# Concept of duality for projective spaces and manifolds

• A
SVN
I firstly learned about duality in context of differentiable manifolds. Here, we have tangent vectors populating the tangent space and differential forms in its co-tangent counterpart. Acting upon each other a vector and a form produce a scalar (contraction operation).

Later, I run into the term of «duality» reading textbook on projective geometry. But here it is used somewhat differently, one talks about duality between points and straight lines in projective plane (point and planes in the case of projective space). Point and lines in ℙ reverse their roles when mapped into ℙ*.

I don't quite understand how this two definitions of duality are related. Can they be formulated in similar words and notions (preferably using «language of manifolds»)? How can a point of projective space be equivalent for tangent vector and line for form? What is the equivalent of manifold itself in case of projective geometry?

## Answers and Replies

Homework Helper
Gold Member
I can't quite pin it down yet. But maybe this is helpful...
It seems that....
tangent vectors and points can be represented by column vectors,
whereas covectors and lines can be represented by row vectors.
The metric tensor provides a metric-dual correspondence between vectors and covectors.
A conic provides a pole-polar correspondence between points and lines.

(Corrections, clarifications, and illuminations are welcome...)

SVN
Yes, you are right!

Metric tensor mapping vectors to one-forms can be visualised as a circle in Euclidean space (I read about this in «Spacetime, geometry, cosmology» by W. L. Burke). The way to recover one-form from vector (arrow) and tensor (circle) is the same procedure as construction of polar by given pole and conic (the latter, obviously, takes place of circle when considering projective plane).

However, one puzzle remains. Vector acts upon one-form producing scalar, which represents value for directional derivative in that point of manifold.

The equation for for a straight line belonging to projective plane in homogeneous coordinates looks like u1x1 + u2x2 + u3x3 = 0, where xi and ui — coordinates of the point and the line respectively. Just like you stated it can be thought of as matrix multiplication for a column and a raw vectors. But why is it always equal to zero? OK, we may say the zero to be an artefact of using homogeneous coordinates. They are nice, but inhomogeneous coordinates are not nixed. The equation turns into u1x + u2y + u3 = 0 or u1x + u2y = -u3, where x = x1/x3 and y = x2/x3. But again what is the meaning for this value of -u3? And what is the equivalent of «manifold» tangent space of which is our projective space?

Homework Helper
Gold Member
That book by Burke was my first introduction to projective geometry, although he only mentioned it in passing.
Over time, I had to make my way over to math books on the subject.
It hasn't been easy... but I am making slow but steady progress into understanding how those ideas
might show up in physics (likely behind the scenes... hiding in the mathematical foundations of various topics
[like tensors, special-relativity, and quantum mechanics]).

Here's an interesting book
https://www.amazon.com/dp/3527405674/?tag=pfamazon01-20

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SVN
It looks like we are sailing on parallel courses here.

I received no special education in mathematics and sincerely believed projective geometry to be some set of formal rules used in doing technical or architecture drawing, something that albeit being practicallly important just thrives on tedium (in my personal opinion, of course). Currently, I am enamoured with elegance and generality of this branch of geometry.

I agree, projective geometry seems to lie in basement of many ideas and theories in physics. Although, one normally finds only scattered and scanty allusions to it in monographs on physics. So thank you for recommending that book, judged by content it indeed looks very interesting, I will certainly check it out.

Homework Helper
the duality in linear algebra and hence also differential geometry is exactly the same as that in projective geometry. in differential geometry and linear algebra, the dual of a vector space (such as a tangent space) is the space of linear functions defined on the first vector space. now a projective plane is defined as the set of lines through the origin of a given 3 diml vector space V. thus the projective plane based on its dual space V*, is the set of lines of linear functions on the first vector space. such a "line" of linear functions, all scalar multiples of one another, have a common kernel, namely a plane in the first vector space. This is precisely a line in the first projective space. So a "point" of the dual projective space P(V*), given by a linear function on V, defines precisely a line in the projective space P(V).

In higher dimensions of course the duality for an n dmensional projective space P(V), where V is an n+1 dimensional vector space, pairs a point of P(V*) i.e. a pencil of linear functions on V with a common kernel, with that kernel, which is not a line in P(V) but a hyperplane, i.e. an n-1 dimensional projective subspace of P(V).

In general dualty often means essentially the same thing, i.e. there is a duality between maps into the field of scalars and maps out of it. Given a vector space V, a map from the reals into V defines a line in V. a map out of V, into R, is defined up to a scalar by its kernel, a hyperplane of V. Algebraically, a duality is defined by a pairing, such as the canonical pairing VxV*-->R given by evaluation of a function on its argument, say a covector on a vector. Since evaluation is a natural pairing, there is a natural duality between V and V*, hence also between P(V) and P(V*). a chopice of an inner product on a vector space V, i.e. a pairing VxV-->R, defines a notion of perpendicularity, hence a duality between lines and their orthogonal complements, namely hyperplanes.

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SVN
the duality in linear algebra and hence also differential geometry is exactly the same as that in projective geometry.
Thank you for such a thorough explanation. The whole picture clarified considerably. But one point still escapes me.

Let's say we have a differentiable (smooth) manifold. It «generates» tangent and co-tangent linear (vector!) spaces, that are dual to each other. The question is whether those generated spaces can, in principle, be projective spaces (or affine spaces for that matter)? If yes, what kind of properties or characteristics should the original manifold (or maybe the proper term for this entity is not «manifold», but «topological space»?) have, in order to «generate» tangent and co-tangent projective spaces? Do such manifolds (or topological spaces) have any special name?

And the second related question. Let's say a manifold «generates» standard linear (vector) tangent space. We can create projective space based on this tangent space. How to describe relation between original manifold and this projective space? Any term for that? The «tangency» should be preserved in some sense, should it? But if the manifold was, say, 2D, than the projective space will be 1D. Can it still be called «tangent» than?

Homework Helper
yes of course one can make out of the tangent bundle on a manifold, the projective tangent bundle. This is useful in some topological constructions but has no more information than the tangent bundle in some sense. yes of course the projective bundle has n-1 diml fibers. one advantage it has for some arguments is that it is compact (when the manifold is) whereas the tangent bundle is not.

Your question on the relation between the topology of the projective bundle made from the tangent bundle and the topology of the original manifold is quite profound. Indeed the cohomology of P(T) (where T is the bundle of tangent spaces to the manifold M), and the cohomology of M itself, describes the basic invariant of the manifold, namely its chern classes. If we consider the case of a complex manifold and a complex tangent bundle, then the cohomology of P(T) is a free module over the cohomology of M, with certain specific generators, and one natural relation gives rise to coefficients which define the chern classes of M. This is explained in the book Differential forms in algebraic topology, of Bott-Tu, pp.269-270, in the chapter on characteristic classes. This book is highly recommended.

back in classical projective geometry,
the geometric duality in projective space is quite rich. we are familiar with the fact that for finite dimensional vector spaces, the dual of the dual is naturally isomorphic to the original space, i. V** ≈ V. Hence this holds also for projective space P(V) ≈ P(V**) = P(V)**. But the correspondence goes further. Given a curve C in the projective plane P(V), we can consider for each point of the curve C, its tangent line, which is apoint of P(V)*. Thus we get a curve C* in P(V)* the dual of the curve C. As you perhaps guessed, the dual of this curve C** in P(V)** ≈ P(V), is isomorphic to C again.

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• lavinia
Gold Member
[QUOTE="SVN, post: 5488556, member: 564947"
But if the manifold was, say, 2D, than the projective space will be 1D. Can it still be called «tangent» than?[/QUOTE]

Tangency is not preserved. For instance if one projectivises the tangent bundle of the 2 sphere one gets a different space.
One can see this by considering the projective bundle is the quotient of the unit circle bundle by multiplication by -1 on each fiber. While the tangent circle bundle has fundamental group isomorphic to Z/2, the projectivised bundle has fundamental group Z/4 so they are not homeomorphic.

I think that this works in the opposite direction as well. The tangent circle bundle of a Riemann surface of genus greater than 1 is a projectivised circle bundle. But the covering bundle is not isomorphic to the tangent bundle.

As Mathwonk pointed out, a key invariant of complex vector bundles is their Chern classes. For orientable surfaces all of the Chern classes are zero except the first. For each integer there is a different complex vector bundle whose first Chern class equals that integer (when integrated over the surface). This Chern class uniquely distinguishes the complex vector bundle. If one successively projectivises the tangent circle bundle of the sphere I think you get a sequence of vector bundles each with double the Chern class of the previous one. I am not 100% sure of this though.

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