@MaximilianJanisch Hold on. I was mistaking. I've been to Lausanne (not Lucerne) :) I've also been to Zurich and Geneve. In general, Switzerland is very beautiful, unless you hate mountains :)
@MaximilianJanisch Not so with stochastic DEs. Problem is, number one, they're random, and number two, any particular instance of a solution is Weierstrass-like in that it's continuous and (almost) nowhere differentiable.
a revealing line from the paper I cited: "The weak-noise theory assumes that $\epsilon$ is a small parameter. The stochastic problem for Eq. (2) can be formulated as a functional integral which, in the limit of $\epsilon \ll 1$, admits a “semi-classical” saddle-point evaluation."
What I hunted for last time was an approximation to the analytic PDF, but that still requires running sample numerical solutions ever-how-many times and then PDFing that.
@BalarkaSen if $U$ is a nonempty open subset of $\Bbb A^n$ then it contains $D_f$ for some $f \in \Bbb C[X_1, \cdots, X_n]$ and we know that $D_f = V(tf-1)$
@LeakyNun If $x^n - 1$ is separable over any field $K$, the roots would generate a subgroup of $E^\times$ where $E$ is the splitting field, but that's cyclic, so the group is a cyclic group of order $n$. This means $\text{Gal}(E/K)$ injects into $\text{Aut}(\Bbb Z_n) \cong \Bbb Z_n^\times$
@LeakyNun You have a well-defined aut $\Bbb Q(\zeta) \to \Bbb Q(\zeta^m)$ if $(m, n) = 1$, right? Then extend to $\overline{\Bbb Q}$ by machinery, right?
@BalarkaSen let $f$ be the minpoly of $\zeta$, so $X^n-1 = fg$ for some $g \in \Bbb Z[X]$. we want to show that $f(\zeta^p) = 0$, so assume $g(\zeta^p) = 0$, so $\zeta$ is a root of $g(X^p)$, so $f(X) \mid g(X^p)$. now for the magic trick, it's the fact that $\overline{g(X^p)} = \overline{g(X)}^p \in \Bbb F_p[X]$, so $\overline{f} \mid \overline{g}^p$, so $X^n - 1 = \overline{f} \overline{g}$ is not separable, which is absurd
@MikeM: Knowest thou offhand. If we take the restriction of the tautological bundle on $\tilde G(2,4)$ to the $S^2$ of self-dual oriented $2$-planes, what's $\chi$?
@skill: It's not rehab. It's a desperate attempt not to degenerate further.
@LeakyNun I think $\text{Gal}(\Bbb F_p(\zeta_n)/\Bbb F_p)$ is going to be $p(\Bbb Z_n)^\times$, because the Galois group is generated by the Frobenius endomorphism, which takes everything to the subgroup generated by $p$.
Well, when $(n, p) = 1$
Otherwise $x^n - 1$ is not separable from the beginning
Well, one way is to realize that the Plücker equation $x_1x_6-x_2x_5+x_3x_4=0$ intersect the unit sphere is $S^2\times S^2$.
Alternatively, if you decompose $\omega\in\Lambda^2$ into self-dual + anti-self-dual, $\omega = \sigma+\tau$, the equation $\omega\wedge\omega = 0$ turns into $|\sigma|^2 - |\tau|^2 = 0$.
> The first use of “whoops-a-daisy” per se is around 1925, in a New Yorker cartoon. It’s an expression of surprise or dismay, specifically upon discovering one’s own error.
The unoriented Grassmannian is just $SO(4)/SO(2) \times SO(2)$, no? But once you oriented it that's tantamount to factoring out by the center in $SO(4)$
