User daniel briggs - MathOverflow

Homotopy Transfer Theorem for Differential Graded Associative Algebras

2012-12-13T19:18:26Z

As in Algebra+Homotopy=Operad by Bruno Vallette, let $A$ with multiplication $\nu$ be a differential graded associative algebra equipped with degree +1 map $h$ and let $H$ be a chain complex such that there exist chain maps $i$ and $p$ such that

and I work in characteristic 2 to make everything easier. Define

$$\mu_2=p\circ\nu\circ(i\otimes i):H\otimes H\to H,$$

and in general,

is a degree $+(n-2)$ map, where $PBT_n$ means binary trees with $n$ nodes and Vallette's summand on the right is an example of a summand for $n=5.$ For $f\in\hom(H^{\otimes n},H),$ define $\partial f=d\circ f + f\circ d_{H^{\otimes n}}$; remember that we are in characteristic 2.

One way to visualize $\partial\mu_n$ is that the one term decorates the leaves with $d$'s, as $d$ is a derivation for the raw tensor product, and the other puts a $d$ at the root, which then propagates upwards, as $d$ is a derivation for $\nu.$

The Homotopy Transfer Theorem for Differential Graded Associative Algebras is that $H$ equipped with the $\mu_n$ is an $A_\infty$ algebra, which means precisely that

All images have been directly screencapped from Vallette's paper. He writes that it should be an "easy and pedagogical" exercise to prove this theorem, but I'm getting caught in the thicket even in this characteristic 2 case where there are far fewer $\pm$'s to keep track of. I was wondering if anyone could provide me with any insights as to how to proceed without trees popping up all over the place occluding the forest.

Answer by Daniel Briggs for The Surreal numbers satisfy all the field axioms except that its elements constitute a proper class. Is it safe to call it a field?

2012-12-07T23:56:23Z

In ZFC+I, where I asserts the existence of a strongly inaccessible cardinal, we can create a Grothendieck universe out of that inaccessible cardinal and have a set that satisfies all the properties of the surreal numbers, where quantification is taken over elements of that universe. From the perspective of the outside universe, this would be a field. Since no one has found a contradiction in ZFC+I, it's safe to say at present that the field theorems apply to surreal numbers and that an algebraist cannot detect any essential difference, only relative differences that you probably know such as "This Field contains all totally ordered fields."

In the quaternions, "any imaginary unit may be called i"

2011-01-30T20:56:09Z

Introduction

Suppose we are trying to prove that $\rm PSO_3\times PSO_3$ is isomorphic with $\rm PSO_4,$ and we catch on to the idea of using the quaternions to do so. We realize (as in Conway & Smith's On Quaternions and Octonions, whence the quotation) that we can encapsulate $\rm PSO_3$ as the set of all maps $x\mapsto\bar qxq$ for a unit quaternion $q$ operating on imaginary quaternions $x,$ and go on to trying to understand why $\rm PSO_4$ is the set of all maps $\pm(x\mapsto\bar lxr)$ for unit quaternions $l,r$ operating on quaternions $x.$

To show that all maps $\pm(x\mapsto\bar lxr)$ really are elements of $\rm PSO_4,$ we begin by showing this in the case $\bar l=\cos\theta+isin\theta,r=1.$ This is simple, since this operation rotates the plane spanned by $1$ and $i$ through an angle of $\theta,$ and rotates the plane spanned by $j$ and $k$ through an angle of $\theta$ at the same time.

But at this point, we really are done proving that all maps $\pm(x\mapsto\bar lxr)$ are elements of $\rm PSO_4,$ since, as in the book, "any imaginary unit may be called $i,$ and perpendicular one $j,$ and their product $k$" (although this was said at a different point), right-multiplication has the same geometric properties as left-multiplication, and the composition of any two elements of $\rm PSO_4$ is an element of $\rm PSO_4.$

Idea

It makes all the intuitive sense in the world to me that "any imaginary unit may be called $i,$" and I can really visualize this geometrically. Furthermore, I could go back through a proof, change all the $i$'s to $u$'s for an arbitrary imaginary unit quaternion $u,$ etc. But suppose I had a collection of proofs written by someone who wasn't very careful, and she/he had used $i, j,$ and $k$ for simplicity and computed examples, stating at the end of each one that it generalizes to all quaternions. Suppose I pored through these proofs and discovered that about a third of them were careless to the point of being false, because of some lack of care in going back/forth between general ($u,v,w$) and specific ($i,j,k$) contexts. To formalize this imaginary formalization attempt, suppose I had a computer that understood category theory really well and wanted to scan these proofs in for it and get it to check whether this person's proofs really proved whatever facts from geometry they purported to prove.

Question

In the specific example, the notion that any imaginary unit may be called $i$ and left-multiplication is like right-multiplication can be dealt with at a first approximation using the notion of automorphisms. There is a ring automorphism sending $u$ to $i$ and vice versa, and there is a group automorphism between the multiplicative group of the quaternions and its opposite group. But I wonder if it follows directly enough that geometric facts can be proved "by example."

Is there a category-theoretic context in which certain ring/group automorphisms are natural and in which their being natural is biconditional with their preserving geometric properties?

To explain why the notion of an automorphism by itself might not be enough, we can imagine $\mathbb{Q}[\sqrt{2}]$ acting on $\mathbb{R}$ by multiplication. Multiplication by $\sqrt{2}$ preserves order, but multiplication by $-\sqrt{2}$ reverses it, so in the context of orderings, it would not be accurate to say "any square root of 2 may be called $\sqrt{2}.$"

What I'm envisioning is a category $\mathcal{C}$ with an object $\mathbb{H},$ as well as a morphism for each automorphism of $\mathbb{H},$ an object $\mathbb{H}^\star,$ and a morphism for each group automorphism of $\mathbb{H}^\star,$ perhaps geometric objects or morphisms as well, and whenever we speak of "an instance of the quaternions" we are really speaking of a functor $\mathcal{C}\to\mathcal{C}$ at least one existing for each possibility of an imaginary unit being called $i,$ a perpendicular one $j,$ and their product $k.$ I know this doesn't work out as stated, because the identity morphism doesn't go to the identity morphism, but perhaps there's a way to fix that. Then an automorphism of the quaternions can be viewed as a natural transformation between functors $\mathcal{C}\to\mathcal{C},$ one preserving the hidden (say, setwise) structure of the quaternions, and another preserving their apparent ($i,j,k$) structure and for some reason, because it's natural, the geometry comes out alright.

Answer by Daniel Briggs for What is the simplest, most elementary proof that a particular number is transcendental?

2010-12-14T06:40:05Z

The original Liouville's number is probably the easiest, but most of the proofs tend to invoke calculus (because why not?), so let me try to show it in a more 7th-grade friendly way. I'll call this the swaths-of-zero approach.

So we know that Liouville's number $L$ looks like this: .1100010000000000000000010... with a 1 in the $n!$ places.

When we square it, we get this: .012100220001000000000000220002...

What happens is that in the $2n!$ places we get a 1, and in the $p!+q!$ places we get a 2. (The great thing about this is that it can be explained using the elementary-school algorithm, the one they are all familiar with, for multiplication.)

If we multiply $L$ by an integer and write down the answer, the value of that integer will be "laid bare" as we go deeply enough into $L$'s decimal expansion, as eventually the 1s are far enough away to become that integer without stepping on each other.

Similarly, if we multiply $L^2$ by an integer, we will see that integer in some places, and 2 times that integer in others. For large enough $n,$ if we look between the $n!$ place and the $(n+1)!$ place, the last thing we'll see is that integer written at the $2n!$ place.

Thus the swaths of zero in the multiple of $L$ are, $n!-(n-1)!=(n-1)(n-1)!$ long (minus a constant), whereas the widest swaths of zero in the multiple of $L^2$ are $n!-2(n-1)!=(n-2)(n-1)!$ (minus a constant) long, which is shorter, so there is no way to add positive multiples of $L$ and $L^2$ together to clear everything after the decimal point, or find positive multiples of each so that everything after the decimal point is equal.

More generally:

Suppose $a_jL^j+...$ and $a_kL^k+...$ are integer polynomials in $L,$ where $j>k.$ We show that their values cannot match up fully past the decimal point. The swaths of zero in the first polynomial, moving back from the $n!$ spot, are a constant away from $(n-j)(n-1)!$ long (the constant being the length of the sum of the coefficients), whereas in the second they are a constant away from $(n-k)(n-1)!$ long, in the same place (moving back from the $n!$ spot).

I don't know if this explanation holds up to the standards of rigor you like to maintain when teaching them, but I think they will find it fascinating.

Answer by Daniel Briggs for Choice Function on the Powerset of the Reals

2010-11-08T20:22:18Z

It seems as if you're interested in the question: "Does there exist a model of ZF in which there is a choice function on the reals but in which the axiom of choice is false?" I don't know how to prove it precisely but there does.

The axiom of global choice is an axiom of Gödel-Bernays (NBG) set theory (sets & classes) which states that there exist choice functions on proper classes as well as sets. NBG canonically uses the axiom of limitation of size instead, which implies global choice.

Now a Grothendieck universe is a set in the context of ZF whose rank is an inaccessible cardinal (the existence thereof is independent of ZF). I believe ZF plus the assertion that such a cardinal exists is formally equivalent to NBG minus global choice. Thus, we would by using a Grothendieck universe have a model of ZF onto which we could add the stipulation that there exist choice functions for all sets of rank less than that of the given inaccessible cardinal; choice for the sets of larger rank would not be guaranteed; although I don't know how to prove this, either.

Comment by Daniel Briggs

2011-12-18T17:33:11Z

Is it possible, for a given $n$ and $K=\sigma_1(n),$ for the statement "$e^\gamma n\log\log n=K$" to be undecidable? Our arbitrary-precision approximations just showing values closer and closer to $K$ and that's all we know?

Comment by Daniel Briggs

2011-04-08T13:47:47Z

One can construct nothing (but random lengths, if one allows randomness) with just a straightedge. But perhaps we should start with a picture of the x- and y-axes plus the unit circle?

Comment by Daniel Briggs

2011-01-31T17:10:35Z

@Mark Sapir: No, I guess that does about wrap it up, without a need to change to the context of category theory. I will accept Emerton's answer later today.

Comment by Daniel Briggs

2010-12-15T14:20:43Z

Or, increasing from 2(n-1)!, the first factorial sum to be seen is n!+1!. (And decreasing from 2(n-1)!, it's (n-1)!+(n-2)!, which is very far away, and this shows that the factorial sums can't conspire to make "magic" 0s.) Similarly, with mL^3, all the products involving at least one 1 from the n! place on make less than 3m 10^-n! L^2 (the 3 is from choosing the 1 to be in the n! place in the first, second, third L, and the "less than" from the microscopic overcounting); moving left from here, the first thing we see is m at the 3(n-1)! place, which is sooner than in a multiple of L^2.

Comment by Daniel Briggs

2010-12-15T13:53:03Z

Given mL^2 for an integer m, let's go back from the n! spot towards the 2(n-1)! spot. Near the n! spot there can be contributions of the form 2m 10^-n!10^-k! for small k, but notice that the effect of the 10^-k! is to move right, rather than left, and the smallest values k! can take on are 1, 2, 6, 24, ... . So .2m+.02m+.000002m+... will be seen here, but that makes at most one more positive digit left than .2m does, and it's the same near the n! spot for any n. Once you get past it, there's nothing going back to 2(n-1)!, and then there's something (if m has final 0s, once we get past them).

Comment by Daniel Briggs

2010-12-14T23:15:17Z

It would be interesting to see what portion of Liouville numbers can be taken care of without much trouble by using an appropriate base for expansion and discussing the number in terms of the base (it seems that (1) q must be able to be chosen in a relatively uniform way, such as powers of the base, and (2) the places with the positive digits would have to become sparse enough so that the number wouldn't get muddy: would this requirement be equivalent to the Liouville criterion? Or stronger?)

Comment by Daniel Briggs

2010-12-14T06:55:10Z

@David Feldman: Agreed! It was, for me, a very thought-provoking question.

Comment by Daniel Briggs

2010-11-09T16:36:41Z

I took a look back at "Set Theory for Category Theory" by Michael Shulman and realized I had misremembered what I'd read. "One should think of the classes in NBG and MK as corresponding to the large sets of rank (or cardinality) $≤\kappa.$ In fact, one can make this precise. If $\kappa$ is inaccessible, we obtain a model of MK (and hence NBG) by taking $V_\kappa$ for the sets and $V_{\kappa+1} = \mathcal{P}V_\kappa$ for the classes. (Note that this means we can prove Con(MK) in ZFC+I, so ZFC+I is strictly stronger than mk.)" In particular, I had forgot what was in the parentheses.