no issue with it being defined on $C^{\infty}(\mathbb{R}^n)$
issue is with the $C^{\infty}_p$ stuff, how is $\phi : T_p(\mathbb{R}^n)\rightarrow D_p(\mathbb{R}^n)$ above a map between the two spaces? maps in $D_p$ are derivations $D: C^{\infty}_p\rightarrow \mathbb{R}$
Lee is not correct in defining derivations as a map from $C^\infty$. It happens to not be a problem because in the smooth world you have bump functions; in the algebraic world it is a map from the ring of germs.
(In the holomorphic setup you run into problems because not all holomorphic germs come from holomorphic functions)
$f$ there extends to a smooth function on the whole manifold $M$ because of bump functions. So you can equivalently define it to be equivalence classes of smooth functions on $M$, equivalence being "agreeing on some open nbhd of $p$"
Identity theorem. The simplest example of a bump function is something like a smooth function $f : \Bbb R \to \Bbb R$ such that $f \equiv 1$ on the open interval $(-1, 1)$ and $f \equiv 0$ outside the open interval $(-2, 2)$. You will learn to construct such things.
There is no holomorphic function $f : \Bbb C \to \Bbb C$ such that $f \equiv 1$ on the open disk $D_1(0)$ and $f \equiv 0$ outside the open disk $D_2(0)$.
If a holomorphic function on $\Bbb C$ is constant on a large enough set it is just constant.
Globally so
So there are more holomorphic functions on an open neighborhood of the origin in $\Bbb C$ than on all of $\Bbb C$, roughly speaking. Local is different from global.
Let $f : X \to Y$ be a continuous map between locally compact spaces, and $f_! : Sh/X \to Sh/Y$ be direct image with compact support. I will show that that this has no right adjoint. Suppose $f^! : Sh/Y \to Sh/X$ was right adjoint, so that the formula $\mathbf{Hom}(f_! \mathscr{F}, \mathscr{G}) \cong f_* \mathbf{Hom}(\mathscr{F}, f^! \mathscr{G})$ holds.
$i : U \hookrightarrow X$ be inclusion of an open subset. $(i^! \mathscr{G})(U) = \text{Hom}(\Bbb Z_U; i^! \mathscr{G})$ which is section over $U$ of the sheaf $i_* \mathbf{Hom}(\Bbb Z_U; i^!\mathscr{G})$ on $X$, which by above formula is isomorphic to $\mathbf{Hom}(i_! \Bbb Z_U; \mathscr{G})$
Sections over $U$ of that is just $\text{Hom}(\Bbb Z_U, i^* \mathscr{G}) = i^* \mathscr{G}(U)$, so $i^! = i^*$ for open inclusions
Let $f : X \to Y$ be a continuous map, $i : U \hookrightarrow X$ be an open inclusion and $g = fi$, $\mathscr{G}$ a sheaf on $U$. Then $(f^! \mathscr{G})(U) = \text{Hom}(\Bbb Z_U, i^* f^! \mathscr{G})$, and $i^* f^! = i^! f^! = g^!$, so we are computing section of $\mathbf{Hom}(\Bbb Z_U, g^! \mathscr{G})$ over $U$
So confusing
Ok, this is the same as global sections of $g_* \mathbf{Hom}(\Bbb Z_U, g^! \mathscr{G})$ over $Y$. By formula this sheaf is $\mathbf{Hom}(g_! \Bbb Z_U, \mathscr{G})$.
So $f^! \mathscr{G}$ is just the sheaf $U \mapsto \text{Hom}(f_! (\Bbb Z_U)^X, \mathscr{G})$ on $X$, where $(\Bbb Z_U)^X = i_! \Bbb Z_U$ is just the skyscraper sheaf on $U$ over $X$
The contradiction is that this is not a sheaf lmao
@loch Someone asked me what is the meaning of Grothendieck-Riemann-Roch theorem, a question I am obviously ill-equipped to answer. In my mind it says there's some map $ch : K_0(X)\otimes \Bbb Q \to \text{Chow}(X; \Bbb Q)$ related to Chern classes that gives an isomorphism - topologically that doesn't sound too far-fetched to me, because the complex $K$-theory spectrum is $BU \times \Bbb Z$, and $H^*(BU; \Bbb Q)$ is a polynomial algebra on the universal Chern classes.
is there any sense to this vague ass explanation or is it just nonsense?
for inclusions that's my understanding yeah. in general it's some complicated definition like $f_! \mathscr{F} = \{s \in \mathscr{F}(f^{-1}(U)) : s|\text{supp}(s) : \text{supp}(s) \hookrightarrow f^{-1}(U) \to U \; is\; proper\}$ but i am not sure what this means
im trying to learn that business rn haha. i think why i care is it gives something called Verdier duality, which is some duality involving these many functors -- for example, Poincare duality is a direct corollary of this somehow
it seems people just mess around with these six functors and get some identity which gives a fact about cohomology theories somehow
you define support of a section $s$ of $F(U)$ for a sheaf $F$ on $X$ to be the set of points $x \in U$ such that $s_x = 0$, i.e., there is some open nbhd of $x$ such that $s \equiv 0$ on that nbhd
This sits inside $\varinjlim\limits_{U \supset y} F(f^{-1}(U))$ and in the directed system of neighborhoods of $f^{-1}(y)$, the saturated ones form a cofinal subset
Well, if $f$ is closed
This I don't know
Anyway assume this so that it sits canonically inside $\varinjlim\limits_{V \supset f^{-1}(y)} F(V) = F(f^{-1}(y))$
So for closed $f$ at least we have a map $(f_! F)_y \to \Gamma_c(f^{-1}(y); F)$
So I confirmed with Muesnter that I accept the doctoral position they offered. I just got an email which is basically "that's great. Attached you will find the forms you need to fill for the contract". With 20 attachements
@Thorgott Ah it's related to the remainder theorem, I see. I recall learning about Taylor series, then I came across the remainder theorem, which actually sounded fucking cool by description. I looked at the proof, and I gave up... :')
Let $X$ be a topological space and $\mathbb{U}$ an open cover of $X$.
Suppose we are given a basis for each $U\in \mathbb{U}$, show that the union of all these bases is a basis for $X$.
My attempt:
Suppose each $U\in \mathbb{U}$ has a bases, $\mathbb{B}_U$. Claim: $\mathbb{B}=\bigcup_{U\in \mathbb{U}}\mathbb{B}_U$ is a basis for $X$. If $P\in \mathbb{B}$, then it is a basis member for some $U$, and thus open
in $U$. Since $U$ is open in $X$, $P$ is open in $X$. Let $V$ be open in $X$, I must show that it is the union of some collection of members in $\mathbb{B}$. Let $x\in V$. Since $\mathbb{U}$ coverse $X$, $x\in V\cap U$. Since $V\cap U$ is open in $U$ , there exists a $B\in \mathbb{B}_U$ such that that $x\in B\subseteq V\cap U \subseteq V$. $V$ is the union of all such possible $B$.
I don't know whether it's a right platform but anyway my question is - How do you guys motivate yourself while learning new stuff when you don't know the importance at the very start? For instance, I am learning about subspaces and I don't see why we need it. I asked my professor, he said that I will see the importance of it in the future like in eigenvectors or orthogonality but in this scenario what do you do when you don't know at the very start?
@abhas_RewCie Thank you Abhas, but I think I will undertake that gory proof once more 😂 Let's see if I can do it :3 But of course I'll hit you later if I can't understand it.
Also, Taylor series in the second chapter of a book? *chuckles* I'm in danger.
unfortunately my topology is too abysmal for me to decide if what you're saying makes sense :(
but anyway -- i think of GRR as just a relative version of Riemann-Roch (taking Y to be a point recovers Hirzebruch-Riemann-Roch). this 'makes sense', except one might wonder what does a 'relative version of Riemann-Roch' mean ---
so maybe it's worth staring at Hirzebruch Riemann-Roch for a second. tl;dr - it says that the holomorphic euler characteristic of a vector bundle can essentially be computed by integrating some chern classes of the vector bundle (and the todd class, which is data from …
small "dumb" question, I have that $u_\theta=r^{3/2}\dot{\theta}$ and $u_r=r^{1/2}\dot{r}$ moreover $-V(\theta)=\frac{u_r^2+u_\theta^2}{2}$, how can I simplify further this expression $-u_ru_\theta/2 - \frac{d \ V(\theta)}{d\theta}$?
obviously $u_r$ does not depend on $\theta$ so the derivative wrt $\theta$ is 0
but i'm getting confused for some reason for the $\frac12u_\theta^2$ part, I end up with $-\frac{1}{2} r^{3/2}\frac{d\theta}{dt}\frac{d}{d\theta}\left[\frac{d\theta}{dt}\right]$ which doesn't make sense (btw the dot from earlier is derivative wrt $t$)
it allows people to do topology without seeing the topology for example your various long exact sequences in topology actually come from morphisms you get from adjunctions of functors...
but more seriously somehow people figured out that sheaves are very powerful in topology - eg like balarka mentioned its good for singular spaces
and singular spaces really do show up quite often in geometry, esp. 'relatively' - eg fibers of a morphism between two smooth spaces can be singular. in topology there's leray spectral sequence where when you have a fibration you can relate the topology of the b…
so i guess step 1 of the story is - convince yourself that it's ok to do topology with sheaves (e.g. singular cohomology = sheaf cohomology of constant sheaf etc.)
and that sheaves come with a bunch of abstract nonsense so you can apply them to topology
@loch i learnt half a day ago that if $f : X \to Y$ is a map (continuous between LCH spaces in my setup), $F^\bullet$ is a complex of sheaves on $X$, then there's an isomorphism in hypercohomologies $H^\bullet(X; F^\bullet) \cong H^\bullet(Y; R^\bullet f_* F^\bullet)$. For $F^\bullet$ to be constant sheaf on $X$ concentrated in degree 0, this says singular cohomology of $X$ can be recovered from hypercohomology of the "fiber cohomology sheaves" on $Y$
if $f$ is fibration this gives me Leray-Serre SS I believe
so it seems believable these kind of things are useful in algeo where you have a sufficiently nice family of schemes (eg flat family?) instead of a bundle
$Rf_*$, $Rf_!$ etc are actually just relative versions of cohomology (w compact support respectively) which I think is a very algeo perspective of things which magically happens to be useful in topology
thats what their stalks keep track of, cohomology of fibers
Hello. Quick question: let's say that $A=\{A_1,A_2,\dots\}$ is an infinite-countable collection of sets. Then, is the set of all intersections of $A_i$ still countable?
It would be the set containing all the $A_i\cap A_j$.
If $W$ is the set of all intersections $A_i\cap A_j$, I think that the map $f:W\to \mathbb{Q}$ can be defined as $f(A_i\cap A_j) = i/j$. Doesn't this define a bijection? And $\mathbb{Q}$ is countable.
We know that the relation in $\Bbb{Z}$ given by $xRy\iff5\mid x-y$ is an equivalence relation, where $[x]=\{x+5k\mid x\in\{0,1,2,3,4]\}$, so we have 4 equivalence classes
But now I am wondering if this other set: $\{x+5k\}$ (which I think we assume $x\in\Bbb{Z}$) is also the quotient set
I am asking how do we justify that $\Bbb{Z}/R\neq\{\ldots,\overline{-2},\overline{-1},\overline{0},\overline{1},\overline{2},\overline{3},\overline{4},\overline{5},\overline{6},\overline{7},\ldots\}$
