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age 28
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Math PhD. student and functional programming enthusiast.


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.
Dec
18
comment Using quaternions to represent an affine transformation?
hm, not sure where I have the idea from. One source is definitely reading about K-theory. To classify rank $k$ vector bundles on $S^n$, you can consider maps from the equator (which is $S^{n-1}$) to $GL(k)$. This tells you how to glue the north and south portions of the bundle (each of which is a trivial bundle because disk is contractible). So you need to find the homotopy type of $GL(n)$ and it turns out it's $O(n)$ resp. $U(n)$ for reals resp. complexes (by Gramm-Schmidt). So in the end one is studying homotopy groups of classical groups.
Dec
18
comment Using quaternions to represent an affine transformation?
thank you for your kind words. Although I'm not as fluent as I'd like to be, I'm beginning to see the light here and there :) By the way, one piece of analysis was missing: we always stay in the positive-determinant component of $GL(2)$. That's why it's enough to work with $SO(2)$. Otherwise there would be a further ambiguity of adding a reflection.