Ted I finally referenced simultaneous diagonalizability in a pset problem lol
So take the standard representation of $GL(n,\mathbb{F})$ on $\mathbb{F}^n$
Here $\mathbb{F}$ is $\mathbb{R}$ or $\mathbb{C}$. This guy is irreducible, in fact it's irreducible if we restrict to $SL(n)$ or $O(n)$ ($U(n)$ for $\mathbb{C}$)
But we wanna show it's not irreducible, or even completely reducible, for the group of upper triangular matrices $B$
I basically split into cases, the $n=2$ case is "upper triangular matrices aren't all simultaneously diagonalizable"
Maybe I give myself too much credit. Basically you have the first standard basis vector $e_1$ as an invariant subspace. Let $W$ be a complementary subspace. Okay if $n=2$ that's a common eigenvector but nope. If $n > 2$ it contains two linearly independent vectors $v,w$.
Choose some linear combination $\lambda v + \mu w$ which kills the first slot. Then you choose a very specific matrix which basically amounts to adding $e_1$ to that guy.
I'm not quite sure what you're hinting at here. Beyond the standard fact that commuting operators preserve each other's eigenspaces, so they share a common eigenvector over C
it says "A simple example of a variety is a (complex) affine subspace, that corresponds to the vanishing of a finite collection of affine polynomials."
Take an example. Imagine you have five questions : each one has options $A,B,C,D$. Then for an example, we can say $1:A,2:B, 3:A,4:D, 5:C$. Your situation is different :it's like four questions, each having five options. You will have to play around a little bit to see the difference. Don't worry if you take some time, lot of people get stuck here.
Five questions, each having four choices. First one four choices, second one four choices, .... fifth one four choices. So 4 multiplied with itself 5 times.
@user15072279 Practice makes perfect. Take your time. One more thing : CURED has a different purpose, so if you post your query there people will not respond. Ideally, if you just post a query here there are enough people to help you out. So don't post these on CURED. But you can come to CURED and read the discussions and participate on that basis.
Let $F$ be a free group and $\alpha,\beta\in F$ be two primitive elements of $F$ then $\langle \alpha\rangle =\langle \beta\rangle$ or $\langle \alpha\rangle \cap\langle \beta\rangle=\varnothing$.
In a group $G$ an element $x$ is said to be primitive if $x$ can $\textbf{not}$ be written as $x=y^n$ for some $y\in G$ and $|n|>1$.
Any help?
Here, $\varnothing$ means intersection has the trivial element.
if $G$ is a group, and $N$ is the normal subgroup of $G$ generated by some subset $R$ of $G$, and $\{x_{\alpha} \} \subset G$, then if $\pi : G \rightarrow G / N$ is the quotient map, is it true that $(G/N) / K$ where $K $ is the normal subgroup in $G/N$ generated by $\{\pi(x_{\alpha}) \}$, is isomorphic to $G / <R \cup \{x_{\alpha} \}>$ where $<...>$ is the normal subgroup generated by ...?
Theorem: Let $\pi:E\rightarrow M$ be a vector bundle. Suppose for each $p\in M$ there exists a linear subspace $F_p\subseteq \pi^{-1}(p)$.
$F=\coprod_{p\in M}F_p$ is a subbundle of $E$ if and only if for each $p\in M$ there exists a neighborhood $U$ on which there are local sections $s_1,..........
Doing a measure theory course right now, and the proofs for additivity of the measures of disjoint sets are kind of throwing me for a loop. There are some kind of nuances I'm not picking up.
Like, the step where you go from $\mu(\bigcup_{n=1}^kA_n)=\sum_{n=1}^k\mu(A_n)$ to $\mu(\bigcup_{n=1}^\infty A_n)=\sum_{n=1}^\infty\mu(A_n)$
I want to just say "Yeah, $d(1_{\bigcup_{n=1}^kA_n},1_{\bigcup_{n=1}^\infty A_n})\leq\varepsilon$ and so $\vert\mu(\bigcup_{n=1}^kA_n)-\mu(\bigcup_{n=1}^\infty A_n)\vert\leq\varepsilon$ and our result immediately follows."
@Thorgott If you start with an outer measure and restrict it to measurable sets additivity is something you need to prove, maybe that's what's happening here
So I've been looking at more alg. geo. things and I was wondering about a definition I can't seem to find so consistently- Is an algebraic set $Z$ Zariski dense whenever $V(I(Z)) = Z$?
ok come to think of it I think this was a kind of early fact in Hartshorne- there was an example where nonempty open subsets of an irreducible space are irreducible and dense
@Astyx I am going to have to learn schemes probably so this should make sense soon
$\varphi(s)=\sum_{n\ge1} e^{-n^s}.$ I don't understand: "what are the function values for s<0 that you would like to extend? The series, as written, certainly diverges there, so you surely meant something different from what you wrote"
If you look at the affine plane, Spec k[x,y], then we've already discussed that you get more "types" of points (ie dim 0 points (x-a, y-b), dim 1 points (f) and the dim 0 point, the generic point, (0))
When you localize at a dim 0 point (x-a, y-b), you get the space of rational functions that have no pole at (a,b)
When you localize at (f), you get rational functions where the denominator is not a multiple of f. However the denominator can still cancel at points in the zero set of f
Right... so when you localize at a dim 1 "point" (f) you get... $C_f$, the coordinate ring of $f$, Which I know by Hartshorne is $\mathcal{O}_{(f)}$ or something
About 6 months ago I asked for an analytic continuation of $\varphi(s)=\sum_{n\ge1} e^{-n^s}.$
From the answers and feedback I got, we know that an analytic continuation exists for $0<s<1$ and there is no analytic continuation for $s>1.$
I'd like to understand whether or not there is an analytic...
The point is that functions on V(f) are Spec k[x,y]/(f) (assuming you're working with reduced schemes), and you don't want to "invert 0" so when localizing at (f) you don't have an inverse for f
And this leads to the fact that (f) is the generic point of Spec k[x,y]/(f) (trivially because (f) = (0) in that ring) but it brings some insight as to what quotienting does
About 6 months ago I asked for an analytic continuation of $\varphi(s)=\sum_{n\ge1} e^{-n^s}.$
From the answers and feedback I got, we know that an analytic continuation exists for $0<s<1$ and there is no analytic continuation for $s>1.$
What's the maximal analytic continuation of $\varphi(s)?$
...
quotienting is like making the "denominator" go to $0$, and localizing usually leaves you with a local thing, like you get $A_M$ and $M A_M$ is a (the) maximal ideal
