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Jun
3
comment Looking For a Neat Proof of the Fact that the Grassmannian Manifold is Hausdorff
Yes. ${}{}{}{}$
Jun
3
answered Looking For a Neat Proof of the Fact that the Grassmannian Manifold is Hausdorff
Jun
3
comment Finding the Jordan Canonical form of a $6 \times 6$ matrix
$A-I$ is nilpotent, it will most certainly not have the same rank as its square! You'll find that $(A-I)^2$ has all its coefficients zero, except for its last line which equals $$(\;4\quad 0\quad 0\quad 0\quad 0\quad 0\;)$$
Jun
1
comment Prove that $G = \langle x,y\ |\ x^2=y^2 \rangle $ is torsion free.
@Kaster $G$ won't be an abelian group, it's basically the set of words in $x,x^{-1},y,y^{-1}$ where each occurrence of $x^2$ may be replaced by $y^2$, and vice versa. Multiplication is by juxtaposition of words.
Jun
1
comment Curves With Known Arc Length
You can calculate the arclength of the exponential function, it takes a few changes of variable, but it's doable.
May
31
revised Maximal order of elements of $\textrm{SL}(n, \mathbb{Z})$
added 312 characters in body
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.