Now let's take a step back: injectivity of $\bar{v}$ tells you that $v$ is surjective and exactness at $\operatorname{Hom}{(M,{-})}$ tells you that $v$ is the cokernel of $u$.
@Ilya: Also, how can $\beta_{ij}$ range over $i=1,\dots,n$ and $j=1,\dots,n$ if $\beta\in\mathbb{R}^n$? That could just be the OP forgetting to make it $\in\mathbb{R}^{n\times n}$ though.
@BenjaminLim so, that amounts to saying that $v$ is a (the) cokernel of $u$, and in turn this is saying that $M' \to M \to M'' \to 0$ is exact. Nothing deep here, just rephrasing.
I do have a good collection (though slashed into a small fraction; I once hid a hard drive so well that no one would find it if I ever forgot its whereabouts. well, I forgot its whereabouts...) That particular one I just remembered off the top of my head. I think I first saw it like half my lifetime ago!
@Skullpatrol you mean Newton spent there the first period of his life and one of balls was periodically beating his had? That's quite a reasoning why did he come up with mechanics and stuff
@Raphael I had some problems with MathJax on Chrome for the last month - like it started render slower than before, it blinks when I type the text etc.
@Ilya I have had problems too (FF). Braces even topped working without escaping the slash, HTML+CSS renderer wrote across text... It seems as if they were changing some stuff for the worse.
Today I have such a silly question: consider an equation on matrix S in O(n): (M - S)a = 0, where a, Ma are on unit sphere S^{n-1} and a, M are known. Does this equation have a solution for any M?
Heh, i'm so stupid. Of course it has :D
i was wandering about a continious solution when M = M(t), a = const
In 3-d case it is OK, i can find a perpendicular to the plane (a,Ma) and choose S as the rotation of space around this perpendicular. In further dimensions I'm stocked
So S is the unknown? You might as well just forget M and make it b=Sa, in which case of course there's an S: O(n) acts transitively on the sphere S^(n-1).
And the construction in higher dimensions works just the same, you just have higher degrees of freedom in your choice of S.
yes, now I have to specify a continuous solution S(t) of b(t) = S(t)a. In 3d it is easy, but I don't know what to do in further dimensions. Is there some similar result to the Euler's theorem? I
If you choose a chain of $S(t_i)$'s that take you from $b(t_0)$ to $b(t_1)$ to ... to $b(t_n)$, where the $t_i$ partition some interval (u,v) more and more finely (uniformly blah blah), then the $S(t_i)$'s should approach some curve in O(n), right?
I'm not sure I understand the question: but if you intersect the plane spanned by (a,Ma) you get a circle (well, assuming that a and Ma are not antipodal). On this circle it is easy to write down a continuous solution, then take an orthogonal complement of that plane and fix that complement.a
The question is, let a be a fixed vector and let b(t) trace out a continuous path on the sphere. find a continuous function S(t) in the orthogonal group such that b(t)=S(t)a.
If you have the following definition of a subring: "A subset $S$ of a ring $A$ is a *subring* of $A$ if $S$ is closed under addition and multiplication and contains the identity element of $A$." According to Wikipedia, a subring is supposed to contain $0$ (implied by definition of a subring)
Take $A = \mathbb{Z}$ and $S$ the subset containing $1$ (the identity). Now how on earth will this contain $0$? Is Atiyah not requiring a subring to contain $0$ or what?
Because if not I think this is a typo. Unless we completely redefine subring to not actually be a ring itself because it won't contain the additive neutral element.
Matt: That's what I take it to mean. It might be the case Atiyah has a stronger idea of what closure is than what I've seen everywhere, one that includes inverses (or he just forgot to state the latter). Either that or his definition of subring is so loose it allows for multiplicative monoids.
Hmm. Moral from Matt's question: Ever be alert. I had gone through this chapter and thought these were stupid definitions that I knew, In fact I was trying to argue in favour of Atiyah. How sick of me.
@Skullpatrol In my browser (Firefox on Mac) that animation looks funny since the browser uses a steady framerate. The animation actually has a variable framerate to make the animation look more natural.
As of 14:20 UTC, MathJax.org is down, hence all MathJax requests fail.
