What does a complex vector "look" like?

[bsaucer]

I'm trying to picture a complex vector in my mind. A real vector is usually drawn as an arrow pointing in some direction in space and having a positive real magnitude "length". I'm trying to understand a complex vector the same way.
Is it drawn as an arrow pointing in some direction having a complex "length"? Or is it two arrows pointing in different directions having positive real length representing the real and imaginary components? Or is it something else?
Also, if I cover the ordinary sphere with complex vectors, does the hairy ball theorem still hold?

 

[fresh_42]

I'm trying to picture a complex vector in my mind. A real vector is usually drawn as an arrow pointing in some direction in space and having a positive real magnitude "length". I'm trying to understand a complex vector the same way.

It is the same, with the difference, that we cannot imagine the complex numbers as a number line due to the lack of an order.

Is it drawn as an arrow pointing in some direction ...

In principle, yes.

... having a complex "length"?

No. The length is still a real number. But the direction is an arrow in $\mathbb{C}^n.$

Or is it two arrows pointing in different directions having positive real length representing the real and imaginary components?

That's our usual workaround. Two real directions instead of one complex. However, that already comes to an end for $\mathbb{C}^2$ because we don't know what a four-dimensional real space looks like.

Or is it something else?

It is something else because the real plane that we use to illustrate complex numbers does not know the relation $i^2+1=0.$

Also, if I cover the ordinary sphere with complex vectors, does the hairy ball theorem still hold?

Hmm, I first have to think about what a complex sphere is.

 

[bsaucer]

Can you cover a Riemann sphere with complex vectors?

 

[fresh_42]

Can you cover a Riemann sphere with complex vectors?

Yes. The points correspond to complex numbers plus infinity, and a number is a one-dimensional vector. You can construct all kinds of mappings, so it all depends on what you mean by "cover".

 

[bsaucer]

In other words, a Riemann sphere is covered by complex scalars?
Does this mean that a complex vector resides in a complex space having twice as many real dimensions?

 

[fresh_42]

In other words, a Riemann sphere is covered by complex scalars?

Whatever "covered" means. The points on the Riemann sphere can be identified with complex numbers, except the north pole, which corresponds to infinity. Whether this can be described as "covered" is another question.

Does this mean that a complex vector resides in a complex space having twice as many real dimensions?

Yes, you can identify complex vectors with real vectors of twice as many components. However, you cannot identify complex numbers with two real numbers. This needs an additional rule: $(0,1)\cdot (0,1)=(-1,0)$ which is more than a simple identification. It characterizes complex numbers and makes them different from $\mathbb{R}^2.$ But if you only consider addition and real scalar multiplication as in the case of real vector spaces, i.e. no multiplications, then you can make this identification.

You see, we must be careful with what we say. A single word can change everything. That's why "covered" isn't a well defined word in this context.

 

[jbergman]

I'm trying to picture a complex vector in my mind. A real vector is usually drawn as an arrow pointing in some direction in space and having a positive real magnitude "length". I'm trying to understand a complex vector the same way.
Is it drawn as an arrow pointing in some direction having a complex "length"? Or is it two arrows pointing in different directions having positive real length representing the real and imaginary components? Or is it something else?
Also, if I cover the ordinary sphere with complex vectors, does the hairy ball theorem still hold?

The main difference as I think of it is that the scalar can rotate your vector. So for instance if we have a 1-d complex vector space with a unit basis vector pointing in the x direction. Then multiplication by a complex scalar can give you any vector in the plane.

 

[mathwonk]

@bsaucer: In one sense the hairy ball theorem is still true for a field of complex tangent vectors to the ordinary sphere. I.e. one constricts the real tangent space to the real sphere by gluing copies of R^2 together using the the derivatives of the smooth coordinate maps. Then it makes sense to speak of smooth tangent vector fields, and any smooth tangent vector field on the sphere must have 2 zeroes, "counted properly". But one can also cover the sphere by open sets whose coordinate maps have not just smooth but holomorphic compositions, hence one can also construct the sphere as a complex holomorphic manifold of complex dimension one, and by gluing copies of C ≈ R^2 together by the derivatives of these holomorphic coordinate maps, one gets the "same" object, i.e. the sphere with a real 2 dimensional vector bundle over it, only this time the fibers of the vector bundle are copies of C, i.e. R^2 with more structure.

Now a holomorphic complex tangent vector field on the complex Riemann sphere, is in particular also a smooth tangent vector field on the underlying real sphere, hence the holomorphic complex hairy ball theorem is implied by the smooth one.

In particular I am interpreting a "covering" of the complex sphere by a family of complex vectors, as a section of the complex tangent bundle, i.e. one vector at each point of the sphere. So here the real smooth 2-diml sphere, and its rank 2 real tangent bundle, have been given the structure of a 1-diml holomorphic complex manifold, with a rank one complex tangent bundle. In relation to jbergman's answer, we have added to the real tangent bundle, a rotation operator on each planar fiber, i.e. a real linear operator J with J^2 = -Id, varying smoothly over the sphere. Such an operator is called an "almost complex structure", on the real sphere. (Having an almost complex structure on the tangent bundle is a necessary, but not sufficient, condition for the real smooth manifold itself to have a complex holomorphic structure. In particular such an almost complex structure already forces the real manifold to have even real dimension, since the only irreducible factor of the minimal polynomial, hence also of the characteristic polynomial, is X^2 +1.)

You can check this result by computing the poles of the one - form dz, at infinity on the Riemann sphere. I.e. a one - form is a section of the dual, or cotangent, bundle, hence should have degree -2 instead of 2 for the tangent bundle, i.e. 2 poles instead of 2 zeroes. In fact dz has no zeroes or poles in the finite plane, and using the coordinate w = 1/z at infinity, we have zw = 1, so zdw + wdz = 0, so dz = -zdw/w = -dw/w^2, a pole of order 2 at w=0.

It is actually enough to check the result for a single one-form, since dividing two one forms defines a meromorphic function, i.e. a holomorphic map to the Riemann sphere, which sends the same number of points to zero as to infinity, i.e. such a map has a well defined degree d. Hence for both forms, #zeroes - #poles = the same number, (which is just d if both forms are holomorphic with distinct zeroes).

The hairy ball theorem for a complex holomorphic Riemann surface of genus g, says that a holomorphic 1-form has 2g-2 zeroes, counted properly, or for a meromorphic form, #zeroes - #poles = 2g-2. The nice part in the holomorphic or meromorphic case, is there are always only a finite number of zeroes and poles, at least for a form that is not identically zero, and their order can be computed from a Taylor series expansion.

In answer to your original question, complex vectors look exactly like real vectors. As jbergman said, a complex vector space is just an even dimensional real vector space plus a real linear rotation operator T. The vectors don't look any different, the only role of the complex structure is that you are given a way to multiply the real vectors not only by real scalars but also by i, i.e. to rotate each one "90 degrees" in a certain real 2 dimensional subspace. I.e. given a real linear operator T with T^2 = -Id, and hence minimal polynomial X^2 + 1 = 0, on a real vector space V, we get an action on V of the field k[X]/(X^2+1) ≈ C, i.e. a structure of complex vector space on V. Any basis for V over this field decomposes V into a direct sum of real 2 dimensional T invariant subspaces, i.e. "complex" subspaces.

 

[Ishika_96_sparkles]

isn't the Argand diagram sort of a pictorial representation of the complex number? How would one define a unit direction vector then? Could we draw an analogy with the usual real space vector algebra i.e.,
$$\hat{a}=\frac{\vec{a}}{|a|}$$
would it be a valid exercise?

 

[fresh_42]

isn't the Argand diagram sort of a pictorial representation of the complex number?

Yes.

How would one define a unit direction vector then?
Could we draw an analogy with the usual real space vector algebra i.e.,
$$\hat{a}=\frac{\vec{a}}{|a|}$$
would it be a valid exercise?

No. A complex vector has complex coordinates. So first we have to decide is, whether our vector is over the real numbers or over the complex numbers. A vector space needs to know its scalar field. Without naming it, it isn't a vector space. This means in return that we can only draw a unit vector of the one-dimensional complex vector space or equivalently the two-dimensional real vector space.

This vector space is the complex numbers. Everything else would already be at least real four-dimensional and that cannot be drawn. And, yes, the complex numbers of length one are the points on the unit circle represented by your formula. However, a number is a poor vector space as it is only a line.

 

[Ishika_96_sparkles]

A complex vector has complex coordinates.

Sure, we have the column vector representation
$$ \begin{bmatrix} z_1 \\ z_2 \\ \vdots \\ z_n \end{bmatrix} $$
which implies $$ \begin{bmatrix} x_1+\iota y_1 \\ x_2+\iota y_2 \\ \vdots \\ x_n + \iota y_n\end{bmatrix} $$
and we defined the field F to be $\mathbb{C}^n$ or the complex coordinate space. This n-tuple represents an n-dimensional vector. If we restrict ourselves to a 2-tuple $$\begin{bmatrix} x_1+\iota y_1 \\ x_2 +\iota y_2 \end{bmatrix}$$
and so these components are along which direction would be set by the kind of basis we have? its difficult to draw a picture on paper as you mentioned before

... we can only draw a unit vector of the one-dimensional complex vector space or equivalently the two-dimensional real vector space...at least real four-dimensional and that cannot be drawn.

the complex numbers of length one are the points on the unit circle represented by your formula.

Now, suppose we have $z=1+\iota$ and we wish to use the usual formula. Then, we would need to find $|z|^2 \implies z\bar{z}$, which in our case would be $\sqrt{2}$ and the form we get would be $\frac{1}{\sqrt{2}} (1+\iota)$. Thus, we just get the desired equivalent output.

Is this what you wanted to describe?

 

[fresh_42]

I am not sure what you want to say but it is more a "no" than it is a "yes". E.g. $\mathbb{C}^n$ isn't a field, it is a vector space of dimension $n$ over the complex numbers and of dimension $2n$ over the reals.

Since we can only draw real vectors, we are restricted to real dimensions $0$ which is a point, $1$ which is a real line, $2$ which is a sheet of paper, or $3$ which is the space surrounding us. So $2n\in \{0,1,2,3\}$ implies $n\in \{0,1\}.$ That means we can only draw a complex point or a complex line. The latter is described by the Argand diagram. It has two real dimensions, but only one complex dimension, a complex "number line". All other directions cannot be drawn. It also shows that we cannot imagine a complex direction since we need an entire plane only to visualize a complex "number line". We have no Archimedean ordering of the complex numbers because $\mathrm{i}^2=-1 <0$ prevents such an ordering, so we cannot arrange the complex numbers along a line. But algebraically, they are a line, namely the vector space $\mathbb{C}\cdot \vec{1}.$ Geometrically, we are restricted to the real world, which makes it a two-dimensional real plane.

 

[Ishika_96_sparkles]

I am not sure what you want to say but it is more a "no" than it is a "yes". E.g. $\mathbb{C}^n$ isn't a field, it is a vector space of dimension $n$ over the complex numbers and of dimension $2n$ over the reals.

So $2n\in \{0,1,2,3\}$ implies $n\in \{0,1\}.$ That means we can only draw a complex point or a complex line. The latter is described by the Argand diagram. It has two real dimensions, but only one complex dimension, a complex "number line". All other directions cannot be drawn. It also shows that we cannot imagine a complex direction since we need an entire plane only to visualize a complex "number line". We have no Archimedean ordering of the complex numbers because $\mathrm{i}^2=-1 <0$ prevents such an ordering, so we cannot arrange the complex numbers along a line. But algebraically, they are a line, namely the vector space $\mathbb{C}\cdot \vec{1}.$ Geometrically, we are restricted to the real world, which makes it a two-dimensional real plane.

Its much clear to me now, although i need to get back to studying the rigorous mathematical expositions of linear algebra to fully appreciate the response. Still, the topic of Archimedean ordering is not clear to me as i never took the real analysis course.

 

[fresh_42]

Archimedean ordering is simple: it means that for every number $z$, there is a natural number (positive integer) $n$ such that $n$ is greater than $z$. That's all. And complex numbers cannot be ordered so that is true. Means: no lines as we know them from the real number line.