Let $X=\begin{bmatrix}
a&b\\0&0
\end{bmatrix}
$ and $Y=\begin{bmatrix}
x&y\\0&0
\end{bmatrix}
$. Furthermore, let the Lie bracket be the matrix commutator: $[X,Y]:=XY-YX$.
We get $XY= \begin{bmatrix}
ax&ay\\0&0
\end{bmatrix}$, but $YX= \begin{bmatrix}
ax&bx\\0&0
\end{b...
It's isomorphic to that one. I think you're misunderstanding what the bracket is. It doesn't spit out the first element. We're defining it by defining it on a basis.
I don't disagree with that. But I think faculty need to pull more weight. It's fine in a large lecture course to have slaves, but I don't approve in a major course.
@Semiclassical Related to what I was talking about earlier is the apparent identity $$\text{Re} \left[K \left(\frac{1-i \sqrt{3}}{2} \right) \right]^{2} = \frac{\sqrt{\pi} \, \Gamma^{2} ( \frac{1}{6})}{2^{11/3} \, 3^{1/2} \, \Gamma (\frac{5}{6})}$$ where, using Mathematica's notation, $K(m)$ is the complete elliptic integral of the first kind with parameter $m=k^2$. Is there any light you can shed on this apparent identity?
Basic category theory question - if we view a monoid as a category, the isomorphisms are precisely the invertible elements. ("Isomorphism" = "arrow with an inverse".) Awodey says that in this case we can't view "isomorphism" as "bijective homomorphism". Why can we not think of a monoid element acting on the monoid by left multiplication, and obtain a bijective homomorphism out of every invertible element that way?
I agree that "bijective hom" is the wrong way to view isomorphisms - for instance, the existence of bijective non-homeomorphism homomorphisms between topological spaces
i'm just not convinced that there are cases where we can't make a concrete definition which makes sense
@anon i read "only the abstract definition makes sense" as "we can't make another definition which makes sense", and in particular "we can't make a concrete definition which makes sense"
so the regular representation is too specific a procedure for it to work as a definition?
is it that we want a definition which is very general, and the representation is not general enough?
the idea of the regular representation is that the monoid is equivalent to a concrete category in which the isomorphisms are bijections, but that's not the same as defining the monoid a priori that way.
we don't define groups as collections of functions for instance, even if every group happens to be equivalent to a concrete symmetry group (via Cayley's). (although indeed stronger things can be said with categories, as categories aren't necessarily concretizable)
@anon I agree that it's extremely useful to have the abstract definition, and the abstract definition is often better
@anon i was just confused by awodey's statement that "only the abstract definition makes sense"
it's like defining a manifold, and saying "only the abstract definition makes sense", when we have theorems that let us embed all manifolds into euclidean space
if the morphisms are not functions, then the bijection definition doesn't make sense. it doesn't matter if there is an equivalent concrete category in which it does, since (again) the original morphisms are not functions.
@anon yep, I agree now, but only because you've shown me a non-concretisable category so I can now believe that there is no well-defined way to turn morphisms, whatever they are, into functions
the original statement wasn't necessarily about our ability to turn morphisms into functions, it was about morphisms being functions in the first place
i would be unhappy with a definition that insisted on casting morphisms into functions to deal with them, but awodey said it was impossible so i tried to see why it was impossible
what does it even mean to say morphisms are "functions"? a category has a bunch of objects, and morphisms between them are arrows. the objects may or may not be sets.
@anon ok, fair enough, but we don't care about the implementations of things usually, do we? like, there's loads of ways i could implement the integers, but i don't care what objects i'm dealing with as long as i can operate with them in some way
@BalarkaSen well, exactly!
@BalarkaSen like, integers are equivalence classes of pairs of naturals, but we have a way of viewing them in a nice way
in the same way, we might have a nice way of viewing morphisms
consider the following statement: "in many cases only the abstract definition makes sense, for instance the integers or fundamental groups." this could accurately refer to the abstract vs. concrete definition of groups. one could identify integers or based loops with functions, but that's not the type of thing that they are
"arrows" are a very abstract thing. one shouldn't confuse it with functions. think of the collection of arrows in between two objects a,b, in a category C as a set hom_C(a, b)
it's just a honset to gosh set, with nothing known about the elements
@anon i'm not very happy with "the type of thing that something is", because we may have many different ways of implementing things, and for each application we just need to be able to find an implementation which works - i'll have to think about that
@BalarkaSen yep, i'm happy with arrows-are-not-naturally-functions - my question was mainly "can we coerce them to be functions in a standard way"
@BalarkaSen whatever objects we are dealing with, we can define them in a set-theoretic way, can't we? it might be as a class, like the class of all sets - not a true set
see, the point is, objects in categories aren't a well-defined thing. it's like the word "point" in euclidean geometry, which you never define, but write down a bunch of axioms about.
my lecturers were very careful to give set-theoretic definitions of everything - a graph is a tuple (V, E) where V is a set and E is a set of tuples (v1, v2) such that v1, v2 are in V, and so on
@BalarkaSen ah, ok
@BalarkaSen i'm still in the "rigour" stage of terry tao's three stages, i like having these things precisely stated :P
it's a kind of "synthetic" thing. very hard to understand for logical minds, but it was surprisingly the thing which came first, if you trace down history.
maybe, maybe not. but that's not the point of category theory. you start with something which is not defined, and write down a bunch of axioms about it so that you can work with that thing. always trying to find a functor which takes your non-concrete category to a "concrete" category makes it hard for one to work with it.
@PatrickStevens yes, but V is the set of all the vertices. a single vertex here is not a set.
@Anon If I want to find the commutation relations for $-\frac{i}{2}\lambda_i$ where $\lambda_i$ are the Gell-mann matrices, will I need to manually compute all of $[u_i,u_k],i,k\in\{1,2,\cdots\}$ or is there some trick to find all of these quickly?
@LieAlgebra automatically you can cut the calculations from 8*8=64 to 8*7/2=28 brackets. there's probably something special that can be done beyond that but I am not familiar with them.
@robjohn hey. Related to the problem we discussed the days, the one in AMM, simple Taylor series reveals a form of this type $$\frac{1423813 x^{10}}{518400}+\frac{816049 x^9}{362880}+\frac{36139 x^8}{13440}+\frac{7183 x^7}{5040}+\frac{1411 x^6}{720}+\frac{67 x^5}{40}+\frac{49 x^4}{24}+\frac{7 x^3}{6}+\frac{3 x^2}{2}+x+1$$
@robjohn excepting the first 2 terms, there is a perfect alternance between the coefficients with odd and even powers. This should make us think of limsup and liminf, and numerically it is confirmed that one of the sequence is below $1/\log(2)-1$ and the other one above.
@robjohn I think once we make the split of the coefficients with odd and even powers, we might possibly use some simple inequalities and related the whole thing to something that gives us the desired result.
For the first coefficients we have $1, 1, 1.5, 1.16, 2.04, 1.67, 1.95, 1.42, 2.68, 2.24$, that shows the alternating values I was talking about.
@robjohn the author could have simply asked: calculate $$\lim_{n\to\infty} \frac{\log(a_n)}{\log(n)}$$ but he specified that we have sup and inf and this is due to the different magnitude of the coefficients of the even and odd powers.
It's not the first time when I met such problems, but the last part seems the most difficult one. Using some clever inequalities to relate some numbers about OEIS says anything about to some numbers that plugged in the stated limits yield the desired results.
The coefficients with odd powers are clearly below $1/\log(2)-1$.
@Chris'ssistheartist we don't need the precise limits, but we need to show the trends continue and the limits don't drop too low or rise too high at some point.
@RandomVariable hmm, I'll think about it. one thing I do notice is that, since $m=\frac{1}{2}+i \text{ Im }m$, one has $m'=1-m=\frac{1}{2}-i \text{ Im }m=\overline{m}$. So the LHS can be expressed in terms of $K(m)$ and $K(m')=K'(m)$ rather than in terms of real parts
@DanielFischer The Riesz-Dunford Functional Calculus definition of a mapping $f: \mathcal{A} \to \mathcal{A}$, from a Banach algebra to itself is as follows:
$$f(a) := \frac{1}{2 \pi i} \int_{\Gamma}f(z)(z-a)^{-1}dx$$ where the $f$ in the integrand is an analytic function on some open $U \subset \mathbb{C}$. Is there a reason why he calls the function on the left $f$ when there is already a $f$ in the integrand, would it not be better to call it $\Phi(f)$? Is there a specific reason for this way of writing the integral?
@Moses It may be cleaner in some respects to denote it $\Phi(f)$, but from a certain abstract point of view it is indeed the same function. If we take a polynomial, you are probably used to viewing polynomials as abstract algebraic entities, and that you can evaluate polynomials at elements of any algebra over the base ring. So writing $P(a)$ when $P$ is a polynomial is not surprising.
Analytic functions have many things in common with polynomials. In simple situations, you can identify the function with its power series, and a power series with elements of a Banach algebra also makes immediate sense (if convergent). The integral is written in that way because it is basically the Cauchy integral formula.
@DanielFischer I get your point regarding the polynomials but there the relation between $P(a)$ and $P$ seems clear and not also included in the same equation. I just don't see how in this case it is equivalent. But I get your point about the analogy to Cauchy Integral Formula. Is the reason why you say they are equivalent quite technical?
@Chris'ssistheartist I am stuck with a last piece of the puzzle .. I guess we could estimate the gap $\displaystyle \lim\sup_{n \to \infty} \frac{\log a_n}{\log n} - \lim\inf_{n \to \infty} \frac{\log a_n}{\log n}$ as well with a bit of knowledge about subseries of $\sum \frac{1}{m!}$ and it's powers :)
user147690
There is a sample questions list for the exam, with 16 questions on it, and written at the top is:
Four of the questions listed below will appear on Quiz 1 and make up 66 marks out of a total of 100.
user147690
@BalarkaSen Advanced alg I have fallen a little behind on, but otherwise everything seems good
user147690
I'll be doing some representation theory soon it looks like
@r9m you suggest, in other words, that $$\lim_{n\to\infty} \frac{\log(a_n)}{\log(n)}=1/\log(2)-1$$, because if so, I'm not convinced of that. I don't say it is not so, but I'm not sure of that.
user147690
Just finished group theory revision I guess, but I haven't got it all down yet
@Chris'ssistheartist nope ... $a_n$ has an error term that oscillates between a range of value .. I'm only saying we could work out an estimate of the difference between limsup and liminf with a bit of luck ;)
@r9m These days I don't have a great focusing power, my condition is not OK. But it's nice to taste it bit with bit like a very good chocolate, no hurry to finish it. ;)
user147690
I'm going to shower now, and then sleep probably(or come back), so goodnight I suppose
@Chris'ssistheartist proving the liminf < 1/log 2 - 1, is easy enough .. limsup > 1/log 2 - 1 desires miraculous observation which I still don't see :)
@Chris'ssistheartist it'd be my pleasure to re-enact the dance I had with my chair after I finished 11832 :P (I had to pay damage charges for that later .. )
@r9m Yeah, and they also produce a terrible lack of energy. Anyway, I think all this comes from overwork since for long periods of time I pushed myself very much and slept less.
@Chris'ssistheartist allergy medicines are to be discontinued only after the symptoms have left you .. don't stop abruptly from a med course (for allergy treatments the results could be aggravating) ..
@RandomVariable it's been a while, but i seem to remember the generalized leibniz rule for derivatives (i.e. $D^n(u(x)v(x))= \sum_{k=0}^n \binom{n}{k} u^{(k)}(x)v^{(n-k)}(x)$) being handy for that kind of calculation
@RandomVariable would be terribly tedious if you try to do that by hand. i'd expand by taylor series, quotient and look at the coefficient of $z^{-1}$, though.
that's not lacking in tedium either, though, because the power of $(z-i)^5$ in the denominator means you're in effect looking for the coefficient of $(z-i)^4$ in $\log(1-i z)/(z(z+i)^5)$
@DanielFischer Is the following reasoning correct: For an integral $\int_{\Gamma}\frac{g(w)}{w-z}dw$, where $g$ is analytic and where $z$ is outside the contour $\Gamma$, the integral is zero. My idea is that if $z$ is outside $\Gamma$ then $w-z \neq 0$ hence $\frac{g(w)}{w-z}$ is still analytic hence by Cauchy Integral Formula we have that $\int_{\Gamma}\frac{g(w)}{w-z}dw = 0$.
@Moses You assume that $g$ is analytic on an open set containing $\Gamma$ and all points $z_0$ with $n(\Gamma,z_0) \neq 0$? And "$z$ outside $\Gamma$" means $n(\Gamma,z) = 0$? Then Cauchy's integral theorem says the integral is $0$ (since $\Gamma$ is nullhomologous in the domain where $\frac{g(w)}{w-z}$ is holomorphic).
@TobiasKildetoft Hey, I'm studying for quals now and have a small question. I'm trying to show the following. Let $G$ be a finite group, $\pi : G \to GL(V)$ an irrep of $G$. Let $\chi$ be the character of $\pi$. If for an element $g \in G$, we have $|\chi(g)| = \dim V$, then there exists $c \in \Bbb{C}$ with $\pi(g)v = cv$ for all $v \in V$.
Here's what I've tried. I know that $\pi(g)$ is diagonalizable with eigenvalues $\lambda_1,\ldots,\lambda_n$, since $G$ is a finite group. Then, the hypothesis and the triangle inequality that all the $\lambda_i$'s have the same complex argument. However, I'm having trouble concluding that all the $\lambda_i$'s are equal to some constant $c$. I haven't used anything about irreducibility yet, and I'm struggling to see how to use it.
@RandomVariable Actually, checking numerically I think there's a typo. the identity is false if one squares the real part of that elliptic integral, but it's true if one takes the real part of the square. so the exponent of 2 should be inside the square brackets.
@DanielFischer If we consider the product of two integrals then you can rename it so that $\big(\int f(x) dx\big)\big(\int g(x) dx\big)= \big(\int f(x) dx\big)\big(\int g(y) dy\big)$ hence you take it into the integral and get $\int \big(\int g(y)dy) f(x) dx$. You could also switch the order of the integration. Is this a special case of Fubini's Theorem?
@SohamChowdhury there were technical things, but the gist is, she talked about what we can know about the variety from looking at it's coordinate ring.
it's fascinating stuff. you can even tell about "smoothness" of a variety from looking at it's sheaf of rings.
I mean that lagarnage theorem is only true for finite groups. So is there any way to find the number of left cosets of H ( A subgroup of G) in G for an infinite set G?
If $|G| = \kappa$, and $H$ is a subgroup of $G$, then $|H| \eta = \kappa$ for some cardinal $\eta$. Assuming the axiom of choice and that $|G|$ is not finite, this just says "$|H| \leq |G|$". Not very exciting.
That's not true. The conclusion is that infinite groups are harder to deal with in this context.
Quotients go mad.
For example, there are infinite groups $G$ with subgroup $H$ such that $G/H \cong G$ even though $H$ is not trivial. This is not quite relevant though.
@MikeMiller I pondered on your problem. My general idea for the less harder problem was a map $K(G, 1) \to X$, where $X$ is simply connected. To get this to work for homology + homotopy, one needs to choose a group $G$ such that $H_n(G)$ vanish for $n > 1$. I am not sure if I know of an acyclic group, yet.
Don't reveal the problem, though. I have not put all of my concentration on it yet.
and playing around in that vein, i come up with $K'(m)=-i m K(m)$. (alas, mathematica recognizes that only at the numerical level)
and from that, i get an identity for $K(m)$ itself, namely that $K(m)=Re^{i\pi/12}$ where $R$ is a positive real constant related to the one you gave in yours
@Semiclassical The identity is related to the definite integral $\int_{0}^{\infty} J_{0}(x)^{3} \, dx$ where $J_{0}(x)$ is the Bessel function of the first kind of order zero.
@Semiclassical There is an generalization of that integral on the main site, but the evaluation relies heavily on Mathematica. What does Mathematica return for that integral?
Consider the ring $F[t, t^{-1}]$ (the polynomials in $t$ and $t^{-1}$ with coefficients in the field $F$).
Theorem 1.
Assume that the characteristic of $F$ is zero. Then the existential theory of $F[t, t^{-1}]$ in the language $\{+, \cdot , 0, 1, t\}$ is unecidable.
Theorem 2.
Assu...
1) So is Fraktur. 2) The point is ease of communication, which includes speed of writing. If you have to learn to read Fraktur, it doesn't take long to learn to read Kurrent.
Every algebraist I've seen actually write things that are normally mathfrak on the board either writes in Kurrent or just writes in standard Roman script.
And for things like $\mathfrak p$? Writing in Roman script can be confusing.
Let $X=\begin{bmatrix}
a&b\\0&0
\end{bmatrix}
$ and $Y=\begin{bmatrix}
x&y\\0&0
\end{bmatrix}
$. Furthermore, let the Lie bracket be the matrix commutator: $[X,Y]:=XY-YX$.
We get $XY= \begin{bmatrix}
ax&ay\\0&0
\end{bmatrix}$, but $YX= \begin{bmatrix}
ax&bx\\0&0
\end{b...
Consider the ring $F[t, t^{-1}]$ (the polynomials in $t$ and $t^{-1}$ with coefficients in the field $F$).
Theorem 1.
Assume that the characteristic of $F$ is zero. Then the existential theory of $F[t, t^{-1}]$ in the language $\{+, \cdot , 0, 1, t\}$ is unecidable.
Theorem 2.
Assu...
