# Category Archives: math analysis

this studies generalizations of calculus

## wedge product example

When bivectors are defined by

$\beta^i\wedge\beta^j:=\beta^i\otimes\beta^j-\beta^j\otimes\beta^i$,

so, for two generic covectors

$\theta=a\beta^1+b\beta^2+c\beta^3$ and $\phi=d\beta^1+e\beta^2+f\beta^3$,

we have the bivector

$\theta\wedge\phi=(bf-ce)\beta^2\wedge\beta^3+(cd-af)\beta^3\wedge\beta^1+(ad-be)\beta^1\wedge\beta^2$.

Otherwise,

Cf. this with the data $\left(\begin{array}{c}a\\b\\c\end{array}\right)$ and $\left(\begin{array}{c}d\\e\\f\end{array}\right)$ to construct the famous

$\left(\begin{array}{c}a\\b\\c\end{array}\right)\times\left(\begin{array}{c}d\\e\\f\end{array}\right)=\left(\begin{array}{c}bf-ce\\cd-af\\ad-be\end{array}\right)$

So, nobody should be confused about the uses of the symbol $\wedge$ dans le calcul vectoriel XD

## double coset counting formula

the double coset counting formula is a relation inter double cosets $HaK$, where $a\in G$ and $H,K$ subgroups in $G$. This is:

$\#(HaK)=\frac{|H||K|}{|H\cap aKa^{-1}|}$

and

$\#(G/K)=\sum_a[H;H\cap aKa^{-1}]$

The proof is easy.

One is to be bounded to the study of the natural map $H\times K\stackrel{\phi_a}\to HaK$. And it uses the second abstraction lemma.

The formula allows you to see the kinds of subgroups of arbitrary $H$ versus $K$ a $p-SS$ of $G$, $p-SS$ for the set of the $p$– Sylow subgroups.

Or, you can see that through the action $H\times G/K\to G/K$ via $h\cdot aK=haK$ you can get:

• ${\rm Orb}_H(aK)=\{haK\}$ which comply the equi-partition
• $HaK=aK\sqcup haK\sqcup...\sqcup h_taK$, so $\#(HaK)=m|K|$, for some $m\in \mathbb{N}$
• ${\rm St}_H(aK)=H\cap aKa^{-1}$

then you can deduce:

$|G|=\sum_a\frac{|H||K|}{|H\cap aKa^{-1}|}$

Now, let us use those ideas to prove the next statement:

Let $G$ be a finite group, with cardinal $|G|=q_1^{n_1}q_2^{n_2}\cdots q_t^{n_t}$, where each $q_i$ are primes with $q_1 and $n_i$ positive integers.

Let $H$ be a subgroup of $|G|$ of index $[G:H]=q_1$.

Then, $H$ is normal.

Proof:

By employing $K=H$ in the double coset partition, one get the decomposition:

$G=HeH\sqcup Ha_1H\sqcup...\sqcup Ha_tH$

So by the double coset counting formula you arrive to:

$|G/H|=1+[H:H\cap a_1Ha_1^{-1}]+\cdots+[H:H\cap a_tHa_t^{-1}]$

i.e.

$q_1=1+\frac{|H|}{|H\cap a_1Ha_1^{-1}|}+\cdots+\frac{|H|}{|H\cap a_tHa_t^{-1}|}$

From this, we get $\frac{|H|}{|H\cap a_iHa_i^{-1}|}.

But $|G|=q_1|H|$ as well $|H|=|H\cap a_iHa_i^{-1}|[H:H\cap a_iHa_i^{-1}]$ so

$|G|=q_1|H\cap a_iHa_i^{-1}|[H:H\cap a_iHa_i^{-1}]$, i.e.

$[H:H\cap a_iHa_i^{-1}]$ divides $|G|$

Then $[H:H\cap a_iHa_i^{-1}]=1$. So $|H|=|H\cap a_iHa_i^{-1}|$ for each $a_i$.

This implies $H=H\cap a_iHa_i^{-1}$ and so $H=a_iHa_i^{-1}$ for all the posible $a_i$, hence, $H$ is normal.

QED.

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## puntos críticos de una función suave en el círculo

En esta breve nota demostraremos que cada función $f:S^1\to{\mathbb{R}}^1$ que tenga un punto crítico aislado debe de tener otro.

Entonces supongamos que existe un punto $p$ en $S^1$ talque ${\rm grad}f(p)=\left[\frac{\partial f}{\partial x}|_p,\frac{\partial f}{\partial y}|_p\right]=\vec{0}$, pero si elegimos la parametrización $\phi:\ ]0,2\pi[\longrightarrow S^1$ dada por $t\longmapsto\left(\begin{array}{c}\cos(t)\\ \\ \sin(t)\end{array}\right)$, entonces tenemos una función $g=f\circ\phi$ para la cual, la regla de la cadena implica que $g'=f'(\phi)\phi'$ satisface

$\frac{d g}{dt}|_{t_0}=\left[\frac{\partial f}{\partial x}|_p,\frac{\partial f}{\partial y}|_p\right]\left(\begin{array}{c}-\sin\\ \\ \cos\end{array}\right)_{|_{t_0}}$

i.e.

$\frac{d g(t_0)}{dt}=-\frac{\partial f(p)}{\partial x}\sin(t_0)+\frac{\partial f(p)}{\partial y}\cos(t_0)$

entonces si ${\rm grad}f(p)=\vec{0}$ tendremos $\frac{d g(t_0)}{dt}=0$, en otras palabras $g$ tiene puntos críticos en $t_0$ y en $t_0+2\pi$.

Pero además $g(t_0)=f\circ\phi(t_0)=f(p)$ tanto como

$g(t_0+2\pi)=f\circ\phi(t_0+2\pi)=f\circ\phi(t_0)=f(p)$

es decir $g(t_0)=g(t_0+2\pi)$ y entonces –por el teorema de Rolle– existe $t_1$ en el intervalo abierto $]t_0,t_0+2\pi[$ talque $\frac{d g(t_1)}{dt}=0$.

Pero si nos restringimos a $S^1\setminus\{p\}$ entonces $f=g\circ\phi^{-1}$,
y así (también por la regla de la cadena) tenemos ${\rm grad}f=\frac{dg}{dt}\ {\rm grad}\ \phi^{-1}$ i.e.

$\left[\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}\right]=\frac{dg}{dt}\left[\frac{\partial\phi^{-1}}{\partial x},\frac{\partial\phi^{-1}}{\partial y}\right]$

que evaluando en $t_1$ implica

$\frac{\partial f(q)}{\partial x}=\frac{dg(t_1)}{dt}\frac{\partial\phi^{-1}(q)}{\partial x}=0$

tanto como

$\frac{\partial f(q)}{\partial y}=\frac{dg(t_1)}{dt}\frac{\partial\phi^{-1}(q)}{\partial y}=0$

por lo tanto ${\rm grad}f(q)=\vec{0}$, donde $q\neq p$ $\Box$

critical points of functions on the circle

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## Multilinear Algebra

álgebra multilineal es como un cálculo vectorial dos o álgebra lineal tres

entonces para poder hacer cálculos en otras geometrías, inclusive muy diferentes a $\mathbb{R}^n$ vamos viendo hacia donde tenemos que caminar: ver  (un post previo con estas ideas en mente).

RE-ENGINEERING LINEAR ALGEBRA

RE-ENGINEERING VECTOR CALCULUS

## words-length and words in a group

the group is $\mathbb{Z}_2*\mathbb{Z}_3=\langle a,\ b\mid a^2,\ b^3\rangle$, and the picture:

Correctly predicts the next one. It is A164001 in the OEIS data-base. Dubbed “Spiral of triangles around a hexagon“. It has the generating function $-(x+1)+\frac{x^2+2x+1}{1-x^2-x^3}$, Why would it be? :) . This another is A000931.

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Filed under free group, math, sum of reciprocals

## multilinear algebra 1, a synoptic view

what is math? let us discuss:

 Baby Abstract Multilinear Algebra Baby Multilinear Algebra  of Inner Product Spaces Reciprocal basis Metric tensor, lenght, area, volumen Bilinear transformations Musical isomorphisms Change of basis Calculus in $\mathbb{R}^n$ Partial derivatives Taylor series Jacobians Chain’s rule Directional derivatives Covariant derivative and Gauss equation Coordinated changes Differential forms with exterior derivatives the $\mathbb{R}^3$ de Rham’s complex Covariant gradient little Stokes’ theorems: Green, Gauss. Algebraic Differential Geometry Parameterizations: curves and surfaces Tangent vectors, tangent space, tangent bundle Curves in $\mathbb{R}^2$ and $\mathbb{R}^3$ and on surfaces in $\mathbb{R}^3$ Surfaces in $\mathbb{R}^3$ all classical surfaces rendered tangent space change of basis vector fields and tensor fields Christoffel’s symbols (connection coefficients) Curvatures (Gaussian, Mean, Principals, Normal and Geodesic) Vector Fields, Covector Fields, Tensor Fields Integration: Gauss-Bonnet, Stokes Baby Manifolds (topological, differential, analytic, anti-analytic, aritmetic,…) Examples: Lie groups and Fiber bundles

## duality rules!

in this post I am gonna again insist about the cruel confusion spawn by people who, I think, misunderstand certain strategic points in the grasping of tensors.

Vectors in $\mathbb{R}^n$ are linear combinations of

$e_1=\left(\begin{array}{c}1\\ 0\\ 0\\\vdots\\ 0\end{array}\right)$ , $e_2=\left(\begin{array}{c}0\\ 1\\ 0\\\vdots\\ 0\end{array}\right)$ , … , $e_n=\left(\begin{array}{c}0\\\vdots\\ 0\\ 0\\ 1\end{array}\right)$ ,

that is, vectors are expressions as $v=v^1e_1+v^2e_2+\cdots+v^ne_n$, or employing the Einstein-Penrose convention: $v=v^se_s$.

But, covectors are linear combinations of

$e^1=[1,0,0,...,0]$ , $e^2=[0,1,0,...,0]$ , … , $e^n=[0,...,0,1]$ ,

which are also dubbed “projectors” due its role as a linear maps ${\mathbb{R}}^n\to{\mathbb{R}}$.

So a general covector is a linear combination $f_se^s$, using the Einstein-Penrose Convention again.

It is regrettable how many authors are merciless in the misuses of the rows and columns concepts for vectors and covectors respectively. It seems that sacrificing the difference by writing vectors as rows to gain space in the texts of linear algebra books is a big cost in the habits of understand well the matter.

This is perhaps a simple reason why it is so difficult to everybody to grasp the idea of tensors at its first steps. Well, not everybody … hehehe.

We say that duality rules ‘cuz

$e^i(e_j)={\delta^i}_j$

becomes a simple device to serve as a guide for a (apparently) complex calculations that arise when one consider tensors of higher rank.

Once recognized this hurdle, the constructions of a calculus à la  Cartan, using the wedge product and the exterior derivative among tensor fields is an easy cake. These techniques will allow you to free of the ugly vector calculus that rules in each bachelor science schools nowadays.

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