for the limit, you go in the other direction. So the map $\Bbb Z/p\Bbb Z \to \Bbb Z/p^2\Bbb Z$ is given by $\overline{n} \mapsto \overline{pn}$, that's not even a ring homomorphism
So basically you have $0 \to \Bbb Z/p\Bbb Z = p\Bbb Z/p^2 \Bbb Z \hookrightarrow \Bbb Z/p^2\Bbb Z = p \Bbb Z/p^3 \Bbb Z \dots$
the colimit is basically just a union
You take the union over $\Bbb Z/p^n\Bbb Z$ for all $n$. This makes actually sense, e.g. if you realize that every $\Bbb Z/p^n \Bbb Z$ embeds uniquely into $\Bbb C^\times$
ah I meant for the colimit you go in the other direction
terminology is confusing: inverse limit = projective limit "=" limit, direct limit = injective limit "=" colimit
@TobiasKildetoft so I wanted to talk about Galois representations, I think some of this stuff is even interesting for finite groups
Suppose $G$ is a profinite group that satisfies some finiteness condition: for every $n$, there are only finitely many open subgroups of index $n$ (don't want to go into much detail, but sadly you can't take $\operatorname{Gal}(\overline{\Bbb Q}/\Bbb Q)$ because it's too big. But you can take the Galois group of a local field or the Galois group of the maximal extension of some number field that is unramified except at some finite fixed set of primes)
Let $k$ be a finite field and suppose we fix some continuous representation $\overline{\rho}: G \to \operatorname{GL}_n(k)$, where we put the discrete topology on $k$
(I think some of this also works when $k$ is not finite, but yeah)
Suppose $A$ is a complete Noetherian local commutative ring such that the residue field of $A$ is isomorphic to $k$. Then we call a continuous representation $\rho: G \to \operatorname{GL}_n(A)$ a deformation of $\overline{\rho}$ if it reduces to $\overline{\rho}$ modulo the maximal ideal
Now some technicality: we say that two representations $\rho_1,\rho_2: G \to \operatorname{GL}_n(A)$ are properly equivalent if they are conjugate by some matrix that reduces to the identity modulo the maximal ideal
Now consider a kind of technical category $\hat{C}_k$ which consists of all complete Noetherian local commutative rings together with a fixed isomorphism of the residue field to $k$ and morphisms are local ring homomorphisms (which is equivalent to continuous in the topologies induced by the maximal ideals) such that induced morphism on the residue field is the identity (if we identify the residue fields under the fixed isomorphisms)
We have a full subcategory $C_k$ consisting of those rings which are also Artinian (note: Artian local rings are automatically complete, since the maximal ideal is nilpotent)
Essentially by definition of complete local rings, every element in $\hat{C}_k$ is a projective limit (indexed by $\Bbb N$ actually) of elements in $C_k$
Now one can show that the assignment which sends each $A \in \hat{C}_k$ to the set of proper equivalence classes of deformations of $\overline{\rho}$ is a functor $\textrm{Deform}(\overline{\rho}):\hat{C}_k \to \mathbf{Set}$
this is the deformation functor associated to $\overline{\rho}$
basically we put those weird conditions on the morphisms so that this works
of course, we just apply our ring homomorphism to the coefficients
Now the main result is that if we restrict the deformation functor $\textrm{Deform}(\overline{\rho})$ to $C_k$, then it is not quite representable, but prorepresentable, which means that there is an object $R$ in the larger category $\hat{C}_k$ such that $\operatorname{Hom}_{\hat{C}_k}(R,A) \cong \textrm{Deform}(\overline{\rho})(A)$ for all $A \in C_k$ naturally
this follows from some general result about prorepresentability about functors $C_k \to \mathbf{Set}$, but the conditions and proofs are super technical
As with general representability stuff, there's also a bit of abstract non-sense involved
But what this basically means is that there exists some "universal deformation ring" $R$ of $\overline{\rho}$ and a universal deformation of $\overline{\rho}$ such that every deformation for any ring in $C_k$ is just given by composing the universal deformation with a ring homomorphism (satisfying some conditions)
you can say something about $R$ with the Cohen structure theorem and the Witt vector construction, if you want
The notation is just $R_{\overline{\rho}}$
you can show that the Krull dimension of $R_{\overline{\rho}}$ is equal the dimension of the cohomology group $H^1(G, \mathrm{ad} \overline{\rho})$ over $k$
The idea behing Wiles proof of FLT is very roughly that if you take the Galois representation coming from an elliptic curve over $\Bbb Q$, you reduce it mod $p$, then you get a universal deformation ring $R_{\overline{\rho}}$. Then also have some kind of Hecke algebra which happens to be an element of $\hat{C}_k$ which turns out to classify those representations coming from modular forms. So if one can show that the Hecke algebra is isomorphic to the universal deformation ring, then every deformation must come from a modular form
but for finite fields, we have that the ring of Witt vectors of $\Bbb F_p$ is $\Bbb Z_p$ and the ring of Witt vectors of $\Bbb F_{p^n}$ is the ring of integers in the unique unramified degree $n$ extension of $\Bbb Q_p$
@TobiasKildetoft apparently a lot can be said when you construct your representation from an elliptic curve that has good reduction mod $p$, but we don't go into that in the course, because alg geo/elliptic curves is not a prerequisite
@TobiasKildetoft I just thought that lifiting to characteristic $0$ is also something you do in modular representation theory
Hi, how could I argue that the boundary of the rotational body is a Lebesgue null-set?
With rotational body I mean $K_f = \left \lbrace {(x,y,z)^t \in \mathbb {R}^3} \mid {a \leq x \leq b,\ 0 \leq \sqrt {y^2+z^2} \leq f(x)}\right \rbrace$ for a continuous function $f: [a,b] \to \Bbb R_{\ge 0}$.
I want to argue that $K_f$ is Jordan measurable. This is equivalent to saying it is bounded and $\partial K_f$ is a Lebesgue null-set. $K_f$ is bounded since $f$ is bounded, but is its boundary also a Lebesgue null-set?
I am working on a review for MathSciNet. I am considering summarising the results as seen from two different points of view. One from the point of view of "vector product algebras are interesting to study" and one from "algebraic groups are interesting and vector product algebras turn out to be useful in this paper"
Maybe start with the "vector product algebras turn out to be useful" PoV and then why they are interesting on their own after that? That seems to be the logical order, since being useful is a strong argument for being interesting
Because what they do is consider morphisms between tensor powers of a vector product algebra which commute with the action of its automorphism group
And to me that just means that they take a certain representation of the automorphism group and consider those tensor powers of it and the homomorphisms between them as representations
where the fact that there is a structure of a cross product algebra is just a convenient thing that allows for a nice description of the these morphisms
Also, I just realized that I have been calling them vector product algebra when they are really called cross product algebras
don't quote me on this, but I think if you take $k=\Bbb F_p$ and $G=\operatorname{Gal}(\overline{\Bbb Q_p}/\Bbb Q_p)$ and let $\overline{\rho}$ be the trivial $1$-dimensional representation, you get that $R_{\overline{\rho}}=\Bbb Z_p$ and the universal deformation can be identified with the restriction map $G \to \operatorname{Gal}(\Bbb Q_p(\mu_{p^\infty})/\Bbb Q_p)$ to the field where adjoin all $p^n$-th roots of unity under the identification $\operatorname{Gal}(\Bbb Q_p(\mu_{p^\infty})/\Bbb Q_p) \cong \Bbb Z_p^\times = \operatorname{GL}_1(\Bbb Z_p)$
wait, no, not the trivial reprentation
We have $\operatorname{Gal}(\overline{\Bbb Q_p}/\Bbb Q_p) \to \operatorname{Gal}(\Bbb Q_p(\mu_{p^\infty})/\Bbb Q_p) \cong \Bbb Z_p^\times \to (\Bbb Z/p\Bbb Z)^\times$
or to put it simply, this is just how $\operatorname{Gal}(\overline{\Bbb Q_p}/\Bbb Q_p)$ acts on the $p$-th roots of unity
Hmm, so if $V$ is of dimension $3$ (let's say over $\mathbb{C}$ for now) and $b$ is a non-degenerate symmetric bilinear form on $V$, then $SO(V,b)$ is isomorphic to $SL_2(\mathbb{C})$, right? And $V$ becomes the adjoint representation?
Hmm, actually I don't see why $V$ should be irreducible. It should have a $1$-dimensional anisotropic subspace, and it seems like this should be a subrepresentation.
ohh, actually, it is not clear that it should be a subrepresentation
Shouldn't every subspace have an orthobasis basis and $SO(V,b)$ acts transitively on the orthogonal bases (modulo permutations to get the determinant right)?
@TobiasKildetoft sorry for being so random, but don't you agree that it's weird that the map $\Bbb Z[T] \to \Bbb Z[T,T^{-1}]$ is not an epimorphism in the category of not-necessarily commutative rings?
From abstract nonsense, one can show that $f:A \to B$ in any category is an epimorphism iff this diagram is a pushout square: $\require{AMScd} \begin{CD} A @>{f}>>B \\ @V{f}VV @VV{\operatorname{id}_B}V\\ B @>>{\operatorname{id}_B}> B \end{CD}$
Ah no wait
I took the wrong pushout
nevermind
I took the pushout over $\Bbb Z$, not over $\Bbb Z[T]$
@Waiting There was no need for motivating bounty! If I had the patience to sit through that calculation I would have already ... :) lemme know if you'd want the bounty points back .. I can't promise I'll will work on that monster any time soon :P
@r9m hehe, no! No hurry with that, you can post a solution anytime you want to (in the following months, years). But even if you don't post one, the answer deserves a bounty, so no need to return any bounty points back.
@r9m If you plan to post a solution in the future, please do it only when you have a full solution. :D
It's a well-known fact that
$$\lim_{n\to\infty} (H_n-\log(n))=\gamma$$
Now, if I change things a bit and use the fact that $\displaystyle \Gamma \left( \displaystyle \frac{1}{ n}\right) \approx n$ when $n$ is large, then I wonder
if it's possible to compute the following limit in a closed-form...
Straight integration seems pretty tedious and difficult, and I suppose that the symmetry might possibly open some new ways of which I'm not aware. What would your idea be?
$$\int_0^1 \int_0^1 \int_0^1 \frac{x^2}{\sqrt{x^2+1} \left(x^2-y^2\right) \left(x^2-z^2\right)}+\frac{y^2}{\sqrt{y^2+1} \lef...
so since all the integration limits are identical, it is indeed as mickep observed it is 3 times one of these integrals. So that means we should focus on:
Why are the values of sine and cosine in the bottom left quadrant of a unit circle diagram negative? Shouldnt they be positive as x and y are both negatve?
Let $V$ be vector space of polynomials of degree less than or equal to $2$. Let: $S=\{x^2+x+1, x^2+2x+2, x^2+3\}$. Then $S$ spans $V$. Is this correct?
I checked that $x^2,x,1$ are all in span of $V$. Is that enough? @Daminark
hi, $Fr$ is the frobenius transform($x\to x \ ^p$) i showed that $<Fr> \le Gal(\overline{F_p} / F_p)$ and that it is infinite. also i showed that $\overline{F_p} \ ^{<Fr>} = F_p$. now i need to show that $Gal(\overline{F_p}/F_p)$ is abelian. someone can help?
I am trying to understand a proof where there is the following:
$$\int_0^{\infty}\rho'(t)\left [\int_a^bu\, dx\right ]\, dt+\int_0^{\infty}\rho (t)\left [A(u(a,t))-A(u(b,t))\right ]\, dt=0$$ Since $\rho (t)$ is arbitrary we have $$-\frac{d}{dt}\left [\int_a^bu\, dx\right ]+A(u(a,t))-A(u(b,t))=0$...
I am trying to understand that what could be the possible outcomes of this following expression as $k \to \infty$ and as $|t| \to \infty$ where $k,t \in \mathbb{Z}$ .
$$\sum_{k=1}^{\infty}\dfrac{(\prod_{m=1}^{t}k)^{-i}}{\sqrt{k}}$$
Considering the numerator first:
$$(\prod_{m=1}^{t}k)^{-i}=\pro...
@JakeRose you know from the fundamental theorem of algebra that there are only two roots for a quadratic, so you only need to find 2 numbers from the quadratic formula
Imagine a rod moving in a $2$-dimensional space where the endpoints move with velocities $v_1, v_2$ (assume the velocities are directed perpendicular to the rod). How would you describe the rod's rotation around its center of mass?
But if you remember that $H_n\sim \ln n$, you can infer that $a_n=c\ln n$ is probably a good choice (with $c$ as a fudge factor) and all you need prove is that $c\ln(1+1/n)>1/n$ for some choice of $c$
I have a quick question regarding dynamical systems. Particularly to do with stability/manifolds and Lyapounov functions. It's only a quick q so I didnt want to makea thread about it.
But if a linearized system has a non empty unstable manifold at the origin (crit point) then so does the non linear system by stable manifold theorem. Does that mean that the origin can not be Lyapounov stable? Because there will be some values who are arbitrarily close at t=0 but as t -> infinity if one of the points is on the unstable manifold then it will just shoot off?
If
$$
h(k) = \begin{cases} 1 & -l \leq k \leq l \\0 & \text{otherwise}\end{cases}
$$
Where $l \geq 0$ is some integer.
I've done some computation and the summation
$$
F[h](\omega)=\sum_{k=-\infty}^{+\infty} h(k)e^{j\omega k} = e^{-j\omega} \frac{\sin(\omega(l-1/2))}{\sin(\omega/2)} = H(\omega...
Problem: Let $V$ and $W$ be finite dimensional vector spaces over a division ring $D$ with the same dimension. Let $f : V \to W$ be a $D$-linear map. Then if $f$ is surjective, it is also injective...I already showed the converse. Basically, it comes down to showing that $ V \rightarrow W \rightarrow C$ is an exact sequence, where $C$ is the cokernel of $f$, and using the fact that $\dim W = \dim V + \dim C$ to show $C=0$. But I am having trouble showing surjectivity implies injectivity.
My book says that the "surjectivity implies injectivity" implication is similar, but I don't see it. What relevant exact sequence will show that the kernel of $f$ is trivial?
@eurocoder To define the square root of a complex number, you ned to pick a branch, which will invariably require a branch cut. The domain of the branch of the square root will be the complex plane, minus some line or curve originating at zero. The square root from this domain to some subset of $\mathbb{C}$ will be a continuous map with continuous inverse (i.e. a homomorphism), but not of the entire complex plane.
Oh... wait... you said homomorphism, not homeomorphism.
In that case, homomorphism of what?
What structure are you trying to preserve?
Presumably, you want $\sqrt{x+y} = \sqrt{x} + \sqrt{y}$, which is not generally true.
@Semiclassical hi some bohmian stuff in my latest blog, hope to get your feedback at some pt & maybe discuss further, was influenced by/ building on quite a few of your leads discussed in hbar :) + lots of deep math for anyone else in here too vzn1.wordpress.com/2018/05/25/fluid-paradigm-shift-2018
If
$$
h(k) = \begin{cases} \frac{1}{2l+1} & -l \leq k \leq l \\0 & \text{otherwise}\end{cases}
$$
Where $l \geq 0$ is some integer.
I've done some computation and the summation
$$
F[h](\omega)=\sum_{k=-\infty}^{+\infty} h(k)e^{j\omega k} = \frac{1}{2l+1}e^{-j\omega} \frac{\sin(\omega(l+1/2))}...
I am trying to understand that what could be the possible outcomes of this following expression as $k \to \infty$ and as $|t| \to \infty$ where $k,t \in \mathbb{Z}$ .
$$\sum_{k=1}^{\infty}\dfrac{(\prod_{m=1}^{t}k)^{-i}}{\sqrt{k}}$$
Considering the numerator first:
$$(\prod_{m=1}^{t}k)^{-i}=\pro...
If
$$
h(k) = \begin{cases} 1 & -l \leq k \leq l \\0 & \text{otherwise}\end{cases}
$$
Where $l \geq 0$ is some integer.
I've done some computation and the summation
$$
F[h](\omega)=\sum_{k=-\infty}^{+\infty} h(k)e^{j\omega k} = e^{-j\omega} \frac{\sin(\omega(l-1/2))}{\sin(\omega/2)} = H(\omega...
