@KannappanSampath the signum map $\operatorname{sign}: S_n \to \{\pm 1\}$ can also be implemented as the canonical morphism $S_n \to S_n/[S_n,S_n] \cong \mathbb{Z}/2\mathbb{Z}$.
@KannappanSampath well, since the commutator subgroup is a characteristic subgroup, all automorphisms must preserve it. Now if you define parity (the signum) as the element $\pm 1$ in the abelianization, it follows tautologically that automorphisms preserve parity.
It seems to me that that is enough. However this seems a bit more difficult than the other direction (you also need to know that $A_n$ is generated by $3$-cycles).
@Jeff one thing you should check: if $g: [a,b] \to [c,d]$ is an increasing continuously differentiable function then $\int_{c}^d |\dot{\gamma}(t)|\,dt = \int_{a,b} |(\gamma \circ g)'(t)|\,dt$. "The arc length integral does not depend on the parameterization".
(of course, I assume that $g(a) = c$ and $g(b) = d$ and $\gamma: [c,d] \to \mathbb{C}$ is a $C^1$-parameterization)
I've been trying to pick random points on a unit sphere, then I found this cool article: http://mathworld.wolfram.com/SpherePointPicking.html in the end, there is a method that uses "Gaussian random variables". Could someone tell me what are they?
If I take 3 uniformly distributed random numbers from -n to n and build a vector (x, y, z) = V, is it quick to see that $V \over |V|$ is not uniformly distributed over the sphere? Because I don't know why it isn't...
But, basically, this is what I want to say, why not diagonal matrix with some $\lambda_1$'s some $\lambda_2$'s and so on. and the nilpotent part as all 1's that come on top of that?
You can write $A = D + N$ where $D$ is diagonalizable and $N$ is nilpotent. To see this, note that $A = SJS^{-1}$ and write $J = D' + N'$ (diagonal + nilpotent part). Then $D = SD'S^{-1}$ is diagonalizable (but not diagonal in general) and $N = SN'S^{-1}$ is nilpotent
@tb Does it not seem clear that if we have a family of submodules $M_i$ such that for each pair $i,j$ there is $k$ such that $M_i + M_j \subseteq M_k$, $i \leq j$ to mean $M_i \subseteq M_j$ and $\mu_{ij} M_i \to M_j$ to mean just inclusion then $\varinjlim M_i = \bigcup M_i$?
Because what are even the maps $\tau_i$ out of $M_i \to \bigcup M_i$, they are just inclusion maps no?
@tb well the $\tau_i$ are all well defined clearly
suppose we take $x_i \in M_i$
then $\tau_i(x_i)$ is just viewing $x_i$ as an element in $\bigcup M_i$
But then $\tau_j (\mu_{ij}(x_i)$ is first viewing $x_i$ as an element of $M_j$, then as an element of $\bigcup M_i$
is this not almost a tautology?
Perhaps we may need to check that $\bigcup M_i$ is an $A$ - module
the only non-trivial thing to check is closure under addition
but then if we take $y$ and $z$ in $\bigcup M_i$, then $y \in M_k$ for some $k$ and $z \in M_j$ for some $j$ Then $y+z \in M_j + M_k$. But we are guaranteed the existence of some $l\geq j,k$ such that $M_j + M_k \subseteq M_l$. Hence $y+z \in M_l \subset \bigcup M_i$.
much later, i realized that i,j and k unit vectors must have been hold-overs from a hamiltonian view of things...quaternions fell out of style rather quickly, i'm afraid
they might have had a better chance of surviving if Hamilton hadn't decided to build a whole religion out of it. The opposition to that religion was very strong.
@KannappanSampath I often vote too localized when a guy just cranks out one homework question after the other. The rationale being the day after tomorrow (or later today) when he's handed in the copied solution he won't give ... about the answer.
whatever you want to call a space and I suppressed the functor from the category with one object and one morphism to "Spaces" sending the object to $\ast$.
yes, the functor on the right is just the identity functor.
Ben's got a family $M_i$ of submodules of a fixed module $M$ such that for each $i,j$ there exists $k$ such that $M_i + M_j \subset M_k$. He wants to check that $\bigcup M_i$ is the colimit of $M_i$.
@tb The direct limit of a directed system $\{M_i, \mu_{ij}\}$ is the unique module up to isomorphism $\varinjlim M_i$ satisfying the following universal mapping property:
yes. You agree that there's a bit more to it than just "because each $\sigma_i$ is a homomorphism"? You use that the system is directed and that you've already shown that $L$ is well-defined as a map of sets.
yep. Just a bit of beautification: what exactly does the subtraction do here? Nothing. If $x \in \bigcup M_i$ then there's $j$ such that $x \in M_j$ but then $L\circ \tau_j (x) = \sigma_j(x) = \varphi \circ \tau_j(x)$, so $L$ and $\varphi$ agree on $M_j$. Since $x$ was arbitrary, we're done.
I love the fact that my country has a good education system, a good healthcare system, good social welfare, a good police force, good roads and so on. This is a consequence, in a sense, of the existence of the tax service. So maybe I don't love the tax service itself, but I love its effects.
