Marek
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 Dec 20 comment What is the importance of conformal vector fields on Riemannian manifolds? Flow is just a family of mappings, one for each $t$ in some interval. So it preserves something when all of those mappings do. I do not understand your second question. What do you mean by physical? I'll give you one example, perhaps it will help. In complex plane any holomorphic map is conformal. So a conformal flow here is just a smooth family of holomorphic maps, e.g. $z + te^z$. Dec 19 comment Inverse image of a PID is a PID @rschwieb: I had something bit more non-trivial in mind. For example, considering $S = {\bf C}$, which $R$ mapping onto it are PIDs? And similarly for other choices of $S$ or even whole classes of $S$. Dec 19 comment Multidimensional complex integral of a holomorphic function with no poles I deleted my answer because as Daniel pointed out (thank you), the proof by the Stokes' theorem theorem only works when $2n - 1 = n$, i.e. when $n=1$ because we'd like to integrate an $n$-form over the boundary of an $2n$-dimensional domain. The reason we need an $n$-form is that there are $n$ independent conditions of holomorphy in $n$ dimensions that we need to use. So it's impossible to even set up a reasonable integral when $n > 1$. Dec 19 comment Inverse image of a PID is a PID What lead you to conjecture this? There's hardly any map to PIDs that would have the property you mention. In fact, it would be fun to post another question: find some conditions on a ring so that your statement holds :) Dec 19 comment How can I get the two identities math.stackexchange.com/questions/81203/… here is proved your first identity. Dec 19 comment How can I get the two identities It might help you if you used consistent notation. What you write as $D$ and as $\nabla_0$ is the same thing, the covariant derivative. Then it's quite obvious that $\Delta_0$ is actually a trace of the Hessian. Dec 19 comment Find all positive integers $n$ s.t. $3^n + 5^n$ is divisible by $n^2 - 1$ @Tim: okay, I should officially go to sleep :D For some reason I thought the problem asked for equality now. Duh... Dec 19 comment Find all positive integers $n$ s.t. $3^n + 5^n$ is divisible by $n^2 - 1$ @Tim: $3^n + 5^n$ is always even whereas $n^2 - 1$ would be odd. Dec 19 comment Find all positive integers $n$ s.t. $3^n + 5^n$ is divisible by $n^2 - 1$ @Tim: even $n$ can't be a solution. Dec 19 comment Find all positive integers $n$ s.t. $3^n + 5^n$ is divisible by $n^2 - 1$ @Tim, Ian: thank you both, I forgot that $n$ must be prime, although it's obvious in retrospect -- for general $n$ one needs to use $\phi(n)$ to satisfy the equation at every prime dividing $n$. Perhaps there's some way to leverage that here although I do not see it yet. Dec 19 revised Find all positive integers $n$ s.t. $3^n + 5^n$ is divisible by $n^2 - 1$ added explanation Dec 19 answered Find all positive integers $n$ s.t. $3^n + 5^n$ is divisible by $n^2 - 1$ Dec 18 answered How many degrees of freedom does a metric have on a psuedo-Riemannian manifold? Dec 18 comment $\pi_n(SU(2)/Z_N)\simeq?$, $\pi_n(SO(3)/Z_N)\simeq?$, $\pi_n(U(1)/Z_N)\simeq?$ @Idear: did you read the answer? It's written explicitly in the second paragraph. Dec 18 comment $\pi_n(SU(2)/Z_N)\simeq?$, $\pi_n(SO(3)/Z_N)\simeq?$, $\pi_n(U(1)/Z_N)\simeq?$ Ah, all right, I didn't understand that you are leaving the general case to the OP :) Dec 18 comment $\pi_n(SU(2)/Z_N)\simeq?$, $\pi_n(SO(3)/Z_N)\simeq?$, $\pi_n(U(1)/Z_N)\simeq?$ Incidentally, there is a some information on $\pi_n(S^3)$ (which is isomorphic to $\pi_n(S^2)$) in this old question of mine: math.stackexchange.com/questions/50377/homotopy-groups-of-s2 Dec 18 comment $\pi_n(SU(2)/Z_N)\simeq?$, $\pi_n(SO(3)/Z_N)\simeq?$, $\pi_n(U(1)/Z_N)\simeq?$ Is there really an extension problem here? $SU(2)$ is simply connected, $U(1)$ has free abelian homotopy group and $SO(3)$ is a quotient of $SU(2)$. Where could the problem possibly arise? Dec 18 answered $\pi_n(SU(2)/Z_N)\simeq?$, $\pi_n(SO(3)/Z_N)\simeq?$, $\pi_n(U(1)/Z_N)\simeq?$ Dec 18 comment Fixed point of a shrinking map. Proof. If you check out the proof in the first link from the first comment, you'll see that the sought for fixed-point can be obtained as a limit of the sequence $f^n(x)$ for any $x \in X$. In other words, limit of $f^n(X)$ is a single point. So your $A$ is actually a one-point set. It seems very hard (if not impossible) to prove anything about $A$ (like your $A \subset f(A)$ without using that information. Dec 18 comment Fixed point of a shrinking map. Proof. In (5) you say 'by continuity'. But of what? Notice that on the LHS you have $f^n$ which depends on $n$, not a single function.