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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.
Dec
18
comment Using quaternions to represent an affine transformation?
I know I read about some parts of the above on wikipedia and in Hatcher's book on Vector Bundles and K-theory (there are many other fantastic insights in that book). But I got much of my intuition from Bott and Tu's Differential Forms in Algebraic Topology. That one is a gem.