# Tag Archives: tensor

## covariant derivative of covectors

How do you think that the covariant derivative in $\mathbb{R}^3$ is extended over covector fields defined over a surface $\Phi:{\mathbb{R}}^2\hookrightarrow\Sigma\subset{\mathbb{R}}^3$?

We use the Riesz Representation’s Lemma, so if

$dx^k(\quad)=\langle\partial^k,\quad\rangle=\langle g^{sk}\partial_s,\quad\rangle$

then

$\nabla_{\partial_i}dx^k(\quad)=\langle \nabla_{\partial_i}\partial^k,\quad\rangle=\langle -{\Gamma^k}_{is}\partial^s,\quad\rangle$

This implies that we have:

$\nabla_{\partial_i}dx^k=-{\Gamma^k}_{is}dx^s$

This contrast nicely with $\nabla_{\partial_i}\partial_k={\Gamma^s}_{ik}\partial_s$

For a general $w=w_sdx^s$, we use the Leibniz’s rule to get

$\nabla_{\partial_i}w=({w_s}_{,i}-{w_t\Gamma^t}_{si})dx^s$

and

$\nabla_{\partial_i}(w\otimes\theta)=(\nabla_{\partial_i}w)\otimes\theta+w\otimes\nabla_{\partial_i}\theta$

The proof that $\nabla_{\partial_i}\partial^k=\nabla_{\partial_i}(g^{sk}\partial_s)=-{\Gamma^k}_{is}\partial^s$ is very fun!

You gotta remember firmly that the $\partial^k=g^{sk}\partial_s$ form the reciprocal coordinated basis, still tangent vectors but representing (à la Riesz) the coordinated covectors $dx^k$.

Filed under differential geometry, multilinear algebra

## cambio de coordenadas, cambio de base y cambio de componentes

dos GPS como exchange informazione inter sus mediciones

Filed under math

## tangent space duality of a surface

let us describe how the euclidean duality is carried to the tangent bundle of a surface.

We know how euclidean duality is: For each euclidean space, $\mathbb{R}^n$, the coordinated (rectangular)  functions: $x^i(p)=p^i$ are linear, so their derivatives, $Jx^i$, are equal themselves i.e the gradients obey $Jx^i=x^i$, they are dubbed $dx^i$ and they satisfy duality:

$dx^i(e_k)={\delta^i}_k$

Now, if $\Phi:\Omega\to\mathbb{R}^3$  is a parameterization of the surface, $\Sigma\subset\mathbb{R}^3$, and $f:\Sigma\to\mathbb{R}$ is a “measure” then, doing calculus in the surface means do calculus to $f\circ\Phi$. Let $g=f\circ\Phi$.

Let us name $\xi^1,\xi^2,\xi^3$ the coordinated (rectangular) functions on $\mathbb{R}^3$.

So by the chain rule we have: $Jg=Jf\cdot J\Phi$ that is, in terms of gradients:

${\rm grad}(g)={\rm grad}(f)\!\cdot\!J\Phi$

or

$[\begin{array}{cc}\frac{\partial g}{\partial x^1}&\frac{\partial g}{\partial x^2}\end{array}]=[\begin{array}{ccc}\frac{\partial f}{\partial\xi^1}&\frac{\partial f}{\partial\xi^2}&\frac{\partial f}{\partial\xi^3}\end{array}]\!\cdot\!\left(\begin{array}{cc}\frac{\partial \xi^1}{\partial x^1}&\frac{\partial \xi^1}{\partial x^2}\\\frac{\partial \xi^2}{\partial x^1}&\frac{\partial \xi^2}{\partial x^2}\\\frac{\partial \xi^3}{\partial x^1}&\frac{\partial \xi^3}{\partial x^2}\end{array}\right)$

So for the functions $u^i=x^i\circ\Phi^{-1}$ we get $x^i=u^i\circ\Phi$ and by the same rule just above

$dx^i=du^i\!\cdot\!J\Phi$

where evaluating at the basis $e_1,e_2$ of $\mathbb{R}^2$ give

$du^i(J\Phi e_k)=dx^i(e_k)={\delta^i}_k$

but since it is known that the $J\Phi e_1,J\Phi e_2$ generate $T_p\Sigma$, then both: $du^1,du^2$ generate $T_p\Sigma^*$, the co-tangent space at $p\in\Sigma$,… wanna see a picture?

## multilineal lección six

### tensor fields on a surface and its covariant derivatives

the collection $T\Sigma$ of all the tangent spaces $T_p\Sigma$ is called the tangent bundle of the surface, i.e.

$T\Sigma=\bigsqcup_pT_p\Sigma$

A vector field in a surface is a mapping $X:\Sigma\to T\Sigma$ with the condition $p\mapsto X\in T_p\Sigma$ and since $\partial_1,\partial_2$ span $T_p\Sigma$ then

$X=X^s\partial_s$

This construction determines a contravariant tensor field of rank one, which is taken as the base to ask how other tensor fields -of any rank and any variance- vary…

wanna know more?

Filed under differential geometry, fiber bundle, math, multilinear algebra

## real multilinear algebra

el álgebra multilineal sobre los números reales, $\mathbb{R}$, incluye a las formas diferenciales euclideas, ahí uno estudia la amalgama producida por el álgebra lineal y el cálculo en varias variables. Pero además si uno dispone del lenguage elemental del álgebra tensorial de espacios vectoriales sobre los reales, es decir la categoría ${\rm{Vect}}_{\mathbb{R}}$, entonces uno puede incluir los principios de la geometría diferencial (de curvas y de superficies en $\mathbb{R}^3$) para obtener un curso realmente útil y moderno. En el mero corazón de esta teoría está el complejo de de Rham que permite construir los módulos cohomológicos del álgebra de Grassmann (módulo-$C^{\infty}$), de un conjunto abierto euclídeo.

Otra cosa es la complex multilinear algebra…

## wedge complements in finite dimension

in the next counting experiment we are going  to calculate the dimensions of some linear subspaces of the Grassmann algebra of a vector space of dimension $n$.

We should  be using the intuition granted by $\Lambda(\mathbb{R}^n)$

The definitions are:

• $C_{dx}=\{\alpha\mid \alpha\wedge dx=0\}$
• $M_{dx}={C_{dx}}^{\top}$
• $C_{dx\wedge dy}=\{\alpha\mid \alpha\wedge dx\wedge dy=0\}$
• $M_{dx\wedge dy}={C_{dx\wedge dy}}^{\top}$
• $C_{dx\wedge dy\wedge dz}$
• $M_{dx\wedge dy\wedge dz}$

doesn’t anybody know the name of the result?…  ‘cuz if it hasn’t, I will claim mine : )

Meanwhile, let me refrain the definition  that says:

$\Lambda(\Omega)$

is the $C^{\infty}(\Omega)$module over the symbols $dx^1,dx^2,...,dx^n$ and over an open set $\Omega\subseteq\mathbb{R}^n$

stay tune…