@Gato Quelle propertion d'élèves de Lycée savent redémontrer le th. de Pythagore ? Quelle proportion infime sait comment se partitionne les aires des carrés correspondants, et comprennent donc ainsi en profondeur le théorème ? :(
@TedShifrin Well I can always apologize if you took all of that badly but my global point of view hasn't changed. We's rather not talk about that now anyway or it's likely to end up like the other time
but that wouldn't prove that center is subgroup of centralizer however I know how to prove it its just some set theory inclusion but I was wondering can we get it directly?
ok that is what I don't understand @TedShifrin the center is yes trivial map since we have all kernel(G) is all of G but this proves that center is subgroup of G however not of $C_G(A)$. Maybe I am misunderstanding something
well yeah I guess so @TedShifrin because one we will have that kernel of the map is all of G and the other we will get a subset of G so the map itself has less in the kernel has more elements so that is why the center is subgroup of the centralizer let me know what do you think of this reasoning when you come also you @BalarkaSen
@copper.hat I don't think it was either. I'm not sure how it's relevant (or how you deciphered that vowel-less notation), but it certainly wasn't rude - merely a fact being stated.
@Vrouvrou Look, if the $L_{p^\ast}^1$ norms of $u_n$ were bounded, the $L_{p^\ast}^1$ norm of $u$ would be finite. So the $L_{p^\ast}^1$ norms of $u_n$ must be unbounded.
Monotone convergence says that since the $u_n$ increase to $u$, if $\|u_n\|_{L_{p^\ast}^1}$ increases to a finite value, $\|u\|_{L_{p^\ast}^1}$ will be equal to that finite value. Since $\|u\|_{L_{p^\ast}^1}=\infty$, we have that $\|u_n\|_{L_{p^\ast}^1}$ increases without bound.
@Vrouvrou NO... $\alpha\lt0$. Remember? That is what all of this is about.
@KarimMansour I'm not sure I follow your reasoning above. Is the question just why the center of a group $G$ is a subgroup of the centralizer of $A \subseteq G$?
so I am using $N_G(A)$ act on set G via conjugation and consider the kernel to get the center and for centralizer I consider S to be A some subset of G.
under same action
this will indeed get that they are subgroup of G but I am trying to reason as I reasoned above center is subgroup of centralizer
@Karim So $N_G(A)$ is acting on $A$ by conjugation? I'm seeing a little more, but I would still be pretty surprised if you could prove this without... well, the basic definition, basically.
yeah but now we have a map given by group action one map that has the kernel of the action as the whole group G and the other has the kernel of the action as something smaller that is subset of G so the other map must be more surjective right?
@Vrouvrou Given an $x\ne0$, by the Archimedian property of the reals, there must be an $n_x$ so that $|x|\ge\frac1{n_x}$. For any $n\ge n_x$, we have $u_n(x)=u(x)$. By the definition
Geometrically, why do line bundles have inverses with respect to the tensor product?
Here my thoughts on the problem so far, please excuse their scatteredness.
I know algebraically, it is just because they are locally modules generated by $1$ element. Basically, it is just the fact that if ...
I'm sorry I didn't say that it always reduces the distance by a scale that's bounded above by some constant less than 1 because it seemed like that would probably make the concept harder to understand!
Geometrically, why do line bundles have inverses with respect to the tensor product?
Here my thoughts on the problem so far, please excuse their scatteredness.
I know algebraically, it is just because they are locally modules generated by $1$ element. Basically, it is just the fact that if ...
