@Thorgott There are various notions of the dimension of a topological space which don't require too much in the way of setup which distinguish R^n from R^{infinite set}. You can also compute using local homology that they are not locally homeomorphic.
Note that if V is a vector space, (V + R) - 0 has the homotopy type of the suspension of V - 0. Since an infinite dimensional vector space has V + R ~ V, it follows that if V is an infinite dimensional vector space the local homology groups are zero.
For every $n$, there is a point $x$ and an open $U$ containing $x$ such that for all open $V$ containing $x$ and whose closure is contained in $U$, the boundary has inductive dimension $n$?
For each of the following $R$-modules $M$ determine whether they are free. If free find a basis and rank: $\Bbb{Z}_{10}$-module $(2,1,2) \Bbb{Z}_{10} + (4,1,0) \Bbb{Z}_{10} + (0,0,1) \Bbb{Z}_{10}$...Question: Should I interpret those tuples as living in $\Bbb{Z}_{10}^3$?
Learning generating functions but occasionally come up with two expressions for the same sequence which aren't equivalent. Would appreciate help detangling it!
Eg. the sequence $1,1,1,2,3,4,5\dots$ seems like it could be either $\frac{x^3}{1-x}+\frac{1}{1-x}$ or $1+x+\frac{x^2}{1-x}$.
Maybe I've been doing this stuff for too long tonight, but let's consider the following... let $X_1, \dots, X_n$ be numbers such that $0 < X_i \leq 1$, and assume $X = \sum_{i=1}^{n}X_i$ satisfies $0 < X \leq 1$ as well. How do I know that $X_j \leq \dfrac{1}{n}$ for some $j$?
@MikeMiller is there a name for what you get by identifying the hemispheres of a sphere, but with a twist? Like say you had an "L" on the north pole, and you drag it to the south pole, then take its mirror image. There is a glide reflection that preserves this pattern, so I want to know if its orbifold (quotient under the action of this single symemtry) has a name. It looks like a cross-cap sort of thing, but "flattened" out?
or rather, I think I'm going to be completely überfordert in this seminar and I'm trying to get as far ahead as I can before the semester starts so I have some buffer space in case I don't understand shit
Ahhh yes the "i'll try to get some buffer space" tactic where you get hardly any buffer space but in exchange you have a semester that starts way earlier
I do that almost every semester ans never feel like I've bought much time but as I do it every time, I dont know how hard the semester would be if I didnt
the proof that smooth irreducible rep of G has countable dimension?
for it to work you need to assume that $G/K$ is countable for any compact open subgroup $K$, which holds for all the groups that are relevant in the book
and then $V$ is spanned by $\lbrace \pi(g)v : g \in G/K\rbrace$ lol
let's say I want to make a $C^k$ function that is not $C^{k+1}$, the idea is to integrate a function that's not $C^k$ but is $C^{k-1},...,C^1$, does anyone have any idea on how to do this?
theorem 2.34 the problem is first he write very sloppy $V_q$ and $W_q$ are neighborhood of p and q but I think why book means is $V_q$ is neighborhood of $p$ and vice versa now problem with $1/2d(p,q)$ so does it means this neighborhood V_q and W_q be less than this $0.5d(p,q)$? I find it notation abuse if my first statement is true they use q in both V_q and W_q Now first Why V is defined that way? What is purpose of intersection ? and why V and W intersection is empty? how does V being subset of K^c which means, not K is is open set?
I have tried drawing these thing but I can't grasp the theorem
For each of the following $R$-modules $M$ determine whether they are free. If free find a basis and rank: $\Bbb{Z}_{10}$-module $(2,1,2) \Bbb{Z}_{10} + (4,1,0) \Bbb{Z}_{10} + (0,0,1) \Bbb{Z}_{10}$...Question: Since $\Bbb{Z}_{10}$ is a PID, and the module generated by those three vectors is a submodule of the free module $\Bbb{Z}_{10}^3$, that submodule should also be free, right?
@anakhro I don't understand the description. Can you give me a formula?
It sounds like this is supposed to be an involution, but every involution on the 2-sphere is equivalent to (x,y,z) mapsto (pm x, pm y, pm z) for some choices of plus or minus.
When they're all + the quotient is the identity. When 1 is - the quotient is a disc (the hemisphere). When two are - it's the suspension of RP^1, so it looks sort of like a football in my mind. And when all three are - you get RP^2.
pick $V$ and $U$ as $\mathbb{F}$-vector spaces. If $\{v_i\}_{i\in I}$ and $\{u_i\}_{i \in I}$ are bases, then choose the linear map that gets $v_i \mapsto u_i$ and extend linearly. this is a bijective homomorphism, i.e., isomorphism
Do $(4,1,0)$ and $(0,0,1)$ span $(2,1,2)$ in $\Bbb{Z}_{10}^3$, where the coefficients are coming from $\Bbb{Z}_{10}$?
I don't think they do, and here is my reasoning (but correct me if I am wrong): If there were $n_1,n_2 \in \Bbb{Z}_{10}$ such that $(2,1,2) = n_1 (4,1,0) + n_2 (0,0,1)$, then we'd have $n_1 = 1$, so $4n_1 = 4$ which is not equal to 2 in $\Bbb{Z}_{10}$.
I am trying to find a basis for the $\Bbb{Z}_{10}$-module $(2,1,2) \Bbb{Z}_{10} + (4,1,0) \Bbb{Z}_{10} + (0,0,1) \Bbb{Z}_{10}$. I know that $(4,1,0)$ and $(0,0,1)$ are linearly independent over $\Bbb{Z}_{10}$, but I don't think they span the module for the reason I gave above. I could use help finding a basis.
I'm trying to show that if $F$ is a free module, then $Hom_{R}(F,F)$ is also a free module. If $F$ is free, then it has a basis $\{m_i \mid i \in \}$. I was thinking that the functions $f_i : F \to F$ defined by $f_i(m_j) = m_j$ if $i = i$ and $f_i(m_j) = 0$ if $i \neq j$ would form a basis for $Hom_{R}(F,F)$, but I think this forms a basis only in the case that $F$ is a free module of finite rank.
But Thorgott and some other person said it would be...
Hmm...I guess I need to rethink my proof of the fact that if $P$ is finitely generated projective module, then $Hom_{R}(P,P)$ is projective too.
I wanted to say that if $P$ is finitely generated projective, there exists a module $Q$ such that $P \oplus Q$ is free, then $Hom_{R}(P \oplus Q, P \oplus Q)$ is free, and then use this to argue that $Hom_{R}(P,P)$ is projective.
If $\{p_1,...,p_n\}$ generate $P$, I guess the most naive thing would be to take $Q$ to be the free module generated by the $p_i$ are something like that...But would $P \oplus Q$ be free...hmmm....
Okay. I think I have a proof (sorry it took so long...had to attend to other things). Let $P$ be generated by $\{p_1,...,p_n\}$, consider $R^n$ with standard basis $\{e_1,...,e_n\}$, and define $g : R^n \to P$ as the $R$-linear extension of $e_i \mapsto p_i$. Then the sequence $0 \rightarrow \ker g \rightarrow R^n \rightarrow P \rightarrow 0$, where the first map is inclusion and the second map is $g$, is exact and therefore $R^n \cong P \oplus \ker g$.
The vector space of $2\times 2$ traceless Hermitian matrices is isomorphic to $\Bbb R^3$ right? This relates to proving the double-cover relationship between $SU(2)$ and $SO(3)$
And we define the inner product on $V$ as $\frac{1}{2}\text{Tr}(X_1X_2)\forall X_1,X_2\in V$ which, if the above is true makes them isomorphic as inner product spaces
It is generated as a $\Bbb R$-vector space by $\begin{pmatrix}1&0\\0&-1\end{pmatrix}$, $\begin{pmatrix}0&i\\-i&0\end{pmatrix}$ and $\begin{pmatrix}0&1\\1&0\end{pmatrix}$
Is this an appropriate use of Cauchy-Schwarz or am I missing something? $$\sum\limits_{n=1}^\infty\big\vert\frac{x_n}{2^n}\big\vert^2\leq \sum\limits_{n=1}^\infty\big\vert\frac{1}{2^n}\big\vert^2\sum\limits_{n=1}^\infty\big\vert x_n\big\vert^2$$
So if $f:A\rightarrow B$ is a constant map then the subsets of $B$ that have the entire domain as a preimage are $f(A)$ and those that have the empty set are $B-f(A)$
Okay. I am trying to understand the hom functor. For example, I want to understand one of the form $Hom_{R}(_,R)$ and how it "acts" on sequences. If I have a morphism $f : A \to B$, how does this induce a map from $Hom_{R}(A,R) \to Hom_{R}(B,R)$?
on the other hand, $\operatorname{Hom}$ is covariant in the second variable, a morphism $f\colon A\rightarrow B$ induces a map $\operatorname{Hom}(R,A)\rightarrow\operatorname{Hom}(R,B)$, you can also guess what that one looks like
Ah, I see. Thanks. Okay, so I am trying to show that $Hom(,\Bbb{Z})$ is generally not exact. Is the following a counterexample? Consider the sequence $0 \rightarrow \Bbb{Z} \rightarrow \Bbb{Q} \rightarrow \Bbb{Q}/\Bbb{Z} \rightarrow 0$, where the first map is the inclusion map, and the second is the canonical projection. Applying the Hom functor we get the sequence $0 \rightarrow Hom(\Bbb{Q}/\Bbb{Z}) \rightarrow Hom( \Bbb{Q},\Bbb{Z}) \rightarrow Hom(\Bbb{Z},\Bbb{Z})$.
But this cannot be an exact sequence because the middle term is $0$ and the last term is $\Bbb{Z}$, and there cannot be any surjection from $\{0\} \to \Bbb{Z}$. Does this sound right?
the short gander I took at Kolar-Michor-Slovak suggested otherwise
$M\mapsto C(M,\mathbb{R})$ is a fully faithful contravariant embedding of the category of differentiable manifolds into the category of commutative $\mathbb{R}$-algebras
So you mean if we consider $g(\delta x)$ as a sequence of function that converges to $g(x)$, then as $g$ is continuous and have compact support, it's bounded so by DCT, $\int |g(\delta x)-g(x)|\to 0$ as $\delta\to 1$?
suppose $M$ is a locally Euclidean Hausdorff space, show that $M$ is second countable if and only if it is paracompact and has countably many components.
This is Problem 2-15 p.59 (or 1-5 p.30 in the new version) in: Introduction to smooth manifolds by John M. Lee.
In the hint he said if $M$ is...
@BalarkaSen ok, if I assume that then the noncompact riemann surfaces are stein manifolds from wikipedia page, so $H^1(M,O)$ for that is zero. But is it zero for compact riemann surfaces as well ?
How do people compute them ? I only know the exact sequence $0 \mapsto H^0(M,Z) \mapsto H^0(M, O) \mapsto H^0(M,O^*) \mapsto H^1(M,Z) \cdots$ and the fact that $H^n(M,Z)$ is same as $n$-th simplicial homology group
But I don't know any easy way to compute either of $H^q(M,O)$ or $H^q(M,O^*)$
OK, so I see $H^0(M,Z) = H^2(M,Z) = Z$, $H^0(M,O) = \mathbb{C}$ and $H^0(M,O^*) = \mathbb{C}^*$, and $H^1(M, Z) = Z^d$ for some $d$
I think you can do a sneak computation of $H^1(X, O)$ for a torus $X$. Run the exponential sequence $0 \to O^*(X) \to H^1(X, \Bbb Z) \to H^1(X, O) \to \text{Pic}(X) \to H^2(X, \Bbb Z) = \Bbb Z \to 0$
Also, where I'm going wrong here: Since $0 \mapsto O \mapsto M \mapsto PP \mapsto 0$ is exact, a data $\tau$ for Mittag Leffler is solvable iff $\delta \tau = 0$ in $H^1(M, O)$. For $M = \mathbb{CP}^{1}$, $H^1(M,O) = 0$ so this then implies any mittag leffler problem is solvable, which is very wrong
I would recommend looking at $H^1(M,\mathscr O)\otimes H^0(M,\Omega^1)\to H^1(M,\Omega^1) \cong H^{1,1}(M)$. Both the other two spaces have explicit generators.
