Challenge Math Challenge - July 2021

[Twigg]

Thanks for posting these, @fresh_42!

Spoiler: Problem 5

Use $x+y+z = 0$ to suggest a new coordinate system $(X,Y,Z)$ over $\mathbb{R}^3$ where $Z = x+y+z$. I choose to define $X = x - y$ and $Y = y - z$. These coordinates are orthogonal, since $\nabla X \times \nabla Y = \nabla Z$.

Since this is a linear transformation, we can write it in matrix form:
$$\left( \begin{array}{cc} X \\ Y \\ Z \end{array} \right) = \left( \begin{array}{cc} 1 & -1 & 0 \\ 0 & 1 & -1 \\ 1 & 1 & 1 \end{array} \right) \left( \begin{array}{cc} x \\ y \\ z \end{array} \right)$$
and we can invert that
$$\left( \begin{array}{cc} x \\ y \\ z \end{array} \right) = \frac{1}{3} \left( \begin{array}{cc} 2 & 1 & 1 \\ -1 & 1 & 1 \\ -1 & -2 & 1 \end{array} \right) \left( \begin{array}{cc} X \\ Y \\ Z \end{array} \right)$$

Now we can write the second constraint in matrix form $$ 9 = \left( \begin{array}{cc} x & y & z \\ \end{array} \right) \left( \begin{array}{cc} 1 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 1 \\ \end{array} \right) \left(\begin{array}{cc} x \\ y \\ z \\ \end{array} \right)$$
Then we can transform that to the $(X,Y,Z)$ coordinate system:
$$\frac{1}{9} \left( \begin{array}{cc} 2 & -1 & -1 \\ 1 & 1 & -2 \\ 1 & 1 & 1 \end{array} \right) \left( \begin{array}{cc} 1 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 1 \\ \end{array} \right) \left( \begin{array}{cc} 2 & 1 & 1 \\ -1 & 1 & 1 \\ -1 & -2 & 1 \end{array} \right) = \left( \begin{array}{cc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array} \right)$$
So, finally, we have $$X^2 + Y^2 = 9$$ so the space is a submersion (or is it immersion? I can never remember the difference) of $\mathbb{S}^1$ and thus a manifold.

...I'll get back to the the tangent and normal spaces at point p later. My partner is giving me the evil eye for spending Saturday afternoon doing physics. I think I'm in trouble

 

[fresh_42]

... so the space is a submersion (or is it immersion? I can never remember the difference) of $\mathbb{S}^1$ and thus a manifold.

Both are primarily mappings, submersions have an everywhere surjective derivative, immersions an everywhere injective derivative.

 

[StoneTemplePython]

second time's the charm?

Spoiler: 2

Suppose $X$ has $m$ orbits and let $X$ be an ordered set such that the first $r_1$ elements belong to orbit $X_1$ (where $\vert X_1\vert =r_1$), the next $r_2$ elements belong to orbit $X_2$ and so on.

Then $\phi_i$ is the homomorphism for the standard permutation matrix representation (with matrices in $\mathbb C^{r_i\times r_i}$) of $G$'s action on $X_i$.

lemma:
$T_i:= \frac{1}{\vert G\vert}\sum_{g\in G}\phi_i(g)$
then $\text{trace}\big(T_i\big)=1$

Spoiler

*proof:*
(i) $T_i$ is idempotent
$T_i^2=\Big(\frac{1}{\vert G\vert}\sum_{g\in G}\phi(g)\Big)\Big(\frac{1}{\vert G\vert}\sum_{g'\in G}\phi_i(g')\Big)=\frac{1}{\vert G\vert^2}\sum_{g\in G}\Big(\phi_i(g)\sum_{g'\in G}\phi_i(g')\Big)=\frac{1}{\vert G\vert^2}\sum_{g\in G}\vert G\vert \cdot T_i=\frac{1}{\vert G\vert^2}\cdot \vert G\vert^2 \cdot T_i=T_i$

missing detail:
$\phi_i(g)\sum_{g'\in G}\phi_i(g')=\sum_{g'\in G}\phi_i(g)\phi_i(g')=\sum_{g'\in G}\phi_i(gg')=\sum_{g''\in G}\phi_i(g'')= \sum_{g\in G}\phi_i(g)=\vert G\vert \cdot T_i$

(ii) $T_i$ is a positive matrix
$T_i$ is a sum (re-scaled by a positive number) of real non-negative matrices and hence is real non-negative. If component $(k,j)$ of $T_i$ is zero that implies there is no $g\in G$ in that maps the kth element of $X_i$ to the $j$th element -- but $G$ acts transitively on $X_i$ so this cannot be true hence $T_i$ is a positive matrix.

(iii) $T_i$ has one eigenvalue of 1 and all else are zero, hence it has trace 1
$T_i$ is a convex combination of (doubly) stochastic matrices hence it is (doubly) stochastic. Since $T_i$ is positive, we may apply Perron Theory for stochastic matrices which tells us that $\mathbf 1$ is an eigenvector with eigenvalue 1, which is *simple*. And since $T_i$ is idempotent, all other ($r_i -1$) eigenvalues must be zero. Thus $\text{trace}\big(T_i\big)=1$,

(Since $T_i$ is doubly stochastic this actually tells us that it is $\propto \mathbf{11}^*$-- which implies, in terms of irreducibles, that $\phi_i(g)$ includes exactly one trivial representation-- though as is often the case with representations, we only need the trace.)

the main case
now that we've looked at individual orbits, consider $G$'s action on $X$ as a whole. This gives a standard permutation representation of, for $g\in G$

$\phi(g\big) = \left[\begin{matrix}\phi_1(g) & \mathbf 0 & \cdots & \mathbf 0\\ \mathbf 0 & \phi_2(g) & \cdots & \mathbf 0\\ \vdots & \vdots & \ddots & \vdots \\ \mathbf 0 & \mathbf 0 & \cdots & \phi_{m}(g)\end{matrix}\right]$
(i.e. the orbits allow us to decompose $G$s action on $X$ into a direct sum of its action on individual orbits)
so we have

$\text{# of orbits}=m = \sum_{i=1}^m \text{trace}\big(T_i\big)= \frac{1}{\vert G\vert}\sum_{i=1}^m \sum_{g\in G}\text{trace}\big(\phi_i(g) \big)$
$=\frac{1}{\vert G\vert}\sum_{g\in G}\sum_{i=1}^m\text{trace}\big( \phi_i(g) \big)=\frac{1}{\vert G\vert} \sum_{g\in G}\text{trace}\big(\phi(g) \big)=\frac{1}{\vert G\vert}\sum_{g\in G}|X^g|$

 

[Twigg]

Annnd back to finish Problem 5

Spoiler: Problem 5, con't:

$p = (2,-1,-1)$ implies $X_p = 3$, $Y_p = 0$, $Z_p = 0$. The tangent plane $T_p M$ points along the Y-axis since $p$ is on the X-axis, so $$T_p M = \{ t \left( \begin{array}{cc} 0 \\ 1 \\ -1 \\ \end{array} \right) : \forall t \in \mathbb{R} \}$$ since $(0,1,-1)$ points along the Y-axis.

For the normal plane, it's the same deal except that the normal vector space spans the X-axis, since the Y-axis is the tangent plane. That means $$N_p M = \{ t \left( \begin{array}{cc} 1 \\ -1 \\ 0 \\ \end{array} \right) : \forall t \in \mathbb{R} \}$$

As far as showing that the space is a manifold, I feel like just showing it's a circle in the transformed coordinates is sufficient. But if you feel like you need more formal proof, I could write you an atlas with stereographic projection onto the X-axis.

 

[fresh_42]

Annnd back to finish Problem 5

Spoiler: Problem 5, con't:

$p = (2,-1,-1)$ implies $X_p = 3$, $Y_p = 0$, $Z_p = 0$. The tangent plane $T_p M$ points along the Y-axis since $p$ is on the X-axis, so $$T_p M = \{ t \left( \begin{array}{cc} 0 \\ 1 \\ -1 \\ \end{array} \right) : \forall t \in \mathbb{R} \}$$ since $(0,1,-1)$ points along the Y-axis.

For the normal plane, it's the same deal except that the normal vector space spans the X-axis, since the Y-axis is the tangent plane. That means $$N_p M = \{ t \left( \begin{array}{cc} 1 \\ -1 \\ 0 \\ \end{array} \right) : \forall t \in \mathbb{R} \}$$

As far as showing that the space is a manifold, I feel like just showing it's a circle in the transformed coordinates is sufficient. But if you feel like you need more formal proof, I could write you an atlas with stereographic projection onto the X-axis.

I have a different tangent vector, but it's impossible to tell where you went wrong. And the normal space to a one-dimensional subspace in a three-dimensional Euclidean space is two-dimensional.

 

[fresh_42]

second time's the charm?

Spoiler: 2

Suppose $X$ has $m$ orbits and let $X$ be an ordered set such that the first $r_1$ elements belong to orbit $X_1$ (where $\vert X_1\vert =r_1$), the next $r_2$ elements belong to orbit $X_2$ and so on.

Then $\phi_i$ is the homomorphism for the standard permutation matrix representation (with matrices in $\mathbb C^{r_i\times r_i}$) of $G$'s action on $X_i$.

lemma:
$T_i:= \frac{1}{\vert G\vert}\sum_{g\in G}\phi_i(g)$
then $\text{trace}\big(T_i\big)=1$

Spoiler

*proof:*
(i) $T_i$ is idempotent
$T_i^2=\Big(\frac{1}{\vert G\vert}\sum_{g\in G}\phi(g)\Big)\Big(\frac{1}{\vert G\vert}\sum_{g'\in G}\phi_i(g')\Big)=\frac{1}{\vert G\vert^2}\sum_{g\in G}\Big(\phi_i(g)\sum_{g'\in G}\phi_i(g')\Big)=\frac{1}{\vert G\vert^2}\sum_{g\in G}\vert G\vert \cdot T_i=\frac{1}{\vert G\vert^2}\cdot \vert G\vert^2 \cdot T_i=T_i$

missing detail:
$\phi_i(g)\sum_{g'\in G}\phi_i(g')=\sum_{g'\in G}\phi_i(g)\phi_i(g')=\sum_{g'\in G}\phi_i(gg')=\sum_{g''\in G}\phi_i(g'')= \sum_{g\in G}\phi_i(g)=\vert G\vert \cdot T_i$

(ii) $T_i$ is a positive matrix
$T_i$ is a sum (re-scaled by a positive number) of real non-negative matrices and hence is real non-negative. If component $(k,j)$ of $T_i$ is zero that implies there is no $g\in G$ in that maps the kth element of $X_i$ to the $j$th element -- but $G$ acts transitively on $X_i$ so this cannot be true hence $T_i$ is a positive matrix.

(iii) $T_i$ has one eigenvalue of 1 and all else are zero, hence it has trace 1
$T_i$ is a convex combination of (doubly) stochastic matrices hence it is (doubly) stochastic. Since $T_i$ is positive, we may apply Perron Theory for stochastic matrices which tells us that $\mathbf 1$ is an eigenvector with eigenvalue 1, which is *simple*. And since $T_i$ is idempotent, all other ($r_i -1$) eigenvalues must be zero. Thus $\text{trace}\big(T_i\big)=1$,

(Since $T_i$ is doubly stochastic this actually tells us that it is $\propto \mathbf{11}^*$-- which implies, in terms of irreducibles, that $\phi_i(g)$ includes exactly one trivial representation-- though as is often the case with representations, we only need the trace.)

the main case
now that we've looked at individual orbits, consider $G$'s action on $X$ as a whole. This gives a standard permutation representation of, for $g\in G$

$\phi(g\big) = \left[\begin{matrix}\phi_1(g) & \mathbf 0 & \cdots & \mathbf 0\\ \mathbf 0 & \phi_2(g) & \cdots & \mathbf 0\\ \vdots & \vdots & \ddots & \vdots \\ \mathbf 0 & \mathbf 0 & \cdots & \phi_{m}(g)\end{matrix}\right]$
(i.e. the orbits allow us to decompose $G$s action on $X$ into a direct sum of its action on individual orbits)
so we have

$\text{# of orbits}=m = \sum_{i=1}^m \text{trace}\big(T_i\big)= \frac{1}{\vert G\vert}\sum_{i=1}^m \sum_{g\in G}\text{trace}\big(\phi_i(g) \big)$
$=\frac{1}{\vert G\vert}\sum_{g\in G}\sum_{i=1}^m\text{trace}\big( \phi_i(g) \big)=\frac{1}{\vert G\vert} \sum_{g\in G}\text{trace}\big(\phi(g) \big)=\frac{1}{\vert G\vert}\sum_{g\in G}|X^g|$

A beautiful solution.

As @Office_Shredder already mentioned, there is still an elementary proof possible.
(Note to all who want to try with the orbit-stabilizer formula.)

The statement is commonly known as Burnside's Lemma. Burnside wrote this formula down around 1900. Historians of mathematics, however, found this formula already from Cauchy (1845) and Frobenius (1887). Therefore the formula is sometimes referred to as The Lemma which is not from Burnside.

 

[julian]

Problem #2:

The set: $\{ g \cdot x : g \in G \}$ is the orbit of $x$ under the action of $G$ which we denote $\text{Orb}(x)$. We can define an equivalence relation on $X$: $x \sim y$ iff $y = g \cdot x$ for some $g \in G$. We check: $x \sim x$ as $e \cdot x = x$. If $y = g \cdot x$ then $x = g^{-1} \cdot y$ so that $x \sim y$ implies $y \sim x$. If $y = g \cdot x$ and $z = h \cdot y$ then $z = h g \cdot x$ so that $x \sim y$ and $y \sim z$ implies that $x \sim z$. As this is an equivalence relation on $X$, the set $X$ is partitioned into disjoint equivalence classes and with each $x$ appearing in one of these equivalences classes.

We have

\begin{align*}
\sum_{g \in G} |X^g| = \sum_{g \in G} \sum_{x \in X} \delta_{g\cdot x, x} = \sum_{x \in X} \sum_{g \in G} \delta_{g\cdot x, x} = \sum_{x \in X} |G_x|
\end{align*}

The elements of $G_x$ are easily verified to form a group called the stabalizer subgroup of $G$ with respect to $x$.

We define a map

\begin{align*}
\phi : G \rightarrow \text{Orb} (x)
\end{align*}

by

\begin{align*}
\phi (g) = g \cdot x
\end{align*}

It is clear that $\phi$ is surjective, because the definition of the orbit of $x$ is $G \cdot x$. Now

\begin{align*}
\phi (g) = \phi (h) \Longleftrightarrow g \cdot x = h \cdot x \Longleftrightarrow g^{-1} h \cdot x = x \Longleftrightarrow g^{-1} h \in G_x .
\end{align*}

That is, $\phi (g) = \phi (h)$ if and only if $g$ and $h$ are in the same coset for the stabalizer subgroup $G_x$. Thus there is a well-defined bijection:

\begin{align*}
G / G_x \rightarrow \text{Orb} (x)
\end{align*}

So $\text{Orb}(x)$ has the same number of elements as $G / G_x$. This together with Lagrange's theorem says that

\begin{align*}
|\text{Orb}(x)| = |G| / |G_x|
\end{align*}

Recall we have that $X$ is the disjoint union of all its orbits. Let $x_i$ be a representative point of the $i$th orbit, then:

\begin{align*}
\frac{1}{|G|} \sum_{g \in G} |X^g| & = \frac{1}{|G|} \sum_{x \in X} |G_x|
\nonumber \\
&= \sum_{x \in X} \frac{1}{|\text{Orb}(x)|}
\nonumber \\
& = \sum_{i = 1}^{|X / G|} \sum_{x \in \text{Orb}(x_i)} \frac{1}{|\text{Orb}(x_i)|}
\nonumber \\
& = \sum_{i = 1}^{|X / G|} 1
\nonumber \\
& = |X / G| .
\end{align*}

 

[martinbn]

5. Show that
$$
M:=\{\,x\in \mathbb{R}^3\,|\,x_1+x_2+x_3=0\, , \,x_1^2+2x_2^2+x_3^2-2x_2(x_1+x_3)=9\,\} \subseteq \mathbb{R}^3
$$
is a manifold, and determine the tangent space $T_pM$ and the normal space $N_pM$ at $p=(2,-1,-1)\in M\,.$

The second equation can be written as $(x_1-x_2)^2+(x_2-x_3)^2=9$. Make the change of variables
$$
x=x_1-x_2, \; y=x_2-x_3, \; z=x_1+x_2+x_3
$$
It is invertable, and in the new variables the set $M$ is given as the solution to the equations
$$
x^2+y^2=9, \; z=0
$$
Obviously a manifold. The point $p$ in the new coordinates is $(3,0,0)$, so the tangent space is $x=3, z=0$ or in the original coordiantes $x_1-x_2=0, x_1+x_2+x_3=0$.

 

[fresh_42]

The second equation can be written as $(x_1-x_2)^2+(x_2-x_3)^2=9$. Make the change of variables
$$
x=x_1-x_2, \; y=x_2-x_3, \; z=x_1+x_2+x_3
$$
It is invertable, and in the new variables the set $M$ is given as the solution to the equations
$$
x^2+y^2=9, \; z=0
$$
Obviously a manifold. The point $p$ in the new coordinates is $(3,0,0)$, so the tangent space is $x=3, z=0$ or in the original coordiantes $x_1-x_2=0, x_1+x_2+x_3=0$.

... which leads to $T_pM=\ldots$ and $N_pM= \ldots$

 

[julian]

Hi @fresh_42. So when you hit 15,000 posts do you get a set of steak knives or something?

 

[martinbn]

... which leads to $T_pM=\ldots$ and $N_pM= \ldots$

Right, forgot the normal one. It is the plane $x_2=x_3$.

 

[fresh_42]

Hi @fresh_42. So when you hit 15,000 posts do you get a set of steak knives or something?

... maybe in the back ...

 

[julian]

You deserve a special award off Greg.

 

[fresh_42]

The second equation can be written as $(x_1-x_2)^2+(x_2-x_3)^2=9$. Make the change of variables
$$
x=x_1-x_2, \; y=x_2-x_3, \; z=x_1+x_2+x_3
$$
It is invertable, and in the new variables the set $M$ is given as the solution to the equations
$$
x^2+y^2=9, \; z=0
$$
Obviously a manifold. The point $p$ in the new coordinates is $(3,0,0)$, so the tangent space is $x=3, z=0$ or in the original coordiantes $x_1-x_2=0, x_1+x_2+x_3=0$.

Right, forgot the normal one. It is the plane $x_2=x_3$.

The easier the problem, the lazier the solver ...
For all viewers, here is the long version:

We consider the function $f\, : \,\mathbb{R}^3\longrightarrow \mathbb{R}^2$ defined by
$$
\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix} \stackrel{f}{\longrightarrow} \begin{bmatrix}x_1+x_2+x_3 \\ x_1^2+2x_2^2+x_3^2-2x_2(x_1+x_3)-9\end{bmatrix}
$$
such that $M=f^{-1}(\{(0,0)\}).$ Its Jacobi matrix is
$$
J_xf = \begin{bmatrix}1&1&1\\2(x_1-x_2)&2(2x_2-x_1-x_3)&2(x_3-x_2)\end{bmatrix}
$$
$\operatorname{rk}J_xf=1$ if $x_1-x_2=2x_2-x_1-x_3=x_3-x_2$ or $x_1=x_2=x_3\,.$ Since $f(t,t,t)=(3t,-9)\neq 0$, $(0,0)$ is a regular value of $f$ and $M$ a submanifold of dimension $3-1-1=1\,.$

For $p=(2,-1,-1)$ we have $J_pf=\begin{bmatrix}1&1&1\\ 6&-6&0\end{bmatrix}$, hence $T_pM=\ker D_pf=\mathbb{R}\cdot \begin{bmatrix} 1&1&-2 \end{bmatrix}$ and $N_pM=(T_pM)^\perp=\mathbb{R}\cdot \begin{bmatrix}1\\1\\1\end{bmatrix}+ \mathbb{R}\cdot \begin{bmatrix}1\\-1\\0\end{bmatrix}$, which is the row space of $J_pf\,.$

 

[martinbn]

For 3. The kernel of the adjoint map is the centre. Semisimple Lie algebras have trivial centre. For the second part, take a 1d algebra, then the general linear algebra over it is also 1d algebra, hense they are isomorphic. And the abelian algebra is not semisimple.

 

[fresh_42]

For 3. The kernel of the adjoint map is the centre. Semisimple Lie algebras have trivial centre. For the second part, take a 1d algebra, then the general linear algebra over it is also 1d algebra, hense they are isomorphic. And the abelian algebra is not semisimple.

Yes, and a bit more detailed:

A semisimple Lie algebra has no Abelian ideals. Its center, however, is an Abelian ideal. Thus we have
\begin{align*}
\mathfrak{Z(g)}&=\{\,Z\in \mathfrak{g}\,|\,[X,Z]=0\;\forall \,X\in \mathfrak{g}\,\}\\&=\bigcap_{X\in \mathfrak{g}}\operatorname{ker}\mathfrak{ad}X = \operatorname{ker}\mathfrak{ad} = \{\,0\,\}
\end{align*}
This means that $\mathfrak{ad}\, : \,\mathfrak{g}\longrightarrow \mathfrak{gl(g)}$ is a monomorphism of Lie algebras and
$$
\mathfrak{g} \cong \mathfrak{ad(g)} \cong \mathfrak{Der(g)} \subseteq \mathfrak{gl(g)}
$$
The adjoint representation cannot be onto, since the center of $\mathfrak{gl(g)}$ are all multiples of the identity matrix

If we consider the non Abelian two dimensional Lie algebra defined by $[X,Y]=Y$, which is the Borel subalgebra of the simple Lie algebra $\mathfrak{sl}(2)$, or the Lie algebra of matrices $\begin{bmatrix}*&*\\0&0\end{bmatrix}$, then we have a solvable and therewith no semisimple Lie algebra which has only a trivial center, too. Hence the condition of semisimplicity is not necessary.

As already mentioned says Ado's theorem, that any finite-dimensional Lie algebra over a field of characteristic zero can be seen as a subalgebra of $\mathfrak{gl}(n)$ for sufficiently large $n.$

 

[suremarc]

I'm pretty unsure about this solution. Linear algebraic groups over finite fields is new territory to me, but I think I managed to leverage some of my abstract algebra knowledge.

Spoiler: Problem 7

We start over the general finite field $\mathbb{F}_q$:
$$|\mathrm{GL}(2,q)|=(q^2-q)(q^2-1)$$ a well known formula which can be deduced by fixing a nonzero vector for the first column (of which there are $q^2-1$), and counting the number of linearly independent vectors as possible values for the second column (of which there are $q^2-q$ for each vector in the first column).

From here on out we assume that $-1$ is a nonresidue modulo $q$, so that $\mathbb{F}_q{[}i{]}$ is a quadratic field extension with $i^2=-1$. ($q=19$ happens to satisfy this property.) We next make use of a representation of $\mathrm{GL}(2,q)$ over $\mathbb{F}_{q^2}$. Choose the following basis for $M(2,q)$: $$\mathbf{1} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix},\quad \mathbf{i} = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix},\quad \mathbf{j} = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix},\quad \mathbf{k} = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$$ Also known as the split-quaternions, we have a 2-dimensional representation over $\mathbb{F}_{q^2}$ via the following isomorphism:
$$w\mathbf{1}+x\mathbf{i}+y\mathbf{j}+z\mathbf{k}\Rightarrow\begin{bmatrix} w + xi & y + zi \\ y - zi & w - xi \end{bmatrix} = \begin{bmatrix} u & v \\ v^* & u^* \end{bmatrix}$$ The center of the split-quaternions consists of scalar matrices over $\mathbb{F}_{q^2}$, of which there are $q^2-1$. Hence, we have $\mathrm{GL}(2,q)\triangleright\mathrm{PU}(Q,q^2)$. The latter group is the 2-dimensional projective unitary group with respect to the Hermitian form $Q=\begin{bmatrix} 1 & 0 \\ 0 & -1\end{bmatrix}$, the form derived from the norm on the split quaternions $|u|^2-|v|^2$. Its order is $q(q-1)$.

We restate the following of the Sylow theorems for clarity: the number of conjugates of the Sylow $p$-subgroup of a group $G$, $n_p$, satisfies $n_p\equiv 1\, (\mathrm{modulo}\,p)$ and $n_p|[P : G]$ where $P$ is any Sylow $p$-subgroup.

The Sylow $p$-subgroup of order $q$ can be generated by the element $$1+\frac{1}{2}(\mathbf{i}+\mathbf{j})$$which maps to the matrix $\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ in the original representation over $\mathbb{F}_q$. Observe that the Sylow $q$-subgroup must be normal: $n_q$ divides $q-1$ and must have remainder 1 when divided by $q$, implying the subgroup only has one conjugate. Since $q$ is prime, the Sylow $q$-subgroup is isomorphic to $C_q$.

Now we restrict our attention to the case $q=19$. We have $\mathrm{PU}(Q,q^2)\cong C_q\times G$, where $|G|=19-1=2\cdot 3^2$. By inspection, we see that the Sylow $3$-subgroup must be normal, since $n_3$ must divide $2$ and have remainder 1 when divided by $3$. Note that the equivalence class of the element $5+10\mathbf{j}$ in $\mathrm{PU}(Q,19^2)$ has order $9$ (since the smallest $n$ such that $(5+10\mathbf{j})^n$ is scalar is $9$), generating a cyclic subgroup of order 9, thus the Sylow $3$-subgroup in $G$ is isomorphic to $C_9$.

In total, we have $\mathrm{GL}(2,19)\cong\mathbb{F}_{19^2}^\times\rtimes\mathrm{PU}(Q,19^2)$, and $\mathrm{PU}(Q,19^2)\cong C_{19}\rtimes (C_9\rtimes C_2).$ The composition factors are thus: $$\mathrm{GL}(2,19)\cong (C_{8}\times C_{9}\times C_5)\rtimes(C_{19}\rtimes (C_9\rtimes C_2)).$$

 

[fresh_42]

I'm pretty unsure about this solution. Linear algebraic groups over finite fields is new territory to me, but I think I managed to leverage some of my abstract algebra knowledge.

Spoiler: Problem 7

We start over the general finite field $\mathbb{F}_q$:
$$|\mathrm{GL}(2,q)|=(q^2-q)(q^2-1)$$ a well known formula which can be deduced by fixing a nonzero vector for the first column (of which there are $q^2-1$), and counting the number of linearly independent vectors as possible values for the second column (of which there are $q^2-q$ for each vector in the first column).

From here on out we assume that $-1$ is a nonresidue modulo $q$, so that $\mathbb{F}_q{[}i{]}$ is a quadratic field extension with $i^2=-1$. ($q=19$ happens to satisfy this property.) We next make use of a representation of $\mathrm{GL}(2,q)$ over $\mathbb{F}_{q^2}$. Choose the following basis for $M(2,q)$: $$\mathbf{1} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix},\quad \mathbf{i} = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix},\quad \mathbf{j} = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix},\quad \mathbf{k} = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$$ Also known as the split-quaternions, we have a 2-dimensional representation over $\mathbb{F}_{q^2}$ via the following isomorphism:
$$w\mathbf{1}+x\mathbf{i}+y\mathbf{j}+z\mathbf{k}\Rightarrow\begin{bmatrix} w + xi & y + zi \\ y - zi & w - xi \end{bmatrix} = \begin{bmatrix} u & v \\ v^* & u^* \end{bmatrix}$$ The center of the split-quaternions consists of scalar matrices over $\mathbb{F}_{q^2}$, of which there are $q^2-1$. Hence, we have $\mathrm{GL}(2,q)\triangleright\mathrm{PU}(Q,q^2)$. The latter group is the 2-dimensional projective unitary group with respect to the Hermitian form $Q=\begin{bmatrix} 1 & 0 \\ 0 & -1\end{bmatrix}$, the form derived from the norm on the split quaternions $|u|^2-|v|^2$. Its order is $q(q-1)$.

We restate the following of the Sylow theorems for clarity: the number of conjugates of the Sylow $p$-subgroup of a group $G$, $n_p$, satisfies $n_p\equiv 1\, (\mathrm{modulo}\,p)$ and $n_p|[P : G]$ where $P$ is any Sylow $p$-subgroup.

The Sylow $p$-subgroup of order $q$ can be generated by the element $$1+\frac{1}{2}(\mathbf{i}+\mathbf{j})$$which maps to the matrix $\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ in the original representation over $\mathbb{F}_q$. Observe that the Sylow $q$-subgroup must be normal: $n_q$ divides $q-1$ and must have remainder 1 when divided by $q$, implying the subgroup only has one conjugate. Since $q$ is prime, the Sylow $q$-subgroup is isomorphic to $C_q$.

Now we restrict our attention to the case $q=19$. We have $\mathrm{PU}(Q,q^2)\cong C_q\times G$, where $|G|=19-1=2\cdot 3^2$. By inspection, we see that the Sylow $3$-subgroup must be normal, since $n_3$ must divide $2$ and have remainder 1 when divided by $3$. Note that the equivalence class of the element $5+10\mathbf{j}$ in $\mathrm{PU}(Q,19^2)$ has order $9$ (since the smallest $n$ such that $(5+10\mathbf{j})^n$ is scalar is $9$), generating a cyclic subgroup of order 9, thus the Sylow $3$-subgroup in $G$ is isomorphic to $C_9$.

In total, we have $\mathrm{GL}(2,19)\cong\mathbb{F}_{19^2}^\times\rtimes\mathrm{PU}(Q,19^2)$, and $\mathrm{PU}(Q,19^2)\cong C_{19}\rtimes (C_9\rtimes C_2).$ The composition factors are thus: $$\mathrm{GL}(2,19)\cong (C_{8}\times C_{9}\times C_5)\rtimes(C_{19}\rtimes (C_9\rtimes C_2)).$$

This is wrong. A short way to the answer is to factor out obvious normal subgroups, and deal with the rest.

 

[suremarc]

This is wrong. A short way to the answer is to factor out obvious normal subgroups, and deal with the rest.

I think I've now figured out how I'm supposed to do it, but I seem to have "proven" in my original proof that $\mathrm{PSL}(2,q)$ is not simple, which contradicts what I'm reading. I will take a closer look.

 

[mathwonk]

A relevant result for prob. #7 seems to be Thm 8.3, chapter 8, page 296, of Mike Artin's Algebra. that PSL2(Z/19Z) is simple.

 

[fresh_42]

A relevant result for prob. #7 seems to be Thm 8.3, chapter 8, page 296, of Mike Artin's Algebra. that PSL2(Z/19Z) is simple.

... in which case I want to know the name of the criterion for this result.

 

[mathwonk]

not sure how to answer. mike just proves directly that any non trivial normal sub group is everything, by showing it must contain elements that generate the group. so,.."bare hands" criterion? there are 6 lemmas in the proof. He proves it for any field of characteristic not equal to 2, and containing at least 7 elements.

 

[Shreya]

Hello, I am Shreya Anish, Class X, India
This is my solution for question 15 for high schoolers

Please correct me if I am wrong

 

[fresh_42]

Hello, I am Shreya Anish, Class X, India
This is my solution for question 15 for high schoolers

Please correct me if I am wrong
View attachment 286166

Here is the reason why the formula is correct:

The evacuated Magdeburg hemispheres are affected by the difference of external and internal air pressure $\Delta p$ which presses them together. To calculate the total force on one of the two hemispheres, we consider a surface element $dA$. The ambient air exerts a force $d\vec{F}$ on this area that is perpendicular to the surface element and of an amount $dF=\Delta p\,dA\,.$ However, we are only interested in the horizontal part of this force $\vec{F}_{\|}$ which is parallel to the direction to which the horses pull, i.e. parallel to the horizontal symmetry axis of the hemisphere. The perpendicular components $\vec{F}_{\perp}$ cancel themselves out. If we denote the angle $\varphi$ between the normal to the surface and the direction of pull, then the parallel component has an amount of
$$
dF_{\|} = dF\cdot \cos \varphi = \Delta p \;dA\cdot \cos \varphi =:\Delta p\; dA'=:dF'_{\|}
$$
The quantity $dA'=dA\cdot\cos \varphi$ can be viewed as parallel projection ot the surface area $dA$ onto a cylinder (see the figure).

The parallel component $dF'_{\|}$ of the force which the air pressure exerts onto the projected surface area element $dA'$ is thus of the same amount as the parallel component $dF_{\|}$ of the original force exerts on the original surface element $dA\,.$

The total amount of force exerted by the air pressure onto the hemisphere is the sum of all forces over the surface elements which compose the hemisphere. Since $dF_{\|}=dF'_{\|}$ we have for the total amount $F_{air}=F'_{air}$, the force onto the projection. So the two hemispheres are pressed together as two cylinders were, whose diameters correspond to the section of the hemispheres: $R$. The force of air pressure on a cylinder is easy to calculate. It's simply the product of pressure and circle area: $$F_{air}=F'_{air}=\Delta p \;A_{\circ}=\pi R^2 \Delta p$$
The example then calculates to a force of
$$
F_{air}=\pi \cdot (0.3)^2 (1.013-0.1)\;10^{5}\;N\approx 26\;kN
$$
which each group of horses has to come up with in order to separate the hemispheres. For comparison: One horsepower is approximately $735.5$ Watt, so $30$ horses would produce $22\;kW$. A horse pulls with approximately $10-12\,\%$ of its weight, i.e. with ca. $700\;N$ or $21\;kN$ for $30$ horses.

 

[Shreya]

Here is the reason why the formula is

Thanks a Lot, fresh_42 ! That was a really beautiful problem. Where can I get more such physics questions with interesting math?

 

[fresh_42]

Good question. We will have 5 of those next month, however, I don't even remember where I got the one above from. I would look for physics olympiad questions, or google for mechanics exams or similar.

 

[Shreya]

For question Number 13, I think x=y=z=c/3

 

[Not anonymous]

13. Maximize $f(x, y, z) = 4x^2y^2 - (x^2 + y^2 - z^2)^2$ under the conditions $x + y + z = c$ and $x, y, z > 0$.

Spoiler: Question 13

Since $z= c - (x+y)$, $f(x, y, z)$ can be expressed as a function of just the 2 variables, $x, y$, i.e. as $g(x, y) = 4x^2y^2 - (x^2 + y^2 - (c-x-y)^2)^2 = 4x^2y^2 - (c^2 + 2xy - 2cx - 2cy)^2$. To find the maximum or minimum value of $g(x,y)$ for a fixed value of $y$, we equate the partial derivative of the function w.r.t. $x$ to 0 and solve for $x$. $$
\frac {\partial g} {\partial x} = 8xy^2 -2(c^2+2xy-2cx-2cy)(2y-2c) = 16cxy - 8c^2x - 12c^2y + 8cy^2 + 4c^3
$$

Equating the partial derivative to zero yields $$
16cxy - 8c^2x - 12c^2y + 8cy^2 + 4c^3 = 0 \Rightarrow x = \dfrac {12c^2y - 8cy^2 - 4c^3} {16cy - 8c^2} =
\dfrac {3cy - 2y^2 - c^2} {4y - 2c}
$$

$\Rightarrow x_m = \dfrac {(2y - c)(y-c)} {-2(2y - c)} = \dfrac {c-y}{2}$ where $x_m$ is the value of $x$ that yields minimum or maximum of $g(x, y)$ for a fixed value of $y$. To find whether it is the minimum or maximum point, we need to consider the 2nd order partial derivative w.r.t. $x$.

$\frac{\partial^2 g} {\partial x^2} = 16cy - 8c^2 = 8c(2y-c)$. Since $c > 0$, this expression is positive if $y > \frac{c}{2}$ and negative if $y < \frac{c}{2}$. Therefore, if $x_m$ would correspond to a maximum (for a fixed value of $y$) if $y < \frac{c}{2}$. (Observation 1)

Now $g(x_m, y)$ can be viewed as a function of just a single variable, $y$.

$$
h(y) \equiv g(x_m, y) = f(\dfrac {c-y}{2}, y, c - y - \dfrac {c-y}{2}) = 4{\dfrac {c-y}{2}}^2 - y^4 = c^2y^2 - 2cy^3
$$

At the maxima and minima of $h(y)$, the first order derivative would be zero, i.e. $$h'(y) = 0
\Rightarrow 2c^2y - 6cy^2 = 0 \Rightarrow 2cy(c - 3y) = 0
$$

Since $c, y > 0$, the above equation implies that we must have $c -3y = 0$ as the only solution for $h'(y)=0$, i.e. $y = \dfrac{c}{3}$.

Now $h''(y) = 2c^2 - 12cy = 2c(c - 6y)$ and this expression takes a negative value when $y = \dfrac{c}{3}$, hence $y= \dfrac{c}{3}$ must correspond to a maximal point, not a minimum. We also note that as per (Observation 1) too, $x_m$ will correspond to a maximum point of $g(x, y)$ (for a fixed $y$) if $y < \dfrac{c}{2}$ and $y = \dfrac{c}{3}$ meets this condition too. Hence, $f(x, y, z)$ is maximized with $y = y_m = \dfrac{c}{3}$ and $x = x_m = \dfrac{c-y_m}{2}$ and $z$ derived using $c - x - y$. In other words, the maximum is achieved with $x = y = z = \dfrac{c}{3}$ and this maximum value is $f(\dfrac{c}{3}, \dfrac{c}{3}, \dfrac{c}{3}) = 3\left(\dfrac{c}{3}\right)^4 = \dfrac{c^4}{27}$

 

[fresh_42]

Spoiler: Question 13

Since $z= c - (x+y)$, $f(x, y, z)$ can be expressed as a function of just the 2 variables, $x, y$, i.e. as $g(x, y) = 4x^2y^2 - (x^2 + y^2 - (c-x-y)^2)^2 = 4x^2y^2 - (c^2 + 2xy - 2cx - 2cy)^2$. To find the maximum or minimum value of $g(x,y)$ for a fixed value of $y$, we equate the partial derivative of the function w.r.t. $x$ to 0 and solve for $x$. $$
\frac {\partial g} {\partial x} = 8xy^2 -2(c^2+2xy-2cx-2cy)(2y-2c) = 16cxy - 8c^2x - 12c^2y + 8cy^2 + 4c^3
$$

Equating the partial derivative to zero yields $$
16cxy - 8c^2x - 12c^2y + 8cy^2 + 4c^3 = 0 \Rightarrow x = \dfrac {12c^2y - 8cy^2 - 4c^3} {16cy - 8c^2} =
\dfrac {3cy - 2y^2 - c^2} {4y - 2c}
$$

$\Rightarrow x_m = \dfrac {(2y - c)(y-c)} {-2(2y - c)} = \dfrac {c-y}{2}$ where $x_m$ is the value of $x$ that yields minimum or maximum of $g(x, y)$ for a fixed value of $y$. To find whether it is the minimum or maximum point, we need to consider the 2nd order partial derivative w.r.t. $x$.

$\frac{\partial^2 g} {\partial x^2} = 16cy - 8c^2 = 8c(2y-c)$. Since $c > 0$, this expression is positive if $y > \frac{c}{2}$ and negative if $y < \frac{c}{2}$. Therefore, if $x_m$ would correspond to a maximum (for a fixed value of $y$) if $y < \frac{c}{2}$. (Observation 1)

Now $g(x_m, y)$ can be viewed as a function of just a single variable, $y$.

$$
h(y) \equiv g(x_m, y) = f(\dfrac {c-y}{2}, y, c - y - \dfrac {c-y}{2}) = 4{\dfrac {c-y}{2}}^2 - y^4 = c^2y^2 - 2cy^3
$$

At the maxima and minima of $h(y)$, the first order derivative would be zero, i.e. $$h'(y) = 0
\Rightarrow 2c^2y - 6cy^2 = 0 \Rightarrow 2cy(c - 3y) = 0
$$

Since $c, y > 0$, the above equation implies that we must have $c -3y = 0$ as the only solution for $h'(y)=0$, i.e. $y = \dfrac{c}{3}$.

Now $h''(y) = 2c^2 - 12cy = 2c(c - 6y)$ and this expression takes a negative value when $y = \dfrac{c}{3}$, hence $y= \dfrac{c}{3}$ must correspond to a maximal point, not a minimum. We also note that as per (Observation 1) too, $x_m$ will correspond to a maximum point of $g(x, y)$ (for a fixed $y$) if $y < \dfrac{c}{2}$ and $y = \dfrac{c}{3}$ meets this condition too. Hence, $f(x, y, z)$ is maximized with $y = y_m = \dfrac{c}{3}$ and $x = x_m = \dfrac{c-y_m}{2}$ and $z$ derived using $c - x - y$. In other words, the maximum is achieved with $x = y = z = \dfrac{c}{3}$ and this maximum value is $f(\dfrac{c}{3}, \dfrac{c}{3}, \dfrac{c}{3}) = 3\left(\dfrac{c}{3}\right)^4 = \dfrac{c^4}{27}$

This is true in case $x,y,z$ are the side lengths of a triangle. The case $c>z\geq x+y > y \geq x>0$ is a bit more difficult (see discussion at the beginning of the thread).

 

[Not anonymous]

This is true in case $x,y,z$ are the side lengths of a triangle. The case $c>z\geq x+y > y \geq x>0$ is a bit more difficult (see discussion at the beginning of the thread).

In the proof I gave, I did not assume that $x, y, z$ must obey the triangle inequality. The only assumptions made are what are given in the question. I do not understand why the derivative-based solution will not help find the maximum if $c>z\geq x+y > y \geq x>0$, mentioned above as the more difficult case. Was there an add-on to the original question which is mentioned in some post? Or is a simpler, more intuitive solution expected?