- #71
martinbn
Science Advisor
- 3,413
- 1,716
Right, forgot the normal one. It is the plane ## x_2=x_3##.fresh_42 said:... 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 said:... which leads to ##T_pM=\ldots ## and ##N_pM= \ldots ##
... maybe in the back ...julian said:Hi @fresh_42. So when you hit 15,000 posts do you get a set of steak knives or something?
martinbn said: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##.
martinbn said:Right, forgot the normal one. It is the plane ## x_2=x_3##.
Yes, and a bit more detailed:martinbn said: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.
This is wrong. A short way to the answer is to factor out obvious normal subgroups, and deal with the rest.suremarc said: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.
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: $$\mathrm{GL}(2,19)\cong (C_{8}\times C_{9}\times C_5)\rtimes(C_{19}\rtimes (C_9\rtimes C_2)).$$
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.fresh_42 said:This is wrong. A short way to the answer is to factor out obvious normal subgroups, and deal with the rest.
... in which case I want to know the name of the criterion for this result.mathwonk said: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.
Here is the reason why the formula is correct:Shreya said: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
Thanks a Lot, fresh_42 ! That was a really beautiful problem. Where can I get more such physics questions with interesting math?fresh_42 said:Here is the reason why the formula is
fresh_42 said: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##.
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 said: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 said: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).
There is no maximum if the boundary is excluded, under the assumption I didn't make a mistake.Not anonymous said: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?
fresh_42 said:There is no maximum if the boundary is excluded, under the assumption I didn't make a mistake.
##c>z\geq x+y > y \geq x>0## makes ##f(x,y)\leq 0## and ##x=y=z=c/3## isn't a solution anymore because ##z=c/3< 2c/3 =x+y.##
fresh_42 said:11. Assume we have put a Cartesian coordinate system on France and got the following positions:
Paris ##(0, 0)##, Lyon ##(3, -8)## and Marseille ##(4, -12)##. Look up the definitions and calculate the distance between Lyon and Marseille according to
- the Euclidean metric.
- the maximum metric.
- the French railway metric.
- the Manhattan metric.
- the discrete metric.