@Ted I was thinking about this concrete example; say I have the holomorophic function $\sqrt{z}$ away from $0$. I take the germ $g$ at some point $p \neq 0$ and try to analytically continue it; that's done by taking the component $X$ of $|\mathscr{O}(\Bbb C - 0)|$ that contains the germ $g$ so $p : X \to \Bbb C - \{0\}$ is a covering map and the function $f : X \to \Bbb C$ is given by $f(\xi) = \xi(p(\xi))$.
$p$ should be the double cover of $\Bbb C - \{0\}$, right?
It's not immediately clear to me why
I guess the point is if I take a circular loop on the base from $p$ that winds once around the origin, then the germ returns back to the germ of $-\sqrt{z}$? So that should guarantee $p$ is a two-fold cover?
This makes me wonder what $f$ is after I identify $X$ with $\Bbb C - 0$. Should just be the coordinate function $z$, shouldn't it?
(I guess a better way to parse the two-fold thing is that monodromy of $p$ is $\Bbb Z/2\Bbb Z$-valued)
@EricSilva Well in this case where the measure is induced integrating a function, I can just divide the space into $A=f^{-1}([-\infty,0))$ and $B=f^{-1}([0,\infty])$, all subsets of the first have negative measure and all subsets of the second have positive measure
Right, @Balarka. You can make it explicit by putting in the $e^{i\theta}$ and seeing that you get $e^{i\theta/2}$ so that you get $e^{i\pi} = -1$ when you make one circuit around.
The European style is to write $AB$ for $A\cdot B$ and $A^2$ for $\|A\|^2$, but I won't allow it, @Semiclassic. It leads to serious errors. I would take off.
Just to clarify, $f$ is the analytic continuation of the germ of $\sqrt{z}$ to the cover $p : \Bbb C - 0 \to \Bbb C - 0, p(z) = z^2$. You're claiming $f$ is the squaring map?
@Ted: Right, OK, I am not confused anymore. All we have to check is that $\text{germ}_{\sqrt{p}}(f)$ pushforwards to the germ of $\sqrt{z}$ at $p$ by the (inverse of the) isomorphism of stalks $\mathcal{O}_{\Bbb C\setminus 0, p} \to \mathcal{O}_{\Bbb C\setminus 0, \sqrt{p}}$
@Semiclassical in my opinion, this is wrong and if it is too trivial to deduct points I would make a frowny face writing why it is wrong. If they had written more things, like wlog assume $A$ is real, then that would be okay
@Semiclassical I can understand that sentiment, I did physics tutorials for a while before I gave up and stuck to math ones. There I can make all the frowny faces I want without feeling guilty :)
define $\phi (gxg^{-1})=g C_{G}(x)$, where $g$ and $x$ belong in $G$ and $C_{G}(x)$ is the conjugacy class of $x$. I was able to show that this map is well defined and 1-1, but i am stuck at proving onto. what should i write exactly?
No, @Kasmir, I already told you that you should be fine reading groups. It will occasionally refer to stuff you haven't done, but don't worry about it.
@TedShifrin yes i noticed that , you said stuff like , where quotion in rings allways work, we need extra condtion on the subgroups , in case of groups
The way these problems are set up the coefficients are always chosen to be real in physics; You always write $\Psi=Ae^{ikx}+B e^{-ikx}$ and use boundary conditions to eliminate $B$
Q : The number of elements in the conjugacy class of $x$ in $G$ is $[G : C_{G}(x)]$. for this i defined the map $\phi$ as is. @TedShifrin is my thinking correct?
quoting from a footnote in the text: "Incidentally, normalization only fixes the modulus of $A$; the phase remains undetermined. However, as we shall see, the latter carries no physical significance anyways."
For instance, some people made the problem harder than intended and took $$\Psi(x,t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \phi(k)e^{i (k x-\omega t)}\,dk$$
now, if they did this and wrote out the current $$J(x,t)=\frac{i\hbar}{2m}\left(\Psi \frac{\partial \Psi^*}{\partial x}-\Psi^* \frac{\partial \Psi}{\partial x}\right)$$
in a correct if tedious way, then I'm giving full credit.
yep. as math it's nonsense, since $k$ is an integration variable. and as physics it's no better, since it leads to the final answer containing a $k$ which has no determinate meaning (the wavefunction in the above case is a superposition of waves of different $k$)
How can we show that the sign of the permutation $\pi: (1, \ldots , k, k+1, \ldots , k+\ell) \mapsto (k+1, \ldots , k+\ell, 1, \ldots , k)$ is $(-1)^{k\ell}$ ?
