bio | website | |
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location | ||
age | ||
visits | member for | 4 years, 2 months |
seen | Sep 19 '13 at 17:36 | |
stats | profile views | 181 |
Jun 25 |
awarded | Citizen Patrol |
Jan 30 |
awarded | Yearling |
Jan 16 |
awarded | Nice Question |
Nov 9 |
answered | Algebraic analogue of the Moebius bundle over the circle |
Oct 22 |
comment |
Morphisms between representations
I don't know what is a "block-diagonal matrix representing a direct product of orthogonal groups", but there exists one fixed matrix Q which simultaneously block-diagonalizes all matrices of G. Sizes of the blocks are just dimensions of irreducible representations of G. In particular, if G=P then you are talking about a regular representation of $S_n$. |
Oct 13 |
comment |
can KL divergence go to 0, but E[log(p/q)^2] diverge in certain cases?
Second moment does not necessarily exist in the first place. So yes, you can find such p and q with arbitrary large A. For example take p to be geometric upto some large $N$ and then $1\over n^3$. $q$ is geometric. Thus as $N\to\infty$ $KL(p||q_\epsilon)\to0$, but the second moment is always $+\infty$ (if you don't like this, cut $p$ at $M\gg N$. |
Sep 20 |
comment |
estimate the error term in CLT
Also instead of Berry-Esseen it is better to apply non-uniform Berry-Esseen, i.e. something like (do not remember the constants exactly) $$ |F_m(x) - F(x)| \le {\mathrm{const}\over\sqrt{m}} {1\over 1+x^3} $$ |
Sep 7 |
accepted | Alternating forms as skew-symmetric tensors: some inconsistency? |
Sep 7 |
comment |
Alternating forms as skew-symmetric tensors: some inconsistency?
Ok, Andrew I get it: your stance is that $\mathrm{Alt}^k$ is the mother object. I might not agree (think of alien civilization: I am sure our definition of the $\otimes$ will agree in the tensor algebra $\otimes^*(X)$, and thus in its quotient $\Lambda^* X$, whereas in $\mathrm{Alt}^k$ they might have defined the product not via $\mathrm{det}$ but via ${1\over k!} \mathrm{det}$, who knows), but I agree all your explanations make complete sense from that viewpoint. |
Sep 7 |
comment |
Is every real vector bundle over the circle necessarily trivial?
For example, for (real) line bundles this is very easy: without loss of generality we may assume that local trivilialization charts are $(0,1)\times \mathbb{R}^1$ and that the transition functions are $\pm 1$. Then by going over the circle counterclockwise we may "fix" all transition functions to be 1 (by changing the local trivialization frame $e\to-e$ if need be), except possibly for the last one -- thus there are only two line bundles. |
Sep 2 |
comment |
Alternating forms as skew-symmetric tensors: some inconsistency?
To elaborate on 2: if you change the top-row map then it is natural to change the definition of $\wedge$ in $\mathrm{Alt}^* V$ to preserve the isomorphism of algebras. For this reason I said before that the choice of $(\Lambda^k V)^* \rightarrow \Lambda^k V^*$ implies the choice of $\wedge$ in $\mathrm{Alt}^* V$. BTW, I just realized that convention (**) also affects differentials (i.e. d(x dy) is now a different 2-form) and volumes of Riemannian manifolds (so that the volume of unit 2-sphere becomes 2\pi). |
Sep 2 |
comment |
Alternating forms as skew-symmetric tensors: some inconsistency?
Andrew, sorry for being equivocal. 1. By the map $(\Lambda^k V)^* \rightarrow \Lambda^k V^*$ I mean the inverse of your $\mathrm{det}$ map (which exist since I assume my $V$ to be finite dimensional). 2. You are right that algebra $\mathrm{Alt}^* V$ has a product, but the reason we denote it $\wedge$ is because the map $\mathrm{Alt}^* V$ in the top-row of my diagram is an isomorphism of graded algebras. 3. I am not sure what you mean $\mathrm{Sk}_{V^*}$ and $p^*_V$ go in the opposite directions (and are the inverses of each other). Again, thanks for clarifying discussion! |
Sep 1 |
revised |
Alternating forms as skew-symmetric tensors: some inconsistency?
deleted 21 characters in body |
Sep 1 |
comment |
Alternating forms as skew-symmetric tensors: some inconsistency?
Georges, thanks! BTW, Birkhoff-MacLane in Section XVI.10 where they discuss $\mathrm{Sk}$ never use the fact that $\mathrm{char} F=0$ and all their statements hold true without ${1\over k!}$ and under the assumption $\mathrm{char} F \neq 2$ (in particular, they never state that $\mathrm{Sk}$ is the right inverse of the projection). So I think they also hiddenly agree with Schwartz! |
Sep 1 |
revised |
Alternating forms as skew-symmetric tensors: some inconsistency?
added 1428 characters in body |
Sep 1 |
comment |
Alternating forms as skew-symmetric tensors: some inconsistency?
Andrew, thanks for your very elaborate answer! To summarize: you show that if I insist on keeping the definition of Sk then the top-right horizontal map should be changed to ${1\over k!} \mathrm{det}$ instead of the (more usual?) $\mathrm{det}$. Similarly, George shows I should modify Sk to remove ${1 \over k!}$ and keep the top-right arrow. I have edited the question to explain both approaches. |
Sep 1 |
revised |
Alternating forms as skew-symmetric tensors: some inconsistency?
added 19 characters in body; edited body |
Sep 1 |
asked | Alternating forms as skew-symmetric tensors: some inconsistency? |
Jul 2 |
comment |
Demystifying complex numbers
Your example has a hidden assumption that a student actually admits the importance of calculating F.S. of $\ln\left|\sin{x\over 2}\right|$, which I find dubious. The examples with an oscillator's ODE is more convincing, IMO. |
May 15 |
comment |
Efficiently sampling points uniformly from the surface of an n-sphere
BTW, normal is the only distribution that upon taking $n$-fold product induces a rotationaly invariant distribution. So in some sense, this is the minimal method (i.e. you only need to build a single machine: one for outputting normals). |