Supppose there are integers $a_1,a_2,dots$ and a polynomial $p$ so that the integers $p(a_1),p(a_2)...$ satisfy some linear recurrence, i.e. $sum p(a_i)x^i$ is a rational function of $x$. Must integers $b_iin p^{-1}(p(a_i))$ so that $sum b_ix^i$ is a rational function, necessarily exist?
(The answer is no if we ask for the function $sum a_i x^i$ to be rational, as can be seen when $p(t)=t^2$ and $a_i$ being a random sequence of $pm1$)
Matthias' polemics are funny at points but also misleading in several respects:
ZFC also has enormous length and depth of deductions for trivial material. According to Norman Megill's metamath page, "complete proof of 2 + 2 = 4 involves 2,452 subtheorems including the 150 [depth of the proof tree] above. ... These have a total of 25,933 steps — this is how many steps you would have to examine if you wanted to verify the proof by hand in complete detail all the way back to the axioms." Megill's system is based on a formalism for substitutions so there may be an enormous savings here compared to the way in which Matthias performs the counts (i.e., the full expanded size in symbols) for Bourbaki's system. If I correctly recall other information from Megill about the proof length he estimated for various results in ZFC, the number of symbols required can be orders of magnitude larger and this is what should be compared to Matthias' numbers.
The proof sizes are enormously implementation dependent. Bourbaki proof length could be a matter of inessential design decisions. Matthias claims at the end of the article that there is a problem using Hilbert epsilon-notation for incomplete or undecidable systems, but he gives no indication that this or any other problem is insurmountable in the Bourbaki approach.
Indeed, Matthias himself appears to have surmounted the problem in his other papers, by expressing Bourbaki set theory as a subsystem of ZFC. So either he has demonstrated that some reasonably powerful subsystems of ZFC have proofs and definitions that get radically shorter upon adding Replacement, or that the enormous "term" he attributes to the Theorie des Ensembles shrinks to a more ZFC-like size when implemented in a different framework.
EDIT. A search for Norman Megill's calculations of proof lengths in ZFC found the following:
"even trivial proofs require an
astonishing number of steps directly from axioms. Existence of the
empty set can be proved with 11,225,997 steps and transfinite recursion
can be proved with 11,777,866,897,976 steps."
and
"The proofs exist only in principle, of course, but their
lengths were backcomputed from what would result from more traditional
proofs were they fully expanded. ..... In the current version of my proof
database which has been reorganized somewhat, the numbers are:
empty set = 6,175,677 steps
transf. rec. = 24,326,750,185,446 steps"
That's only the number of steps. The number of symbols would be much, much higher.
Yesterday, in the short course on model theory I am currently teaching, I gave the following nice application of downward Lowenheim-Skolem which I found in W. Hodges A Shorter Model Theory:
Thm: Let $G$ be an infinite simple group, and let $kappa$ be an infinite cardinal with $kappa leq |G|$. Then there exists a simple subgroup $H subset G$ with $|H| = kappa$.
(The proof, which is short but rather clever, is reproduced on p. 10 of http://www.math.uga.edu/~pete/modeltheory2010Chapter2.pdf.)
This example led both the students and I (and, course mechanics aside, I am certainly still a student of model theory) to ask some questions:
$1$. The theorem is certainly striking, but to guarantee content we need to see an uncountable simple group without, say, an obvious countable simple subgroup. I don't know that many uncountable simple groups. The most familiar examples are linear algebraic groups like $operatorname{PSL}_n(F)$ for $F$ an uncountable field like $mathbb{R}$ or $mathbb{C}$. But this doesn't help, an infinite field has infinite subfields of all infinite cardinalities -- as one does not need Lowenheim-Skolem to see! (I also mentioned the case of a simple Lie group with trivial center, although how different this is from the previous example I'm not sure.) The one good example I know is supplied by the Schreier-Ulam-Baer theorem: let $X$ be an infinite set. Then the quotient of $operatorname{Sym}(X)$ by the normal subgroup of all permutations moving less than $|X|$ elements is a simple group of cardinality $2^{|X|}$. (Hmm -- at least it is when $X$ is countably infinite. I'm getting a little nervous about the cardinality of the normal subgroup in the general case. Maybe I want an inaccessible cardinal or somesuch, but I'm getting a little out of my depth.) So:
Are there there other nice examples of uncountable simple groups?
$2$. At the beginning of the proof of the theorem, I remarked that straightforward application of Lowenheim-Skolem to produce a subgroup $H$ of cardinality $kappa$ which is elementarily embedded in $G$ is not enough, because it is not clear whether the class of simple groups, or its negation, is elementary. Afterwards I wrote this on a sideboard as a question:
Is the class of simple groups (or the class of nonsimple groups) an elementary class?
Someone asked me what techniques one could apply to try to answer a problem like this. Good question!
$3$. The way I stated Hodges' result above is the way it is in my lecture notes. But when I wrote it on the board, for no particular reason I decided to write $kappa < |G|$ instead of $kappa leq |G|$. I got asked about this, and was ready with my defense: $G$ itself is a simple subgroup of $G$ of cardinality $|G|$. But then we mutually remarked that in the case of $kappa = |G|$ we could ask for a proper simple subgroup $H$ of $G$ of cardinality $|G|$. My response was: well, let's see whether the proof gives us this stronger result. It doesn't. Thus:
Let $G$ be an infinite simple group. Must there exist a proper simple subgroup $H$ of $G$ with $|H| = |G|$?
Wait, I just remembered about the existence of Tarski monsters. So the answer is no. But what if we require $G$ to be uncountable?
Eric Broug in his book Islamic Geometric Patterns gives
straightedge and compass construction of some simpler patterns.
It is clear his techniques will provide constructions for many
Islamic patterns.
Looking at formal constructibility, the Wikipedia pages gives Gauss' result that
7, 9, 11, 13, 14, 18... etc sided polygons are not constructible. Hence the pattern
http://tilingsearch.org/HTML/data160/J43C.html
is not constructible since it contains a regular
9-pointed star polygon.
I have over 800 Islamic patterns on my web site but
I use a computer and trigonometry to produce my images. It seems that
about 40 Islamic patterns on my site are not constructible.
Given an Islamic pattern that is not excluded from construction by Gauss'
result, it is almost certainly constructible if the following is true:
Given two points on the plane, a polygon ($n$ sides) can be constructed with the two points
as an edge, provided $n$ is not equal to 7, 9, 11, 13, 14, 18... etc.
This result would allow patterns to be built up piece-by-piece.
EDIT, Will Jagy: from his profile page, the OP's website, in this address preset to display the tilings in a slideshow format on a web browser, is at
In number theory, here are four applications of techniques or results in first-year calculus.
(1) Finding equations of tangent lines by first-semester calculus methods lets us add points on elliptic curves using the Weierstrass equation for the curve. This is more algebraic geometry than number theory, so I'll add that the methods show if the Weierstrass equation has rational coefficients then the sum of two rational points is again a rational point.
(2) The recursion in Newton's method from differential calculus is the basic idea behind Hensel's lemma in $p$-adic analysis (or, more simply, lifting solutions of congruences from modulus $p$ to modulus $p^k$ for all $k geq 1$).
(3) The infinitude of the primes can be derived from the divergence of the harmonic series (the zeta-function at 1), which is based on a bound involving the definition of the natural logarithm as an integral.
(4) Unique factorization in the Gaussian integers can be derived from the Leibniz formula
$$
frac{pi}{4} = 1 - frac{1}{3} + frac{1}{5} - frac{1}{7} + frac{1}{9} - cdots = sum_{n geq 0} frac{(-1)^n}{2n+1}
$$
by interpreting it as a case of Dirichlet's class number formula $2pi h/(wsqrt{|D|}) = L(1,chi_D)$ for $chi_D$ the primitive quadratic character associated to ${mathbf Q}(sqrt{D})$ where $D$ is a negative fundamental discriminant, $h$ is the class number of ${mathbf Q}(sqrt{D})$ and $w$ is the number of roots of unity in ${mathbf Q}(sqrt{D})$. Taking $D = -4$ turns the left side into $2pi h/(4sqrt{4}) = (pi/4)h$, so the Leibniz formula is equivalent to $h = 1$, which is another way of saying $mathbf Z[i]$ is a PID or equivalently (for Dedekind domains) a UFD.
Here are two more applications, not in number theory directly.
(5) Gerry Edgar mentions in his answer Niven's proof of the irrationality of $pi$, which is available in Spivak's calculus book. The same ideas imply irrationality of $e^a$ for every positive integer $a$, which in turns easily implies irrationality of $e^r$ for nonzero rational $r$ and thus also irrationality of $log r$ for positive rational $r not= 1$. The calculus fact in the proof of irrationality of the numbers $e^a$ is that for all positive integers $n$ the polynomial
$$
frac{x^n(1-x)^n}{n!}
$$
and all of its higher derivatives take integer values at $0$ and $1$. That implies a certain expression involving a definite integral is a positive integer, and then with the fundamental theorem of calculus that same expression turns out to be less than 1 for large $n$ (where "large" depends on the hypothetical denominator of a rational formula for $e^a$), and that is a contradiction.
(6) Prove that if $f$ is a smooth function (= infinitely differentiable) on the real line and $f(0) = 0$ then $f(x) = xg(x)$ where $g$ is a smooth function on the real line. There is no difficulty in defining what $g(x)$ has to be if it exists at all, namely
$$
g(x) = begin{cases}
f(x)/x, & text{ if } x not= 0, \
f'(0), & text{ if } x = 0.
end{cases}
$$
And easily the function defined this way is continuous on the real line and satisfies $f(x) = xg(x)$. But why is this function smooth at $x = 0$ (smoothness away from $x = 0$ is easy)? You can try to do it using progressively messier formulas for higher derivatives of $g$ at 0 by taking limits, but a much slicker technique is to use the fundamental theorem of calculus to write
$$
f(x) = f(x) - f(0) = int_0^x f'(t),dt = xint_0^1 f'(xu),du,
$$
which leads to a different formula for $g(x)$ that doesn't involve cases:
$$
g(x) = int_0^1 f'(xu),du.
$$
If you're willing to accept differentiation under the integral sign (maybe that's not in the first-year calculus curriculum, but we used first-year calculus to get the slick formula for $g(x)$) then the right side is easily checked to be a smooth function of $x$ from $f$ being smooth.
If you want to teach something intriguing, you should do something that introduces a new geometric idea while also involving algebra in an essential way. I recommend that you give an introduction to the projective plane, showing the other students that it is a natural extension of ordinary space which makes some geometric properties more uniform (such as intersection properties of curves), gives a fruitful new way to think about old topics (like asymptotes), and lets you do things that are impossible to conceive without it (reducing rational points mod $p$). There should be substantial interplay between algebra and geometry, but make sure to draw pictures to emphasize the geometric aspects.
In algebra, we can conceive of the quadratic formula in a uniform manner, but the ancient Greeks [Edit: Babylonians, not Greeks] couldn't do this because they didn't have the idea of negative numbers. So they had several quadratic formulas on account of not being able to write something as simple as $ax^2 + bx + c = 0$ at one stroke (for any signs on $a, b$, and $c$, with $a$ nonzero). Our extended skill at algebra lets us work with one case where the ancients had to take multiple cases. We can also say with complex numbers that any quadratic equation has two roots, allowing for a double root to count as one root with multiplicity two.
The thrust of what comes next is to extend the plane so that geometric properties become nicer in a similar way the algebra is becoming nicer when we use more general number systems.
Consider the intersection properties of lines in the plane. There is a dichotomy: usually two lines in the plane meet in one point, but some pairs of lines (the parallel ones) meet in no points. Let's see what this looks like under stereographic projection. Lines in the plane become circles through the north pole, but not including the north pole itself. It's natural to close up the image and take that whole circle as a substitute for the original line. So we can see that lines in the plane naturally close up into circles through the north pole. Under stereographic projection, the old dichotomy between parallel and non-parallel lines takes on a new appearance: a pair of non-parallel lines corresponds under stereographic projection to a pair of circles intersecting in two different points, one of which is the north pole, while a pair of parallel lines corresponds under stereographic projection to a pair of circles which are tangent at the north pole. It is natural to think of two tangent circles as having their point of tangency be an intersection point of multiplicity two, much like a quadratic polynomial can have a double root. So after stereographic projection we can "see" two points of intersection for any pair of lines. This geometric construction is something like the algebraic use of more general number systems to find roots to all quadratic equations. The moral to take from this example is that in a larger space, curves that used to not intersect may now intersect (or rather, their natural closures in the new space intersect) with a uniform count of the number of intersection points. If the students agree that enlarging number systems to create solutions to polynomial equations is good, they should agree that enlarging space to make intersection properties more uniform is good too. Another important feature is that the sphere, like the plane, is a homogeneous object: we can transform (rotate) the space to carry one point to any other point. On the sphere as a space in its own right, there is truly nothing special about the north pole.
An even better geometric extension of the plane is the projective plane, although at first it will feel unfamiliar and strange because you can't see it all at once.
You should introduce it in a uniform manner as points described with homogeneous coordinates $[x,y,z]$ where $x$, $y$, and $z$ are not all 0 and, say,
$$
[3,6,2] = [1,2,2/3] = [1/2,1,1/3] = [3/2,3,1] text{ and } [0,5,0] = [0,1,0].
$$
Although it is impossible to see the whole projective plane at once, we can get glimpses of large parts of it using three different charts: $U_0$ is the points where $x not= 0$, $U_1$ is the points where $y not= 0$ and $U_2$ is the points where $z not= 0$. These three charts together cover the projective plane. Any nonzero coordinate can be scaled to 1 and that fixes the other two homogeneous coordinates of the point, e.g., $[x,y,1] = [x',y',1]$ if and only if $x = x'$ and $y = y'$. This means we can identify each of $U_0$, $U_1$, and $U_2$ with the usual plane (e.g., identify $U_2$ with ${mathbf R}^2$ by identifying $[x,y,1]$ with $(x,y)$). This means the projective plane locally looks like the plane, much like the sphere does, except we can't see all of it at the same time as we can with the sphere.
(In case you want to show students that the projective plane is a really natural model of something they have known in another context, think about nonzero ideals in ${mathbf R}[x]$. Any ideal has a generator, but the polynomial generator is only defined up to a nonzero scaling factor. Usually we normalize the generator to be monic, but if we don't want to insist on a particular choice of generator then the right model for the generator is a point in projective space. In particular, for any nonzero ideal $(f(x))$ where $deg f(x) leq 2$, write $f(x) = ax^2 + bx + c$; the coefficients $a, b, c$ are only defined up to an overall scaling factor, so the point $[a,b,c]$ is one way to think about that ideal.)
Next introduce curves in the projective plane as solutions to homogeneous polynomial equations in $x$, $y$, and $z$ and explain what the algebraic process of homogenization and dehomogenization of polynomials is, e.g., it makes $y = 2x + 1$ into $y = 2x + z$ or $x^2 - y^2 = x+ 1$ into $x^2 - y^2 = xz + z^2$. In particular a line in the projective plane is the solution set to any equation $ax + by + cz = 0$ where the coefficients are not all 0.
Now let's look at what a point on a specific curve in the projective plane looks like in each of the three standard charts, carry out the same kind of calculus computation in each chart, and compare the results with each other. We will use the curve $C : x^2 + y^2 = z^2$ in the projective plane (not to be confused with a surface in 3-space given by the same equation) and the points $P = [3,4,5]$ and $Q = [1,0,1]$ which lie on $C$. How do $C$, $P$, and $Q$ appear in each of the charts $U_0$, $U_1$, and $U_2$?
a) In $U_0$, which is identified with the plane by $[x,y,z] mapsto (y/x,z/x)$, $C$ becomes the hyperbola $z^2 - y^2 = 1$, $P$ becomes $(4/3,5/3)$, and $Q$ becomes $(0,1)$. Here we identify $U_0$ with the usual $yz$-plane. By calculus, the tangent line to $z^2 - y^2 = 1$ at the point $(4/3,5/3)$ is $z = (4/5)y + 3/5$ and the tangent line at $(0,1)$ is $z = 1$.
Note that we actually miss two points from $C$ when we look at the intersection of it with $U_0$: $[0,1,pm 1]$.
b) In $U_1$, $C$ becomes the hyperbola $z^2 - x^2 = 1$ in the $xz$-plane, $P$ becomes the point $(3/4,5/4)$ with tangent line $z = (3/5)x + 4/5$, and $Q$ doesn't actually live in this chart (kind of like the north pole under stereographic projection not going to anything the in the plane). Here two points from $C$ are missing: $[1,0,pm 1]$.
c) In $U_2$, $C$ becomes the circle $x^2 + y^2 = 1$, $P$ becomes $(3/5,4/5)$ with tangent line $y = (-3/4)x + 5/4$, and $Q$ becomes $(1,0)$ with tangent line $x = 1$. Every point from $C$ lies in $U_2$, so no points are missing here. We see the "complete" curve in this chart.
It is essential to draw three pictures here (of the $yz$-plane, $xz$-plane, and $xy$-plane) and mark $P$ and $Q$ in each (except you don't see $Q$ in the $xz$-plane).
Now comes the beautiful comparison step: in all three charts the homogenization of the tangent line at $P$ is exactly the same equation: $3x + 4y = 5z$. The tangent line at $Q$ in $U_0$ and $U_2$ homogenizes in both cases back to $x = z$. This suggests there should be an intrinsic concept of tangent line in the projective plane to the curve $C$ at the points $P$ and $Q$, and you can compute the tangent line by looking at any chart containing the relevant point of interest, doing calculus there, and then homogenizing back. The homogenization of your ordinary linear equation to a homogenuous linear equation will always be the same, and its solutions in the projective plane define the tangent line to the projective curve at that point.
As further evidence of the consistency of this new space and the geometry in it, look at the intersections of the two tangent lines at $P$ and $Q$: in $U_0$ -- the $yz$-plane -- the tangent lines meet in $(1/2,1)$ while in $U_2$ -- the $xy$-plane -- the tangent lines meet in $(1,1/2)$. These points both homogenize back to the same point $[2,1,2]$, which is the unique (!) point in the projective plane satisfying $3x + 4y = 5z$ and $x = z$.
Remember that $Q$ went missing in the chart $U_1$? Well, its tangent line did not go missing: the projective line $x = z$ in the projective plane meets the chart $U_1$ in the ordinary line $x = z$ of the $xz$-plane, which is an asymptote to the piece of $C$ we can see in $U_1$. This is really amazing: asymptotes to (algebraic) curves in the usual plane are "really" the tangent lines to missing points on the complete version of that curve in the projective plane. To see this from another point of view, move around $C$ clockwise in the chart $U_2$ (where it's a circle) and figure out the corresponding motion along the piece of $C$ in the chart $U_0$ (where it's a hyperbola): as you pass through the point $Q = (1,0)$ in $U_2$, what happens in the chart $U_0$ is that you jump off one branch of the hyperbola onto the other branch by skipping through an asymptote, sort of. (There is a second point on $C$ in $U_2$ that you don't see in $U_0$ -- the point $R = [-1,0,1]$ is $(-1,0)$ in $U_2$ -- and paying attention to that point may help here.)
The conic sections -- parabolas, hyperbolas, and ellipses -- which look quite different in ${mathbf R}^2$, simplify in the projective plane because they all look like the same kind of curve (once you close them up): $y = x^2$ becomes $yz = x^2$, $xy = 1$ becomes $xy = z^2$, and $x^2 + y^2 = 1$ becomes $x^2 + y^2 = z^2$, which is the same as $x^2 = (z-y)(x+y) = z'y'$, where $z' = z-y$ and $y' = z+y$. I like to think about this as a fancy analogue of the Greek [Edit: Babylonian] use of many forms of the quadratic formula because they didn't have the right algebraic technique to realize there is one quadratic formula. Using the projective plane we see there is really one conic section.
You might want to show by examples the nicer intersection properties of lines in the projective plane: any two lines in the projective plane meet in exactly one point. This is just a glimpse of the fact that curves in the projective plane have nicer intersection properties than in the ordinary plane, but to get the correct theorem in that direction for curves other than lines, you need to (a) work over the complex numbers and (b) introduce an appropriate concept of intersection multiplicity for intersection points of curves, somewhat like the idea of tangent circles intersecting in a point of multiplicity two which I mentioned earlier. The relevant theorem here is Bezout's theorem, but to state it correctly is complicated precisely because it is technical to give a good definition of what the intersection multiplicity is for two curves meeting at a common point.
For the student who wants to be a number theorist, compare reduction mod $p$ in the usual plane and the projective plane. In the study of Diophantine equations (e.g., to show $y^2 = x^3 - 5$ has no integral solutions), it is very useful to reduce mod $p$, and there is a natural way to reduce a point in ${mathbf Z}^2$ modulo $p$ However, there's no reasonable way to reduce all points in ${mathbf Q}^2$ modulo $p$: when the rational numbers have denominator divisible by $p$, you can't make sense of them mod $p$: we can reduce $(-7/4,51/8)$ mod 5, for example, but not mod 2. In the projective plane, however, we can reduce rational points mod $p$ by the idea of choosing a set of primitive integral coordinates, where the homogeneous coordinates are relatively prime. For example, $[-7/4,51/8,1] = [-14,51,4]$ in ${mathbf P}^2({mathbf Q})$, and this can be reduced mod $p$ for any $p$ at all. For example, in ${mathbf P}^2({mathbf F}_2)$ it becomes $[0,1,0]$.
(There is another primitive set of homogeneous coordinates for the point, namely $[14,-51,-4]$, but that reduces mod $p$ to the same thing as before, so this reduction mod $p$ process is well-defined.) This suggests that the projective plane has better mapping properties than the usual plane, in some sense.
Any pure mathematician will from time to time discuss, or think about, the question of why we care about proofs, or to put the question in a more precise form, why we seem to be so much happier with statements that have proofs than we are with statements that lack proofs but for which the evidence is so overwhelming that it is not reasonable to doubt them.
That is not the question I am asking here, though it is definitely relevant. What I am looking for is good examples where the difference between being pretty well certain that a result is true and actually having a proof turned out to be very important, and why. I am looking for reasons that go beyond replacing 99% certainty with 100% certainty. The reason I'm asking the question is that it occurred to me that I don't have a good stock of examples myself.
The best outcome I can think of for this question, though whether it will actually happen is another matter, is that in a few months' time if somebody suggests that proofs aren't all that important one can refer them to this page for lots of convincing examples that show that they are.
Added after 13 answers: Interestingly, the focus so far has been almost entirely on the "You can't be sure if you don't have a proof" justification of proofs. But what if a physicist were to say, "OK I can't be 100% sure, and, yes, we sometimes get it wrong. But by and large our arguments get the right answer and that's good enough for me." To counter that, we would want to use one of the other reasons, such as the "Having a proof gives more insight into the problem" justification. It would be great to see some good examples of that. (There are one or two below, but it would be good to see more.)
Further addition: It occurs to me that my question as phrased is open to misinterpretation, so I would like to have another go at asking it. I think almost all people here would agree that proofs are important: they provide a level of certainty that we value, they often (but not always) tell us not just that a theorem is true but why it is true, they often lead us towards generalizations and related results that we would not have otherwise discovered, and so on and so forth. Now imagine a situation in which somebody says, "I can't understand why you pure mathematicians are so hung up on rigour. Surely if a statement is obviously true, that's good enough." One way of countering such an argument would be to give justifications such as the ones that I've just briefly sketched. But those are a bit abstract and will not be convincing if you can't back them up with some examples. So I'm looking for some good examples.
What I hadn't spotted was that an example of a statement that was widely believed to be true but turned out to be false is, indirectly, an example of the importance of proof, and so a legitimate answer to the question as I phrased it. But I was, and am, more interested in good examples of cases where a proof of a statement that was widely believed to be true and was true gave us much more than just a certificate of truth. There are a few below. The more the merrier.
Consider the evaluation map $S^1 times [S^1,K(Z,n)] to K(Z,n)$. Since $S^1$ is a $K(Z,1)$, and $[S^1, K(Z,n)]=Omega^1(K(Z,n))$ is a $K(Z,n-1)$, up to homotopy we get a map $K(Z,1)times K(Z,n-1) to K(Z,n)$. I'm not actually sure if this induces the product on cohomology. If it does, there is a natural generalization:
Consider the space of pointed maps $[K(A,n),K(A,m+n)]$. Then $pi_k([K(A,n),K(A,m+n)])=0$ for $k>m$, and $=A$ for $k=m$. To see this, note
that (all maps are pointed)
$$[S^k, [K(A,n),K(A,m+n)]] = [K(A,n), Omega^k(K(A,m+n))]$$
$$ =[K(A,n),K(A,m+n-k)]= {check H}^{m+n-k}(K(A,n),A) = 0 $$
if $k>m$, and $=A$ if $k=m$ (by Hurewicz).
Thus, we have a map $i: K(A,m) to [K(A,n), K(A,m+n)]$ by obstruction theory sending $pi_m(K(A,m))to pi_m([K(A,n),K(A,m+n)])$ isomorphically, which of course is equivalent to a map (by evaluation)
$$ K(A,m) times K(A,n) to K(A,m+n).$$
Maybe someone could explain to me if this gives the correct cohomology operation?
In the book "Toposes, Triples and Theories", Barr and Wells study the question when a particular endofunctor admits a free monad. This is the case if the underlying category is complete and cocomplete and if the endofunctor preserves filtered colimits (we say that such a functor is finitary).
The question remains whether the resulting adjunction between monads on $C$ and endofunctors on $C$ is monadic. If $C$ is locally finitely presentable (lfp), this is true: Steve Lack showed in
"On the monadicity of finitary monads", Journal of Pure and Applied Algebra, Volume 140, Number 1, 21 July 1999 , pp. 65-73(9) (available here)
that the forgetful functor $mathrm{Mnd}_f(C) rightarrow mathrm{End}_f(C)$ from finitary monads on a lfp category $C$ to finitary endofunctors is monadic. Note that both these categories are again lfp categories, and we don't need universes to make sense of them, essentially because a finitary endofunctor is determined by what it does on a set of objects.
The result remains true if we consider categories and functors enriched in a complete and cocomplete symmetric monoidal closed category $V$ which is lfp as a closed category.
This is a follow-up to the question on the Ramanujan constant.
Can one find an asymptotic family of shockingly near-integers obtained by raising e to some simple algebraic number, in the spirit of Ramanujan's
$|exp(pi sqrt{163}) - text{ some integer }| le 10^{-12}$? Single numerical coincidences don't impress me (and for good reason too ...).
Is there a sequence of such near-integers, whose nearness to integers is closer than what you know exists by pigeonholing exp(alpha) for all alpha of bounded "height" or "complexity"?
This is a (far too long) comment on Buzzard's comment about Hida's remark.
I think I can guess what Hida was saying. He was probably talking about non-vanishing of L-functions of Hecke eigenforms of level one and weight $k equiv 0$ (mod 4). This is a long-standing (folklore? ) conjecture in its own right, well-known among analytic number-theorists.
Here is how such a thing can be proven using Maeda's conjecture. There is a result of Shimura that says that the Galois group acts nicely on the central values (in fact any critical value) L-function of eigenforms. In particular, if one of them is zero then all the Galois twists are also zero and hence their sum is also zero. Now, even though it may be difficult to show that an L-function doesn't vanish at the centre, it is often easy to show that the sum of the central values of L-functions in a family is non-zero (see, for example, the work of Rohrlich and Rodriguez-Villegas on non-vanishing of L-functions of Hecke characters).
In the case in question, Maeda's conjecture will imply that if one central L-value is zero then the sum of all the central L-values over the whole basis must be zero and I think a contradiction will ensue after one uses the approximate functional equation to write the central value in terms of the Fourier coefficients and then using the Petersson formula ( I need to check this up).
Note 1: There is an article by Conrey and Farmer titled "Hecke operators and nonvanishing of L-functions" (Ahlgreen et al. (eds.), Topics in Number Theory, 1999) where they prove the above mentioned result along a different line.
Note 2: I think the following is easier. One can think of $frightarrow L(f,k/2)$ as a linear functional on the space of cusp forms $S_k(Gamma(1))$ and indeed it is possible to explicitly write a function $G$ such that
$L(f,k/2)=langle f,G rangle$
for all Hecke eigenform $f$ in $S_k(Gamma(1))$. Now Maeda + Shimura's result will imply that $G$ is orthogonal to the whole space and therefore zero. So it is just a matter of checking that $G$ is not identically zero, which shouldn't be too hard.
I'm not a number theorist, so apologies if this is trivial or obvious.
From what I understand of the results of Green-Tao-Ziegler on additive combinatorics in the primes, the main new technical tool is the "dense model theorem," which -- informally speaking -- is as follows:
If a set of integers $S subset N$ is a dense subset of another "pseudorandom" set of integers, then there's another set of integers $S' subset N$ such that $S'$ has positive density in the integers and $S, S'$ are "indistinguishable" by a certain class of test functions.
They then use some work of Goldston and Yildirim to show that the primes satisfy the given hypothesis, and note that if the primes failed to contain long arithmetic/polynomial progressions and $S'$ did, they'd be distinguishable by the class of functions. Applying Szemeredi's theorem, the proof is complete.
Obviously I'm skimming over a great deal of technical detail, but I'm led to believe that this is a reasonably accurate high-level view of the basic approach.
My question(s), then: Can one use a similar approach to obtain "Goldbach-type" results, stating that every sufficiently large integer is the sum of at most k primes? Is this already implicit in the Goldston-Yildirim "black box?" If we can't get Goldbach-type theorems by using dense models, what's the central obstacle to doing so?
I think that the results are "folklore" in the sense that if you have no concern at all for numerical stability, using small translates of the Gaussian to obtain derivatives is the first thing that you'd think to do. Likewise with the trig functions with frequency $omega ll 1$, you can just directly differentiate many times with respect to $omega$ and then evaluate at $0$. You get polynomials, which are dense by the Weierstrass approximation theorem. You say several times in the paper that the results are surprising, but in my opinion the qualitative results are not all that surprising.
These results are at least similar to popular results that certain sets of wavelets are bases, and certain other sets span but are overcomplete. I don't have a reference, I just remember hearing about a set of wavelets on $mathbb{R}$ that comes from a general lattice in $mathbb{C}$. Whether the wavelets span or are overcomplete depends on the determinant of the lattice. This is not the same result, but it is similar.
Your paper also analyzes the numerical stability of your approximations. That seems like the more original contribution to me. Maybe the referees would be happier if you shifted the emphasis of the paper to the numerical stability issue.
For the record, I do NOT find this question off-topic.
Math is moving away from the pen and paper (or chalk and chalkboard) days. Computers are everywhere, and, unlike days of old when computers were notoriously BAD at math, modern languages, such as Python, have built in representations of fractions, complex numbers, positive and negative infinity, etc.
Matrices and vectors, for example, are used in everything from the latest shoot-em-up to bank transactions, to simulation software and drawing programs.
It is about time that we had a real solution for entering and working with real equations on a computer, rather than having to slog it out with --> area = (PI * (r^2)).
Would I buy a keyboard that had all the standard mathematical symbols built into it?
Absolutely.
And while, yes, I am aware that I can remap a standard keyboard, and while yes, I touch type... some of these symbols would be used often enough to want on the keyboard, yet rarely enough to have no idea where you put them. (Where the %&@# did I put phi? CTRL-P? Bah, PI... Alt P? Rho? Whats that doing there?)... a quick look down on the proper keyboard could solve this easily.
And yes... I came to this board after doing a Google search for just such a beast... if this question had not been asked here, I never would have wandered to your site.
Me
Over real-closed fields such as $langle mathbb{R}, +, *, -, <, 0, 1 rangle$, there is an interesting simple answer: every polynomial inequality is equivalent to a projected equation. E.g.,
Given $p_1, p_2 in mathbb{Q}[vec{x}]$ we have $left( p_1 > p_2 iff exists z text{ s.t. } z^2(p_1 - p_2) - 1 = 0 right),$ and $left( p_1 geq p_2 iff exists z text{ s.t. } p_1 - p_2 - z^2 = 0 right).$
Geometrically, this is the simple observation that every semialgebraic set defined as the set of $n-$dimensional real vectors satisfying an inequality is the projection of an $n+1$-dimensional real-algebraic variety defined by a single equation. Semialgebraic sets defined by boolean combinations of equations and inequalities can be similarly encoded as the set of satisfying real vectors of (an) equation(s) by using the Rabinowitsch encoding $(p_1 = 0 vee p_2 = 0 iff p_1p_2 = 0)$ and $(p_1 = 0 wedge p_2 = 0 iff p_1^2 + p_2^2 = 0).$
Combining the above two observations, one obtains the fact that every semi-algebraic set $S subseteq mathbb{R}^n$ is the projection of a real algebraic variety $V subseteq mathbb{R}^{n+k}$, where $k$ is the number of inequality symbols appearing in the defining Tarski formula for $S$. In fact, due to a construction of Motzkin [``The Real Solution Set of a System of Algebraic Inequalities is the Projection of a Hypersurface in One More Dimension,'' Inequalities II, O. Shisha, ed., 251-254, Academic Press (1970)], it is known that every such $S$ is in fact the projection of a real-algebraic variety in $mathbb{R}^{n+1}$.
The answer is: no there isn't such a thing. Here is a rough argument (a full proof would deserve a little more care).
Using the main result of
S. Schwede, The stable homotopy category is rigid, Annals of Mathematics 166 (2007), 837-863
your question is equivalent to the following: does there exist a model category $C$, which is additive, and such that $C$ is Quillen equivalent to the usual model category of spectra?
In particular, we might ask: does there exist an additive category $C$, endowed with a Quillen stable model category structure, such that the corresponding stable $(infty,1)$-category is equivalent to the stable $(infty,1)$-category of spectra?
Replacing $C$ by its full subcategory of cofibrant objects, your question might be reformulated as: does there exist a category of cofibrant objects $C$ (in the sense of
Ken Brown), with small sums (and such that weak equivalences are closed under small sums), and such that the corresponding $(infty,1)$-category (obtained by inverting weak equivalence of $C$ in the sense of $(infty,1)$-categories) is equivalent to the stable $(infty,1)$-category of spectra? If the answer is no, then there will be no additive model category $C$ such that $Ho(C)$ is (equivalent to) the category of spectra (as a triangulated category).
So, assume there is an additive category of cofibrant objects $C$, with small sums, such that $Ho(C)$ is (equivalent to) the category $S$ of spectra (as a triangulated category). Let $C_f$ be the full subcategory of $C$ spanned by the objects which correspond to finite spectra in $S$. Then $Ho(C_f)simeq S_f$, where, by abuse of notations, $Ho(C_f)$ is the $(infty,1)$-category obtained from $C_f$ by inverting weak equivalences, while $S_f$ stands for the stable $(infty,1)$-category of finite spectra (essentially the Spanier-Whitehead category of finite CW-complexes). Given any (essentially) small additive category $A$ denote by $K(A)$ the "derived $(infty,1)$-category of $A$" (that is the $(infty,1)$-category obtained from the category of bounded complexes of $A$, by inverting the chain homotopy equivalences). Then, the canonical functor $Ato K(A)$ (which sends an object $X$ to itself, seen as a complex concentrated in degree $0$), has the following universal property: given a stable $(infty,1)$-category $T$, any functor $Ato T$ which sends split short exact sequences of $A$ to distinguished triangles (aka homotopy cofiber sequences) in $T$ extends uniquely into a finite colimit preserving functor $K(A)to T$. In particular, the functor $C_fto Ho(C_f)simeq S_f$ extends uniquely to a finite colimit preserving functor $F:K(C_f)to S_f$. Let $Ker(F)$ be the full $(infty,1)$-subcategory of $K(C_f)$ spanned by objects which are sent to zero in $S_f$. Then the induced functor
$$K(C_f)/Ker(F)to S_f$$
is an equivalence of (stable) $(infty,1)$-categories (to see this, you may use the universal property of $S_f$: given a stable $(infty,1)$-category $T$, a finite colimit preserving functor $S_fto T$ is the same as an object of $T$; see Corollary 10.16 in DAG I). This implies that, for any object $X$ of $S_f$, if $X/n$ denotes the cone of the map $n:Xto X$ (multiplication by an integer $n$), then $n.X/nsimeq 0$ (see Proposition 1 in Schwede's paper Algebraic versus topological triangulated categories). But such a property is known to fail whenever $X$ is a finite spectrum for $n=2$ (see Proposition 2 in loc. cit.). Hence there isn't such a $C$...
Background/motivation
It is a classical fact that we have a natural isomorphism $Sym^n (V^*) cong Sym^n (V) ^ast$ for vector spaces $V$ over a field $k$ of characteristic 0. One way to see this is the following.
On the one hand elements of $Sym^n (V^*)$ are symmetric powers of degree n of linear forms on $V$, so they can be identified with homogeneous polynomials of degree n on $V$. On the other hand elements of $Sym^n (V) ^ast$ are linear functionals on $Sym^n V$; by the universal property of $Sym^n V$ these correspond to n-multilinear symmetric forms on $V$. The isomorphism is then as follows.
An n-multilinear symmetric form $phi$ corresponds to the homogeneous polynomial $p(v) = phi(v, dots, v)$. In the other direction to a polynomial $p(v)$ we attach the multinear form obtained by polarization $phi(v_1, dots, v_n) = frac{1}{n!}sum_{I subset [n]} (-1)^{n - sharp I} p(sum_{i in I} v_i)$. Here $[n]$ is the set $lbrace 1, dots, n rbrace$.
Problem
Of course this will not work for $n$ greater than the characteristic of $k$ if the latter is positive.
One can expect that an isomorphism $Sym^n (V^*) cong Sym^n (V) ^ast$ holds also in positive characteristic, and that this should be trivially true by using the universal properties of the symmetric powers. The problem is that if I try to define a natural map between the two spaces using the universal properties I have at some point to divide by $n!$ anyway.
Still there may be some natural isomorphism that I cannot see. Or maybe there is not a natural isomorphism, but I don't know how to prove this.
Is there a natural isomoprhism $Sym^n (V^*) cong Sym^n (V)^ast$ in positive characteristic?
The classic handshake puzzle goes something like this:
Its common variations are:
Let's say N is the number of other people (patients/women...etc) and K is the number of gloves (or condoms). The above case of N=3 and K=2 is not hard (and its solution readily available on the net).
QUESTION 1: In general, what can we say about the feasible N's and K's? It seems like (2K >= N+1) is a necessary condition (K gloves has 2K sides and there are a total of N+1 people involved). Is this also sufficient?
While researching on Google, I came across a posting that claimed the generalization of this similar puzzle is an open problem:
QUESTION 2: I assume the general form of the question would study the feasibility of N couples and K condoms. What is known about the general problem? Is it still open?
(Qiaochu Yuan:) Based on the downvotes, which I would guess are directed at the way in which the problem is stated rather than its content, here is a "cleaned up" version appropriate for mathematicians:
You have a collection of $K$ tokens which have two sides, each of which can be marked. There are two families of marks, $N$ of which are of the first family and $M$ of which are of the second family. For each pair $(i, j)$ of a mark of the first family and the second family, attempt the following:
For which values of $K, N, M$ is this possible?
What is the relationship between $G_infty$ (homotopy Gerstenhaber) and $B_infty$ algebras?
In Getzler & Jones "Operads, homotopy algebra, and iterated integrals for double loop spaces" (a paper I don't well understand) a $B_infty$ algebra is defined to be a graded vector space $V$ together with a dg-bialgebra structure on $BV = oplus_{i geq 0} (V[1])^{otimes i}$, that is a square-zero, degree one coderivation $delta$ of the canonical coalgebra structure (stopping here, we have defined an $A_infty$ algebra) and an associative multiplication $m:BV otimes BV to BV$ that is a morphism of coalgebras and such that $delta$ is a derivation of $m$.
A $G_infty$ algebra is more complicated. The $G_infty$ operad is a dg-operad whose underlying graded operad is free and such that its cohomology is the operad controlling Gerstenhaber algebras. I believe that the operad of chains on the little 2-discs operad is a model for the $G_infty$ operad. Yes?
It is now known (the famous Deligne conjecture) that the Hochschild cochain complex of an associative algebra carries the structure of a $G_infty$ algebra. It also carries the structure of a $B_infty$ algebra. Some articles discuss the $G_infty$ structure while others discuss the $B_infty$ structure. So I wonder: How are these structures related in this case? In general?
In 1965 or so, Glen Bredon defined ordinary equivariant cohomology, ordinary meaning
that it satisfies the dimension axiom: For each coefficient system $M$ (contravariant
functor from the orbit category of G to the category of Abelian groups), there is a
unique cohomology theory $H^*_G(-;M)$ such that, when restricted to the orbit category,
it spits out the functor $M$. Just as in the nonequivariant world, it can be defined
using either singular or cellular cochains, the latter defined using $G$-CW complexes.
This works as stated for any topological group $G$.
For an abelian group $A$, Borel cohomology with coefficients in $A$, $H^*(EGtimes_G X;A)$
is the extremely special case in which one takes $M$ to be the constant coefficient system
$underline{A}$ at the group $A$ and replaces $X$ by $EGtimes X$. That is,
$$ H^*_G(EGtimes X; underline{A}) = H^*(EGtimes_G X;A) $$
This is not a complete answer by any means, but is intended to get the ball rolling.
First of all, it need not be the case that $H^1(X_{fl},mu_n) = 0.$ Rather, what follows
from the vanishing of $H^1(X_{fl},{mathbb G}_m)$ is that
$H^1(X_{fl},mu_n) = R^{times}/(R^{times})^n.$
(This is not always trivial; imagine
e.g. that $R$ is a non-algebraically closed field. You might have been thinking of the case
when $X$ is projective and smooth over an algebraically closed field, when the $H^0$-part of the exact sequence is itself exact, and so can be omitted from consideration. That is not the case here.)
Secondly, this doesn't hurt your arguments, because the same consideration of $H^0$-terms has to be made for the cohomology of $U$. Since $X$ is three dimensional and a complete intersection, restriction induces an isomorphism
$H^0(X,mathcal O) cong H^0(U,mathcal O)$, and so also an isomorphism
$H^0(X,mathcal O^{times})cong H^0(U,mathcal O^{times}),$ and so also isomorphisms
$H^0(X_{fl},mu_n) cong H^0(U_{fl},mu_n)$ and
$H^0(X_{fl},{mathbb G}_m)cong H^0(U_{fl},{mathbb G}_m).$ Thus in fact one finds that
the $n$-torsion in Pic$(U)$ is equal to the cokernel of the injection
$H^1(X_{fl},mu_n) hookrightarrow H^1(U_{fl},mu_n).$ And as your analysis shows, this cokernel embeds into $H^2_{{m}}(X_{fl},mu_n)$, with the cokernel of that embedding itself embedding into $H^2(X_{fl},mu_n).$
So what can be said about this latter cohomology group?
Since $H^1(X_{fl},{mathbb G}_m)$ vanishes, as you observed, one finds that $H^2(X_{fl}, mu_n)$ coincides with the $n$-torsion in the cohomological Brauer group $H^2(X_{fl},{mathbb G}_m).$ (Here I am using the fact that since ${mathbb G}_m$ is smooth, flat and etale cohomology coincide, so $H^2(X_{fl},{mathbb G}_m) = H^2(X_{et}, {mathbb G}_m).$) So it seems that one wants to kill off the torsion in this Brauer group.
I don't see why this need be true, but what one actually needs is that $H^2(X_{fl},mu_n)
rightarrow H^2(U_{fl},mu_n)$ is injective. Since $H^2(X_{fl},mu_n)$ embeds into
$H^2(X_{fl},{mathbb G}_m)$, it would be enough to show that the restriction
$H^2(X_{fl},{mathbb G}_m) to H^2(U_{fl},{mathbb G}_m)$ induces an injection on torsion. Might this be some kind of purity result on Brauer groups of the kind Gabber discusses in his abstract? It would be related to a vanishing of (torsion in) $H^2_{{m}}(X_{fl}, {mathbb G}_m)$. Somewhere (maybe here?) one presumably has to make use of the dimension and lci hypotheses.
P.S. You may well just want to email Gabber to ask him about this. If you do, and you get an answer, please share it!
EDIT: This is an excerpt from the email referred to in Hai Long's comment below:
To learn about these kinds of arguments, my advice is to
do just what you are doing. One works with the exact sequence
linking $mu_n$ and ${mathbb G}_m$, as you did.
Number theorists (at least of a certain stripe) have some advantages
with this, because the case $X =$ Spec $K$ ($K$ a field) comes up a lot
under the name of Kummer theory, and also Mazur in one of his
famous papers uses a lot of flat cohomology. But in the end, the
formalism is just the one you used in your question.
Then, typically, one has to inject something additional that
is less formal. My suggestion would be to look at de Jong's proof
of Gabber's result showing $Br'(X) = Br(X)$ discussed in his abstract.
(There is a write-up on de Jong's web-page.)
Reading the proof of a result like this might give some insight
into how to work with Brauer groups in a less formal way.
I call a set X of positive integers strongly lcm-closed if a,b ∈ X if and only if lcm(a,b) ∈ X. In the finite case X is the set of divisors of lcmx ∈ Xx. But in the infinite case it is more interesting, for example, ${a geq 1: a notequiv 0 pmod p}$ and ${p^a:a geq 0}$ for any prime p, are strongly lcm-closed sets.
Which sets are strongly lcm-closed sets?
This question arose in my Ph.D. thesis (p.107) where strongly lcm-closed sets describe where autotopisms of Latin squares give rise to subsquares.
As a side question:
Is there a common name for strongly lcm-closed sets?
If there is a group G and an injective endomorphism $sigma$ of G, there is a group K containing G such that $sigma$ extends to an inner automorphism of K.
Even more generally, if two subgroups of a group are isomorphic, the group can be embedded in a bigger group where those two subgroups are conjugate via a conjugation map that extends the isomorphism. The construction used is called a HNN-extension, and it basically adjoins elements to the group that act by conjugation as the isomorphism. (This generalizes even further: given any number of isomorphisms between pairs of subgroups of a group, there is a group containing the group such that all these isomorphisms become conjugations in that bigger group).
Thus, to find examples that answer your question, it is enough to find examples of a group that is isomorphic to a proper subgroup. For instance, if we consider the example of the group of integers isomorphic to the subgroup of even integers, the corresponding HNN-extension is the Baumslag-Solitar group mentioned above.
Incidentally, the statement above (that any two isomorphic subgroups of a group become conjugate in some bigger group) is also true when we restrict to finite groups, though this does not give any examples for the question you are interested in because no finite subgroup can be isomorphic to a proper subgroup.
See this and http://groupprops.subwiki.org/wiki/Isomorphic_iff_potentially_conjugate_in"finite">this for more notes on these.
I'd start with Lee's Introduction to Smooth Manifolds.
It covers the basics in a modern, clear and rigorous manner.
Topics covered include the basics of smooth manifolds, smooth
vector bundles, submersions, immersions, embeddings, Whitney's
embedding theorem, differential forms, de Rham cohomology, Lie
derivatives, integration on manifolds, Lie groups, and Lie algebras.
After finishing with Lee, I'd move on to Hirsch's Differential
Topology. This is more advanced then Lee and leans more
towards topology. Also, the proofs are much more brief then
those of Lee and Hirsch contains many more typos than Lee.
The topics covered include the basics of smooth manifolds,
function spaces (odd but welcome for books of this class),
transversality, vector bundles, tubular neighborhoods, collars,
map degree, intersection numbers, Morse theory, cobordisms,
isotopies, and classification of two dimensional surfaces.
These two should get you through the basics. However, if that
is not enough, I'd move on to Kosinski's Differential Manifolds
which covers the basics of smooth manifolds, submersions, immersions,
embeddings, normal bundles, tubular neighborhoods, transversality,
foliations, handle presentation theorem, h-cobordism theorem,
framed manifolds, and surgery on manifolds.
What do You mean by word analogy here? From wikipedia we have ( among others):
The word analogy can also refer to the
relation between the source and the
target themselves, which is often,
though not necessarily, a similarity
So You see similarity in differential equation versus difference equation, but this is mostly matter of aesthetic. In practice if You need discrete equation for continues one, You have to put usually a large amount of work in order to make this analogy working. Of course in principle there is relation among differential and difference equation. But what is important here is not what is similar, but what is a gap between them.
When You say, that discrete case may approximate continues one, in fact You take many assumptions, for example about criteria which constitutes what is that mean approximation.
Say what is analogy of holomorphic function? Is discrete complex function on lattice of Gauss integers, good approximation for some complex analytical function? In what meaning? What are criteria? Are all properties of holomorphic function shared by "discrete analogy" and vice versa?
For example, it is not true that whole
theory of differential equations may
be deduced from difference
equations. We have
several equations when we cannot
find correct approximations, for
example Navier-Stokes equation has
no discrete model, at least
till now. You may say: but chaos is
analogous to turbulence. Why?
Because is similar? Why do You may
say that? Is that someone think two
things are similar enough to say
that they are?
Then analogy is so broad in meaning word, that I may say, I can see analogy between every things You may point. It may be very useful as inspiration, sometimes it lead us to great discoveries. For every thing You say is analogous to some continues case, we may have differences between them which allows us to distinguish this cases. They nearly almost are non equivalent even in approximate meaning. They are never the same. It is a matter of criteria, if You may say two things are in analogy.
This question arose while I was trying to work out examples for the second question of this thread: Reconstruction Conjecture: Group theoretic formulation?
In the beginning, I considered some computable properties of groups and wondered whether two groups of the same order having equal value for that computable property would necessarily be isomorphic. For instance, take centers of groups and it is not difficult to find many specific examples where two groups have the same order and isomorphic centers but then the two groups are not necessarily isomorphic. Considering lattice of groups, Scott Carnahan has already given a counterexample there.
Are there any two finite groups of the same order that have the same number of subgroups?
Let A be a C*-algebra, let H be a Hilbert space, and let $T:Arightarrow B(H)$ be a completely bounded (cb) map (that is, the dilations to maps $M_n(A)rightarrow M_n(B(H))$ are uniformly bounded). We can write T has $T_1-T_2+iT_3-iT_4$ where each $T_i$ is completely positive. If $T$ is hermitian in that $T(x^*)^* = T(x)$ for all $xin A$, then $T=T_1-T_2$. We can order the hermitian cb maps $Arightarrow B(H)$ by saying that $Tgeq S$ if $T-S$ is completely positive.
I'm interested in criteria by which we can recognise that $Tgeq S$. Even special cases would be good (for example, I'm happy to assume that $T$ is completely positive).
An old paper of Arveson ("Subalgebras of C*-algebras") shows that if T and S are both completely positive, and T has the minimal Stinespring dilation $T(x) = V^*pi(x)V$, then $Tgeq Sgeq 0$ if and only if $S(x) = V^*pi(x)AV$ where $0leq Aleq1$ is a positive operator in the commutant of $pi(A)$. This is nice, but suppose all I know is that $T(x)=V^*pi(x)V$ and $S(x) = U^*pi(x)U$ (notice that the representation $pi$ is the same). Can I "see" if $Tgeq S$ by looking at $U$ and $V$? What if S is only cb, so $S(x)=Api(x)B$? Maybe that's too much to hope for, but anything vaguely in this direction would be interesting.
Your question seems to be to identify the Clifford algebras $Cl(0,1)$ and $Cl(0,2)$ in the usual mathematics notation of, say, Spin Geometry by Lawson and Michelsohn. It is very easy to show that
$$Cl(0,1) cong mathbb{R} oplus mathbb{R}$$
and
$$Cl(0,2) cong mathbb{R}(2),$$
where $mathbb{R}(2)$ is the algebra of $2times 2$ real matrices.
There are two inequivalent one-dimensional irreducible representations of $Cl(0,1)$, where the gamma matrix (here a real number) is $Gamma^1=pm 1$.
There is a unique two-dimensional irreducible representation of $Cl(0,2)$, where the two gamma matrices can be given by
$$Gamma^1 = begin{pmatrix} 1 & 0 cr 0 & -1 end{pmatrix}$$
and
$$Gamma^2 = begin{pmatrix} 0 & 1 cr 1 & 0 end{pmatrix}.$$
I am interested in theorems with unexpected conclusions. I don't mean
an unintuitive result (like the existence of a space-filling curve), but
rather a result whose conclusion seems disconnected from the
hypotheses. My favorite is the following. Let $f(n)$ be the number of
ways to write the nonnegative integer $n$ as a sum of powers of 2, if
no power of 2 can be used more than twice. For instance, $f(6)=3$
since we can write 6 as $4+2=4+1+1=2+2+1+1$. We have
$(f(0),f(1),dots) = $ $(1,1,2,1,3,2,3,1,4,3,5,2,5,3,4,dots)$. The
conclusion is that the numbers $f(n)/f(n+1)$ run through all the
reduced positive rational numbers exactly once each. See A002487 in
the On-Line Encyclopedia of Integer Sequences for more
information. What are other nice examples of "unexpected conclusions"?
These functions like the Cantor function and the continuous-but-not-differentiable function are all well and good, but contrived - the only place you ever see them is as counterexamples. Here is a function that has many uses in Number Theory, and still manages to have a strange property or two. Let
$x=h/k$ with $h$ and $k$ integers, $k>0$. Define $$s(x)=sum_{c=1}^{k-1}((c/k))((ch/k))$$ where $((y))=0$ if $y$ is an integer, $((y))=lbrace yrbrace-1/2$ otherwise. It is easily proved that the sum depends only on the ratio of $h$ and $k$, not on their individual values, so $s$ is a well-defined function from the rationals to the rationals. It is known as the Dedekind sum; it came up originally in Dedekind's study of the transformation formula of the Dedekind $eta$-function.
Now for the strange properties.
Hickerson, Continued fractions and density results for Dedekind sums, J Reine Angew Math 290 (1977) 113-116, MR 55 #12611, proved that the graph of $s$ is dense in the plane.
With Nick Phillips, I proved (Lines full of Dedekind sums, Bull London Math Soc 36 (2004) 547-552, MR 2005m:11075) that, with the exception of the line $y=x/12$, every line through the origin with rational slope passes through infinitely many points on the graph of $s$. We suspect that the points are dense on those lines, though we could only prove it for the line $y=x$.
One profitable thing to look at might be geometric Satake:
Roughly, one can categorify the symmetric polynomials acting on all polynomials as perverse sheaves on $GL(n,mathbb{C}[[t]])setminus GL(n,mathbb{C}((t)))/GL(n,mathbb{C}[[t]])$ acting on perverse sheaves on $GL(n,mathbb{C}[[t]])setminus GL(n,mathbb{C}((t)))/I$ where $I$ is the Iwahori (matrices in $GL(n,mathbb{C}[[t]])$ which are upper-triangular mod $t$).
The maps to polynomials are take a sheaf and send it to the sum over sequences $mathbf{a}$ of $n$ integers of the Euler characteristic of its stalk at the diagonal matrix $t^{mathbf{a}}$ times the monomial $x^{mathbf{a}}$.
One nice thing that happens in this picture is the filtration of polynomials by the invariants of Young subgroups appears as a filtration of categories. Thus, one can take quotient categories and get a nice basis, with lots of good positivity, for the isotypic components.
For the multiplicity space, one might be able to do some trick using cells. It's not immediately clear to me how.
If what you need is a simple practical method to do the interpolation,
then just multiply the "time domain" samples by the linear-phase
signal ${expleft(alphacdot i 2pi k /Nright)}_{k=0}^{N-1}$, where
$alphain(0,1)$ is the sub-sample shift in the "frequency domain."
(I'm not sure if an additional constant of absolute value 1 is
required here). As
already noted in previous answers, the motivation for this is the
assumption that your samples vector ${x_k}$ comes from sampling some
continuous-time signal ${X(t)}_{tin mathbb{R}}$, as in
$x_k=X(kT_s)$ for $k=0,ldots, N-1$, where $T_sin mathbb{R}_{++}$ is
the sampling period. If the continuous-time signal has a compactly
supported Fourier transform, than by Shannon's sampling theorem $X$ is determined by the infinite sequence ${X(kT_s)}_{kin
mathbb{Z}}$ for small enough $T_s$. Since we only have $N$ entries
of the infinite sequence, we loose some information on $X$. But if
$X$ decays rapidly (rather than having a compact support), then at
least intuitively we don't loose much (I know that this is a dangerous
and imprecise statement :) ).
For example, try the following Matlab lines:
ttt=-32:31;
x_time=exp(-ttt.^2/10);
figure;plot(x_time);
x_freq = fft(x_time);
figure;plot(fftshift(abs(x_freq)));
figure
for alpha=-4:0.5:4
x_time_2 = x_time .* exp(2 * pi * i * (0:63)/64 * alpha);
x_freq=fft(x_time_2);
plot(fftshift(abs(x_freq)));
axis([26,37,0,10]);
grid on;
pause(1);
end
There is a category of "sequential spaces" in which objects are spaces defined by their convergent sequences and morphisms are those maps which send convergent sequences to convergent sequences.
As stated above, all metric spaces are sequential spaces, but so are all manifolds, all finite topological spaces, and all CW-complexes.
To build this category, one actually just needs to look at the category of right $M$-sets for a certain monoid $M$. Consider first the "convergent sequence space" $S:=${$frac{1}{n}|nin{mathbb N}cup${$infty$}}$subset {mathbb R}$. In other words $S$ is a countable set of points converging to 0, and including $0$. Let $M$ be the monoid of continuous maps $Sto S$ with composition. Then an $M$-set is a "set of convergent sequences" closed under taking subsequences.
The category of $M$-sets is a topos, so it has limits, colimits, function spaces, etc. And every $M$-set has a topological realization which is a sequential space.
This is my first question; I hope it worthy of this awesome forum and its members.
Let $R$ be a commutative ring, perhaps with unit, perhaps not. As usual let $R[x]$
denote the ring of polynomials over $R$, and let $I$ be an ideal in $R[x]$. Let
$D(I)$ be the subset of $I$ consisting of all $f(x) in I$ of minimal degree. It is
well known from elementary algebra that in the event that $R$ is a field, $D(I)$
consists of essentially one element $d(x)$ (up to multiplication by a unit of $R$),
and that $I$ is the principal ideal generated by $d(x)$: $I = (d(x)) = Rd(x)$.
In the case of general commutative $R$, does $D(I)$ generate $I$ in the sense that
any $f(x) in I$ may be written $f(x) = sum_{i = 1}^{m} f_{i}(x)d_i(x)$ with $d_{i}(x)in D(I)$ and $f_{i}(x) in R[x]$? Does the existence of $1 in R$ make any difference?