i looked at you differential geometry notes. they look great. i am a teacher too. i love to be in the classroom with the kids. but the administration is horrible.
At the university level, I haven't had so much problem with administration (I was associate department head for 8 years). But I find that I can no longer motivate students to excel the way I could in past years.
incurrence, you are doing really well. it is nice to see so many young people interested and competent in math. i almost feel like i want to learn algebraic topology!
But the power of the inverse function theorem is absolutely amazing ... truly deep theorems that undergraduates can follow, rather than needing second-semester algebraic topology.
@TedShifrin few days ago my brother told me about an amazing new topic in mathematics . Incidence combinatorics!!!!!!!! in this we use algebraic topology and algebraic geometry to solve combinatoric problems ......Feels so amazing
@Balarka i had this very weird question to ask i dont understand when you and mike solve algebraic topology question why do you draw those weird shapes??
I have a group $G$ defined inductively $G^0$ is the group of upper triangular matrices, with $1$'s on the diagonal, and $G^1 = [G^0,G^0], G^{i+1}=[G^0,G^i]$
@Incurrence $PA$ is messes up the diagonal of $A$ by permuting rows. $(PA)P^{-1}$ messes up the diagonal of $PA$ by permuting columns. Show that these messes cancel out.
That's sort of standard, @Incurrence. I haven't thought about it in years. But can't you see that each step you take puts another "superdiagonal" of zeroes in there?
@Rememberme Patna is in south bihar and relatively well built ... the north part was more unlucky ... very undeveloped, had a tropical storm just 2 days ago and then two quakes in short time.
I just labelled then in reference to their existence(what I mean is, I started the count on when they started to exist) oh I just realised this is stupid as hell, but I'll show you the picture for a chuckle
@TedShifrin well the example is the Union of a convergent sequence without it Limit (here $\{\frac{1}{n} : n \in \mathbb{N}\}$) and some open Intervall minus one Point, a singleton, a closed set and some set such that both the set and its complement is dense in some intervall
@Rememberme Before you ask you question, you should do some research to make you question more specific, also you might want to consider restricting to classes of group you are actually interested in (like, maybe finite groups, or whatever). here and here are some places to start.
I am looking at the proof that $G$ has an Euler tour iff in-degree($v$)=out-degree($v$), that I found at this site: www.cs.duke.edu/courses/fall09/cps230/hws/hw3/headsol.pdf (Problem 2)
A simple cycle is a path in a graph that starts and ends at the same vertex without passing through the same ...
@evinda that Theorem doesn't Sound that spectacular, some example for some complex circle that isn't s simple one is going from 2 to 1 to 2 to 3 and back to 2
and if your graph is a complex circle it surely must be connected to be an euler tour
if there is no complex circle reaching all Points, it can't be euler
and for the sake of easiness you say you take the longest simple circle in the complex one and this one doesn't pass all Vertices because if it would do, this cycle would be an euler tour
@DominicMichaelis So at the first part do we consider the Euler tour as complex cycle and use only the fact that we can decompose a complex cycle to a set of simple cycles and for simple cycles it holds that in-degree(v)=out-degree(v) for all vertices v?
@Incurrence: Here's what you should prove as a lemma. If the first $k$ diagonals of $C$ and $D$ agree, then so do the first $k$ diagonals of $C^{-1}$ and $D^{-1}$.
@DominicMichaelis Could you explain me why from the fact that the graph is connected we deduce that there must be some complex cycle which passes trough every vertex?
@DominicMichaelis So always when we have a graph and it holds that in-degree(v)=out-degree(v) for all the vertices, does this mean that there is a complex cycle that contains all the vertices?
@DominicMichaelis Could you explain me further this part of the proof?
But if those degrees aren't all $0$ we could add some cycle to $C$ which would result in some complex cycle with more edges, which is some contradiction to our assumption, that $C$ does have the most edges.
@DominicMichaelis I am a little confused now.. G is the graph with the set of vertices V and the set of edges E, right? And we said that C is a complex cycle which passes trough every vertex, right? So will G' contain any vertices?
@DominicMichaelis A ok.. You said that: But if those degrees aren't all 0 we could add some cycle to C . Why would we know that G-C would contain a cycle?
While playing around with Mathematica I noticed that most polynomials with real coefficients seem to have most complex zeroes very near the unit circle. For instance, if we plot all the roots of a polynomial of degree 300 with coefficients chosen randomly from the interval $[27, 42]$, we get some...
@DominicMichaelis I also have an other question. There is this proposition: Since each edge in a complex cycle, and therefore in an Euler tour, is part of one or more simple cycles it will have in-degree(v)=out-degree(v).
How can an edge be a part of more than one simle cycles? Wouldn't that mean that we pass at an edge more than once?
Is a “connected Lie group” defined as simply a Lie group that is connected as a topological space? And does “groupe de Lie connexe” in French mean the same?
@DominicMichaelis I think that I have understood it but I will think about it again... I also want to describe an algorithm that runs in time O(E) and finds an Euler tour of G, if it exists. (Hint: Merge edge-disjoint cycles.) If we apply DFS, we will get a set of cycles formed by disjoint sets of edges, right? But how can we know if it holds that in-degree(v)=out-degree(v), for all vertices in V?
@DominicMichaelis The complexity of DFS is $O(|E|+|V|)$, but it is also $O(|E|)$. Indeed, DFS explores only the connected component to which the starting vertex belongs, and in a connected graph $|V|\le |E|+1$ (equality holds for a tree). Therefore, $O(|V|+|E|)=O(|E|)$.
@MikeMiller I think I saw a question on MSE that asks to show that existence of the Mayer-Vietoris long exact sequence implies homology excision. I don't think this is true, but I don't know of such a homology theory which satisfies the MV-sequence but not the exicison axiom.
@Ted: I don't quite understand. Here is the context. Q: 5 balls will be distributed into 4 labeled boxes (equal probability of going into each box). What is the expected value of number of balls in boxes A and B?
It's just the weighted average, @Jeff, but since all four boxes are equally likely, if you perform the experiment lots of times, on average there will be 5/4 balls in any particular box.
Right, @Jeff, it's the average of all possible outcomes, weighted by their probabilities. You can think of that as the expected average if you repeat the experiment a lot of independent times.
I am looking at the proof that $G$ has an Euler tour iff in-degree($v$)=out-degree($v$), that I found at this site: www.cs.duke.edu/courses/fall09/cps230/hws/hw3/headsol.pdf (Problem 2)
A simple cycle is a path in a graph that starts and ends at the same vertex without passing through the same ...
