My sister didn't appreciate all the math I tried to instill in her ... plus, if you make it so he's totally bored in school, that doesn't help. I dunno.
Not really, @CaptainAmerica. I am finding that working out stuff I used to know in my sleep to answer questions on main takes more time than I expect ...
@CaptainAmerica: I'll be happy to discuss this a little bit when you are doing your research. You should think seriously about whether you want smaller college or larger university. Very different experiences.
You can mollify them by doing some computer science (minor, maybe), which I would recommend even if you want to do pure math research.
@TedShifrin I'd appreciate that. It's been a lot. I can't think of any other option other than double majoring. I'm focusing on colleges that allow that.
Well, I may actually buy a present for one person this year. And I will take out my friend in Michigan for a fancy dinner for his Xmas/birthday present.
If only there were a spider lemma. Then there could be some pretty good Harry Potter jokes about follow the Spiders? Why can't we just use Zassenhaus' lemma?
@TedShifrin if the function is $C_2$ then the above sequence just converges to the parabola, and presumably you mean two distinct functions. I can think of some more pathological functions, but I doubt that's what you're getting at.
I had one question regarding induction. If $B_n = \cup_{k=1}^{n} A_k$ satisfy some property $P \forall n \in \mathbb{N}$. Does it mean that $\cup_{k \in \mathbb{N}} A_k$ satisfy $P$?
Hi umm, can someone guide me on how to start solving this equation: $$i\dfrac{\partial f(x,t)}{\partial t} = \dfrac{\hbar}{2m}\cdot \dfrac{\partial f(x,t)}{\partial x^2}$$ ?
Or any good resource where I can learn to solve these kinds of equations?
@MUH Not always, no. Consider $A_k = [-k,k]$ and let $P$ be the property "$B_n$ is compact". Clearly this holds for all natural numbers $n$, but $\Bbb R$ is not compact.
Induction says "the statement is true for all natural numbers"---you cannot necessarily conclude anything about an infinite case from normal induction.
Let $T$ be some finite dimensional operator. Is it true that the size of the largest Jordan block corresponding to the eigenvalue $\lambda$ is equal to the dimension of the generalized eigenspace associated to $\lambda$?
polytope insanity of the day (so far): I'm running code to get the nonnegative part of a high-dimensional cone. To make things easier for me to track, I'm doing it coordinate-by-coordinate i.e. restrict to have the first coordinate be nonnegative, then the second, etc
@BalarkaSen that's the elementary proof. you can also argue via dimension shifting on Tate cohomology that it is enough to check this on $\hat{H}^0(G;M)$, but that's $M^G/N_GM$ and for that the statement is obvious
Ya, there's a couple proof. You can write down the transfer homomorphism $H^0(1; M) \to H^0(G; M)$ sending $m$ to $\sum_{g \in G} g \cdot m$ at the cochain level. This extends to a homomorphism $\tau : H^n(1; M) \to H^n(G; M)$ by the derived functor formalism, which you can check satisfies $\tau \circ \iota = |G|$ where $\iota : H^n(G; M) \to H^n(1; M)$ is induced from the inclusion map.
Because that's what it does at degree 0
So multiplication by $|G|$ factors through $\iota$, which kills everything
this reminds me: if $H \subset G$ is a finite index subgroup, then $\mathrm{Cor} \circ \mathrm{Res}$ is multiplication by $[G:H]$. For $H$ the trivial subgroup, this recovers the result
the statement about $\mathrm{Cor} \circ \mathrm{Res}$ has a useful consequence: if $P$ is a $p$-Sylow subgroup of a finite group, then the kernel of $\mathrm{Res}:H^n(G,M) \to H^n(P,M)$ is $p$-torsion free
I think you're spoiling Mike's exercise ("what's the relation between cohomology of a group and cohomology of it's p-Sylows?") though, so don't tell me more :P
@MatheinBoulomenos The easiest way to prove that $H^n(G; \text{Hom}_{\Bbb ZH}(\Bbb ZG, M)) \cong H^n(H; M)$ is to see that $\text{Hom}_{\Bbb ZG}(\Bbb ZG, \text{Hom}_{\Bbb ZH}(\Bbb ZG, M)) \cong \text{Hom}_{\Bbb ZH}(\Bbb ZH, M)$ as functors, right?
Below is a problem that I made up and my attempt at a solution to it. I am hoping that somebody here can help me finish it. I believe there is a unique
answer to the problem.
Thanks,
Bob
Problem:
Let $X$ and $Y$ be uniformly distributed independent variables on the interval $(-1,1)$. Let $K$ b...
I was trying to construct a nowhere zero vector field by hand and the only thing that I could think of was to take gradient field of a Morse function and cancel points of opposite indices
Problem: Prove or disprove that $\Bbb{R}/\Bbb{Z}$ is compact...My conjecture is that $\Bbb{R}/\Bbb{Z}$ is homeomorphic to $S^1$. I know that $\Bbb{R}/\Bbb{Z}$ and $S^1$ are isomorphic as groups, but $\Bbb{R}/\Bbb{Z}$ as a quotient group is different from $\Bbb{R}/\Bbb{Z}$ as a quotient space (equivalence classes are defined differently). Is my suspicion right?
Well, from my understanding, in the case of the quotient space $x \sim y$ if and only if $x,y \in \Bbb{Z}$; in the case of quotient groups, $x \sim y$ if and only if $x-y \in \Bbb{Z}$.
You were thinking of the topological $X/A$ where you collapse the subspace $A$. But I don't believe that's what they mean. You'd have to ask the person who wrote it.
Below is a problem that I made up and my attempt at a solution to it. I am hoping that somebody here can help me finish it. I believe there is a unique
answer to the problem.
Thanks,
Bob
Problem:
Let $X$ and $Y$ be uniformly distributed independent variables on the interval $(-1,1)$. Let $K$ b...
