i have a question: if i roll a ball horizontally with and in some part of its track there is a well, so will the ball come back after falling inside (*collisions are elastic, no kind of resistant force*)
For example, the angular momentum tensor as seen in a different frame of reference with a Lorentz transformation matrix Lambda is given by Lambda M and not inner(Lambda, M) = Lambda^T g M
As a workaround while this request is pending, there exist several client-side workarounds that can be used to enable LaTeX rendering in chat, including:
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An altern...
Like the distinction between points and vectors (which becomes much more evident in curved geometry, but many people equate the 2 because they're working in Cartesian coordinates in R^n)
@lucas It is not the vote that is preventing deletion but the acceptance. Unfortunately we (the mods) are strong encouraged not to delete accepted answers.
Your best bet is probably to improve the post to the point that it is correct. Or to add a caveat to it pointing out the limitation of answer.
@lucas Deleted content isn't removed from the database, it's just not show to most users. The shading indicates that it is deleted. You can see it. Mods and some Stack Overflow staff (Community Mods) can see it. Users with at least 10k rep on the site can see it.
@lucas In my view it says "deleted by owner Apr 22 at 20:02"
@Danu I'm having a stupid problem. It's well-known that the derivatives of the Chern-Simons forms $\omega_{2p-1}$ give $\mathrm{tr}(F^p)$, but, uh...they don't. For instance, $\omega_3 = \mathrm{tr}(\mathrm{d}A\wedge A +\frac{2}{3}A\wedge A\wedge A)$ gives $\mathrm{d}\omega_3 = \mathrm{tr}(\mathrm{d}A\wedge\mathrm{d}A + 2\mathrm{d}A\wedge A \wedge A)$. Where's the $A\wedge A\wedge A\wedge A$ from $F\wedge F$? I must be missing something obvious.
It doesn't help that none of the dozen sources I sought out ever does this computation explicitly. That's why I think I must be missing something rather obvious
@DavidZ Depends on the scope and objective of the project no? I mean sure you'll code faster in Python, but if you'll be doing something computationally intense it will probably pay off
Well, in my experience relatively few projects are computationally intensive enough to make C++ beneficial.
and honestly, the amount of time I've spent tracking down memory errors in C++ is often enough to write a hundred Python programs to do the same thing :-P
I've been ignoring this book for ages because it looked so bad, but when I cracked it open I kept finding at the beginning of many proofs a short summary of what you're actually doing, very cool!
@Danu Sorry, the dual statement is easier, but I think I have it: The resolution $F_\bullet\to A\to 0$ is exact, so $F_\bullet\otimes B\to A\otimes B\to 0$ is still exact. So $\operatorname{im}(F_1\otimes B\to F_0\otimes B) = \ker(F_0\otimes B\to A\otimes B)$and you have $H_0 = F_0\otimes B/\ker(F_0\otimes B\to A\otimes B)$. Moreover, $F_0\otimes B\to A\otimes B$ is surjective, and some isomorphism theorem gives you this quotient is just $A\otimes B$.
This we do by considering that $F_1\otimes B\to F_0\otimes B\to A\otimes B$ is exact, so by exactness, $\mathrm{im}(F_1\otimes B\to F_0\otimes B) = \ker(F_0\otimes B\to A\otimes B)$
To finish my statement, the first isomorphism theorem says that $F_0\otimes B/\ker(\phi) = \mathrm{im}(\phi)$, and $F_0\otimes B\to A\otimes B$ is surjective, so we get that $H_0$ is the original functor.
As a workaround while this request is pending, there exist several client-side workarounds that can be used to enable LaTeX rendering in chat, including:
ChatJax, a set of bookmarklets by robjohn to enable dynamic MathJax support in chat. Commonly used in the Mathematics chat room.
An altern...
$$l=x_A+(x_A-k_1)+k_2+k_3+2\pi R$$
$$k_3=x_B-2R-(x_A-k_1)$$
$$l=x_A+x_B+constant$$
so, we have:
$$0=v_A+v_B$$
and$$a_A=-a_B$$
