TITLE: Proof of The Third Isomorphism Theorem
QUESTION [0 upvotes]: Here's what I'm trying to prove right now:
Let $V$ be a vector space over $\mathbb{F}$. Let $M$ be a linear subspace of $V$ and $N$ be a linear subspace of $M$. Prove that the mapping $x+N \mapsto x+M$ between the quotient spaces $V/N \to V/M$ is linear with kernel $M/N$. Then, deduce that:
$$\frac{V/N}{M/N} \cong V/M$$

Proof Attempt:
We have to show that the given relation $T: V/N \to V/M$ is a linear mapping. We define:
$$\forall x \in V: T(x+N) = x+M$$
This is totally-defined. To show well-definedness, let $x+N = y+N$ where $x,y \in V$. Then, $x-y \in N$. So, $x-y \in M$. Hence:
$$x + M = y+M$$
$$\iff T(x+N) = T(y+N)$$
To prove that it is linear, we need to show additivity and homogeneity.

Proof of Additivity

Let $u,v \in V/N$. Then, $u = x+N$ and $v = y+N$ for some $x,y \in V$. Then:
$$T(u+v) = T((x+N)+(y+N)) = T((x+y)+N) = (x+y)+M = (x+M)+(y+M) = T(u) + T(v)$$
This proves additivity.

Proof of Homogeneity

Let $\alpha \in \mathbb{F}$ and $u \in V/N$. Then, $u = x+N$ for some $x \in V$. So:
$$T(\alpha u) = T(\alpha(x+N)) = T(\alpha x +N) = \alpha x + M = \alpha (x+M) = \alpha T(u)$$
This proves homogeneity. Hence, $T$ is a linear map. To show that the kernel of $T$ is $M/N$, we have:
$$T(x+N) = x+M = \theta_V+M$$
$$\iff x \in M$$
$$\iff x+N \in M/N$$
$$\iff \ker(T) = M/N$$
Now, we notice that $T$ is surjective. By the first isomorphism theorem, it follows that:
$$\frac{V/N}{M/N} \cong V/M$$
That proves the desired result.
Does the proof above work? If it doesn't, why? How can I fix it?

REPLY [1 votes]: Your approach is absolutely right!
I would change the part concerning $\ker T$, writing:
\begin{align}
x + N \in \ker T & \iff T(x+N) = 0 \in V/M \\
 & \iff x+M = 0 \in V/M \\
 & \iff x \in M \\
 & \iff x + N \in M/N
\end{align}
which means $\ker T = M/N$.


TITLE: Is the group-theoretic Grothendieck-Springer resolution Calabi-Yau?
QUESTION [11 upvotes]: Any cotangent bundle is Calabi-Yau (by which I mean the canonical bundle is trivial), so the Springer resolution $T^*(G/B)$ is Calabi-Yau.  I think that the Grothendieck-Springer resolution $\tilde{\mathfrak{g}}$ is Calabi-Yau -- this is because a vector bundle over $X$ with sheaf of sections $\mathcal{E}$ has canonical bundle $\pi^* \omega_X \otimes \Lambda^{top} \pi^* \mathcal{E}^\vee$.  One can check that $\mathcal{E}$ is an extension of the cotangent bundle by trivial bundles.
Is the group-theoretic resolution
$$\tilde{G} = \{(x, g) \in G/B \times G \mid g \in xBx^{-1}\}$$
also Calabi-Yau, and how would one argue it?

REPLY [1 votes]: I think I have an answer.  Let $p: E \rightarrow X$ be a vector bundle over $X$.  Since this is a smooth map, we have a short exact sequence
$$0 \rightarrow p^* \Omega^1_X \rightarrow \Omega^1_E \rightarrow \Omega^1_{E/X} \rightarrow 0$$
and taking top exterior powers,
$$\omega_E \simeq p^* \omega_X \otimes \Lambda^n \Omega^1_{E/X}$$
Checking the definitions gives us $\Omega^1_{E/X} \simeq p^* \mathcal{E}^\vee$.
For the Lie algebra Grothendieck-Springer resolution, we take $\mathcal{E}$ to be a trivial extension of the cotangent sheaf.  Fix this notation.  For the group-theoretic version $p^*: \tilde{G} \rightarrow G/B$ is a $B$-bundle over $G/B$, but it is still smooth, and the short exact sequence still splits.  The result we want follows from the claim that $\Omega^1_{\tilde{G}/(G/B)} \simeq p^* \mathcal{E}^\vee$.  As a warmup, we can compute $\Omega^1_B$; as an equivariant sheaf, this is $\mathcal{O}_B \otimes \mathfrak{b}^*$.  By $G$-equivariance, $\Omega^1_{\tilde{G}/X} = p^*\mathcal{F}^\vee$ where the total bundle of $\mathcal{F}$ is $G \times_B \mathfrak{b}$, but this is just $\mathcal{E}$.


TITLE: Is there an ambiguity between the motion of an aircraft propeller and time dilation in special relativity?
QUESTION [0 upvotes]: We know that the faster a propeller rotates, the faster the attached vehicle (airplane, submarine, etc.) moves. (Assume that the propeller is normal to the motion direction.)
On the other hand, special relativity asserts that in high-speed linear motions, the clocks slow down. If we consider the propeller as a mechanical clock, it is anticipated that $-$ due to the direct effect of time dilation $-$ it runs very slowly when the airplane travels at a considerable fraction of light speed. I just cannot understand how is it possible to see an airplane moving through the air at, say, $0.9999999999999999c$ (digit 9 is repeated 16 times), while its propellers spin extremely slowly.
Remember that in the airplane rest frame the propeller can rotate at a maximum angular velocity of $\omega=c/r$, where $r$ is the propeller's radius. If we assume that this maximum angular velocity is capable of setting the airplane in motion at $0.9999999999999999c$ as seen by the lab observer at rest with respect to the air, this observer sees that the angular speed of the propeller reduces to $\omega^\prime=\sqrt{1-0.9999999999999999^2}\omega$ $\approx 10^{-8}\omega$. Now if $r=10\space m$, we have $\omega \approx 3×10^{7}\space rad/s$ and $\omega^\prime=0.3\space rad/s$.
I wonder how it can be rational for the lab observer to see an aircraft moving very close to the speed of light, while its propeller rotates very slowly at just $\omega^\prime=0.3\space rad/s$? Does not it violate this sentence that the faster a propeller rotates, the faster the attached vehicle moves?

REPLY [0 votes]: There was a similar problem involving a relativistic traincar with toothed wheels moving on  toothed rail.
Due to time dilation, in the reference frame of the rail rotation of the wheel must slow down as speed of the train approaches that of light, that seems as a paradox. It seems that the faster the train moves, the slower its wheels must rotate.
If the train moves at velocity close to $c$, rotation of the wheels must slow down until almost complete stop. This way the train appears to be “slipping” on the rail in the reference frame of the rail, though it is impossible, since “smooth rotation” is an absolute effect and it cannot depend on chosen reference frame.
Resolution of that „paradox“ was in relativistic kinematics. The rim of the wheel Lorentz – contracts as velocity of the train increases.
The rest length of the rim of the wheel must remain constant. This means that the rim Lorentz contracts, and that the radial extension of the wheel contracts accordingly. The result is that the wheel becomes infinitely small in the limit that the train moves with the velocity of light.
If $v$ is velocity on the rim in the rest frame $K$ of the wheel, we have $\Omega=v/R$, where $R=R_0/\gamma$ is the contracted radius of the rotating wheel, and $R_0$ is their radius when they are at rest. The angular velocity of the rotating wheel is then
$$\Omega = \gamma v /R_0$$
Hence, in this case the angular velocity $\Omega$ must approach an infinitely great value in $K$ when the speed of the train approaches that of light.
Through a gear wheel rotation can be transmitted from the wheel to the "propeller"; apparently in the reference system of the traincar angular velocity  of this "propeller" must also approach infinitely large value as speed of the train comes close to that of light.


TITLE: Lagrange interpolation of Galois field functions
QUESTION [3 upvotes]: I'm trying to understand how to apply Lagrangian interpolation for a function $F : GF(2^k) \to GF(2^k)$, but am having some difficulty. Suppose I have a function $F : GF(2^2) \to GF(2^2)$ given by the following mapping $F$ (where every element is in the field $GF(2^2)$ defined by the only irreducible polynomial $p(x) = x^2 + x + 1$):
$0 \to 1$
$1 \to x$
$x \to x + 1$
$x + 1 \to 0$
Now suppose I want to find some polynomial $q(x)$ such that $q(x) = F(x)$. Do I simply apply the interpolation formula: $q(x) = \sum_{i = 1}^4y_i\prod_{1 \leq j \leq 4, j \not= i}^{4} \frac{x - x_j}{x_i - x_j}$? If so, can someone try and walk me through how this is done? My calculations did not yield a valid polynomial $q(x)$.

REPLY [2 votes]: Let's denote $GF(4)=\{0,1,\alpha,\alpha+1\}$ with $\alpha^2=\alpha+1$, so that $x$ is freed as an indeterminate.
Let $z\in GF(4)$. The Lagrange interpolation needs polynomials $f_z(x)$ with the property $f_z(z)=1$ and $f_z(y)=0$, if $y\in GF(4)$ is $\neq z$. Let's find $f_0(x)$. To that end we first construct the polynomial
$$
g(x)=(x-1)(x-\alpha)(x-(\alpha+1))
$$
that vanishes whenever $x\in GF(4)\setminus\{0\}$. Then $f_0(x)=g(x)/g(0)$ will work. But here
$$
g(0)=(0-1)(0-\alpha)(0-(\alpha+1))=1\cdot\alpha\cdot(\alpha+1)=\alpha^2+\alpha=1,
$$
so we actually get $f_0(x)=g(x)$. Expanding the polynomial $g(x)$ actually gives something simple (leaving this to you as an exercise), namely
$$
g(x)=1+x^3.
$$
With a little bit of experience you could have seen this right away. After all, all the non-zero elements of $GF(4)$ are third roots of unity so the polynomial $x^3$ takes value $1$ whenever $x\in GF(4)^*$. Anyway, we know that $f_0(x)=1+x^3$.
What about the other polynomials $f_z(x)$, $z\neq0$. We can either do the same thing, or we can use linear substitutions. The latter approach gives us the recipe
$$
f_z(x)=f_0(x-z).
$$
Why does this work? If we plug in $x=z$ we get $f_z(z)=f_0(z-z)=f_0(0)=1$. OTOH if we plug in $x=y\in GF(4),y\neq z$ we get $f_z(y)=f_0(y-z)=0$, because here $y-z\neq0$ and $f_0$ vanishes at the non-zero elements of $GF(4)$.
Finally we are well placed to do Lagrange interpolation. Let $F:GF(4)\to GF(4)$ be any function. We can write it as a polynomial
$$
q(x)=F(0)f_0(x)+F(1)f_1(x)+F(\alpha) f_\alpha(x)+F(\alpha+1) f_{\alpha+1}(x).
$$
All that is left to do is to plug in the values and expand it out.


TITLE: Blow up of an ideal in $\Bbb C^2$
QUESTION [3 upvotes]: As described in these notes, I am trying to compute the blow up of $\mathbb C^2=\text{ Maxspec }\mathbb C[x,y]$ along the subvariety corresponding to the ideal $\langle\ x^2,y\ \rangle$ but I am having trouble with the steps.
Now the blow up is given by the closure of the image of the map $f$ defined as follows - $$f:\mathbb C^2\setminus\mathcal Z(x^2,y)\to \mathbb C^2\times \mathbb P^1\quad\text{is given by}\quad (a,b)\mapsto(a,b,[a^2:b])$$
Now if $[x_0:x_1]$ are the homogeneous coordinates of $\mathbb P^1$ then  $\mathbb {C^2\times P^1}$ is covered by $\mathbb C^2\times U_0$ and $\mathbb C^2\times U_1$ where $U_i=\{\ [a_0:a_1]\ |\ a_i\neq0\ \}$.
Let $A_0=\overline{\text{ Im }f}\cap(\mathbb C^2\times U_0)$ and $A_1=\overline{\text{ Im }f}\cap(\mathbb C^2\times U_1)$ then the blow up is obtained by appropriately glueing $A_0$ and $A_1$. However I am unable to explicitely write this "appropriate glueing" without a proper description of $A_0$ and $A_1$. By that I mean I want to write the $A_i$ as the vanishing of some ideals in $\mathbb {C^2\times P^1}$. This way I can look at glueing maps at the coordinate ring level.
Could someone hepl me describe $A_0$ and $A_1$?
Thank you.

REPLY [1 votes]: In more generality, there's the following result:
Let $X \subset \mathbb{A}^n$ be an affine variety and $Z\subset X$ be a closed smooth subvariety, which is defined by the vanishing of the polynomials ${f_1, ..., f_k}$ in $\mathbb{A}^n$. Then the blow-up with respect to the subvariety $Z$, denoted $Bl_Z\mathbb{A}^n$, in $\mathbb{A}^n \times {\mathbb{P}}^{k-1}$ (note the "degree of ${\mathbb{P}}$" corresponds to the number of defining polynomials of $Z$) is given by $\{ y_if_j=y_jf_i \vert i,j=1,...,k \}$.
So in your case, the blow-up is the set $\{ (x,y),[a_0,a_1] \in \mathbb{C}^2 \times \mathbb{P}^1 \text{ } \vert \text{ }a_0f_2(x,y)=a_1f_1(x,y) \}$ where $f_1(x,y)=x^2$ and $f_2(x,y)=y$. So your $A_0$ simply becomes $A_0=\{(x,y),[1:a_1]\text{ } \vert \text{ } y=a_1x^2 \}=\{(x,a_1x^2),[1:a_1]\} \cong \mathbb{C}^2_{x,a_1}$. Similarly with $A_1$.
By the way, do you notice what subvariety you are blowing up?


TITLE: Moment Generating Function - Poisson?
QUESTION [0 upvotes]: Problem: If  $MX(t)=e^{−5(1−et)}$ , find $V(X)$.
Also find: $P(X=3)$
My first intuition is that this would follow the Poisson rule: $M(t)=e^{\lambda(e^t-1)}$
This would mean that $\lambda=E(X)=-5?$
I am not sure how to get $E(X^2)$ so that I can finish the equation $V(X)=E(X^2)-(E(X))^2$.
For the $P(X=3)$, it would appear that This is the MGF of the following discrete $rv$ is as follows:
$$X =
\begin{cases}
1,  & \text{if $x=-5$ } \\
0,  & \text{if $x=0$ }
\end{cases}
$$
So I don't know that I'm working through this part correctly either.

REPLY [1 votes]: The MGF of the Poisson as you have written would imply that $\lambda = 5$. The MGF is exactly what it sounds like, a generating function for the moments. $M^{(k)}(0) = E(X^k)$. Thus $M''(0) - (M'(0))^2 = \text{var}(X)$. You can show that this implies that $\text{var}(X) = 5$.
To find $P(X = 3)$, it is better to use the distribution itself
$$ P(X = x) = \frac{e^{-5}5^x}{x!}. $$


text:
\section{Union of Class is Subset implies Class is Transitive}
Tags: Class is Transitive iff Union is Subset

\begin{theorem}
Let $A$ be a [[Definition:Class (Class Theory)|class]].
Let $\displaystyle \bigcup A$ denote the [[Definition:Union of Class|union]] of $A$.
Let:
:$\displaystyle \bigcup A \subseteq A$
Then $A$ is [[Definition:Transitive Class|transitive]].
\end{theorem}

\begin{proof}
Let $\displaystyle \bigcup A \subseteq A$.
Let $x \in \displaystyle \bigcup A$.
Then by definition:
:$\exists y \in A: x \in y$
By definition of [[Definition:Subclass|subclass]]:
:$x \in A$
Thus we have that:
:$x \in y \land y \in A \implies x \in A$
It follows by definition that $A$ is a [[Definition:Transitive Class|transitive class]].
\end{proof}

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