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2h
comment simplicial homology definition
the simplex $\Delta^n$ lives in $\mathbb{R^{n+1}}$ so when he says $e_{\alpha}^n$ is an $n$-simplex of $X$ this is really the canonical map $\sigma_\alpha: \Delta^{n}\to X$
2h
comment A consequence of the inequality $\pi(x)+\pi(y)\ge\pi(x+y)$
that's what i'm saying. Your "proof" doesn't work by my comments above
18h
comment A consequence of the inequality $\pi(x)+\pi(y)\ge\pi(x+y)$
Also, note that it's possible that $\pi(p_n-p_m+1) = \pi(p_n - p_m) + 1,$ so then the left hand side becomes $\pi(p_m) + \pi(p_n-p_m),$ which doesn't tell you anything new.
18h
comment A consequence of the inequality $\pi(x)+\pi(y)\ge\pi(x+y)$
That's not strong enough to give strict inequality. Try, for example, $m=2$ so $p_m = 3$ (in particular then your inequality is not true from the hypothesis). Then the question becomes to determine when $\pi(p_n-3) > \pi(p_n) - 2.$ If $p_n - 2$ and $p_n$ are twin primes, then you're in trouble.
18h
comment Tangent bundle of manifold with no odd dimensional sub-bundles
@JoeS the orientable double cover of a manifold is connected only when the manifold is not orientable, otherwise you get two disjoint copies. I'll think about what you wrote more closely but that's what i meant
May
18
answered A consequence of the inequality $\pi(x)+\pi(y)\ge\pi(x+y)$
May
14
comment What are the limitations /shortcomings of Fourier Transform and Fourier Series?
Often times calculations using fourier coefficients are provably the "worst" possible (I can't find a reference for this at the moment). As Mikhail says, wavelets are often used instead due to the nice properties that come from being localized.
May
14
comment If $q$ is a prime, $gcd(x(x+2),q\#)=1$ and $q < x < q^2$, doesn't it follow that $x,x+2$ are twin primes?
This is true; are you asking for a proof?
May
13
comment If A,B are real symmetric positive definite matrices, then $B^{-1} AB$ is symmetric positive definite.
no, $(B^{-1}AB)^T = B^T A^T (B^{-1})^T = BAB^{-1}$, which is not equal to $B^{-1}AB$ unless you happen to have $B^{-1} = B$
May
13
answered Cute convergence problem. Proving convergence of sequence regarding reciprocals of least common multiple converges.
May
13
comment Quadrilateral $APBQ$.
The easiest way to see this is to take $X$ to $P, Q,$ and the midpoint of $PQ.$ This immediately tells you that the center of the circle is the midpoint of $AO,$ where $O$ is the center of the circle. Then complex numbers gives a rather straightforward solution. Alternatively one can compute the lengths directly using Stewart's theorem, though this can get a bit messy.
May
13
comment Quadrilateral $APBQ$.
see discussion at: artofproblemsolving.com/community/…
May
13
comment If $\sum (a_n)^2$ converges and $\sum (b_n)^2$ converges, does $\sum (a_n)(b_n)$ converge?
using cauchy will require knowing either 1) $l^2$ is an inner product space or 2) monotone convergence for real numbers, which seems slightly complicated for such a simple problem (not to say I don't approve!)
May
7
accepted $G_n(\mathbb{R}^{\infty})$ as a subspace $G_n(\mathbb{C}^{\infty})$
May
7
comment $G_n(\mathbb{R}^{\infty})$ as a subspace $G_n(\mathbb{C}^{\infty})$
After a google search, I understand it now. Thanks!
May
7
comment $G_n(\mathbb{R}^{\infty})$ as a subspace $G_n(\mathbb{C}^{\infty})$
Sorry, could you explain more explicitly how the tensor product works (I'm not very well-versed in this). Are we tensoring over $\mathbb{R}$?
May
7
comment $G_n(\mathbb{R}^{\infty})$ as a subspace $G_n(\mathbb{C}^{\infty})$
So for each $n$-plane (thought of as a vector space), we tensor it with $\mathbb{C}$ and this gives a complex $n$-plane?
May
7
asked $G_n(\mathbb{R}^{\infty})$ as a subspace $G_n(\mathbb{C}^{\infty})$
May
6
accepted Stiefel-Whitney Numbers of $\mathbb{R}P^2\times \mathbb{R}P^2$
May
6
comment Stiefel-Whitney Numbers of $\mathbb{R}P^2\times \mathbb{R}P^2$
I see; interestingly enough, this implies that $\mathbb{RP}^2\times \mathbb{RP}^2$ is cobordant to $\mathbb{CP}^2$ (as a real manifold), since $w(\mathbb{CP}^2) = 1 + a^2 + a^4$ (reducing the Chern class mod 2, essentially) gives the same Stiefel-Whitney numbers. (This popped up while I was showing that $\mathfrak{N}_4$ has order at least $4,$ and since I couldn't do $\mathbb{RP}^2\times\mathbb{RP}^2,$ I looked at $\mathbb{CP}^2$ instead.