Graphical Representation of a Complex Sphere

[Leo Authersh]

@fresh_42 @FactChecker After thinking, I understood that the answer for this question might make the complex numbers comprehensible for me. My question in detail is as follow

Let the equation of a sphere with center at the origin be

$Z1²+Z2²+Z3² = r²$

where Z1 = a+ib, Z2 = c+id, Z3 = s+it

then

How many dimensions do we need to represent this sphere graphically?

 

[fresh_42]

The question about dimension is a question of where. I assume you mean the real numbers as the ones which describe coordinates, because we only can draw real lines. Next you have an object, a sphere, which isn't a vector space, so the term dimension needs to be defined first. Usually one takes the dimension of the tangent space, which is a vector space. But your question seems to be more about the dimension of space, where this sphere can be embedded, i.e. drawn. We have $6$ free real parameters in total, so the answer should be $6$. However, it is a $2$ dimensional complex and a $5$ dimensional real object, because the equation can be used to replace on parameter by an expression of the others and the constant $r$.

An example. Let's consider $z=1+i$. This is a complex number, so $V_\mathbb{C} = z\cdot \mathbb{C}$ is a one dimensional complex vector space spanned by $z$. But with $z \cdot (a+ib) = (a-b) + i(a+b)$ we have $1 \in V_\mathbb{C}$ by choosing $a=-b=\frac{1}{2}$ and also $i \in V_\mathbb{C}$ by choosing $a=b=\frac{1}{2}$, so all complex numbers are in $V_\mathbb{C}$ and as a real vector space $\dim_\mathbb{R}V_\mathbb{C} = 2$ although $\dim_\mathbb{C}V_\mathbb{C} = 1$.

 

[Leo Authersh]

We have $6$ free real parameters in total, so the answer should be $6$. However, it is a $2$ dimensional complex and a $5$ dimensional real object, because the equation can be used to replace on parameter by an expression of the others and the constant $r$.

so all complex numbers are in $V_\mathbb{C}$ and as a real vector space $\dim_\mathbb{R}V_\mathbb{C} = 2$ although $\dim_\mathbb{C}V_\mathbb{C} = 1$.

This explanation makes my comprehension of complex pane better. Thank you for the valuable explanation.

 

[Infrared]

However, it is a $2$ dimensional complex and a $5$ dimensional real object, because the equation can be used to replace on parameter by an expression of the others and the constant $r$.

I think the real dimension should be $4$ since you actually have two constraints (real part and imaginary part both have to be zero). In general, the real dimension should be twice the complex dimension.

 

[Leo Authersh]

I think the real dimension should be $4$ since you actually have two constraints (real part and imaginary part both have to be zero). In general, the real dimension should be twice the complex dimension.

But since this is a complex sphere, which is three dimensional, the real dimensions should be six right? With each complex axis being a superposition of an real axis and an imaginary axis.

 

[fresh_42]

But since this is a complex sphere, which is three dimensional, the real dimensions should be six right? With each complex axis being a superposition of an real axis and an imaginary axis.

Yes. And even if we transform $z_1^2+z_2^2+z_3^2=r^2$ into two real equations to get rid of two real variables, it would be difficult to draw: there are still $4$ free variables (I avoid the term dimension here, as it is no obvious dimension, rather a degree of freedom), and we have expressions $\pm \sqrt{\ldots}$ so uniqueness is lost. In the end we would need all $6$ dimensions for a reasonable embedding, namely $\mathbb{R}^6$. Or the vector space $\mathbb{C}^3$ which is only three dimensional, but still impossible to draw.

 

[WWGD]

@fresh_42 @FactChecker After thinking, I understood that the answer for this question might make the complex numbers comprehensible for me. My question in detail is as follow

Let the equation of a sphere with center at the origin be

$Z1²+Z2²+Z3² = r²$

where Z1 = a+ib, Z2 = c+id, Z3 = s+it

then

How many dimensions do we need to represent this sphere graphically?

Problem is that the equation of a sphere is {$z: |z|=r$}. What you wrote is not equivalent to this. Notice that in $\mathbb R^{n+1}$ , this equation is equivalent to $(x-x_0)^2+...+(x-xn)^2=r^2$, but in the complexes, it is not, i.e., given $w=(z_1, z_2, z_3): |w| \neq z_1^2+z_2^2+z_3^2$..