# 090917qa

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# 2009 September 17th - questions/answers

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#### 1 Mike D

Suppose that is a field that is complete with respect to a non-discrete valuation , and its valuation ring. Complete means that -cauchy sequences convergerge. A -cauchy sequence is a sequence such that for every large in the value group, there exists such that for , .

Let be the completion of at . Is the map a ring isomorphism? Is it a homeomorphism? ( has the -adic topology.)

**FIRST ANSWER** by Critch.

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 . Claim: they can also be divided (by transfinite recursion!), so they're a field. The resulting valuation ring comprises those series with exponents ≥ 0, and comprises those with exponents > 0. But here we have , so the completion is just .

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#### 2 Mike V

Let be an "Scottish flag matrix":

- the th diagonal entry is
- 's on the super-diagonal, 's on the subdiagonal,
- in the bottom left corner, and in the top left corner
- all other entries .

What are its eigenvalues? Do they lie inside a square in the complex plane with corners ?

The eigenvalues do lie in the square.

First, let us prove that the set of eigenvalues is taken to itself under multiplication by . Let G be the group of order generated by x, y and z, subject to , , , .

There is a unique irreducible representation of this group in which acts by the scalar . The matrix in question is the matrix of . There is an (outer) automorphism of the group which sends , . Since fixes , the representation is isomorphic to . In particular, and have the same eigenvalues. But . So the eigenvalues of are permuted amongst themselves by . 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 lie in
But is the desired square.

David Speyer points out that our matrix can be written as where, in the above notation, and . It is easy to see that the eigenvalues of and both lie in , and both and are anti-Hermitian. Greg Kuperberg claims that this is a general phenomenon: if and are Hermitian matrices whose eigenvalues lie in and respectively, then the eigenvalues of lie in the rectangle .

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#### 3 Andrew D

Say is a field of characteristic 2, algebraically closed.

Geometric version: is the zero set of in normal at ?

Algebraic version: if have no common factors, must be squarefree?

**FIRST ANSWER** by Peter M.

Yes. Write , and differentiate both sides wrt x and y (in characteristic 2):

Now f and g have a common factor, p.

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#### 4 Pablo S

If is an integer such that has only 's and 's in its base expansion, must be a power of ?

Scott: Just checked it's true for 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

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#### 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 , 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.

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#### 6 Critch

A finite topological space which should have nontrivial fundamental group:

Let be the unit circle in the complex plane. Identify the open top half of the circle, , to a single point , and the open bottom half to a single point . Let be the resulting 4-point space (the open sets are ).

Can anyone see directly why the map is not contractible, without fancy theorems?

If you want to calculate 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

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#### 7 Dan H-L

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

**FIRST ANSWER** by Hugh Thomas

No. Let be the real plane with the origin removed. Consider the family of circles defined by for positive. Let be a vector space whose flow gives this family of circles, flowing counterclockwise. Let be reflected in the mirror . So, on the line , will have no -component. By controlling the relative speeds of the different orbits defined by , we can make the flow defined by along the ray move towards the missing origin with constant speed.

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