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May
31
comment Maximal order of elements of $\textrm{SL}(n, \mathbb{Z})$
@MarkBell This is the same idea as G. Lewitt uses by the way to obtain his $G$ function. I see though that I made a mistake as there shouldn't be any $\phi$'s in the lowest common multiple part.
May
31
comment Maximal order of elements of $\textrm{SL}(n, \mathbb{Z})$
@MarkBell I think so, because $\phi(6)=2$, so $C_6\in\mathrm{SL}(2,\Bbb Z)$ will do the job. The only thing I'm not sure about is wheter all the companion matrices are in SL rather than in GL...
May
31
revised Maximal order of elements of $\textrm{SL}(n, \mathbb{Z})$
added 93 characters in body
May
31
answered Maximal order of elements of $\textrm{SL}(n, \mathbb{Z})$
May
31
comment Maximal order of elements of $\textrm{SL}(n, \mathbb{Z})$
Regarding the finite order elements of $\mathrm{SL}(2,\Bbb Z)$, there are elementary proofs using only the characteristic polynomial.
May
28
comment Show that $\mathbb{R}^m$ is not homeomorphic to $\mathbb{R}^n$
If two (locally compact Hausdorff) spaces are homeomorphic, their one point compactifications are too (via an extension of the original homeomorphism). The one point compactification of $R^n$ ($n\geq 1$) is $S^n$.
May
28
comment Show that $\mathbb{R}^m$ is not homeomorphic to $\mathbb{R}^n$
Have you heard about one point compactification?
May
28
comment Does the product functor preserve quotient maps?
This math.stackexchange.com/questions/833042/… is related, and talks a little about the categorical part ($-\times Y$ having an adjoint)
May
27
comment Redundance of the Smoothness of the Inversion Map in the Definiton of a Lie Group.
Thanks : ) ${}$
May
27
comment Redundance of the Smoothness of the Inversion Map in the Definiton of a Lie Group.
You don't need smoothness of the inverse map, you only need smoothness of the maps "left multiplication by some fixed $g$" and "right multiplication by some fixed $h$" (of course you'll take $h=g^{-1}$), which follow from the assumption that multiplication is a smooth map.
May
27
revised Redundance of the Smoothness of the Inversion Map in the Definiton of a Lie Group.
added 290 characters in body
May
27
answered Redundance of the Smoothness of the Inversion Map in the Definiton of a Lie Group.
May
27
comment There is no smooth submersion from $S^2$ to $S^1$.
$\pi_1(S^2)=0$.
May
27
comment There is no smooth submersion from $S^2$ to $S^1$.
Another approach to showing that there is no such submersion $f:S^2\to S^1$ would be to endow $S^2$ with its standard riemannian metric, and to pull the vector field $\frac\partial{\partial t}$ back to $S^2$ along the submersion, by selecting the only tangent vector at $m\in S^2$ that projects onto $\frac\partial{\partial t}$ while being orthogonal to $ker(d_mf)$. This contradicts the hairy ball theorem.
May
27
comment There is no smooth submersion from $S^2$ to $S^1$.
The immediate proof that comes to mind lifts such a map to the real numbers, and shows that submersivity is impossible in the presence of a maximum. It's short, did you have this in mind or something else?
May
27
comment Groups with finite automorphism groups.
@ClémentGuérin Yes, it's infinite.
May
25
comment How to prove $\sinh^{-1}x=\ln\left(x+\sqrt{x^2+1}\right)$
Fix some real number $x\in\Bbb R$, you are looking for a formula for $y\in\Bbb R$ such that $\sinh(y)=x$. If you write $Y=e^y$, then you are trying to solve the equation $$\frac{Y-\frac1Y}2=x$$ i.e. $$Y^2-2Yx-1=0$$ and you can apply the quadratic formula to get the (right) $Y$, thus $y$.
May
23
comment Are closed simple curves with this property necessarily circles?
Many curves have this property, for instance ellpises, or squares, or regular polygons drawn from the $(2n)$-th roots of unity...
May
23
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May
19
comment Which other “exotic” permutation-related things exist?
I apologize, this is very basic, and I don't doubt you are aware of it, but it relates to $S_4$. There are 3 ways to cut a 4 element set in two equal halves, hence there exists a (clearly) surjective morphism $S_4\to S_3$.