This would be one part, and the other part would be showing that every vector is a sum of something in each. I think that other part is where $p(T)q(T) = 0$ kicks in
@Daminark you can work that out with group homology purely algebraically if you want. You'll find that $H_n(\Bbb{Z}/p,A)$ is two-periodic and you get $A[p]=\{a \in A \mid pa =0\}$ for even $n \geq 2$ and $A/pA$ for odd $n$. ($A$ is any abelian group)
Might be non-trivially inclined to start pulling out group homology. This is a common theme in my class, no one has any clue how to figure out the non-Hatcher problems until some wacky high powered theorem falls from the sky
the idea for the group cohomology thing is that if $G$ is a cyclic group, you can write down an easy projective resolution of $\Bbb Z$ as a $\Bbb{Z}[G]$-module. You have the norm element $N_G=\sum_{h \in G}h$. And if $g$ is a generator of $G$, then you can write down a LES $\dots \to \Bbb Z[G] \to \Bbb{Z}[G] \to \Bbb{Z} \to 0$. Where the map $\Bbb{Z}[G] \to \Bbb{Z}$ sums over coefficients and the maps $\Bbb{Z}[G] \to \Bbb{Z}[G]$ alternate between multiplication with $g-1$and multiplication with $N_G$
Given a 3-manifold $M$ and a principal $\mathbb{Z}_2\ltimes PSU(4)$-bundle $P$ over $M$ whose isomorphism class is represented by the homotopy class of a map $f:M\to B(\mathbb{Z}_2\ltimes PSU(4))$ where $\mathbb{Z}_2$ acts on $PSU(4)$ by the outer automorphism.
Since there is a (right split) sho...
The context was that we had a problem to show that a compact 3-manifold with finite fundamental group had universal cover homotopy equivalent to S^3. If in turn the fundamental group is abelian, then it's cyclic. First part was ez and then our idea for the second was that you had π_2 = 0, only needed a single 4-cell to kill π_3, and then you can add a bunch of stuff to give yourself $K((\mathbb{Z}/p)^2,1)$.
@BalarkaSen The comment I made about the classification of $O(2)$-bundles is folklore to me, I never did it. I guess it's by explicitly understanding the Postnikov tower, and I guess that's what we should do here
But to do it here I'd need to start by understanding that folklore I think
@Daminark Oh nice, so $H_1(M;\Bbb Z/p)$ and $H_2(M;\Bbb Z/p)$ contradict Poincare duality.
Now we were trying to say that $H_4$ had to be at most dimension 1 which fucks Kunneth. And if we know that either $H_3$ or $H_4$ is $0$ then we're good. Given that there's still stuff left to do I'm slightly inclined to just cite the fact that I totally understand that $H_3(K(\mathbb{Z}/p,1)$ is non-zero, maybe by group homology or smth
Actually in general the natural map $H_2(M;A) \to H_2(\pi_1 M; A)$ is surjective by similar logic, which is why you have the contradiction here for $A = \Bbb Z/p$.
@MikeMiller Weird. If we pass to the double cover corresponding to $w \in H^1(M; \Bbb Z/2)$, the $O(2)$-bundle on $M$ lifts to an $SO(2)$-bundle on $\tilde{M}$, right?
so if $\lambda$ is a cardinal and $f:\lambda \to Ord$ is non-decreasing and $\alpha = \sup_{i \in \lambda} f(i)$ is a successor ordinal then $|\{ i \in \lambda \mid f(i) = \alpha \}| = \lambda$?
Okay so, $H_1(M;\mathbb{Z}/p) = (\mathbb{Z}/p)^2$, then because $H^1$ is the dual group that should be isomorphic, and by Poincare duality we get that for $H_2$
We have this. I suspect, then, that you can simultaneously make your matrices "nearly-upper-triangular", meaning that the diagonal 2 x 2 blocks represent complex scaling and the bottom-left block is zeor.
Simultaneous triangularizability is much easier because triangularization is tantamount to choosing an eigenvector iteratively on smaller and smaller blocks. Commuting matrices have a common eigenvector and the iterative blocks also commute... I suppose that gives a proof
Balarka just gave a counterexample (assuming that's still "this sort of thing") by proving that commuting matrices are simultaneously triangularizable. :)
Anyway, I don't know how to use this to finish Dami's problem
The crucial thing is proving that commuting guys have some common eigenvector, @Balarka. Certainly it's not true of every eigenvector (of one of 'em). Now how do we see this?
Well, if $\Bbb C[x]$ acts on $V$ by having $x\cdot v = T(v)$, you need to know that $xI-T$ gives the relations for the obvious presentation of the module.
Argue by induction. Suppose $V$ is a vector space with two operators $T$ and $S$, which commute. They therefore have a common eigenvector $v$. Then there are induced operators $T'$ and $S'$ on $V/\langle v\rangle$. Inductively, I may choose a linear transformation $L'$ on $V/\langle v\rangle$ to conjugate by to make $T'$ and $S'$ simultaneously triangular. By extending $\{v\}$ to a basis of $V$, we see there is an operator $L$ on $V$ which fixes $v$ and projects to $L'$.
Conjugating by $L$ upstairs we see that we have a new basis so that in the quotient, $T'$ and $S'$ are upper triangular, and $T$ and $S$ have $v$ as an eigenvector. This is what you wanted.
$V$ is f.g. $\Bbb C[X]$-module so $V \cong \bigoplus_{i} \Bbb C[X]/(X-\lambda_i)^{v_i}$ so each $C[X]/(X-\lambda_i)^{v_i}$ should give you a Jordan block: indeed $\{1, X, X^2, \cdots, X^{v_i-1}\}$ is a cyclic basis... I need to rethink my life
You mentioned CRT and I think that's what allows you to convert between different versions of Jordan Form, in particular allowing you to say $\bigoplus_i \mathbb{C}[x]/(x-\lambda_i)^{e_i}$ instead of some random decomposition $\bigoplus_i \mathbb{C}/p_i$
At least I think it does, so in that sense yeah you're right that it kicks in
I think the point was more in even being able to decompose into powers of irreducibles. At least in the case of groups, for example, to be able to go from just saying "a direct sum of cyclic groups" to "a direct sum of cyclic groups of prime power order" is through CRT
the whole Santa thing is so creepy. I wonder why catholics would try to present an image of trust and associate to a creepy old guy visiting the children's bedrooms at night
well naturally the supreme cuckold is a part of the picture too
but the santa thing is just more on the nose, like it's directly suggestive of the idea that there is nothing to fear why some random old guy comes down the chimney in the olden timey equivalent of a mardi gras costume
Take a subset $\mathcal{S}\subset \mathcal{P}(E)$. Now make a set $\mathcal{S}^p\subset \mathcal{p}(\mathcal{S})$ such that any element $B \in \mathcal{S}^p$ is $B \subsetneq E$
if we form a subset from elements in a set $\mathbb S$ by imposing some relational predicate $\mathcal R(\mathbb S)$, provided that that subset is not equal to $\mathbb S$ we are assured that the two subsets formed by imposing $\mathcal R(\mathbb S)$ and $\lnot(\mathcal R(\mathbb S))$ are equivalence classes of $\mathbb S$ with respect to .. and that's as much AA as I can do at any given time or I am pretty sure nose bleeds would occur. anyway tell me what ive said there that is wrong
@Adam This holds when law of excluded middle holds. Also since any$ \mathcal{R}(\Bbb{S}) \subset \mathcal{P}(\Bbb{S})$ and the powerset can be identified to the set of indicator functions, it follows $\mathcal{R}$ and $\neg\mathcal{R}$ partitions $\mathcal{S}$ into two equivalence classes
That I am not sure if I am good enough at logic to demonstrate. For example, say I have some $\mathcal{R}$ defined as "pairs of real numbers whose 10th digit in the decimal expansion are the same. Then for unconputable numbers like the Chaitin's constant, $\mathcal{R}$ is unable to tell you which subset it belongs
So, rock-paper-scissors is a game where two people simultaneously choose one of three possible "throws", and then the winner is determined by a particular tournament graph (en.m.wikipedia.org/wiki/Tournament_(graph_theory)) whose vertices are the throws.
Variants of rock-paper-scissors can be created simply by choosing a different tournament graph.
The original game is "interesting" because it has exactly one Nash equilibrium, which assigns a non-zero probability to each throw.
Also, its tournament graph is vertex-transitive (which means that, essentially, all throws are equivalent to each other).
Suppose I want to create an "interesting" rock-paper-scissors variant (one with exactly one Nash equilibrium, and it assigns a positive probability to each throw) whose tournament graph is asymmetric (which means that, essentially, none of the throws are equivalent to each other).
@CaptainAmerica16 oh that's too bad champ try to be more careful hey you know what is just as fun as spoon and totally risk free? Putting up different license plates on your vehicle and stealing fuel
@LeakyNun your function has the same growth as Gödel's pairing function restricted to the diagonal of $\mathsf{Ord}\times\mathsf{Ord}$. The bound should be sharp for all ordinals with $\beta=\omega^\beta$
Here is a sketch of a start of a definition of category.
It uses definitions found in set so we import that. In that file you'll find $\text{cat} \ \textbf{Set}$ defined and all the basic set operations.
Anytime you compile a diagram and a Node is titled "Definition:..." then that triggers an ...
The vector space structure consists of addition and scalar multiplication. Being linear means that $A(v+w)=Av+Aw$ and $A\lambda v=\lambda Av$, the first part is the preservation of addition, the second the preservation of scalar multiplication
@MatheinBoulomenos I saw yesterday that $H^*(A_5; \Bbb F_2)$ is a domain. I wonder in general how often $H^*(G; \Bbb F_2)$ or $H^{2*}(G; \Bbb F_p)$ (passing to even gradings to ignore the obvious nilpotents).
@MikeMiller so I have been wondering the following: suppose that you have an abelian group $A$, a group $G$ and a central extension $1 \to A \to E \to G \to 1$. I think (or hope) that this should induce an exact sequence on (non-abelian) cohomology $\dots \to H^1(X;A) \to H^1(X;E) \to H^1(X;G) \to H^2(X;A)$ via some Snake-lemma-like diagram chasing. Now suppose the morphism $H^1(X;G) \to H^2(X;A)$ is natural in $X$, then this determines via Yoneda an element of $[K(1,G),K(A,2)]=H^2(G;A)$. Is this the same element in $H^2(G;A)$ that is determined via the standard algebraic construction for t…
when we speak of an equivalence relation on a set $\mathbb S$, the standard convention used is '$\sim$' but how do a formally address a relation composed of multiple '$\sim$'?
The way I would construct that sequence is to construct a fiber sequence $BE \to BX \to B^2A$ classifying the central extension - the construction of such a sequence gives rise to it and vice versa
or more concisely, how do I generalize the notational declaration of a particular composition of equivalence relations, joined by logic operators as I did in the example of $\land$
particular composition of equivalence relations and their necessary negations (necessary for the composite to satisfy the requisites of a standard equivalence relation)*
common on, logic operators and fundamental abstract algebra concepts in the same question, uniting those who are already the best of friends, programmers and mathematical "purists"
> Zhang was identified as one of the top candidates for the Nobel Prize by Thomson Reuters in 2014 for his work on quantum effects, including topological insulators (material that insulate from the inside but allow the flow of electrons on their surface), spintronics, and high temperature superconductivity.
Link should be fixed now
I still have no idea what a topological insulator is
Tomorrow I have my final exam on algorithm. In the previous year paper it was asked - "Is P=NP? Justify your answer." Can anyone tell me how to answer it? :p
The setseelements (0,1) and (1,0) are usually considered as the generators for Zp X Zp. But we can take any two elements of order p as generator set for the cayley graph of Zp X Zp , right?
The set containing the two elements (1,0),(0,1) is usually regarded as the minimal generating set for cayley graph of Zp X Zp
Sorry the first part of the sentence was not properly typed so I sent it again
A generating set with any two elements of order p will give a Cayley graph for Zp X Zp which will look like a square grid similar to the square grid we obtain when considering the generator set {(1,0),(0,1)},right? p is a prime
And if we consider a semidirect product between Zq and G=Zp X Zp then if two elements of order p are also contained in the generator set of cayley graph of the semidirect product then this torus structure will always be created within the cayley graph right?
Here is a sketch of a start of a definition of category.
It uses definitions found in set so we import that. In that file you'll find $\text{cat} \ \textbf{Set}$ defined and all the basic set operations.
Anytime you compile a diagram and a Node is titled "Definition:..." then that triggers an ...
