given two arbitrary sets X and Y with n and m elements accordingly i can have the G(X) and G(Y) free abelian groups generated by X and Y and $ G(X) \oplus G(Y) \cong Z^{n+m} $
so say you have three sets $X,Y,Z$ and the corresponding free abelian groups $A,B,C$, what's a necessary and sufficient condition for $A\oplus B\cong A\oplus C$?
I'm just saying the notation isn't great, but what you mean is the free abelian group of countably infinite rank and that is certainly free abelian, by definition
hi guys morning/evening/afternoon/noon actually i was reading about the graohas of polynomial function $x^3$ and $x^2$ etc and they can easily be traceable ,but if i look a the $z^2$ and $z^3$ does they have any physical significance?I mean if i multiply z by z it onl gives me another complex number if it is not purely real or purely imaginary ,that is contradicting for me if i take $z$ as purely real then $z^24 surely have a physical meaning ,i am confuse in it!
I am trying to show that $f = 0$ a.e. on [0,1] implies \int_M f = 0 for every measurable set in [0,1]. I just let $N = \{ f \neq 0 \}$ then M = (M \cap N) \cup (M \cap N^c) and take the integral and it is $0$ for both, so is that a good enough proof?
the integral of a non-negative function always makes sense
you still have $|f|=0$ a.e.
so that integral is $0$ by the calculation and this, by definition, means $f$ integrable
generally, if two functions agree a.e., then one is measurable iff the other is and if their integrals make sense, they agree
so it should be natural to assume $f$ is automatically measurable and integrable, because it essentially is the $0$ function, which we know to be measurable and integrable
hi guys morning/evening/afternoon/noon actually i was reading about the graohas of polynomial function $x^3$ and $x^2$ etc and they can easily be traceable ,but if i look a the $z^2$ and $z^3$ does they have any physical significance?I mean if i multiply z by z it onl gives me another complex number if it is not purely real or purely imaginary ,that is contradicting for me if i take $z$ as purely real then $z^24 surely have a physical meaning ,i am confuse in it!
Problem Statement: Given $n$ different integers $\{a_1, \cdots , a_n\}$, then there exist a subset $\{a_{j1}, \cdots , a_{jn} \}$ with $1 \leq j_1 \lt \cdots \lt j_m \leq n\}$ such that $n$ divides $a_{j1} + \cdots + a_{jm}$
@Thorgott in the above problem, I don't know what does $$1\leq j_1 \lt \cdots \lt j_m \leq n$$ really mean?
Are they just clarifying that subscripts must be appropriate?
It's funny. I found myself trying the sorts of things students always did when I asked them to prove that if you choose $n+1$ numbers between $1$ and $2n$ (incl.), then there is at least one pair where one divides the other.
They often tried to do it by process of elimination, which doesn't work. I found myself doing the analogous thing here.
@TedShifrin @MikeMiller By the way, it's a boring fact that total space of the tangent bundle of a flat Riemannian manifold is a complex manifold. Under completeness, $M$ is a quotient of $\Bbb R^n$ by a group of affine isometries. So $TM$ is a quotient of $T\Bbb R^n$, which we identify with $\Bbb C^n$, by clearly a group of complex-affine isometries.
So the integrable scenario is boring. It's still curious to me that if $M$ is a Riemannian manifold, $TM$ is an almost-Kahler manifold (almost-complex structure we discussed, metric being Sasaki metric and symplectic form coming from that of $T^* M$ with natural isomorphism $TM \to T^*M$ using Riemannian metric)
Maybe there's a way to deform the almost complex structure to a complex structure on $TM$. Some $h$-principle.
Well, yeah, that's just my throw-away comment about $T^*M$ the other day. Well, it's a good exercise to prove directly that $TS^n$ actually is a complex manifold (you can give it explicitly sitting inside $\Bbb C^{2n}$ or $\Bbb C^{n+1}$—probably the latter).
I think integrability is going to be rare.
Actually, that $TS^n$ exercise might have been on the Guillemin & Pollack exercise sheets I sent you years ago.
(The baby case you know well. $\tilde G(2,4) \cong S^2\times S^2$, but the quadric surface in $\Bbb CP^3$ is, being doubly ruled, of course $\Bbb P^1\times \Bbb P^1$.)
Also, is it immediate how to rearrange for the $TS^n$ thing? $x \cdot y = 0$ is present in the imaginary part, the real part is like $\|x\|^2 - \|y\|^2 = 1$, which, hmm
Not seeing the issue. $z = x \sqrt{1 + \|y\|^2} + i y$ satisfies $z \cdot z = \|x\|^2 (1 + \|y\|^2) - \|y\|^2 + 2 i \sqrt{1 + \|y\|^2} x \cdot y$. Since $\|x\|^2 = 1$, the real part equates to $1$, and the imaginary part is zero since $x \cdot y = 0$
At any rate, that exercise (which I did centuries ago and have assigned numerous times) should have inspired me to think about the issue I just raised the other day.
This shows if you throw away the antidiagonal from $S^n \times S^n$ it's a complex manifold. I wonder if anything at all is known about $S^n \times S^n$.
It's a question in a book about number theory (maybe)
Problem solving strategies-Arthur Engel
It has the solution, but I'm not able to understand it
First solution: The result is $3n^2+3n +1 = (n + 1)^3- n^3$ is striking and allows a geometrical interpretation. One person gets x+y+z = 3n objects with 0 ≤x, y, z≤2n. These are triangular coordinates for an equilateral triangle with altitude 3n.x, y, z can be interpreted as lattice points (make a figure). The hexagon in the figure can be interpreted as the projection of the cube with edge n + 1 from which a cube of edge n is subtracted. This solution is due to Martin Härterich, a gold medallist of the IMO
Second solution. If the first person gets n - p (p € N) objects of the first kind, then the person can get p o 2n objects of the second kind. The remaining ones are objects of the third kind. The sum is
Sum (2n- p + 1) from p=0 to n
p=0 If the first person gets n +q (q€ N) objects of the first kind, then the person gets0 to (2n-q) objects of the second kind, since the person gets 3n objects altogether. The sumis (2n-q+1) The sum altogether is (2n +1)(n +1) +n(2n +1)- n(n +1) = 3n +3n +1
Let $f : \Bbb R \to \Bbb R$ be a smooth even function, i.e., $f(x) = f(-x)$. Prove that there's a smooth function $g : \Bbb R \to \Bbb R$ such that $f(x) = g(x^2)$
I have a question about: https://www.quantamagazine.org/black-holes-prove-that-anti-de-sitter-space-time-is-unstable-20200511/ It says (paragraph starting with "Moschidis imagined standing in the middle of AdS space-time...") that an AdS observer would see a light signal reach the boundary at infinity in finite time because such a wave is traveling at (or near) the speed of light. How does this explain that? Aren't we talking about the observer's perspective?
All odd derivatives vanish at $0$, so you can make a Taylor polynomial argument to generalize what you just did. I think that's what I did in the past.
By Borel's lemma, there is a smooth germ at $0$ whose Taylor polynomial at $0$ is exactly the Taylor polynomial of $g$ at $0$. This says that the negative half-germ of this smooth germ glues to the half-germ we already have
Borel's lemma is something stronger, like if $f_1, f_2, \cdots : \Bbb R^n \to \Bbb R$ are smooth functions defined near $0$ then there is a smooth function $F :\Bbb R^n \times \Bbb R \to \Bbb R$ such that $\partial^k F(x, t)/\partial t^k |_{t = 0} = f_k(x)$
Also saw this in a different formulation in some expository by Michael Weiss; if $L$ is a compact manifold, the map $C^\infty(L \times \Bbb R) \to \prod_{k \geq 0} C^\infty(L)$, $f \mapsto (f(-, 0), \partial_t f|_{t = 0}, \partial^2_t f|_{t = 0}, \cdots)$ has a continuous left-inverse
@TedShifrin So the Malgrange preparation theorem says that if $f : (M, p) \to (N, q)$ is any smooth map, $A$ is a finitely generated module over the local ring $C^\infty_p(M)$ of germs, then under basechange by $f^* : C^\infty_q(N) \to C^\infty_p(M)$, $A$ is a finitely generated module over $C^\infty_p(M)$ iff $A/\mathfrak{m}_q(N)A$ is a finite-dimensional vector space where $\mathfrak{m}_q(N) \subset C^\infty_q(N)$ is the maximal ideal of functions vanishing at $q$.
Let $\phi : \Bbb R \to \Bbb R$, $\phi(x) = x^2$. Consider $C^\infty_0(\Bbb R)$ as a module over itself, and basechange by $\phi$. Then the new $C^\infty_0(\Bbb R)$-module structure on $C^\infty_0(\Bbb R)$ is $f(x) \cdot g(x) = f(x^2) g(x)$.
Before restriction of scalars, $C^\infty_0(\Bbb R)/(x)C^\infty_0(\Bbb R) \cong C^\infty_0(\Bbb R)/(x) = \Bbb R$ was finite-dimensional, so by Malgrange $C^\infty_0(\Bbb R)$ as this weird module after restriction of scalars is also finite-dimensional. But then $C^\infty_0(\Bbb R)/(x) C^\infty_0(\Bbb R) \cong C^\infty_0(\Bbb R)/(x^2) = \Bbb R\langle 1, x\rangle$ is a 2D vector space spanned by $1$ and $x$, so by Nakayama's lemma $\{1, x\}$ is a basis for $C^\infty_0(\Bbb R)$ as this weird module
This means for every $f \in C^\infty_0(\Bbb R)$, $f(x) = g(x) \cdot 1 + h(x) \cdot x$, or, $f(x) = g(x^2) + x h(x^2)$ for some smooth germs $g, h$ at $0$
If $f$ is even $h \equiv 0$, and we have the germinal result
Yeah, basically the idea of "preparation" is elaborated in my proof of this result (originally by Whitney it seems -- what a giant!) above, you reduce Malgrange preparation to the case of Weierstrass preparation (an analytic result) by Borel's theorem
can anyone point me in the direction that puts mathematical rigor to these words? "zooming in, one loses global information about the (object) and zooming out one loses local information about the object
Weierstrass preparation is proved by repeatedly using Cauchy integral formula - a way of reading off the whole Taylor series from integrals
It's not worth reading the proof in detail. Very little payoff.
I still don't quite get it. When is it true that if $f : A \to B$ makes $B$ an $A$-algebra, $M$ is an f.g. $A$-module, then $f^* M$ is an f.g. $B$-module?
What the algebraic content of Malgrange says is if $f$ is a morphism of some special local rings coming from geometry, $f^* M$ is an f.g. $A$-module iff $f^*M/\mathfrak{m}_A f^* M$ is a finite dimensional vector space
One direction is Nakayama, the other direction is usually false
