TITLE: the group of symmetries of the nodes of a cube
QUESTION [2 upvotes]: Can you explain me please how can  I find rotation for a cube. I will attach an image:

Thanks :)

REPLY [2 votes]: started as a comment, but way too long, so posted here as a suggestion to look into
Maybe you have a particular reason to look at permutations of the nodes, but if you want to investigate the rotational symmetry of the cube, it is probably easier to use that its rotational symmetry group is isomorphic to $S_4$. To do this, you won't look at permutations of the vertices, but imagine the 4 diagonals inscribed into the cube, going, eg, from 1 to 7. Each rotational symmetry rotates the 4 diagonals among themselves, and so corresponds to a permutation in $S_4$. All rotations are generated by three types of rotations around axes: (i) through the center of opposing faces (by $\frac{k \pi}{2}$, $k = 0, 1, 2, 3$); (ii) through one of the diagonals (by $\frac{2 \pi}{3}$ and $\frac{4 \pi}{3}$); and (iii) through a line connecting the middle of two opposing edges (eg, middle 12 to middle 67) (by $\pi$). As indicated by a comment by Barto above, the rotation you are particularly interested in is of type (ii), and you can rotate by either $\frac{2 \pi}{3}$ or $\frac{4 \pi}{3}$. This still twists your mind (or at least it does mine, whenever I go through this again), but it is a more parsimonious description of the cube's symmetry.


TITLE: Total number of integral solutions for the given second degree equation!
QUESTION [0 upvotes]: First, the problem statement : "Consider the equation $x^{2}+y^{2}-3z^{2}-3t^{2}=0$. The total number of integral solutions of this equation in the range of the first 10000 numbers, i.e., $1\leq x,y,z,t\leq10000$ is:
(a.) 200 (b.) 55 (c.) 100 (d.) 1 (e.) None of the above"
I know how to find the number of integral solutions for first degree equations,i.e. form $x_{1}+x_{2}+x_{3}+...+x_{n}=0$ but I am unable to start with this equation.
One observation that I made ( a very obvious one ) is that $x^{2}+y^{2}=3\left( z^{3}+t^{2}\right)$ and thus we can further have $\left(x+y\right)^{2}-3\left(z+t\right)^{2}+6zt-2xy=0$ but this doesn't seem to take me anywhere towards the number of integral solutions.
Any insights or reference that I should refer to solve the above problem?
The solution given in key is option (e.)

REPLY [1 votes]: There are no solutions.
Assume the contrary, then by the well-ordering principle there is a solution $(x, y, z, t) $ such that $ x^2 + y^2 $ is minimized. If $ x^2 + y^2 = 3z^2 + 3t^2 $, reducing both sides modulo 3 and noting that the quadratic residues mod 3 are $ 0, 1 $ we have that both $ x $ and $ y $ are divisible by $ 3 $, so we can let $ x = 3x' $ and $ y = 3y' $ to find $ 9x'^2 + 9y'^2 = 3z^2 + 3t^2 $ or $ z^2 + t^2 = 3x'^2 + 3y'^2 $. However, then $ (z, t, x', y') $ is a solution such that $ z^2 + t^2 < x^2 + y^2 $ (assuming that $ x^2 + y^2 > 0 $), contradiction.
The only solution occurs when $ x^2 + y^2 = 0 $ or $ x=y=z=t=0 $, which does not lie in the desired interval.


TITLE: $\sqrt{2}$ as the limit of a decreasing sequence of rational numbers
QUESTION [1 upvotes]: $\sqrt{2}$ is the limit of the sequence $\{1, 1.4, 1.41, 1.414, ...\}$. How are we to express it as the limit of a decreasing sequence of rational numbers?

REPLY [3 votes]: You can simply round up at each decimal point.
We know $\sqrt{2} \approx 1.41421356$.
Hence define the sequence by $\{2, 1.5, 1.42, 1.415, 1.4143, 1.41422, 1.414214,\cdots\}$.


TITLE: Gravitational anomalies and topological order
QUESTION [8 upvotes]: I wonder the relation of gravitational anomaly and topological order. Specifically:

What is the definition of gravitational anomaly here?
How are they related?

REPLY [4 votes]: Let us define gravitational anomaly as low energy effective theory with no UV completion.
Then gravitational anomalies are classified by topological orders in one higher dimension. In other words, all gravitational anomalies can be realized by the boundary of topological orders in one higher dimension. My paper with Liang Kong addresses this issue. http://arxiv.org/abs/1405.5858
Here is a more detailed discussion.
A low energy effective theory (which can be gapped or gapless) described by an
effective action $S_\text{eff}$ is anomalous if such a low energy effective
theory cannot be realized by any well defined local bosonic quantum model in
the same dimension.  A low energy effective theory is anomaly-free if it can be
realized by a well defined local bosonic quantum model in the same dimension.
(This is a UV completion.)
However, different gapped low energy effective theories
may correspond to the same type of gravitational anomaly.
To address this issue,
we can introduce an equivelance relation: $S^{T_1}_\text{eff}$ and
$S^{T_2}_\text{eff}$ are equivalent if there exist anomaly-free low
energy gapped effective theories $S^{C_1}_\text{eff}$ and $S^{C_2}_\text{eff}$ such
that the combined effective theories $S^{T_1}_\text{eff}+S^{C_1}_\text{eff}$
and $S^{T_2}_\text{eff}+S^{C_2}_\text{eff}$ can deform into each other without
encounter phase transitions.
This leads to a notion of
types of gravitational anomalies, which are defined as the
equivalent classes gapped low energy effective theories under the above
equivalence relation.
Conjecture: The types of gravitational anomalies are classified by topological orders in one higher dimension.


TITLE: Radius of convergence of a power series summing from $- \infty$ to $0$.
QUESTION [0 upvotes]: How would one go about computing the radius of convergence of, say, the following power series:
$$\sum_{n=-\infty}^0 n \, 3^{-n} z^n.$$
It is tempting to directly apply the Cauchy-Hadamard theorem here, but the statement is true for power series summing from $n=0$. I tried to make a substitution by realizing the sum above is equal to :
$$\lim_{m \to \infty} \sum_{n=-m}^0 n \, 3^{-n} z^n,$$
and letting $k = n + m$ for $n= -m, -m+1, \ldots, 0$ we have
$$\lim_{m \to \infty} \sum_{k=0}^m (k-m) \, 3^{m-k} z^{k-m}.$$
But this seems like making things worse. Any thoughts on tackling this problem?

REPLY [4 votes]: $$\sum_{n=0}^{\infty}(-n)3^nz^{-n}$$ converges for $|z|>3$.