@Secret Yeah, your conditions basically mean that $a$ is surjective and $b$ is injective (being the right inverse of $a$). This only implies that they are bijective if the set is finite
@Euclid First, you need to find the primitive roots of $p^k$, then primitive roots of $2p^k$ are odd primitive roots of $p^k$ and $p^k+\text{any even primitive root of}~p^k$ if $p$ is an odd prime number.
@Euclid If $p=2$, then $2p^k=2^{k+1}$ just has primitive roots for $k=0,1$.
@euclid You just need to see the proof of several theorems on primitive roots of composite numbers. You can find them probably in any book on elementary number theory. See for example David M. Burton's book.
Ok. I actually don't know the rigorous story; what is the definition of $\Omega_k^{fr}$? Framed cobordism classes of $k$-manifolds inside a sphere of dimension which, exactly?
@user91500 for finding primitive roots of $p^k$ we have to find primitive roots of $p$ and other primitives are $a^m$ that (m,k)=1 when $a$ is a primitive root.this is correct?
and this indeed works modulo homotopy on the left and modulo framed cobordism on the right. it's a bijection between $\pi_{n+k}(S^k)$ and framed cobordism classes of $n$-submanifolds of $S^{n+k}$ actually I believe
It's a pretty neat exercise to compute $\pi_3 S^2$ using that. google told me you can do that for $\pi_4 S^2$ too but I don't know how to compute framed cobordism classes of 2-manifolds. everything should be framed cobordant to the sphere, but.
I actually don't understand how you're thinking of it as nullhomotopy. Normal bundle of a circle in $S^4$ is a 3-bundle on $S^1$. So a framing is a choice of a 3-frame (an elt of SO(3))to each point on $S^1$. the map $S^1 \to SO(3)$ is given by sending each point to the 3-frame at it
@Euclid For example, suppose you want to find the primitive roots of $3^3$. Since $2$ is a primitive root of $3$, you must choose primitive roots of $3^3$ from the numbers $2,5,8,11,14,17,20,23,26$. Note that $p^{k-2}(p-1)=6$ and just for $a=8,17,26$, $a^6\equiv 1\pmod{3^3}$, so primitive roots of $3^3$ are $2,5,11,14,20,23$
You can in fact generalize the map, $J : \pi_k SO(n) \to \pi_{n + k} S^n$, known as the J-homomorphism. It sends an framed $S^k$ inside $S^{n+k}$ to the map $S^{n+k} \to S^n$ according to the bijection mentioned above
the P-T story says it's an isomorphism for $k = 1$ and $k = 2$ (??). I don't know anything about $k =3$, but it fails for $k = 4$. lotta 4-manifolds not even cobordant to the 4-sphere let alone framed
@euclid Since $2$ is a primitive root of $3$ and either $2$ or $2+3$ is a primitive root of $3^2$ by corollary of Lemma 1 in page 160 of Burton's book. See also Lemma 2 in this page cleverly.
@euclid First, find a primitive root for $p$ like $r$ such that $r^{p-1}\not\equiv1\pmod {p^2}$. Then primitive roots of $p^k$ are $r+mp$'s such that $(r+mp)^{p^{k-2}(p-1)}\not\equiv 1\pmod{p^k}$.
@Euclid I don't think such a thing is true but if $r$ is a primitive root of $p$, then the other primitive roots of $p$ are $r^k$'s such that $(k,p-1)=1$.
Hey hey! How many sequences $(a_1,\dots,a_n)\in\mathbb{F}_q^n$ are there such that $a_i\neq a_j$ for all $1\leq i<j\leq n$ and $a_1+a_2+\dots+a_n\neq 0$. Where $p$ is a large prime.
It seems to me that there are $p(p-1)(p-2)\dots(p-n+2)(p-n)$ such sequences. There are $p$ choices for $a_1$, $p-1$ choices for $a_2$ since we can't pick $a_1$, and so on until and including $a_{n-1}$. When picking $a_n$ we can't pick any of the $a_i$ where $i<n$, and also can't pick $-a_1-a_2-\dots-a_{n-1}$, so there are $p-(n-1+1)$ choices.
let be $N(x) = \sum\limits_{d|p - 1} {\mu \left( d \right)} {F_x}(d)$ that $p$ is a prime number and ${F_x}(d) = \frac{{{x^2}}}{{4d}} + O(x{p^{\frac{1}{2}}})$
that $x+1=g(p)$ and $g(p)$ is the least primitive root modulo $p$. Applying Brun's method to $N(x)$ in conjunction with ${F_x}(d)$ in orde...
let be $N(x) = \sum\limits_{d|p - 1} {\mu \left( d \right)} {F_x}(d)$ that $p$ is a prime number and ${F_x}(d) = \frac{{{x^2}}}{{4d}} + O(x{p^{\frac{1}{2}}})$
that $x+1=g(p)$ and $g(p)$ is the least primitive root modulo $p$. Applying Brun's method to $N(x)$ in conjunction with ${F_x}(d)$ in orde...
I have seen the following in the script: $\displaystyle \mathrm{exp}(x) = \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n$ and $\displaystyle \mathrm{exp}(x) = \sum_{k=0}^{\infty} \frac{x^k}{k!}$. It is told that these are the two opportunities to define the exponential function. On the one hand, is that not ordinary to write $f$ for the function, and $f(x)$ for the value of the function at $x$? If so, why these are definitions of the functions and not for some values $x$ only?
@Null Maybe, yes. The review queue for "close votes" is never at $0$ because there's so many people who post questions without a single mention of attempting it themselves.
On the other hand, why is that possible to define functions in that way? How can a function that gives many values be represented as a one limit or a value of one summ?
@Semiclassical it is about the definitions. I was learnt that a function is something like $\mathbb{N} \to \mathbb{R}, x \mapsto x^2$, the thing that orders every $x$ a definitive value under $f$. At the same time I was learnt that $lim$ is a value, one value. So I am asking myself how can a function be described only with one value...
If I had a map $f: \mathbb{Z}-\{0\}\to \mathbb{Q}$ such that $f(x)=\frac{1}{x}$, then it is easy to check $f$ is injective with a fixed point $f(1)$. However it is easy to see that $f$ is not surjective since $\textrm{img}(f) \subset \mathbb{Q}$ despite that $|\mathbb{Q}|=|\textrm{img}(f)|=\aleph_0$. Is this the the correct way to understand it?
I know that for a permutation, it is a bijection that maps from some set S to itself, but is it true even for the uncountably infinite case, that every element in S appears only once in the image of such permutation?
I have been trying to follow the lines in a proof I saw online. You can read the snippet on page 6. The part of the proof I am stuck on can be stated as follows
Let $r$ ble close to $0$ and $p>1$ then for every $z \in D$
$$
\int_{-\pi}^{\pi}\frac{\mathrm{d}u}{\left| 1 - \frac{|z|}{r}e^{iu}...
For permutations in finite set, it is easy to show that every element appear exactly once in the image. For countably infinite case, one have examples such as in $\mathbb{Z}$ where the map is "multiply x by -1", and in general it is also easy to show that every element appeared exactly once via some induction type proof. However I am not very sure about the uncountably infinite case. Is it possible for a permutation in an uncountable set $S$ to map every $x$ to every unique $y$, yet the image is a subset of $S$?
@iwriteonbananas The point is you extend the framing on $S^1 \sqcup S^1$ to $S^1 \times [0, 1] \sqcup S^1 \times [0, 1]$, till the crotch of the pant. Then use the non-manifold level set (wedge of two circles) to modify the 1-frame to the trivial 1-framing on the circle.
can someone please talk with me privately? I feel analysis is way more difficult than my linear algebra course. Maybe I'm just a certain type of student, but maybe that is normal. So how can I make analysis easy again?
We also had one analysis teacher that was really evil. First he writes a proof on the blackboard giving some explanation along the way. Then he writes down a counterexample to what he just proved.
And asks us: which one is wrong, proof or counterexample?
I have been trying to follow the lines in a proof I saw online. You can read the snippet on page 6. The part of the proof I am stuck on can be stated as follows
Let $r$ ble close to $0$ and $p>1$ then for every $z \in D$
$$
\int_{-\pi}^{\pi}\frac{\mathrm{d}u}{\left| 1 - \frac{|z|}{r}e^{iu}...
