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solutions for problems on vector analysis and differential forms

Solutions

Exercise:
Assume that 4 vectors in \(\mathbb{R}^3\) satisfy \(A+B+C+D\)=0.
Simplify
$$ A\times B-B\times C+C \times D-D\times A$$

Solution: By the anti-symmetry \(-B\times C = C\times B\)
and \(-D\times A = A\times D\) we can rewrite the expression as
$$ …

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March 28

  • Consider the \(k\)-form \(\omega \) defined at \(x\in \Omega\subset \mathbb{R}^n\), given by the sum
    $$ \omega(x)= \sum c_I(x) dx_I $$
    where the multi-index
    \( I=\{1\leq i_1\leq i_2 \leq, \ldots, \leq i_k \leq n\} \).
    Each basis element
    \(dx_I:= dx_{i_1} \wedge
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    Exterior Differential forms

    \(\def\pd{\partial}
    \def\Real{\mathbb{R}}
    \)

    • We define a differential \(k\)-form as
      $$\omega(x) = \sum_{I=\{i_1<i_2<\ldots<i_k\}} c_I(x) dx_{i_1}\wedge \cdots\wedge dx_{i_k},$$
      an exterior form with coefficients depending on the positions.

      (We will be using sometimes simplifying notation \(\xi_i = dx_i\) or \(\eta_i=dy_i\).)

    • Pullback maps: Let
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    Exterior Forms

    \(\def\Real{\mathbb{R}}
    \)

    1. To generalize the notion of 1-forms, we need to develop some algebraic apparatus. It is called Exterior Calculus.
    2. Consider a vector space \(V\cong \mathbb{R}^n\).
      We say that the functional
      $$ \omega:\underbrace{V\times \cdots \times V}_{k \text{ times}}\to \mathbb{R} $$
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    Calculus of Exterior and Differential Forms: Motivation

    \(\def\pd{\partial}\def\Real{\mathbb{R}}
    \)

    1. Let \(\Omega \subset \mathbb{R}^n\) be some Euclidean domain, and consider
      a curve \( \gamma:\underbrace{[0,1]}_{I} \to \Omega \) given by
      $$\gamma(t) = \begin{pmatrix} \gamma_1(t), & \ldots &, \gamma_n(t) \end{pmatrix}. $$Given a real-valued multivariate function \(f:\Omega\to \mathbb{R}\) we’ll try to
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    3.14 Vector Analysis And Classical Identities in 3D

    \(\def\pd{\partial}\def\Real{\mathbb{R}}
    \)

    1. There are two main types of products between vectors in \(\mathbb{R}^3\):
      The inner/scalar/dot product
      $$ A\cdot B = A_x+B_x + A_yB_y + A_z Bz \in \mathbb{R} $$
      is commutative, distributive, and homogenenous.
      The vector (cross) product:
      $$ A\times
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    2.28 operators in Hermitian spaces and their spectra.

    \(\def\Real{\mathbb{R}}
    \def\Comp{\mathbb{C}}\def\Rat{\mathbb{Q}}\def\Field{\mathbb{F}}\def\Fun{\mathbf{Fun}}\def\e{\mathbf{e}}
    \def\f{\mathbf{f}}\def\bv{\mathbf{v}}\def\i{\mathbf{i}}
    \def\eye{\left(\begin{array}{cc}1&amp;0\\0&amp;1\end{array}\right)}
    \def\bra#1{\langle #1|}\def\ket#1{|#1\rangle}\def\j{\mathbf{j}}\def\dim{\mathrm{dim}}
    \def\ker{\mathbf{ker}}\def\im{\mathbf{im}}
    \def\tr{\mathrm{tr\,}}
    \def\braket#1#2{\langle #1|#2\rangle}
    \)

    1.
    Given a bilinear form \(Q(\cdot,\cdot)\), or, equivalently, a mapping \(Q:U\to U^*\), one can bake easily new bilinear forms from linear operators \(U\to U\): just take \(Q_A(u,v):=Q(u,Av)\).

    This opens …

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    midterm

    problems and solutions here.…

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    solutions to 2.14

    Exercise:
    Let \(V\) be the space of real polynomial functions of degree at most \(3\). Consider the quadratic form \(q_1:V \to \mathbb{R}\) given by $$ q_1(f):=|f(-1)|^2+|f(0)|^2+|f(1)|^2. $$ Is this form positive definite?
    Consider another form \(q:V\to \mathbb{R}\) given by $$ …

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    2. 14 quadratic forms

    \(\def\Real{\mathbb{R}}\def\Comp{\mathbb{C}}\def\Rat{\mathbb{Q}}\def\Field{\mathbb{F}}\def\Fun{\mathbf{Fun}}\def\e{\mathbf{e}}
    \def\f{\mathbf{f}}\def\bv{\mathbf{v}}\def\i{\mathbf{i}}
    \def\eye{\left(\begin{array}{cc}1&amp;0\\0&amp;1\end{array}\right)}
    \def\bra#1{\langle #1|}\def\ket#1{|#1\rangle}\def\j{\mathbf{j}}\def\dim{\mathrm{dim}}
    \def\ker{\mathbf{ker}}\def\im{\mathbf{im}}
    \def\tr{\mathrm{tr\,}}
    \def\braket#1#2{\langle #1|#2\rangle}
    \)

    1. Bilinear forms are functions \(Q:U\times U\to k \) that depend on each of the arguments linearly.

    Alternatively, one can think of them as the linear operators
    \[
    A:U\to U^*, …

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