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Aug
24
comment Intuition behind the construction of Young Symmetrizer
@anon sorry I have been occupied.
Aug
23
comment Intuition behind the construction of Young Symmetrizer
hep.caltech.edu/~fcp/math/groupTheory/young.pdf
Aug
23
answered Intuition behind the construction of Young Symmetrizer
Aug
20
revised Prove that there exists infinitely many positive integers $n$ such that $\sin^2{(na)}+\sin^2{(nb)}\le \frac{2\pi^2}{n}$
added 195 characters in body
Aug
20
answered Prove that there exists infinitely many positive integers $n$ such that $\sin^2{(na)}+\sin^2{(nb)}\le \frac{2\pi^2}{n}$
Aug
20
awarded  Popular Question
Aug
19
accepted Prove or disprove: $ \sum_{b \vee d = x} \tau(b) \tau(d) = \tau(x)^3$
Aug
19
revised Prove or disprove: $ \sum_{b \vee d = x} \tau(b) \tau(d) = \tau(x)^3$
added 2 characters in body
Aug
19
revised Prove or disprove: $ \sum_{b \vee d = x} \tau(b) \tau(d) = \tau(x)^3$
added 43 characters in body
Aug
19
revised Prove or disprove: $ \sum_{b \vee d = x} \tau(b) \tau(d) = \tau(x)^3$
added 376 characters in body
Aug
19
asked Prove or disprove: $ \sum_{b \vee d = x} \tau(b) \tau(d) = \tau(x)^3$
Aug
16
comment Area of the lattice generated from $(n, n\sqrt{2} \mod 1)$
@LeeMosher I start out with the line $(t, t \sqrt{2}) \in \mathbb{R}^2$ and mod the y-coordinate by $1$, $(x,y) \mapsto (x, y \mod 1)$ so it's wrapping around a cylinder $\mathbb{R}\times S^1$. Additionally $t \in \mathbb{Z}$ so although I started with a line in the plane, it really looks like a 2D lattice on the cylinder.
Aug
16
asked Area of the lattice generated from $(n, n\sqrt{2} \mod 1)$
Aug
16
answered Proof of $\sum_{d|n} {\tau}^3(d)=\left(\sum_{d|n}{\tau}(d)\right)^2$ (not standard proof)
Aug
16
revised Moebius band not homeomorphic to Cylinder.
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Aug
16
comment Moebius band not homeomorphic to Cylinder.
@ThomasAndrews sure I can; I am taking the closure of a subset of the cylinder/Mobius band in the relative topology.
Aug
15
comment Proof of $\sum_{d|n} {\tau}^3(d)=\left(\sum_{d|n}{\tau}(d)\right)^2$ (not standard proof)
Also $\sum k^3 = \left( \sum k \right)^2$. Concidence?
Aug
15
revised Moebius band not homeomorphic to Cylinder.
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Aug
15
comment Moebius band not homeomorphic to Cylinder.
@PyRulez My constructions are fine. If the two spaces where homeomorphic, we could map one meridian circle to the other $S^1 \subset X \leftrightarrow S^1 \subset Y$. Since both spaces fiber over the circle, we can take a closed interval over each point in $S^1$. The result is a closed cylinder on the one hand and a closed Möbius band on the other.
Aug
15
revised Moebius band not homeomorphic to Cylinder.
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