Homomorphisms preserve linear structure, homeomorphisms preserve topological structure. Neither of them have to do with differentiability. That the concepts are different is apparent in the definition. The derivative of a holomorphic function $\mathbb{C}\rightarrow\mathbb{C}$ at a point is a complex number, whereas the derivative of a differentiable function $\mathbb{R}^2\rightarrow\mathbb{R}^2$ at a point is a linear operator.
@N.Maneesh: Yes, when you have diffeomorphic objects, they are "identical" so far as differentiable functions are concerned. If they're $C^k$ diffeomorphic, then we talk about $C^k$ functions. If they're smoothly diffeomorphic, then we talk about smooth functions.
If $A$ be $3\times3$ Matrix with Characteristic polynomial with characteristic polynomial $x^3-3x+a=0$ for what values of $a$, $A$ must be diagonalizable?
$(A) a\in (-2,2)$$ (B)a\in [-10,2] (C)a\in \mathbb R (D) $ No values of $a$
What shall I care about this problem? Should I check the diagonalizability over $\mathbb C$ or $\mathbb R$?
I saw your thing above. I don't want to check it in detail, but you can give a cleaner argument by noting that if $x$ is a unique element of order $2$ in $G$, then $\langle x \rangle$ is normal in $G$.
I want you to do that, instead of listing down the multiplication table by hand or whatever
Is it correct to say that $\lim_{n \rightarrow \infty} a_n = \{x \in \mathbb{Q} \mid \exists n \in \mathbb{N} : \forall m \in \mathbb{N} : m > n \rightarrow a_m > x\}$ ?
Where the right hand side is interpreted as a Dedekind cut
If $a$ be a unique element of order $m$ in $J$, then $a \in Z(J)$.
Proof: Consider the element $jaj^{-1}$.
In this case $x \in Z(G)$. So, for any element $g \in G$
$g\{x,e\}=\{gx,g\}=\{xg,g\}=\{x,e\}g$.
Evidently, $<x>$ is normal in $G$. Here, $Z(G)=<x>$ [since the only possible order of subgroups are $3$ and $2$. If any other element, of order $3$, say $p$ belongs to $Z(G)$, then $p^2=x \implies p^3=e=xp \implies p=x^{-1} $ But, $o(x^{-1}=2$, contradiction ] The quotient group $G/<x>$ exists. But, $o(G/<x>)=o(G)/o(<x>…
Sure, sounds like it. You have proved $\Bbb Z_6$ and $S_3$ are the only groups of order $6$ with very general techniques (eg, you can try to classify groups of order $pq$, $p \neq q$ now)
Note that both $\Bbb Z_6$ and $S_3$ have a normal subgroup isomorphic to $\Bbb Z_3$ which, upon quotienting, gives a $\Bbb Z_2$, which is something you were asking yesterday if I recall correctly.
So the answer to all three are "No". The middle one is the most interesting, though.
Here's some abstraction: Suppose $N, G, H$ are groups and $i : N \to G$ and $q : G \to H$ are homomorphisms which fit in the following "diagram" $$0 \to N \stackrel{i}{\to} G \stackrel{q}{\to} H \to 0$$ This sequence of arrows is called a short exact sequence if kernel of any arrow is image of the arrow before it. This in particular implies $i$ is injective, $q$ is surjective, and $\ker q = \text{im}\, i$.
Note that by first isomorphism theorem, $G/i(N) \cong H$.
Here's a classical problem in group theory: Given $N$ and $H$ in a short exact sequence as above, what groups $G$ can fit in the middle? This is called the "Extension problem"
You gave an example where there are two groups that fit in the middle in $0 \to \Bbb Z_3 \to G \to \Bbb Z_2 \to 0$, namely, either $\Bbb Z_6$ or $S_3$.
There are no other groups of order $6$, you proved, so nothing else can fit. These are the only solutions.
Previous argument. Since, the non identity elements will be of order $p$ and $q$, and the exponent will have an order of $q$ (the order cannot be higher, since it will then become cyclic). Again, $p \nmid q$, but that is a contradiction since in an abelian group the order of all the elements divide the order of the exponent
And Cauchy's theorem ensures that the must exist an element of order $q$. $G$ being abelian
Here's a random detour (but connects back to everything, I promise): Suppose $H, N$ are subgroups of a group $G$ such that $N$ is moreover normal. Consider $HN = \{hn : h \in H, n \in N\}$. Check that $HN$ is a subgroup of $G$.
Suppose $G = HN$, $H \cap N = \{e\}$ and $H$ is also normal in $G$. Then prove that $G \cong H \times N$.
@SubhasisBiswas You're doing pretty good. I'd advice not to evaluate yourself and just enjoy the math though
It's not worth evaluating if you're doing well or not at the end of the day, there's just bound to be things you're good at and things you're bad at - who cares? It's the math that's interesting
If you just collapsed $X \times \{0\} \sqcup X \times \{1\}$ to a point you'd get, in case of $X = S^1$, a sphere with north and south pole identified.
No, $\Bbb Z_2$, right? Mobius \sum Mobius is trivial
@Alessandro Upto isomorphism there are exactly two rank $k$ bundles on $S^1$, one is the trivial guy and another is the twisted guy, exactly like in $k = 1$
Let $M$ be the Moebius line bundle on $S^1$, then the rank $k$ Moebius bundle is isomorphic to $M \oplus \underline{\Bbb R}^{k-1}$
Hello. I am trying to figure out how this formula should be evaluated (which result is vector [w1,w2,w3]) : https://i.imgur.com/ua3Flsv.png RT is a (transposed) matrix I am not asking what the formula does or the meaning each variable (this is OK) What is not clear to me is those values inside round parentheses, should I treat this as matrices (and so multiply several matrices together to evaluate the formula) ?
I checked Hatcher, and apparently what I'm doing is the reduced K-group. $\widetilde{K}^0(X)$ is the group of stable equivalence classes of vector bundles over $X$ equipped with direct sum
Let $f:[0,1]\to\Bbb R$ be a differentiable function such that $f'(x)\ne0$ for all $x\in(0,1)$. Then which of the following statements is false?
a) $f$ monotone. b) $f$ one-to-one. c) $f$ uniformly continuous. d) None of these.
I think d is the answer, right? because e.g. there are monotone as well as non-monotone functions with non-zero derivatives, and so on for b and c as well.
Sorry! It should be c, since because of intermediate value theorem for derivatives, hence either $f'(x)>0$ or $f'(x)<0$ for all $x\in(0,1)$, so monotone, and one-to-one.
To show that $f:(0,1) \to \mathbb{R}$ is not uniformly continuous on $(0,1)$, we pick the sequences $\{1/n\}=(u_n)$ and $\{\frac{1}{n+1}\} =(v_n) $. Now, $\{|u_n-v_n|\}$ converges to zero, but, $|f(u_n)-f(v_n)|=|1| \nrightarrow 0$
$\sum a_n$ is a convergent series of positive terms. What can you say about the convergence of $\sum\frac{ \sqrt{a_n}}{n}$? Since, $\sum a_n$ is a convergent series of positive terms. For $\epsilon=1$, There is a natural number corresponding to it. such that $a_n <1$. So, $\sqrt{a_n}<1$. I am not able to use comparison test. please help me.
@BalarkaSen Going off this post, the answer to the question does turn out to be affirmative as a consequence of Hahn-Banach; math.stackexchange.com/a/2440188/422019
By Serre-Swan $K^0(S^1)=K_0(\mathcal C(S^1))$, so one only needs to classify finitely generated projective $\mathcal C(S^1)$-modules up to isomorphism, which sounds awful
We also have a different definition of $K_0(A)$ for a $C^\ast$-algebra based on stable equivalence of projections in $M_\infty(A)$, which doesn't sound much easier to compute...
@AlessandroCodenotti By Serre-Swan again those correspond to vector bundles over $S^1$ again: If you trust me there are exactly two such things for every rank $k$, then you only have to figure out what the modules corresponding to those are. For the trivial bundle, the corresponding module is free of rank $k$. Now what would be the space of sections of the Mobius bundle?
Let's try $k = 1$. Mobius line bundle on $S^1$.
Is it possible to explicitly write down what the $C(S^1)$-module $\Gamma(S^1, \text{Mob})$ is?
@ÍgjøgnumMeg Oh of course, I always think about the lifting property, but direct summand of a free module is a much nicer definition of projective often enough
$(P \otimes_A B) \oplus (Q \otimes_A B) \cong (P \oplus Q) \otimes_A B$, free again
Great, so the thing above is a f.g. projective $C[0, 1]$-module. Serre-Swan of course tells you all such things are free, but is there a purely algebraic proof?
I doubt. You'd need machinery.
Anyway, do the same thing with the lower hemisphere map $l : C(S^1) \to C[0, 1]$ to get another free $C[0, 1]$-module. Let's call these two things $\Gamma \otimes_u C[0, 1]$ and $\Gamma \otimes_l C[0, 1]$.
There should be a way to recover $\Gamma$ from these two free modules.
This diagram commutes, where the maps $C[0, 1] \to C(\{0, 1\})$ are just restriction to the boundary
$C(S^1)$ is a pullback in CRing, basically.
I have two free $C[0, 1]$-modules over the top-right and bottom-left corners of this diagram, obtained from extending scalars of $\Gamma$ over the top-left corner.
Ok, extend scalars of this free $C[0, 1]$-module to get a free module over $C(\{0, 1\})$. Then we have a corresponding pullback diagram of modules over this pullback diagram, and the whole cube commutes.
Some argument here should tell you $\Gamma$ is uniquely determined by this free module over $C(\{0, 1\})$.
But $C(\{0, 1\})$ is $\Bbb R\oplus \Bbb R$. There are exactly two free modules over rank $k$ over $\Bbb R\oplus \Bbb R$ upto isomorphism, yeah?
One is $\Bbb R^k \oplus \Bbb R^k$ where both copies of $\Bbb R$ in $\Bbb R \oplus \Bbb R$ act by $c \cdot v = cv$ and another is when one acts by $c \cdot v = cv$ and the other acts by $c \cdot v = -cv$. These are not isomorphic as $\Bbb R \oplus \Bbb R$-modules.
Something like that.
@AlessandroCodenotti Modulo details, this tells you there are two f.g. projective modules of rank $k$ over $C(S^1)$ :D
@Alessandro I didn't give you a correct description of the gluing. There is only one module upto isomorphism over $\Bbb R \oplus \Bbb R$ (modules over $A \times B$ are really nothing but pairs $(M, N)$ where $M$ is an $A$-module and $N$ is a $B$-module). That wasn't the right thing to say.
The point I think is two different free rank $k$ modules on $C[0, 1]$ extend by scalars to two different free rank $k$ modules on $\Bbb R \oplus \Bbb R$; they are isomorphic but there's a $\text{GL}_k(\Bbb R) \times \text{GL}_k(\Bbb R)$'s worth of isomorphisms.
If this pair of elements of $\text{GL}_k(\Bbb R)$ have determinants of the same sign, $\Gamma$ if free. Otherwise it's not.
So consider for example $y^2 = x^3 - 54$. You write that as $x^3 = (y+3\sqrt{-6})(y-3\sqrt{-6})$. Well, if we have a prime ideal $\mathfrak{p}$ which divides both $(y+3\sqrt{-6})$ and $(y-3\sqrt{-6})$, then it divides their sum, which contains (and thus divides) $(6\sqrt{-6}) = (\sqrt{-6})^3$. By unique factorization in a Dedekind domain, $\mathfrak{p} = \sqrt{-6}$
In particular, $\mathfrak{p}$ is self-conjugate under the Galois action, so the it divides $(y+3\sqrt{-6})$ with the same order as it does $(y-3\sqrt{-6})$. In particular, that must be a multiple of $3$
But if a prime ideal only divides one of the two, ofc the order is also a multiple of $3$, so $(y+3\sqrt{-6}) = I^3$ for some ideal $I$. But here's the kicker: the class number of $\mathbb{Q}(\sqrt{-6})$ is 2, so the square of any ideal is trivial in the class group. If the cube is also, then the ideal itself is trivial, aka principal
@Daminark Fractional ideals in a Dedekind domain are 1-dimensional projective modules, I think, so the ideal class group is actually the 0-th K-group of the domain
I'll say that even though mornings aren't my friend so I've had trouble going to lectures, the psets alone for my NT class have gotten me excited about NT again. For some time it was like yeah I had fun with it and gotta learn it properly
Also recently I've finally been convinced that AG is cool, day before yesterday Venkatesh gave some lectures about stable homology of symplectic groups over $\mathbb{Z}$. Didn't really understand it too well but it has finally converted me
Actually funny thing along those lines, at one university I was considering there was a guy who, among other things, works on "topological modular forms"
Lol, yeah that's pretty true. Though I guess there seem to be folk who still do something like topology. I've heard Hopkins, also one guy I'm interested in talking to a bit does algebraic topology of varieties
That would be very wrong. Those steps are pulling some $n$'s out of the limit where they stop making sense. Not to mention the number of summands goes to infinity, so that interchange would be non-trivial, if valid. The justification in the answer is given in the "Added:" section.
Well, pick an $x_i$ such that there is a monomial term in $f$ containing $x_i$ with non-zero coefficient. Then keep all the other $x_j,j\neq i$ constant (and non-zero). This restriction yields a non-constant polynomial in $\mathbb{C}[x]$ and you can apply FTA.
Why must there be such a monomial term? What do you mean by "keep all other $x_j$" constant? Do you mean, evaluate them at some nonzero complex number?
If all monomials had a zero coefficient, $f$ would be constant. I'm not sure if evaluation is the correct term in this case, but you can think of it as a kind of partial evaluation, yes.
Oh, slight oversight, I of course mean the non-constant monomials*
Different question: Let $g\colon\mathbb{R}\rightarrow\mathbb{R}$ be continuous and $\delta>0$. Why is there a sequence $(x_n)_{n\in\mathbb{Z}}$ s.t. $x_n\rightarrow\pm\infty$ when $n\rightarrow\pm\infty$ resp. and $|g(x)-g(x_n)|\le\delta$ for all $x_n\le x\le x_{n+1}$ and $n\in\mathbb{N}$ (this essentially says that $g$ can be approximated by step functions with up to countably many steps). I tried using continuity locally and then use that $\mathbb{R}$ is Lindelöf, but it didn't work out
@RyanUnger iunno no one gave me specifics bc i didn’t ask much about ngc but ppl generally told me they’re bad in the way that living in a dorm as an adult is bad
as far as I think, we can modify the Lindelof property by changing the countable sub cover of an open covering into a compact-covering (idk whether the term is valid).
I just had the same idea as well after Leaky's tip. I was considering compact intervals earlier, but even though I considered a limiting construction on increasing compact sets , the much more elegant thought of just splitting up the real line took a second attempt. Thanks :)
Let $f : \mathbb R \to \mathbb R$ be differentiable at $x = a$. Evaluate:
$$ \lim_{n\to \infty}\large[{f(a +\frac{1}{n^2})}+{f(a +\frac{2}{n^2})}+...+{f(a +\frac{n}{n^2})}-nf(a)] $$
Answer: $\ $ $\ \frac{1}{2}f'(a)$
My attempt:
$$ \lim_{n\to \infty}\large[{f(a +\frac{1}{n^2})}-f(a)+{f(a...
