- Re: symplectic group Sp(p,q) and reduction modulo a prime
The "reduction modulo a prime" of an algebraic group is not well-
defined since it depends on the choice of a model over the ring of
integers. In down-to-earth terms, an embedding into GLn defines a
reduction, but a different embedding may give a different reduction.
Models of semisimple groups over rings of integers in local fields
- symplectic group Sp(p,q) and reduction modulo a prime
Hi!
Does anyone know a book or a reference where one can find something
about the reduction modulo a prime of certain algebraic groups, i.p.
f.e. if we have a group G defined over a totally real number field F,
with the help of a quaternion division algebra D/F and a hermitian
form on D^{n+1} s.t.
G(R) ~ Sp(n,1) (\times ...),
- Re: arbitrary factorization as a difference of cubes?
Indeed, it appears that
(#P)! \product_(p \in P) 2p
=? Sum_(Q \subset P) (-1)^#Q ( Sum_(p \in P) PM(p,Q) )^#P,
where PM(p,Q) = -p if p \in Q, p otherwise.
There should be a name for this equation, but I don't know it.
Dan Hoey
haoyuep at aol.com
- Positions at Penn State
Tenure and Tenure Track Faculty Positions
The Department is seeking to fill two or more positions; dependent on
the qualifications and experience of the appointee, these may be at
the assistant, associate or full professor level. One position is
preferably to be filled in the area of probability theory/stochastic
- Re: arbitrary factorization as a difference of cubes?
Along the same lines,
(a+b+c+d)^4 - (a+b+c-d)^4 - (a+b-c+d)^4 - (a-b+c+d)^4 - (-a+b+c+d)^4
+(a+b-c-d)^4 + (a-b+c-d)^4 + (a-b-c+d)^4 = 192 a b c d
- Re: OEIS A004664 related conjecture: n! + n^2 != m^2 for n>=1, m>= 0
FYI - Tony Noe came up with the partial proof for my conjecture (when
"n" is prime)
- see below.
Tony Noe also extended the calculations up to n = 10^5
(my original calculation was up to n = 30,000) and found
that my conjecture holds to be TRUE for entire range.
Thanks,
Best Regards,
Alexander R. Povolotsky
- Re: OEIS A004664 related conjecture: n! + n^2 != m^2 for n>=1, m>= 0
I'm not sure what I was thinking when I wrote the above,
but it's nonsense. My apologies.
- Re: Criterion for SVD Truncation
On Sep 24, 4:00 pm, Robert Israel
There is a discussion of this in "Numerical Recipes", which is
available online. As they point out, it is not often you get to set
infinity equal to zero.
- Re: OEIS A004664 related conjecture: n! + n^2 != m^2 for n>=1, m>= 0
Maria Povolotsky a écrit :
[...]
Yes, but same [extremely classical] proof gives same conclusion --
There is a prime between n and n-squared and so on --
Cheers!
Olivier
- Re: arbitrary factorization as a difference of cubes?
As a trivial aside, it can be written as
24abc = (a+b+c)^3 + (-a-b+c)^3 + (-a+b-c)^3 + (a-b-c)^3.
- Re: OEIS A004664 related conjecture: n! + n^2 != m^2 for n>=1, m>= 0
Hi,
It appears that some responders misunderstood my conjecture ...
Let me reiterate my conjecture once more:
The sum
n! + n^2
for n >=1
doesn't yield the perfect square
I also was advised for the additional clarity to present above in the
TEX format as:
$n! + n^2 \ne m^2$
Also any one could check the OEIS's A004664 to see what is involved
- Re: Does anyone have access to this article?
On Sep 29, 10:30 am, Starblade Enkai
wrote:
I found staff.utia.cas.cz/matus/fmmatp ar.ps
- Re: OEIS A004664 related conjecture: n! + n^2 != m^2 for n>=1, m>= 0
I MEANT sqrt(n). Prime numbers between -n and n is absurd. I do wish
Google would enable you to put symbols in. I know I have said this on
other occaisions.
- Ian Parker
- Re: K-dimensional subspaces of GF(p)^{2k} with zero intersection
I already answered this on sci.math. Consider
V = GF(p)^{2k} as a 2-dimensional vector space over GF(p^k).
The 1-dimensional GF(p^k)-subspaces of V suffice.
Victor Meldrew
"I don't believe it!"
- Does anyone have access to this article?
It's this one:
[link]
Can anyone please explain what "Matroid representations by partitions"
means, and give a little demonstration of the math involved? Thanks!
- Re: K-dimensional subspaces of GF(p)^{2k} with zero intersection
[A complimentary Cc of this posting was sent to
jdm
], who wrote in article :
The case k=2 is trivial. The case k=2 implies the general case via
field extension: consider GF(k^p)^2.
Hope this helps,
Ilya
- Re: arbitrary factorization as a difference of cubes?
For what it's worth, the uses for this identity I've seen
are based on the fact that it provides a way to write the
product ab in terms of additions, subtractions, division
by a simple constant (2), and squaring. Here are a couple of
places that I can think of where this it's used: (1) In numerical
calculations by hand, before electronic computing devices
- Re: OEIS A004664 related conjecture: n! + n^2 != m^2 for n>=1, m>= 0
I think I can oblige.
n^2! + n! = (n^2.(n^2-1).(n^2-2)....(n+1)+ 1)n! Just multiplying out.
Consider n^2.(n^2-1).(n^2-2)....(n+1)+1
It cannot have a factor between n+1 and n^2 for if we divide by such a
number we always obtain a remainder of 1. This number must either be
prime or the product of large primes. If crucially we take the numbers
- K-dimensional subspaces of GF(p)^{2k} with zero intersection
I am trying to construct a set of k-dimensional subspaces of
GF(p)^{2k} such that all of the subspaces are disjoint except for the
zero vector. I know that such a set can always be constructed with
three subspaces contained in it, and have established an upper bound
of p^{k} + 1 subspaces.
I currently believe that this upper bound can always be achived for
- Re: OEIS A004664 related conjecture: n! + n^2 != m^2 for n>=1, m>= 0
Let p be any prime between n and n / 2 (but exceeding n / 2).
Then p^2 divides n^2, p divides n-factorial, p^2 doesn't
divide n-factorial, so p divides left side but p^2 doesn't,
so left side is not a square.
- OEIS A004664 related conjecture: n! + n^2 != m^2 for n>=1, m>= 0
Hi,
In addition to previously posted conjecture
(unfortunately I have not seen any responses yet ;-) )
I came up with OEIS A004664 related conjecture:
n! + n^2 != m^2 for n>=1, m>= 0
I checked using PARI that indeed n! +n^2 doesn't yield perfect square
up to n=30,000
Is this conjecture known (could one on this list point me to the
- Why "Borel" in Borel calculus?
I looked through several "standard books", and could not find an answer:
When one considers Borel calculus of a (bounded) normal/self-adjoint
operator, one can apply any function which is l-infinity w.r.t. the
spectral measure of the operator [*].
So, in principle, one could restrict attention to functions which are
- Mapping Class Group computations
Where can I find a rigorous proof that the (orientation-preserving)
MCG(R^n)={1}?
Also, is MCG(S^4)={1}? What about MCG(S^n), n>1?
- SOCG 2009 Call for Papers
CALL FOR PAPERS, VIDEOS, AND MULTIMEDIA
25th Annual ACM Symposium on Computational Geometry
June 8-10, 2009
Aarhus, Denmark
[link]
In cooperation with ACM SIGACT and SIGGRAPH
The Twenty-fifth Annual Symposium on Computational Geometry will be
held at the University of Aarhus, Denmark, and will be organized by
- Combinatorial (?) assignment problem
Greetz.
I am looking at a problem, probably a simple enough one, for which I don't know
a solution. Here's how it goes. Any help appreciated.
Let s_i(a) and r_i(a) be two groups of functions, that each take one of two
values, 0 or \sigma_i in the case of s_i, and 0 or \rho_i in the case of r_i,
- finiteness of Ext
let R be commutative notherian ring and I, J be ideals of R and M be
an R-module. we know that Ext^i(R/I,M) is finite for evry non-negative
integer i.
1- can we say that Ext^i(R/I,M/JM) is finite for all i?
2- i know with this assumptaion Ext^i(R/I+J,M) is finite, can we say
that Ext^i(R/I,M/JM) is finite?
- connectedness of the omega-limit sets
Consider a diffrential equation dx/dt = f(x)
It is well known that if the omega-limit set omega(x0),issued from an
intial condition x0, is compact, it is also connected. But it is true
that an omega-limit set is also arc-connected ?
- Re: A Conjecture about eigenvalues
Thanks.
The question was answered in sci.math. It can be concluded from the
interlocking eigenvalues lemma used in optimization theory. For
example see p. 300 in David G. Luenberger and Yinyu Y, Linear and
Nonlinear Programming: Third Edition (2008). You can also find the
proof of my version at
[link]
- 2 year Lecturer position in Pure Mathematics at UEA
University of East Anglia
School of Mathematics
Fixed-term Lectureship in Pure Mathematics Ref:ATS311
UK 28,290 to UK 33,780 per annum (pay award pending October 2008).
The University of East Anglia invites applications for a
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- Re: Criterion for SVD Truncation
The usual justification is that the result you get, while not having
Ax = b, has Ax very close to b, or Bx very close to b for some B very
close to A. If the elements of A and/or b are only approximations of some
(possibly unknown) exact values, e.g. come from physical measurements, that is
the best that you can hope for.
- Re: Criterion for SVD Truncation
[A complimentary Cc of this posting was sent to
Junoexpress
], who wrote in article :
*Why* do you solve it?
Hope this helps,
Ilya
- Re: A Conjecture about eigenvalues
[A complimentary Cc of this posting was sent to
Peter B
], who wrote in article :
Look for minimax characterization of eigenvalues of self-adjoint
operators.
Hope this helps,
Ilya
- Re: Minimal Enclosing Triangle
How large is your 'n'? As with many computational geometry
algorithms:
(1) The asymptotic behavior might not be noticeable until n
is much larger than what your application uses.
(2) The effort to implement the algorithm is not worth the time
when a simpler (but asymptotically slower) algorithm is
- Re: Complemented Lattices
In article ,
You are correct...I misunderstood the original question.
- Criterion for SVD Truncation
Hi,
I have a linear system of eqns I am solving of the form:
Ax = b
A is a 32 x 32 matrix with complex entries, having full rank.
It is not a normal matrix, but has, of course, the SVD A = U L V*.
A is somewhat strange in that N of the (2N)^2 entries have a random
component to them.
Also, half of the components of b also have a random component to
- A Conjecture about eigenvalues
I'm looking for a proof of the following Conjecture
Let A be a diagonal p-dim matrix with a_1 >= a_2 >=...>= a_p >=0 on the
diagonal. For an arbitrary column vector v, the matrix (A + v*t(v))
has the eigenvalues b_1>=b_2>=...>=b_p>=0 such that b_i >= a_i for all
i=1,..,p.
Note: t(v) is a transpose of v, that is, a row vector (so, v*t(v) is a
- Re: Combining two second-order linear recurrences
What do you mean by "intersections"?
-- m
- Six papers published by Geometry & Topology Publications
Four papers have been published by Algebraic & Geometric Topology
(1) Algebraic & Geometric Topology 8 (2008) 1523-1565
Organizing volumes of right-angled hyperbolic polyhedra
by Taiyo Inoue
URL: [link]
DOI: 10.2140/agt.2008.8.1523
(2) Algebraic & Geometric Topology 8 (2008) 1567-1579
- Combining two second-order linear recurrences
Hello all,
I'm trying to solve a system of two simultaneous Diophantine
equations. I have found two second-order linear recurrence equations,
A(n) = rA(n-1) + sA(n-2)
B(n) = uB(n-1) + vB(n-2).
I have determined that A(0) = B(0) = 0 and A(2) = B(1), corresponding
to the first two solutions of the original system. Is there a way to
- Re: Complemented Lattices
No. (I see that Tim Chow has given the opposite answer. However, the
theorem he quotes uses a stronger condition - the existence of unique
relative complements.)
R. P. Dilworth proved in 1945 that every lattice can be embedded in a
uniquely complemented lattice. So a uniquely complemented lattice need
- Result of Turpin leads to a contradiction
There's an celebrated result of Philippe Turpin (Philippe Turpin,
Produits tensoriels d‰??espaces vectoriels topologiques, Bulletin de la
Soci?t? Math?matique de France 110 (1982), 3‰??13.) that asserts:
If A and B are F-spaces then there's an explicit formula for an F-norm on
the algebraic tensor product A\otimes B (not the completed tensor
- Re: arbitrary factorization as a difference of cubes?
Hello, all!
Thank you for your responses (both on- and off-list). Sorry I
apparently wasn't any clearer with my "clarification" post than I was
in my original? =)
In any case, I have (in the meantime) derived an identity which is an
example of precisely the kind of thing I was seeking:
ab = F_1(a,b)^3 +- F_2(a,b)^3 +- F_3(a,b)^3 +- ...
- Re: Complemented Lattices
In article ,
Yes; for example Corollary 1 in Chapter IX of Birkhoff's book on lattice
theory says that a lattice is distributive if and only if relative
complements in it are uniquely determined. He proves it via a structure
theorem: If a lattice fails to be distributive then it must contain one of
- How much choice is implied by Urysohn's lemma?
Based on the following 2 links we need some form of choice
to prove Urysohn's lemma:
[link]
[link]
This leads me to wonder, if we assume ZF + "Urysohn's lemma"
can we derive countable choice or dependent choice?
- Re: arbitrary factorization as a difference of cubes?
In article ,
I'm not quite I understand your interpretation of this identity.
I might interpret it as follows: given an integer N we consider
two Diophantine equations to be solved: N = a b and N = x^2 - y^2.
(Or more generally we can play this game in any ring, not just in Z .)
- Minimal Enclosing Triangle
Does anybody have a good C/C++/Java implementation of finding the
minimal enclosing triangle for a (convex) polygon in O(n) time?
I tried implementing the algorithm in "An Optimal Algorithm for
Finding Minimal Enclosing Triangles" (O'Rourke, Aggarwal, Maddila,
Baldwin) in Journal of Algorithms 7 (1986). After grinding my rusty
- Re: radical ideal algorithms
Determining the radical of an ideal in the general case is quite hard.
It goes roughly like this:
1. Find a primary decomposition for the ideal
2. From this, calculate the minimal associated primes of the ideal.
3. Then the radical is the intersection of the minimal associated
primes
The hard part is the primary decomposition.
- Complemented Lattices
Let L be a complemented lattice. That is a lattice with
a top or max element 1 and a bottom or min element 0 and
for all x in L, some y in L with
xy = inf x,y = 0, x + y = sup x,y = 1
x and y with that property are call complements.
If L is a distributive lattice, then each element has
exactly one complement. Is the converse correct,
- Re: arbitrary factorization as a difference of cubes?
Hi Wolfgang,
Thanks for the post. I *do* know that analog for cubes -- allow me to
clarify my question?
Let's write
ab = F_1(a,b)^2 - F_2(a,b)^2
where F_1(a,b) = (a+b)/2 and F_2(a,b) = (a-b)/2. Then I want an
analogous result for cubes, i.e.
ab = F_1(a,b)^3 +- F_2(a,b)^3 +- F_3(a,b)^3 +- ...
where the F_k(a,b) are functions/polynomials in a and b. Since (e.g.)
- Re: arbitrary factorization as a difference of cubes?
hi Kieren,
the above identity is based on a^2 - b^2 = (a+b)*(a-b). The analogon
for cubes is given by
a^3 pm b^3 = (a pm b)*(a^2 mp ab + b^2), where pm means plus/minus and
mp vice versa.
hope this helps, Wolfgang.
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