@PaulPlummer: Do you know if there's an algorithm to calculate the outer automorphism group of a group G? Like if I have a finite presentation for G, is there an algorithm to give me a presentation of Out(G)?
@MikeMiller I am unsure, I highly doubt it, given what @pjs36 what said, and that there is a whole (large) chunk of geometric group theory just studying $Out(F_n)$
They're infinite, but I definitely do care about the computations. I guess I don't really want a presentation. Here's the real story. I have a finitely presented group $\pi$. What I really want to know is what the automorphisms of $\pi$ do on the level of $H = \pi/[\pi,\pi]$. In particular, if I have my favorite ten thousand groups $\pi$, it'd be nice if there was a computationally feasible way of checking whether or not every automorphism of $\pi$ descends to a trivial automorphism of $H$.
In every case I care about $H$ has rank at least 1.
@MikeMiller I was just trying to see if Sage happened to have any "automorphism" group stuff... I know they have a class for finitely-presented groups, and they can handle commutators, etc
I used it quite a lot around two years ago, @mixedmath working with abstract polytopes, but I haven't done much else. I really do like a lot, but I'm not super proficient with many areas
@MikeMiller Still on tablet, but looking around (I don't really know much about automorphisms of groups) but this seems to say no:math.stackexchange.com/questions/825580/…
@PaulPlummer: In practice I want to test this on the fundamental groups of surface bundles over a circle. It's not too terrible to write down a presentation of these groups.
Yes, I know what you mean, @mixedmath. I've wanted to contribute, but I never fully made up my mind to ("didn't know how" and all that). What did you work on, when you contributed?
Not bad, @MikeMiller. I had a very productive day or two this past week. Then I got caught up in some more errands, and squirreled away some time watching the Pre Classic. You?
Ah, not much of a golf(?) guy myself. Gave a talk yesterday that has been contributing to much stress lately, so now things are much more peaceful. Today has been relaxing; time to be productive again tomorrow.
If you're thinking of the high-dimensional topology talk, different talks :) That was for the graduate seminar, which is an informal thing where people just talk about cool math; that was pre-blog post. This was a talk for the topology group's seminar, and much more time consuming to prepare.
@KarimMansour: That's a lot of topics. My introduction to C* algebras was through "C* algebras by example"; its a fields institute book. You need to know some of the fundamentals of functional analysis first.
yeah I am learning these stuff from my honours seminar I am doing project on completely positive maps and some stuff related to it that is EPR paradox etc
There is a reason why it's defined this way: we want to be able to use complex inner-products to define a metric, which has to be real-valued. In other words we want $\langle v, v\rangle \in \Bbb R$.
Well, additively, $\Bbb C^n$ isn't really different than $\Bbb R^{2n}$, and so there's some geometric impetus for wanting to preserve a similar notion of "distance".
When one deals with complex vector spaces, one "visualizes" the entire field $\Bbb C$ as being "a line", but this is only a crude analogy. The formal definitions pretty much take over, at that point.
If we call it $\theta$ for now, I would think $\theta(g_1h_1g_2h_2\cdots g_kh_k) = \psi_G(g_1)\psi_H(h_1)\psi_G(g_2)\psi_H(h_2)\cdots\psi_G(g_k)\psi_H(h_k)$
Well @DavidWheeler since you last talk about maths with me I have really advanced in mathematics...... Now I am thinking of taking a break any takes on how much long my break should be??
For a matrix $X$, a generalized inverse of $X$ is any matrix $Y$ such that $XYX = X$. We use $X^{-}$ to indicate a generalized inverse of $X$.
Suppose $X$ is a matrix. $X^{\prime}$ denotes the transpose of $X$. Then clearly $X^{\prime}X$ is symmetric and has a generalized inverse that is symmetr...
Pretty much this isn't how I like to learn. I prefer to have enough time to sit down and go through the text instead of learning to solve specifically these problems
Suppose $G$, $H$ are generalized inverses of $X^{\prime}X$. Then (i) $XGX^{\prime}X = XHX^{\prime}X = X$ and (ii) $XGX^{\prime} = XHX^{\prime}$.
I have shown that for a particular generalized inverse of $X^{\prime}X$ (say $G$) that $XGX^{\prime}$ is symmetric, but how does this lemma demonstrate that for **any** $G$, $XGX^{\prime}$ is symmetric?
@MikeMiller Them being HNN extentions should make the problem more manageable, although the Baumslag-Solitar group is an HNN extention (which had the infinitely generated automorphism group)
@DavidWheeler From the book: since $X^{\prime}X$ is symmetric, there exists a generalized inverse $G$ that is symmetric (this was from a theorem). For this generalized inverse, $XGX^{\prime}$ is symmetric; so, by the lemma, $XGX^{\prime}$ must be symmetric for any choice of $G$
@DavidWheeler So basically, since $XGX^{\prime}$ is symmetric for a symmetric $G$, since $XGX^{\prime} = XHX^{\prime}$ for any generalized inverse $H$ of $X$, $H$ must be symmetric?
So, $$1,1,2,3,5,8,13,21...$$ Any connection to primes?...it appears not. However, in between the Fibonacci numbers are how much primes? Let's see:
1 and 1 ZERO
1 and 2 NADA
2 and 3 ZILCH
3 and 5 ZIP
5 and 8 1
8 and 13 1
13 and 21 2
21 and 34 3
34 and 55 5
55 and 89 8
89 and 144 13
Hu...
We want to show that $\sup(A)+\sup(B)$ is the least upper bound
of the set $A + B$. First, we need to show that $\sup(A) + \sup(B)$
is an upper bound for the set $A + B$. Indeed, if $z\in A + B$,
then there exists $a\in A$ and $b\in B$ such that $z = a + b$. But by
definition of $\sup(A)$ and $\s...
Because of that reason I dislike sometimes number theory You can pose infinitely many questions in it you don't require any effort to make questions Just find some combinations and Voila!! You get a conjecture @Soham
Even so, that is a fascinating theorem (the diagonalization theorem). And, yes, it doesn't matter for your lemma, because your considering $(X'X)^{-}$ and $X'X$ is certainly symmetric.
Technically, you probably never have to have computations in anything....but....professors and authors like to give examples, which often necessitate them.
@DavidWheeler Let me see if I'm following this logic correctly.
Since $X^{\prime}X$ is symmetric, then it has a generalized inverse that is symmetric. For this particular generalized inverse, say $\left(X^{\prime}X\right)^{-}$, $X\left(X^{\prime}X\right)^{-}X^{\prime}$ is symmetric [verified by simply taking the transpose], so that $X\left(X^{\prime}X\right)^{-}X^{\prime} = XGX^{\prime}$ is symmetric for any generalized inverse $G$ of $X^{\prime}X$. We can say even more here: if $XGX^{\prime} = XG^{\prime}X^{\prime}$ (because we have shown $XGX^{\prime}$ is symmetric)... does it follow tha…
@DavidWheeler Could you explain to me why the book I have says that $G$ is symmetric for any choice of $G$? Maybe I'm missing something else from that lemma
@Paul, are you online? I'm confused about the answer to [this](math.stackexchange.com/questions/1039988/showing-that-the-product-group-of-g-and-h-satisfies-the-universal-property-f). The accepted answer says: $$\sigma\left(g,h\right)=\sigma\left(\left(g,e_{H}\right)\left(e_{G},h\right)\right)\overset{\text{how?}}{=}\sigma\left(g,e_{H}\right)\sigma\left(e_{G},h\right)=\sigma\circ\psi_{G}\left(g\right)\sigma\circ\psi_{H}\left(h\right)=\varphi_{G}\left(g\right)\varphi_{H}\left(h\right)$$.
Isn't the marked step essentially assuming that $\sigma$ is a homomorphism?
Since $X^{\prime}X$ is symmetric, then it has a generalized inverse that is symmetric. For this particular generalized inverse, say $\left(X^{\prime}X\right)^{-}$, $X\left(X^{\prime}X\right)^{-}X^{\prime}$ is symmetric [verified by simply taking the transpose], so that $X\left(X^{\prime}X\right)^{-}X^{\prime} = XGX^{\prime}$ is symmetric for any generalized inverse $G$ of $X^{\prime}X$.
Well, we are used to thinking of $A \times B$ as a "pairing" of $A$ and $B$
In the UMP view, we see it as something that acts as a "go-between" for pairs of maps from a common source, that allows us to consider the pair of maps as a single map.
@MikeMiller I have no idea what is going on with my computer... well I have been flipping through Trees by Serre and Combinatorial Group Theory by Lyndon and Schupp to see if there is anything that pops out
The difference being: in $A\times B$ in the usual view, we have things like $(a,b)$...in the UMP view, we have $f: C \to A$ and $g: C \to B$, and we produce (note the etymological similarity with product) the map $f \times g$.
Oh wait - suppose I could prove that there exists no such function that satisfies these conditions when $a = 0$. Would that be enough to ascertain that no such functions exist in the general case?