Well, at the rate he cranks out answers, this is no surprise. He has already overtaken Didier and it won't be long that he catches up with joriki and Gerry.
user19161
@robjohn You can beat him by trying to publish 7 papers while he is doing that. :-)
Or do it, but keep in mind that you'll have to go through the process of $X$ quits $\Rightarrow$ "$X$ wasn't so bad" $\Rightarrow$ "let's star another message asking $X$ to come back" $\Rightarrow$ "good to have you back, $X$" $\Rightarrow \cdots$.
but seriously, I don't know what the numbers mean, what they represent, so in having no insight I'm left with basically some IQ-ish puzzle checks and pure luck.
When we write something like "the following two are equivalent," we are making a lemma or a proposition, not a definition. For that you just want to say "$Cf(x)$ is the maximum number of times $f$ is differentiable at $x$."
@RajeshD: Alright, I skipped the closure under + section but I think I understand the question now. I might read the given answer / edit the question for readability / tackle the question in the next week sometime.
@Fx I would say that yes. The usual convention is that repeated indices are to be summed over, so more explicitly, your $f$ seems to be defined as $$f(\vec{a}) = \sum_{k=1}^n \sum_{i=1}^n \sum_{j=1}^n S_{ijkk} a_i a_j$$ (where I assumed that all indices run from $1$ to $n$.)
(oh, now I see that your question was answered on physics.SE). Never mind.
@ZhenLin You're not the only one that is confused... There's a lot of silly stuff in the literature. The best definition (rather a sufficient condition) I'm aware of is in the Added bit of my answer here.
You need that at various points in the argument. One place is when you try to look for a replacement of the Horseshoe lemma (for the long exact sequence).
Basically the conditions guarantee that $\mathcal{A}$ is closed under all the constructions you need when verifying that you have a universal $\delta$-functor by defining it using an $\mathcal{A}$-resolution and taking homology.
I am wondering where one would get all the required lifts/extensions needed to define the functor if one doesn't have the lifting/extension property of projectives/injectives...
Yes. Basically the point is that you construct the derived functor by taking the colimit over all $\mathcal{A}$-resolutions. A derived functor is a Kan-extension, after all.
I think the basic idea goes back to Buchsbaum here but I don't know where that is written up in more modern terminology. The derived category approach to that is due to Deligne and is well explained in Keller's notes on derived categories and their uses.
@tb You'll get a carrot if you watch 30 minutes. If by then you are still put off by his face, never mind. But I'm quite sure you'll end up watching it to the end.
Ok. I'll be back later. Have to work on my additive combinatorics thing.
@tb I am very close to proving Sylvester and Jacobi Inertia formula! Thanks for bringing that to my notice. I see that theory of bilinear and quadratic forms is very rich.
This question needs a better title. But I'm not sure if "counting permutations without fixed points" is a good one, since the asker may not understand it.
@ymar Hmm, In the online persona of people, we have no way of gauging someone's hnesty IMO, but as for nice, I might suggest you review your definition.
@leo: re: your comment here. You may be interested in this thread for a derivation of the identity $m(E) = m_\ast(E \cap A) + m^\ast (E \smallsetminus A)$ from scratch without using any topological notions.
Hmph, too much of work. But, before that, let me try hard refresh and cache clear ways of doing things before I reinstall newest chrome, as you say. But, still, Thank you for the suggestion.
Is this your version too: 18.0.1025.142? @AntonioVargas
@KannappanSampath It turns out that I wasn't on the newest version after all. I was running 17.0.963.83m, but when I checked the about window it decided to update for me. I assume I can look forward to MathJax errors too once this update completes?
Apparently, I tried, to instruct the new window, to show up as a new tag; viewing that as a tab works OK. But, we used to get a small screen within the screen.
@DavidWheeler That is known. Everyone has to run the bookmark periodically. For some reason the event that was used to trigger the update does not happen in the same way.
@KannappanSampath That function works fine for me on Firefox
@KannappanSampath once the window pops up you can right-click it in the taskbar and select "maximize" and it will show up. But obviously this is annoying. Please do post it to meta sometime.
@robjohn i am answering a question on another forum about reflections in euclidean n-space. although i have never actually studied this topic, i feel confident in my ability to bluff my way through it
You can always answer those questions with «ok so let R be a root system, let R^+ be a set of positive roots, S the set of simple ones, ...» and by the time you have set up the notation they lost interest.
lol, this was a relatively simple one, if $\phi$ is orthogonal, and $s_\alpha$ the reflection by $\alpha$, show $\phi s_\alpha \phi^{-1} = s_{\phi(\alpha)}$
it seems like the same logic as conjugating a k-cycle in Sn is at work here
Well, just decompose your space into $\phi(\alpha) \perp n$ where $n$ is a normal vector to $\phi(\alpha)$. Then show that $\phi(\alpha)$ is fixed under both maps and $n \mapsto -n$ under both maps.
@KannappanSampath Oh, no, not at all! Sorry, I was just speaking of generally impenetrable error messages software tends to give me. For example, have you ever seen ten pages full of xypic complaints just because you messed up some braces? It is impossible to find that information out from those error logs.
Let $R$ be a unital ring with ideal $L$ such that $L^2\neq L$. I need to prove that $Hom(R/L,\pi)$ is not surjective where $\pi:R\to R/L$ is a standard quotient map
I guess I've used that before. So you want to show that there is a map $f\colon R/L \to R/L$ which is not of the form $\pi \circ g$ for some $g\colon R/L \to R$.
@Norbert Hm. These are supposed to be ring homomorphisms, right? You have some elements $x \in L - L^2$, but it seems hard to me to produce a ring homomorphism out of an element. Modules are much more flexible in this regard.
