As discussed here the complex projective space $\mathbb{C}P^n$ is the set of all lines on $\mathbb{C}^n$ passing through the origin. It would seem natural to assume that any $\mathbb{C}P^n$ can be viewed as a tangent space to a point in an equivalently parametrized space. So for a point $p$ on some manifold $M$ the set of all tangent vectors at $p$ is:


As discussed here, the union of all points on a manifold is written as:

$$TM = \bigcup_{p\in M} T_pM$$ and is known as the tangent bundle.

Also, $\pi$ is used to represent the map of from the tangent bundle to the manifold such that $\pi: TM \rightarrow M$ and for each $p \in M$ :


Where $\pi^{-1}(p)$ is a vector space of dimension $n = \dim M$ (confirming as stated above that the tangent space will have same dimension, or equivalent number of parameters as the underlying manifold).

Ordinary spacetime is often described as $\mathbb{R}^{3,1}$ and it is known that euclidean plane $\mathbb{R}^{2}$ can be described by a complex number in $\mathbb{C}^{1}$, and $\mathbb{R}^{1,1}$ can be described with split-complex numbers in $\mathbb{R}^1\oplus \mathbb{R}^1$. When one considers the Lorentz boost, it is tempting to think of ordinary space as being $\mathbb{C}^{1}\oplus \mathbb{R}^1\oplus \mathbb{R}^1$ and described by two variables $\left( z , \mathring{z} \right)$ where $z$ is a complex number and $\mathring{z}$ is a split complex number.

Since there are exactly three $2$-dimensional unital algebras, complex numbers, split-complex numbers and dual numbers, and the algebra of dual numbers is isomorphic to the exterior algebra of $\mathbb{R}^1$, and there are two separate $\mathbb{R}^1$ in $\mathbb{C}^{1}\oplus \mathbb{R}^1\oplus \mathbb{R}^1$, it is tempting to add an additional set of dual numbers so that we can parameterize space as $$\left( z , \mathring{z} \right)\big| \left( \mathring{v} , \mathring{w} \right)$$ where $ \mathring{v}$ is the dual number for the first $\mathbb{R}^1$ and $ \mathring{w} $ is the dual number for the second $\mathbb{R}^1$. Dual numbers are noteworthy as having "fermionic" directions of "bosonic" directions.

If I wanted to keep my new set of coordinates, $$\left( z , \mathring{z} \right)\big| \left( \mathring{v} , \mathring{w} \right)$$ how best would I describe the tangent space of this "ad hoc" manifold? Would I be able to describe this in $\mathbb{C}P^n$, if not, why not? If only because of compactness, why can't tangent spaces be compact? Could we describe it as pseudo-tangent?

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    $\begingroup$ $\mathbb{C}P^n$ will not be a tangent bundle... it is compact whereas tangent bundles are not. And your description of the tangent bundle is missing crucial features, the way the $T_pM$ "connect" to each other. That all being said, I have no idea what the latter half of your post has to do with the beginning. Please elaborate and just delete the first part about tangent bundles (providing instead a link to Wikipedia that defines it) $\endgroup$ – Chris Gerig Mar 3 '13 at 1:50
  • $\begingroup$ @ChrisGerig thanks for the input, I added a reference, and some additional questions, I will revisit this later for further modifications. $\endgroup$ – user11547 Mar 3 '13 at 2:00
  • $\begingroup$ Comment to the question formulation (v5): What does $\big|$ mean? $\endgroup$ – Qmechanic Mar 4 '13 at 1:09
  • $\begingroup$ @Qmechanic Nothing in particular, however, I wanted to group the coordinates appropriately and so I borrowed $\big|$ from the idea of superspace notation. $\endgroup$ – user11547 Mar 4 '13 at 1:31
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    $\begingroup$ Comments on the question (v5), which has been flagged for being math only. OP writes: Why can't tangent spaces be compact? The answer is: Because tangent spaces are by def. vector spaces. Perhaps OP wants a fiber bundle construction rather than a vector bundle? Is the word pseudo-tangent supposed to be a standard mathematical notion, or is it just supposed to be a yet-to-be-seen-unknown-notion that replaces the word tangent? Why is OP talking about the tangent bundle of Minkowski space, when he really wants to ask about $\mathbb{C}P^n$? Is it possible to make the question more clear? $\endgroup$ – Qmechanic Mar 5 '13 at 15:27

Regarding compactification -- perhaps you are looking for the notion of the "unit tangent bundle", which would be compact provided the manifold is compact. In general, the tangent bundle is a manifold of dimension twice the original and naturally projects onto the original manifold; in other words, if the original manifold is not compact, even the unit tangent bundle is not compact.

You can read more about the unit tangent bundle here.

Regarding changes of coordinates: to be completely frank, I understood neither the motivation nor the technical details. Nevertheless, when dealing with like questions, one has to be careful to not mix the categories that one is working in.

From the point of view of bare set theory, all the spaces you described, and in addition $\mathbb{C}^2$, are equivalent (in the sense that there exists a one-to-one and onto map from any one of the spaces in question to any other). However, in addition to the set structure, our spaces have a much richer structure: a geometric one. Thus, as soon as we pass to the smooth category (think of it, very roughly speaking, as a context of smooth manifolds and smooth maps defined on them), then questions arise as to whether changes of coordinates preserve the structure we're interested in.

To make matters more complicated, the spaces $\mathbb{R}^k$, $\mathbb{C}^k$ and the projective spaces all carry an algebraic structure.

In short, there is no reason to change the coordinates from, say, $\mathbb{R}^2$ to $\mathbb{C}^1$, unless one wants to benefit from the additional structures carried by $\mathbb{C}^1$. In this case, then, one has to make sure that the change of coordinates is canonical (in the sense that whatever is proven in the new coordinates can be carried back to the old coordinates). Thus the transformation that manifests this change of coordinates must possess certain structural integrity (loose language alert!!). This makes the question difficult, especially when it isn't clearl exactly what one is looking for.

Here is a simple example: let us view $\mathbb{R}^2$ as $\mathbb{C}^1$ via the transformation $\Phi: \mathbb{R}^2\rightarrow \mathbb{C}^1$ with $\Phi(a,b) = a + bi$. No tricks here. Moreover, this transformation is continuous with a continuous inverse, so a homeomorphism. Thus all the topological properties are preserved (i.e. whatever one can prove about $\mathbb{C}^1$ and the natural topology thereof, as well as continuous maps defined on them, one can formulate analogous results for $\mathbb{R}^2$).

Notice also that both $\mathbb{R}^2$ and $\mathbb{C}^1$ are manifolds, but here one has to be careful: a manifold is always modeled on a "model manifold" (for example, a smooth surface embedded in $\mathbb{R}^3$ is modeled on $\mathbb{R}^2$ in the sense that it is locally, around any point, essentially a copy of $\mathbb{R}^2$. Now, when we say that both $\mathbb{R}^2$ and $\mathbb{C}^1$ are manifolds, what do we mean by that? Here one has to make a choice. We can either model both on $\mathbb{R}^2$, or model $\mathbb{R}^2$ on $\mathbb{R}^2$, and model $\mathbb{C}^1$ on $\mathbb{C}^1$. In the former case they are equivalent via the change of coordinates $\Phi$; in the latter case, they are not, since there are functions, say $f$, on $\mathbb{R}^2$ which are infinitely differentiable and constant on some open set, while the corresponding "push forward" of this function under $\Phi$, namely $f\circ \Phi$, is not differentiable in $\mathbb{C}^1$. That is, in the analytic category, $\mathbb{R}^2$ and $\mathbb{C}^1$ are not equivalent (analytic category is stronger than continuous; also, being differentiable in $\mathbb{C}^1$ is equivalent to being infinitely differentiable, is equivalent to being analytic).

The latter complication comes from the fact that in addition to a topological structure of $\mathbb{C}^1$ (which, as I've commented above, is the same as that of $\mathbb{R}^2$), $\mathbb{C}^1$ carries an algebraic structure (namely, a field) which is compatible with its topological (even smooth) structure in the sense that the field operations of multiplication and addition are smooth maps (in fact analytic). $\mathbb{R}^2$, on the other hand, as it is usually defined, does not enjoy these properties.

You could argue, of course, that we can "redefine" $\mathbb{R}^2$ to reflect the properties of $\mathbb{C}^1$. Sure we can do that, but then $\mathbb{R}^2$ becomes $\mathbb{C}^1$ that goes by the name of $\mathbb{R}^2$. When we write $\mathbb{R}^2$, we specifically mean the two-dimensional vector space with the underlying field being $\mathbb{R}$. The vector space operations in this case are incompatible with the field operations of $\mathbb{C}^1$.

Long story short: whenever a change of coordinates is introduced, it is usually for the sake of transforming a representation into a simpler form, to simplify computations or to relate to a known problem/solution in the other coordinates. You know you're doing something that doesn't "feel right" when you're changing coordinates for the sake of benefiting from some additional (or principally different) mathematical structure of the new coordinates, because whatever you manage to prove in the new coordinates that requires this additional (or different) structure, you probably won't be able to pull back to the original coordinates (there are some exceptions to this rule, of course!).

Thus, it seems to me, that what you are doing is either (a) ultimately incorrect in the sense of making a category error, or (b) simply "renaming" your spaces without losing or gaining anything.


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