for each $x$ (where does $x$ live? you haven't specified) there is a map $\varphi_x$ defined by the formula $\varphi_x(s) = \zeta(x)^{\frac{1}{\log(x)}}$.
presumably, $x > 0$ is real, and $s\in\mathbb{C}$?
basically, you have a family of functions indexed by $x$
If $X$ is a continuum (compact connected metric space), $p\in X$, and every two points of $X\setminus \{p\}$ are contained in a nowhere dense subcontinuum of $X$, then is there a nowhere dense subcontinuum $N$ with $p\in N$ and $|N|>1$?
hey folks does anyone know an online editor that has a similar feel and functionality to math.se's question bodys? I.E. you write your LaTeX into it and it auto-renders, making line breaks and scrolling nicely
also apologies I have terrible wifi, so my responses will be super choppy and delayed
For all continuous functions it is true that $f:[a,b] \mapsto \mathbb{R}$ (with $a < b$) is bounded from above.
The question is to use the opposite position of that statement as well as to use the Weierstrass Interval technique for a suitable sequence.
Could somebody please provide a nice expla...
Heyo. How does the linear isomorphism $\operatorname{Hom}(H\otimes V\otimes V^*, H)\cong \operatorname{End}(H)\otimes V\otimes V^*$ work? (everything is finite-dimensional vector spaces). From right to left, we can assign to an arbitrary element $f\otimes v\otimes \gamma$ the linear map which does: $h\otimes \tilde{v}\otimes\tilde{\gamma}\mapsto \gamma(\tilde{v})\tilde{\gamma}(v) f(h)$, that is clear. But what is the inverse? I can't figure it out.
Maybe this isn't true in general, so I will tell you the assumptions I withheld: $H$ is a bialgebra, $V$ is an $H$-module under the trivial action, and the Hom-spaces actually consist only of intertwiners.
By Hall's theorem, every p-regular element is contained in some complement of the p-Sylow and by Schur-Zassenhaus all complements are conjugate, so every p-regular element is conjugate to an element in the torus, thus the Brauer characters are determined by the values on the torus
So the composition factors are determined by the restriction to the torus
This approach won't work for the whole group, though
If the minimal polynomial of the matrix is irreducible, it's not conjugate to an element in the maximal torus
this argument works more generally for solvable groups with normal p-Sylow
can someone please help me with finding conditional probability functions?
we toss a regualr coin 50 times and define the following events: x - number of heads obtained totally(out of 50) y - number of heads from the first 20 tossings of the coin
how can i find the conditional probability function of x given that y=j {j=0,1,..,20}
or easier, how can i find the conditional probability function of y given x=i {for 1=0,1,..,50} ? i tried to use $P_{X|Y}(x \mid y) =\frac{P_{XY}{(x,y)}}{p_Y(y)}$
which from what i understand means $P_{X\mid Y}(x\mid y)=P(X≤50\mid Y=y)$
Sanity check: if $K$ is an ordered field and $K_1\supseteq K$ is a real closed field then the relative algebraic closure of $K$ in $K_1$ is real closed, right?
Let $K_o$ be the relative algebraic closure of $K$ in $K_1$. Let $f \in K_o[X]$ be of odd degree. Then, after a whole bunch of conjugation, we get a polynomial in $K[X]$ with odd degree, so it has a root in $K_o$. It is clear that $K_o$ is closed under square root: if $x \in K_o$ is positive, then its square root is algebraic over $K$, so is also in $K_o$
Now by definition $h(J) = E$ and $h(S_{\alpha}) = F$ From the above we can see that $F < E$ thus by definition of initial segment, $F$ is an initial segment of $E$
I think something interesting will happen if we consider $N$ to contain $p$ and a point $y$ that is not $p$, because it means that other point $y$ has to interact with the subcontinuum $Y$ somehow
I still cannot quite visualise what limit points in metric spaces in terms of open balls look like though
both $Y$ and $N$ are nowhere dense, thus it means their interior of their closure is empty. This means the interior of cl(Y), cl(N) cannot be made by union of open balls
so there are no open balls or their unions contained in cl(Y), cl(N)
must every closed ball in a metric space contain an open ball?
If $X$ is a continuum (compact connected metric space), $p\in X$, and every two points of $X\setminus \{p\}$ are contained in a nowhere dense subcontinuum of $X$, then is there a nowhere dense subcontinuum $N$ with $p\in N$ and $|N|>1$?
I am trying to approach it using first principle arguments (and some of the theorems I knew)
@GFauxPas Consider the discrete metric, i.e. $d(x,y)=1$ unless $x=y$ in which case $d(x,y)=0$. Then this induces the discrete topology, i.e. every subset is open and closed. Then if we consider the open ball of radius 1 around $x$, that's just $x$ itself, and we still have only $x$ itself if we take the closure. But the closed ball of radius $1$ is the entire space
(There's a similar, but more advanced example with p-adics)
@Mathei suppose that $K=\Bbb Q(\alpha)$ is a number field such that $\mathcal O_K=\Bbb Z[\alpha]$. I know that all the factorizations of prime ideals in $\mathcal O_K$ have the same shape, but is there an intuitive explanation of why I should expect this to be true?
...hmm I guess I am still not ready for this problem (my intuition on finding limit points for an arbitrary topology is still very poor), I will revisit this later in the future...
So $p\mathcal O_K$ has a factorisation in $\mathcal O_K$ with the same shape as the factorisation of the minimal polynomial of $\alpha$ in $\Bbb Z/p\Bbb Z$, right?
@AlessandroCodenotti note that this also provides intuition for the thing with "primes that ramify are those that divide the discriminant". Since primes p that ramify will be those such that the minimal polynomial has multiple roots, but then the reduction of the minimal polynomial mod p will be zero, which means that p divides the discriminant
Of course the proof in the general case is harder and will likely pass through differents, but that's how I think about that
@AlessandroCodenotti yeah if your extension is not monogenic, then calculations get messy
yeah, cyclotomic and quadratic fields are the two cases where computations are quite nice. Even with cubic fields it can happen that the extension is not monogenic and then factoring prime ideals is kinda annoying
I messed up some stuff in the statement above, I mean that the primes p that ramify are those such that minimal polynomial has multiple roots mod p and then the reduction of the minimal polynomial will have discriminant 0
I think you got it, but I wanted to clarify in case someone else reads that and is confused
I think your bit about non-transversality being a closed condition is better rephrased in terms of ranks of linear maps somehow (rank is upper or lower semicontinuous, one of the two)
Ah, much better formulation, lol. If $f : M \to N$ with $W \subset N$ a submanifold, then $M$ is stratified by open submanifolds consisting of points $p$ with stratification index being the codimension of $df_p(T_pM) + T_p W$ in $T_p N$
@BalarkaSen So it seems to me that there is a map $C^\infty(M, N) \to \text{Closed}(W)$, sending a map $f$ to $f(M) \cap W$. I think there is a natural topology on closed sets in general (Vietoris topology?? maybe is the right name) but Hausdorff metric certainly works
I have missed a step but hadn't seen how to work it in yet. You want more than just closed subsets of $W$
Hm, let me decide what you want
$\text{Closed}(W)$ is not enough data: you want to assign to each point $p \in V \in \text{Closed}(W)$ a quotient space $(T_p N)/(df_q(T_q M) + T_p W)$, and this should be something semicontinuous
So the right space to map to is not $\text{Closed}(W)$, but a space which has a forgetful map to $\text{Closed}(W)$, and it is this final thing that you want to stratify over
And if you set all of this up correctly it should be a matter of showing that the subspace for which that quotient space is always zero is open.
@anakhronizein Are there examples on tori? $T^4$ maybe?
Let's see. I don't follow why you want to stratify $C^\infty(M, N)$ along with the stratification of $M$ that comes from the semicontinuity of the rank of the quotient space you described
I've been skulking around MathOverflow for about a month, reading questions and answers and comments, and I guess it's about time I asked a question myself, so here is one has interested me for a long time.
Suppose $M$ is a compact even dimensional smooth manifold with two symplectic forms $\ome...
@BalarkaSen My thought is just that "the stratification of $M$ that comes from..." doesn't make sense, since to talk about that you need to assume $f(q) = p \in W$ for some fixed $p$.
For all continuous functions it is true that $f:[a,b] \mapsto \mathbb{R}$ (with $a < b$) is bounded from above.
The question is to use the opposite position of that statement as well as to use the Weierstrass Interval technique for a suitable sequence.
Could somebody please provide a nice expla...
