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# 2009 September 17th - questions/answers

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#### 1 Mike D

Suppose that $K$ is a field that is complete with respect to a non-discrete valuation $v$, and $(R,m)$ its valuation ring. Complete means that $v$-cauchy sequences convergerge. A $v$-cauchy sequence is a sequence $f_n$ such that for every large $g$ in the value group, there exists $N$ such that for $i,j > N$, $v(f_i-f_j) > g$.

Let $R_m$ be the completion of $R$ at $m$. Is the map $R \to R_m$ a ring isomorphism? Is it a homeomorphism? ($R_m$ has the $m$-adic topology.)

No. Over some field k, define a "Q-power series" to be a formal power series with whose exponents form a well-ordered subset of the rationals. Because of the well-ordering, these can be multiplied. This ring is also complete with respect to the "smallest exponent" valuation $v$. Claim: they can also be divided (by transfinite recursion!), so they're a field. The resulting valuation ring $(R,m)$ comprises those series with exponents ≥ 0, and $m$ comprises those with exponents > 0. But here we have $m^2=m$, so the completion $R_m$ is just $R/m=k$.

#### 2 Mike V

Let $A$ be an $N\times N$ "Scottish flag matrix":

• the $j$th diagonal entry is $2\sin(2\pi j/N)$
• $1$'s on the super-diagonal, $-1$'s on the subdiagonal,
• $1$ in the bottom left corner, and $-1$ in the top left corner
• all other entries $0$.

What are its eigenvalues? Do they lie inside a square in the complex plane with corners $\pm 2 \pm 2i$?

The eigenvalues do lie in the square.

First, let us prove that the set of eigenvalues is taken to itself under multiplication by $i$. Let G be the group of order $N^3$ generated by x, y and z, subject to $x^N=y^N=z^N=1$, $xy=zyx$, $xz=xz$, $yz=zy$.

There is a unique irreducible representation $V$ of this group in which $z$ acts by the scalar $e^{2 \pi i /N}$. The matrix in question is the matrix of $d := x+iy-x^{-1} - i y^{-1}$. There is an (outer) automorphism $\phi$ of the group which sends $x \to y[itex], [itex]y \to x^{-1}$, $z \to z$. Since $\phi$ fixes $z$, the representation $\phi(V)$ is isomorphic to $V$. In particular, $\phi(d)$ and $d$ have the same eigenvalues. But $\phi(d) = id$. So the eigenvalues of $d$ are permuted amongst themselves by $i$. See Gurevich and Hadani, and the references therein, for much more on this group and its representation theory.

Now, use the Gershgorin circle theorem. This shows that the eigenvalues of $d$ lie in $D := \{ |x|, |y| \leq 2 \} \cup \{ |x+iy-2| \leq 2 \} \cup \{ |x+iy+2| \leq 2 \}$ But $D \cap iD$ is the desired square.

David Speyer points out that our matrix can be written as $X+iY$ where, in the above notation, $X=x-x^{-1}$ and $Y=y-y^{-1}$. It is easy to see that the eigenvalues of $X$ and $Y$ both lie in $[-2,2]$, and both $X$ and $Y$ are anti-Hermitian. Greg Kuperberg claims that this is a general phenomenon: if $X$ and $Y$ are Hermitian matrices whose eigenvalues lie in $[x_1,x_2]$ and $[y_1,y_2]$ respectively, then the eigenvalues of $X+iY$ lie in the rectangle $[x_1,x_2] \times [y_1, y_2]$.

#### 3 Andrew D

Say $K$ is a field of characteristic 2, algebraically closed.

Geometric version: is the zero set of $z(xy-z^2)$ in $A^3$ normal at $(0,0,0)$?

Algebraic version: if $f,g \in A[x,y]$ have no common factors, must $xf^2+yg^2$ be squarefree?

Yes. Write $xf^2+yg^2=p^2h$, and differentiate both sides wrt x and y (in characteristic 2):

$f^2=h_xp^2$
$g^2=h_yp^2$

Now f and g have a common factor, p.

#### 4 Pablo S

If $N$ is an integer such that $N^2$ has only $0$'s and $1$'s in its base $10$ expansion, must $N$ be a power of $10$?

Scott: Just checked it's true for $N$ up to 100 billion...

Greg Kuperberg: I don't have a solution either, but I can check it for N up to a septillion (or more) in Python using Hensel lifting. Since the algorithm works 10-adically, maybe that is a good way to view the problem.

maxmod = 10**24

def check(x):
if not str(x**2).replace('0','').replace('1',''):
print 'Eureka:',x,x**2

def search(x,mod):
x %= mod
if mod == maxmod:
check(x)
check(mod-x)
return
top = -(x**2/mod) % 10
x += (top + top%2)/2 * mod
search(x,mod*10)
search(x + 5*mod,mod*10)

search(1,10)    # Solution is either 1 or 9 mod 10

#### 5 Mike H

The Mandelbrot set is usually defined over C. It can also be defined in the quaternions or hypercomplexes. What does it look like? If we visualize this 4D set as a time-varying 3D set, will it change smoothly transitions, or will it just flicker in and out of existence?

FIRST ANSWER by Greg Kuperberg: Defined how? If you mean defined by the orbit of 0 under $x \mapsto x^2 + c$, then this quaternionic Mandelbrot set is just obtained by revolving the usual complex Mandelbrot set. That's because every quaternion is contained in a copy of the complex numbers. In fact, any reasonable definition of a quaternionic Mandelbrot set would have this symmetry, although it would not necessarily come from the complex Mandelbrot set.

Maybe a better thing to look at is quaternionic Julia sets. If you Google that, you'll find all sorts of fun (but non-rigorous) material.

#### 6 Critch

A finite topological space which should have nontrivial fundamental group:

Let $S$ be the unit circle in the complex plane. Identify the open top half of the circle, $\{z \in S | im(z) > 0\}$, to a single point $T$, and the open bottom half $\{z \in S | 0 > im(z)\}$ to a single point $B$. Let $X=\{+1,T,-1,B\}$ be the resulting 4-point space (the open sets are $\{\}, X, \{B\}, \{T\}, X\setminus \{+1\}, X\setminus\{-1\}$).

Can anyone see directly why the map $S \to X$ is not contractible, without fancy theorems?

If you want to calculate $\pi_1$ the easy way to do this is to find the universal cover of this 4 point space. It consists of three rows of points, the top and bottom open and the middle closed which are staggered so that they go top, middle, bottom, middle, top, etc. The closure of an open point contains the two adjacent closed points. Clearly this is a covering space, so you just need to show that it is contractible. I forget how that argument works, but it isn't hard.

Actually just to answer your question, you don't need that it's contractible, just that it's a covering space. It's clear that your loop lifts to a nontrivial path in this covering space and hence is not null-homotopic.

Noah

#### 7 Dan H-L

If $V,W$ are vector fields on a manifold $M$, each of which defines an "eternal flow" (a flow that is defined for all times $t$), does $V+W$ define an eternal flow?

No. Let $M$ be the real plane with the origin removed. Consider the family of circles defined by $(x-c)^2+y^2=4c^2$ for $c$ positive. Let $V$ be a vector space whose flow gives this family of circles, flowing counterclockwise. Let $W$ be $V$ reflected in the mirror $x=0$. So, on the line $x=0$, $V+W$ will have no $x$-component. By controlling the relative speeds of the different orbits defined by $V$, we can make the flow defined by $V+W$ along the ray $x=0, y>0$ move towards the missing origin with constant speed.