Maybe I'm misunderstanding what the complex tangent bundle means here? I thought it's just $TM$ with fiberwise (almost) complex structure, so the topology doesn't depend on whatever ACS you chose
So consider $M$ with two homotopic ACS's. Then consider the projectivization (as a complex! vector bundle) of $TM$, which of course depends on the ACS you had.
Guys, question about rings; my teacher wrote the following: $\mathbb Z[X]/(X)\cong\mathbb Z$, $\overline x\mapsto 0$. I don’t understand why he wrote $\overline x\mapsto 0$ (and I’m not sure if I copied it correctly). In any case, I don’t know which isomorphism to pick. I initially thought of $\sum_{i=1}^na_iX^i\mapsto\sum_{i=1}^na_i$, but this is of course not injective. So could someone help me?
It just follows from the fact that the sum is in $\langle X\rangle$, which by the definition of the quotient is mapped to $0$, skipping your lemma, so I didn't get sniped.
but if $\displaystyle \sum_{i=1}^n a_i X^i\in\langle X\rangle$, then isn't $\overline X$ just the entirety of $\mathbb Z[X]$? Like, we only have one equivalence class then?
@BalarkaSen by the definition of quotient, which is the set of cosets, in which the coset itself $\langle X\rangle$ is $0+\langle X\rangle$, is naturally sent to $0$, so I didn't get sniped
@BalarkaSen I'm reminding her that if she wants the sum to be able to represent everything in $\Bbb Z[X]$ then it has to start from $0$, so I didn't get sniped
@Balarka Too Western/normal for my taste. I mostly listen to Russian/Balkan music and meme music these days. So Remove Kebab is such a golden combo, I've been listening to it a lot:P
a friend of mine who is like the top at our class said he didn't want to go to Russia for an exchange programme, because he was afraid he would be intimated by the Russians :P (mathematically wise) I'm still not sure how serious he was
So, suppose $f:\Omega\to\Bbb R$ is a function, with $\Omega\subseteq\Bbb R^n$ open and connected (we can also take $\Omega=\Bbb R^n$ if it's easier) and that $\int_{B_r(x)}f(y)\mathrm{d}y=0$ for all balls $B_r(x)$ contained in $\Omega$
I can show that if $f$ is continuous then it must be the $0$ function
Without assuming continuity must it be $0$ almost everywhere?
I'd be very surprised if this turns out to be false, for an arbitrary measure space $(X,\mathcal{A},\mu)$ and a nonnegative function $f:X\to\bar{\Bbb R}$ we have $\int_X f\mathrm{d}\mu=0\implies f=0$ almost everywhere, and $\Bbb R^n$ is much nicer than the general case
Ah, apparently every set of positive measure contains a closed set of positive measure and one can reach a contradiction by assuming that $f$ is $>0$ on a set of positive measure and using that fact
Actually since the Lebesgue measure is Borel regular for every measurable set $E$ of finite measure and every $\varepsilon>0$ there is a closed set $F\subseteq E$ with $\mu(E\setminus F)<\varepsilon$
If $E_n$ is a decreasing sequence of open sets with $\bigcap E_n=E$, when is $\lim\limits_{n\to\infty}\int_{E_n}f\mathrm{d}L=\int_E f\mathrm{d}L$ ($L$ is the Lebesgue measure)
Consider the set $G$ of all functions $f: \mathbb{R}^{n} \to \mathbb{R}$ of $n$ variables such that $f(\overline{x})=c$, where $x \in \mathbb{R}^{n}$ is a fixed point and $c \in \mathbb{R}^{n}$ is a given number.
I need to do the following two things:
Show that the set $G$ is convex.
For whic...
Consider the set $G$ of all functions $f: \mathbb{R}^{n} \to \mathbb{R}$ of $n$ variables such that $f(\overline{x})=c$, where $x \in \mathbb{R}^{n}$ is a fixed point and $c \in \mathbb{R}^{n}$ is a given number.
I need to do the following two things:
Show that the set $G$ is convex.
For whic...
