity.  A singularity (in space or in time) is a sign that there should be some topology change, i.e. a change of manifold.
Since particles collision/creation/annihilation is described by quantum mechanics, the topology change that General Relativity is asking should be a quantum mechanical phenomenon (this isn't surprising :  most of quantum mechanics are about boundaries, i.e. topology !).


TITLE: Why can we set the external momenta to zero while calculating the 1-loop correction to non-abelian theory coupling?
QUESTION [1 upvotes]: Refer to the second diagram in figure 73.2 on page 440 of Srednicki's Quantum Field Theory book. On page 441 he proceeds to calculate the amplitude and argues that since the divergent part is independent of the external momenta, we may set it to zero.
It is obvious that the terms with external momenta on the numerator are convergent (and may be ignored) since, by superficial analysis, the most divergent terms (which have $\ell^2$ on the numerator) are logarithmic divergent.
My doubt is: how can we know that the external momenta appearing in the horrible denominator won't produce external momenta dependence on the divergent terms?
I suspect the answer comes from dimensional analysis but cannot see how.

REPLY [2 votes]: UV-divergences are caused by integrating over a large loop momentum $\ell$. In this region $\ell$ dominates the external momentum $k$ both in numerators but also in denominators. The latter can e.g. be seen by Taylor expansion. The second diagram in Fig. (73.2) is logarithmic UV-divergent, so subdominant ($k$-dependent) contributions are necessarily finite.
References:

M. Srednicki, QFT, 2007; Fig. (73.2)+eq. (73.6). A prepublication draft PDF file is available here.


TITLE: Directly prove using the definition of convergence $\lim\limits_{n \to \infty}\frac{n^2 +1 }{4n -1} = \infty$
QUESTION [0 upvotes]: The definition of convergence is :
$\lim\limits_{n \to \infty}a_n =l \Rightarrow \lim\limits_{n \to \infty}a_n -l=0 $
So $\lim\limits_{n \to \infty}\frac{n^2 +1 }{4n -1} =\infty \Rightarrow \lim\limits_{n \to \infty}\frac{n^2 +1 }{4n -1} -\infty =0$
and we have to show that $ \forall n > X,|\frac{n^2 +1 }{4n -1} -\infty| < \epsilon $.

REPLY [1 votes]: To show that a function f(n) goes to $\infty$ as n goes to $\infty$, we need to show that for any $\alpha$ > 0, there will be a $\delta$ > 0 such that for all n > $\delta$, f(n) > $\alpha$.
You function is f(n) = $\frac{n^2 +1 }{4n -1}$ >$\frac{n^2  }{4n}$ > $\frac{n}{4}$
and we need the f(n) > $\alpha$, which can be satisfied if we have $\frac{n}{4}$ > $\alpha$ i.e. n > 4$\alpha$. So our required $\delta$ = 4$\alpha$ and you have shown the required thing.


TITLE: In the semi linear uniform space
QUESTION [0 upvotes]: In the semi linear uniform space, If  $f$ is a function from $(X ,Γ_X)$ to ($Y,Γ_Y)$ where $f(x_n)$ converges to $f(x)$ whenever $x_n$ converges to $x$,show that $f$ is continuous at $x$.

REPLY [0 votes]: Let $f$ be not continuous at $x$. There is some net $(d_t)_{t\in T}$ with $d_t \to x$ and $f(d_t)\not\to f(x)$. So we can assume there some net $(U_t)_{t\in T}$ of neighborhoods with
$$(\forall t\in T)(f(t)\notin U_t)$$
As $\{U_t\mid t\in T\}$ is a chain, there must be a strictly decreasing sequence ${U_{t_n}}$ such that
$$(\forall n)(f(t_n)\notin U_{t_n})$$
But then by assumption, $d_{t_n}\not\to x$. A contradiction.


TITLE: Show that the spectral radius is an upper semi-continuous function
QUESTION [9 upvotes]: I am stuck in a problem of Conway's A course in a Functional Analysis. Can anyone give me a hint to solve the problem?
The question is "If $A$ is a Banach Algebra, then show that the function $r:A\to R$ defined by $r(a)=$ spectral radius of $a,$ is a upper semi-continuous function."
Recall that spectral radius of $a,$ $r(a):=\sup\{|c|:c\in\mbox{ spectrum of } a\}$ .

REPLY [11 votes]: Assume that $r$ is not upper semi-continuous at $a$. Then we can find a sequence $\{a_n\}\subset A$ converging to $a$ such that $\limsup_{n\to +\infty}r(a_n)>r(a)$. Hence there exists a $\delta>0$ and a subsequence $\{b_k\}$ of $\{a_n\}$ such that $r(b_k)\geq r(a)+2\delta$. So for each $k$, let $\lambda_k$ in the spectrum of $b_k$ such that $|\lambda_k|\geq r(a)+\delta$. As $\{b_k\}$ is bounded, the sequence $\{\lambda_k\}$ of complex numbers is bounded. So we extract a converging subsequence (but we will denote it in the same way for simple).
Then $\{b_k-\lambda_k e\}$ is not invertible and converges to $a-\lambda e$. It's not invertible, as the set of invertible elements of $A$ is open. So $\lambda\in \sigma(a)$. But $|\lambda|\geq r(a)+\delta$, a contradiction.


TITLE: Proving two matrices are equal
QUESTION [5 upvotes]: A friend and I are having some trouble with a linear algebra problem:
Let $A$ and $B$ be square matrices with dimensions $n\times n$
Prove or disprove:
If $A^2=B^2$ then $A=B$ or $A=-B$
It seems to be true but the rest of my class insists it's false - I can't find an example where this isn't the case - can someone shed some light on this?
Thanks!

REPLY [4 votes]: All matrices $E_{ij}$  of the standard basis of $M_n(\mathbf R)$, defined with
$$ a_{kl}=\begin{cases}
1&\text{if }(k,l)=(i,j)\\
0&\text{if }(k,l)\neq(i,j)
\end{cases}$$
satisfy the equation $\;E_{ij}^2=0$ if $i\neq j$.


TITLE: Good conductor of electricity and bad conductor of heat
QUESTION [1 upvotes]: I wonder, how a graphite can be a good conductor of electricity but at the same time, be a bad conductor of heat.?!
As we know, a body conducting electrons are bound to produce heat by resistance, which will, in turn increase the entropy and heat.
So therefore thermal conductivity is directly proportional to electrical conductivity.
Also I guess it is not necessary that a good conductor of heat be also a good conductor of electricity because vibrational energy can also be the cause of it like in the case of diamond.

REPLY [3 votes]: I wonder, how a graphite can be a good conductor of electricity but at the same time, be a bad conductor of heat.?!

You have to be more precise. Can you give values? Graphite is a very anisotropic material. Its thermal conductivity is insanely high in the x-y plane (about 4 times that of copper). But in the z direction, the thermal conductivity is very low, about 2 orders of magnitude less than copper. Something similar happens with its electrical conductivity.

therefore thermal conductivity is directly proportional to electrical conductivity.

Not quite. The so-called Wiedemann-Franz law (which works reasonable well for metals) stipulates that the ratio between the thermal conductivity and electrical conductivity is proportional to temperature (and not merely a constant!). It also assumes that only electrons contribute to heat transfer, not phonons. But keep in mind that this law does not apply "as is" for semiconductors or semimetals.

Also I guess it is not necessary that a good conductor of heat be also a good conductor of electricity because vibrational energy can also be the cause of it like in the case of diamond.

You guessed right. This would hold reasonable well for metals, but not for heavily doped semiconductors (good thermoelectric materials), or insulators. The more the phonons have a non negligible contribution to the heat transfer, the less the statement linking $\kappa$ to $\sigma$ should hold.


TITLE: Principal open sets in graded rings
QUESTION [1 upvotes]: I am interested in Prop II.2.5b in Hartshorne stating that if $D_+ (f) = \{p \in \textrm{Proj } B \mid f \notin p \}$ then there is a canonical homeomorphism $D_+ (f) \cong \textrm{Spec } B_{(f)}$ the latter notation meaning the elements of degree 0 in $B_f$.
Following Hartshorne's proof, there is a natural morphism $D_+ (f) \to \textrm{Spec } B_{(f)}$ by restricting the canonical map $D(f) = \textrm{Spec } B_f \to \textrm{Spec } B_{(f)}$ to $D_+ (f)$. Now Hartshorne claims "the properties of localization show that [this] is bijective."
I would like to be more explicit about surjection so I consulted Liu's book which said the following:

Explicitly, $\theta$ which is the restriction map described above is $p \mapsto (pB_f) \cap B_{(f)}$. I would like to show that $\theta(p) =(pB_f) \cap B_{(f)}= q$. At this point, it seems to be a matter of following the contractions and extensions of ideals with an added twist of having two maps. However, I am having some difficulties following the maps properly. Can someone lend me a hand?

REPLY [1 votes]: I think the problem is complicated mainly by the grading.

Claim: $\theta(p) = pB_f \cap B_{(f)} = q$

$$p = \{x \in B \mid x/1 \in \sqrt{qB_f}\} = \{x \in B \mid \exists m \in \mathbf{Z}^+ x^m / 1 \in qB_f\}$$
so $pB_f$ looks like all the things of the form $xb/f^N$ with $x^m b^m / f^{mN} \in qB_f$. Intersecting with $B_{(f)}$ results in taking
$$ \deg (f) = r, \deg (x) = s, \deg (b) = Nr - s$$
Thus $(xb/f^N)^m \in qB_f \cap B_{(f)} = q$ and $q$ was prime so $xb/f^N \in q$.
Conversely, any element of $q$ is already of degree 0 and satisfies trivially the condition of being in the radical, namely taking $m = 1$.


TITLE: Convergence of sequence of polynomials defined by boundary conditions
QUESTION [4 upvotes]: I'm sorry if my question sounds trivial, but analysis is not my field.
Consider the interval $[a,b]\subset \mathbb{R}$. On $[a,b]$, for every $n\in\mathbb{N}$, $n\ge 3$, I define the polynomials $P_n:=a_{0,n}+a_{1,n}x+\dots +a_{n,n}x^n$ (i.e. $a_{i,n}$ is the coefficient of $x^i$ in $P_n$ and $a_{j,n}=0$ in $P_n$ for $j>n$), the coefficients $a_{j,n}$ are given by the conditions:
$P_n(a)=A,P_n(b)=B,P_n'(a)=C,P_n'(b)=D$, $A,B,C,D\in\mathbb{R}$ for $n=3$ and as $n\ge 3$ I gradually impose that every other derivative of $P_n$ is equal to zero in the two points $a,b$ (for $n=4$ I add the condition $P_4''(a)=0$, for $n=5$ I add $P_5''(a)=0$ and $P_5''(b)=0$, for $n=6$ I add $P_6''(a)=0$, $P_6''(b)=0$ and $P_6'''(a)=0$ and so on).
My questions are:
1) What type of convergence can I expect on the sequence $P_n$ as $n\rightarrow \infty$? Does it converge uniformly?
2) Suppose $P_n$ converges to $P$ in the better possibile way. In addition suppose that $P$ is differentiable, $C\neq D$ and $\frac{B-A}{b-a}\in[min\{C,D\},max\{C,D\}]$. Then is it true $min\{C,D\}\le P'(x)\le max\{C,D\}$ $\forall x\in(a,b)$?
Thank you!

REPLY [1 votes]: You cannot hope for uniform convergence. Consider for instance the case C=D=0. For n=2k+1, you have
$$P_n(x)=A+(B-A)\int_a^x (y-a)^k(b-y)^k\,dy/\int_a^b (y-a)^k(b-y)^k\,dy.$$
For large k, this converges to A+(B-A)H(x-(a+b)/2), where H is the Heaviside function. Since this limit is discontinuous, convergence cannot be uniform.


TITLE: Are there matrices with with non-real elements?
QUESTION [3 upvotes]: I know that the definition of a matrix is a rectangular arrangement of elements, which are real numbers. But does there exist such a thing as a rectangular arrangement of complex numbers? How are these matrices called, if they exist?

REPLY [4 votes]: The entries in a matrix can be anything at all as long as multiplication and addition are things that make sense for it. Formalizing this notion is the concept of a ring, an algebraic structure with two operations, addition and multiplication, under some familiar axioms. Then, one can consider matrices with entries in a ring. This includes real matrices (i.e., matrices in the ring $\mathbb R$), complex matrices (i.e., matrices in the ring $\mathbb C$), but also many other possibilities like matrices whose entries are polynomials, or matrices whose entries are themselves matrices (all of the same size).