0

I don't understand why this statement in lecture is true :

let $\gamma : [a,b] \rightarrow \mathbb{R}^2 \setminus \left \{ p\right \} $ a continuous path, and v any vector in the plane. Moreover, let $\gamma$ +v be the path defined by
$(\gamma + v)(t) = \gamma(t) + v $
it follows that
$W(\gamma +v , p+ v) = W(\gamma, p)$
in this sense we say that winding numbers are invariant under translation.

But why does this follow? Do I have to use some Linear Algebra to get there ? Could someone maybe explain to me why this statement is true ? Thank you very much.

Bünzli Refinej
  • 35
  • 7
  • 1
    $\gamma+v$ intuitivly is just the translation of the curve along a vector. It has no affect to the curvature of $\gamma$. The winding number is given as the total curvature of a closed curve over $2\pi$. Since the curvature remains the same under such translation the total curvature is also the same, hence the winding number is the same aswell. – y_andoni Nov 19 '21 at 08:21
  • thank you very much! this makes sense – Bünzli Refinej Nov 19 '21 at 10:36

1 Answers1

2

Identifying $\mathbb{R}^2$ with $\mathbb{C}$ in the natural way:\begin{eqnarray*}W(\gamma +v , p+ v) &=&\frac1{2\pi i}\int_a^b \frac1{(\gamma+v)t-(p+v)}\frac{d(\gamma+v)}{dt}dt\\ &=&\frac1{2\pi i}\int_a^b \frac1{\gamma(t)-p}\frac{d\gamma}{dt}dt\\ &=&W(\gamma, p)\end{eqnarray*}

tkf
  • 11,563