How are vector spaces viewed as universal algebras?

Geometry: Is geometric inversion in the complex plane curl-free when viewed as a vector field, ? If so, what is the potential function of the gradient?

• Precisely, geometric inversion can be regarded on the complex plane without the origin as complex inversion followed by conjugation. At each point, draw the vector corresponding to its image under the transformation -- this defines the vector field.

• Answer:

  You're considering the function which inverts in the unit circle.  A complex number  [math]x+yi[/math]  is sent to the conjugate of its inverse, [math]\dfrac{x+yi}{x^2+y^2}.[/math]Next we interpret that as a vector field  [math]{\mathbf F}:\mathbb R^2\to\mathbb R^2.[/math]  Then[math]\qquad\qquad\displaystyle{\mathbf F}(x,y)=\left(\frac{x}{x^2+y^2},\frac{y}{x^2+y^2}\right)[/math]Note that this is a radial field.  All it's flow lines are along rays through the origin.Its curl is                     [math]\displaystyle\frac\partial{\partial x}\frac{y}{x^2+y^2}-\frac\partial{\partial y}\frac{x}{x^2+y^2}[/math]which is 0.  So this is an irrotational field.There's a theorem which says If  [math]{\mathbf F}[/math]  is an irrotational vector field defined on a simply connected domain in [math]\mathbb R,[/math] then  [math]{\mathbf F}[/math] is a gradient field, the gradient of some scalar field  [math]f.[/math]  That  [math]f[/math]  is called the potential field for [math]{\mathbf F}.[/math]A simply connected region in the plane is one in which every closed curve in it can be shrunk to a point while staying in the region. In other words, it has no holes.  The domain of our vector field  [math]{\mathbf F}[/math]  is the entire plane minus the origin.  It's got a hole, so it's not a simply connected region.It might still have a potential field, though.  Let's see.  We're looking for a function  [math]f(x,y)[/math]  such that[math]\qquad\qquad\displaystyle\frac\partial{\partial x}=\frac{x}{x^2+y^2}[/math] and                     [math]\displaystyle\frac\partial{\partial x}=\frac{y}{x^2+y^2}[/math] Aha, there is one!                     [math]f(x,y)=\dfrac{\log(x^2+y^2)}2[/math]