Why is it that I have a book which studies triangulations in the homotopy chapter, and calculates fundamental groups using triangulations, yet no actual algebraic topology book I can find even mentions triangulations?
Yeah that's the weird thing, because, say, Hatcher studies simplicial complexes in the Homology chapter not the homotopy chapter, and his whole homotopy stuff is different to the homotopy triangulation stuff I'm reading :\
Hmm, maybe he just took homology calculations and put them in the homotopy chapter?
It is suggested that the next generation wireless technology will be highly directional since this increases the signal gain (MIMO) and decreases interference.
How will the receiver still be able to decode or observe many signals from many directions but still be a directional receiver?
Mathematica vs matlab (comparing the numerical calculation aspect alone (as otherwise it is not a fair comparison) what are the goods and bads of each?
@Danu It is a shame that if and when you would write a scientific book you will be so established that you can have some students doing all that dirty work for you
Take path $\sigma|_{I_i}$ in $B$. Join each $\sigma|_{I_i}$ by straightlines with $x$ (you're in something homeomorphic to the usual $n$-ball, so you can do this). Then slide through those straightlines to make it disjoint from $x$.
@Secret Sorry, I didn't note this message before. A grope is an infinite construction using disks. If you have a map from a 2-disk (circle with interior filled up, yes) to a 4-manifold, it can have a lot of singularities. Now around each singularity, pick loops based at the singular points and similarly attached 2-disks. Those in turn have singularities. Keep attaching them. You'll end up with an infinite tower of singular disks.
Sorry, I have no idea what you mean. If you have a small ball around $x$, you have two paths $\sigma_1, \sigma_2$ inside $B$ with endpoints in $\text{cl}\, B$. You can path homotope rel endpoint each of $\sigma_i$ just fine.
That is precisely what you're doing here. You are in $\Bbb R^n$, you have a path $\sigma$ with endpts $x_0, x_1$, passing through $x$. You want to homotope $\sigma$ to be disjoint from $x$.
I guess your statement is a bit too concise for me to parse. You have some $I_1$ and its endpoints, and you want to homotope those endpoints to $x$ with a straightline homotopy.
Writing down a formula for this would be a wee bit tedious, but I bet you can do it. Join $x$ with $\sigma|_{\text{cl} I_1}(s)$ for every $s$ by a straightline. Now slide $\sigma|_{\cl I_1}$ by sliding each point along the straightline in a continuous fashion, but fixing endpoints.
You can write down the formula for certain. But I'd best not write down a formula when the picture is clear. Pictorial reasoning/logic can be powerful, and if you're doing topology, you have to use that frequently.
I'm still a bit confused... If you join $x$ with $\sigma\big|_{\mathrm{cl} I_1}(s)$ for every $s$, then the straight line going to the point there this curve actually hits $x$ is trivial: It just stays at $x$.
Hence, we cannot move away from it along the straightline homotopy.
Is the map $D \rightarrow A$ where $A$ is any 4-manifold, and D is a 2-disk, injective (because if it is then I don't get why because of how $A$ is a lot larger than $D$ because $A$ has 4 dimensions? but D has only 2, or did I mistaken something here?
nah, curse it. $\sigma_{\cl I_i}$ is a path from $\sigma(a)$ and $\sigma(b)$, where $a, b$ are endpoints of cl $I_i$. take any other path $g$ from $\sigma(a)$ and $\sigma(b)$. they are going to be homotopic by straightline homotopy, as $B$ is homeomorphic to a convex subset of $\Bbb R^n$.
sorry for confusing you on this bit, I wasn't really paying attention + I am sick.
More precisely, we can complete the proof as follows: Given a loop $\ell:[0,1]\to S^n$ such that $\ell(0)=\ell(1)=x_0$, pick a point $x$ and a (small) open $n$-ball that does not contain $x_0$. By continuity, $\ell^{-1}(B)$ is a union of open intervals. Since $\ell^{-1}(x)\subset \mathbb R$ is closed and bounded, it is compact. Hence, it intersects only finitely many such open intervals, which we will call $I_1,I_2,\dots , I_n$.
Consider a corresponding little loop segment $\sigma_{|_{\bar I_i}}$ (note that its end points lie in $\bar B$). It is homotopic to any other curve inside $B$ with the same end points, since $B$ is homeomorphic to a ball in $\mathbb R^n$, which is simply connected.
Take $M$ remove a $D^4$ to make a hole $H$ glue the boundary of the disk $D$ to the boundary of the hole, meanwhile, the interior of $D$ is mapped to points on M that are not in $H$?
@BalarkaSen The physical picture is almost perfect its beauty and simplicity: If the quantity of something contained in a certain region changes in time, then it is flowing out/in of the boundaries. This is precisely reflected by that equation :)
So you mean in $D^2\rightarrow M$ once I map the set of points that is $D^2$ onto $M$, I basically end up having a subset of $M$ that is homeomorphic (is the the correct word) to a disk, and the points that corresponds to the circular boundary of $D^2$ before the mapping is what makes it singular in $M$?
A picture/some thought should convince one that the change in the contained charge is exactly equal to $$\frac{\mathrm d}{\mathrm d t}Q= -\int_{\partial V} \mathrm d^2 x \vec n \cdot \vec I$$
where $\vec n$ is the normal vector to $\partial V$
Hence, vanishinging of the integrant expresses the fact that electric charge cannot simply disappear: For it to leave a volume, it must flow out of the boundary.
(it does give some inconvenience sometimes when more complicated expressions are needed, or when an index has to be repeated more than twice in a single term---something that's not allowed in the ESC)
But for simple expressions such as this it's nice. It makes equations look more accessible :P
I am kind of confused by how exactly long term entropy works, despite I have read many articles here. Would anyone be kind enough to explain that to me?
If my memory serves, in a closed systems, there are many microstates (microscopic degrees of freedom that the time average of the state of the system can be with equal probability each (i.e. assuming ergodicity)) in the system.
By statistical arguments, if there is a larger number of microstates can contribute to the same macrostate (observed quantities of the system such as temperature, volume, energy, alignment of spins etc.), then over time, the system is very likely to be in states with maximum number of microstates, and hence entropy (ln of the number of microstates) is maximised
The above is the statistical mechanics view of entropy. Entropy can also be described in terms of the amount of uncertainty in the information in information theory, and has a simialr formula
However I am not very knowledgeable on infromation theory to say any further
Space filler http://mathworld.wolfram.com/images/eps-gif/Fractal1_1000.gif
https://terrytao.wordpress.com/2008/09/27/what-is-a-gauge/ so basically instad of concerning about one coordainte system and trying to pick which value is a reference level, we generalise it to "a field of coordinate systems?" and fi a gauge by choosing one particular "configuration" as a baseline for this field?
(P.S. Field because it is basically for each point in space you define a coordinate system, I guess...)
@0celo7 True, but I think you cannot talk about whether "Legion" was one of them or not. The Geth are immaterial, they are not bound to physical existence - I think it doesn't make sense to speak of them as if they had any continuity of identity
@0celo7 I think it's more a sign that the Geth are not beyond symbolism - they sent Legion to that collector/reaper thingy where you found him, and they gave him the same (kind of) gun as they used when they first rebelled against their creators.
Implying they see the reapers as masters that must be resisted in the same way as the Quarians
@0celo7 It goes downhill with the "introduction" of Kai Leng, imo. And the whole thing with Cerberus doesn't make a lick of sense, story-wise and logistic-wise (Where do they get all those troops and the equipment? What do they want to accomplish by attacking random worlds?). The reveal in the Asari temple also doesn't hold up to the slightest bit of scrutiny (but I won't talk about that now since you're not there yet, I guess)
@0celo7 Exactly, it makes no sense, especially after what TIM told you on Mars (and what he will tell you near the end) they should be doing some stuff with/about the reapers instead of terrorizing and slaughtering unimportant humans
@0celo7 I meant again story-wise - it's just another weirdly written part of what is the clusterfuck that is the ending, but I'll not spoil the details.
It is suggested that the next generation wireless technology will be highly directional since this increases the signal gain (MIMO) and decreases interference.
How will the receiver still be able to decode or observe many signals from many directions but still be a directional receiver?
I am having trouble finding references for the following identities:
Dirac Operator:
$$
\gamma_a e^{a}_{\mu} D^\mu \gamma_b e^{b}_{\nu} D^\nu =D^\mu D_\mu - \tfrac{1}{4}R
\tag{1}
$$
QED Operator:
$$
\gamma_a e^{a}_{\mu} (D^\mu -iA^\mu ) \gamma_b e^{b}_{\nu} (D^\nu-iA^\mu) = \\(D^\mu -iA^\mu )(D...
