How to understand that RP2 is non-orientable?

[wdlang]

there is a general result that

RPn is orientable if n is odd while non-orientable if n is even

how to prove this or just understand it?

i am a physics student

 

[Jamma]

In terms of understanding it, it just means that you can't have a consistent orientation on the even real projective spaces i.e. if you move some chosen orientation continuously from some point, you can return there again with a different orientation.

In terms of proving it, you can deduce this from the top homology groups of the real projective spaces, it is Z/2 for n even and Z for n odd, I think.

It's easy to see that it's not orientable for the real projective PLANE (i.e. n=2). Just look at the construction by identifying edges of a square. By moving a chosen orientation around the square, off the edge, and back again, it will now by a different orientation.

 

[owlpride]

The square picture Jamma points out is probably the easiest way to see non-orientability in dimension 2.

In general, RP^n is orientable for odd n and non-orientable for even n. That's because the antipodal map on S^n is orientation-preserving in odd dimensions and orientation-reversing in even dimensions. (One way of constructing RP^n is to take S^n and identify all antipodal points x and -x. If the map that takes x to -x is orientation-reversing, there's no chance that the resulting space is orientable because tangent vectors are identified with their flip-images.)

 

[Jamma]

Hmm, that's a nice way of looking at it. How would you prove that the antipodal map is orientation preserving for n-odd and reversing for n even? I remember proving that the degree of the antipodal map is 1 for n odd and -1 for n even (I think), is it to do with this?

 

[quasar987]

Well yea. The antipodal map is 1-1, so deg(a) = ±1 means that at any point x of the sphere, sng(det(a'(x))) = ± (respectively).. i.e. a is orientation preserving (resp. reversing).

 

[lavinia]

on the 2 sphere take a frame at a point and slide it to the antipodal point. This gives a frame at the antipodal point. If projective space were orientable then this frame would map to the original frame under the differential of the projection map. But in fact, it maps to a frame with opposite orientation.

If one slides the frame along a great circle that is tangent to one of the vectors in the frame, then this vector undergoes a 180 degree rotation while the other vector remains constant. In the tangent plane this is the same as a reflection. However the antipodal map rotates the entire frame by 180 degrees in the tangent plane.

BTW: Since the sphere is embedded in Euclidean 3 space these tangent planes can be identified by translating them to the origin.

If projective space were orientable then these two frames would have to have the same orientation. However, one of them differs from the original by a reflection while the other differs from the original by a rotation.

The exact same reasoning applies to all projective spaces. For odd dimensional projective spaces the resulting frames differ by a rotation. For even dimensional projective spaces they differ by a rotation and a reflection.

This methodology applies whenever a smooth manifold is obtained from an orientable manifold via the action of a discrete group. If the group acts by orientation preserving diffeomorphisms, the quotient manifold is orientable. If not, it is non-orientable.

For instance the Klein bottle may be obtained from the plane by the action of the standard lattice plus one other affine motion of the plane which reflects the y-axis then translates by 1/2 in the x direction. Its formula is (x,y) -> (x + 1/2, -y)This last motion is not orientation preserving so neither is the Klein bottle.

If one starts with the frame (1,0) (0,1) at the origin and slides it along the x-axis to the point (1/2,0) , which is the same point in the Klein bottle, then in the Klein bottle this frame differs from the original by a reflection in the y-direction. The argument is exactly the same as for projective space.

 

[quasar987]

For RP², the simplest way is probably this: Realize RP² as the closed 2-disk D² with antipodal boundary points identified. Then consider a diameter of D² and "thicken it" by epsilon. Then in the quotient (i.e., in RP²) this thickened diameter is a Möbius band.

For RP^n I don't know of a comprehensible simple way to see it but the key of the proof is that the antipodal map is orientation reversing iff n is even.

 

[lavinia]

For RP², the simplest way is probably this: Realize RP² as the closed 2-disk D² with antipodal boundary points identified. Then consider a diameter of D² and "thicken it" by epsilon. Then in the quotient (i.e., in RP²) this thickened diameter is a Möbius band.

For RP^n I don't know of a comprehensible simple way to see it but the key of the proof is that the antipodal map is orientation reversing iff n is even.

I do not see why the existence of a embedded Mobius band means that the manifold is not orientable. I can give you a 3 manifold that is orientable that contains an embedded Klein bottle - and the Klein bottle is not homologous to zero either.

 

[lavinia]

there is one last point which you all may understand already but which I didn't until I thought about it a bit.

The antipodal map is orientation reversing on R3 but on the antipodal tangent planes viewed as planes in Euclidean 3 space it is orientation preserving since it just acts by negation on a 2 dimensional plane.

So why then is the antipodal mapping orientation reversing when in fact it preserves the orientation of the tangent planes when viewed as subsets of 3 space? Why should it matter that is also reflects the normal to the sphere?

I thought this was worth thinking about.

 

[quasar987]

I do not see why the existence of a embedded Mobius band means that the manifold is not orientable. I can give you a 3 manifold that is orientable that contains an embedded Klein bottle - and the Klein bottle is not homologous to zero either.

Well of course, any RP^n (n>1) contains an embedded Möbius band, but only the n even are nonorientable.

But for 2-manifolds, the existence of an embedded Möbius band implies non orientability.

 

[lavinia]

Well of course, any RP^n (n>1) contains an embedded Möbius band, but only the n even are nonorientable.

But for 2-manifolds, the existence of an embedded Möbius band implies non orientability.

I understand that it is a special property of 2 manifolds - but... the proof?

 

[Jamma]

Well surely if the manifold is orientable, it is orientable on any subset of it, and clearly that would fail if there was an embedded Möbius band.

 

[quasar987]

For simplicity, assume (without loss of generality) that in the following, by "chart" I mean "connected chart".

By definition, a manifold M is orientable iff you can find a covering of M by coherently oriented charts. Meaning that the determinant of the derivative of the transition functions between overlapping charts are all positive.

Next, note that given two charts (U,f), (V,g) of M of opposite orientation (i.e. the transition function is of negative determinant), the charts (U,f*),(V,g) are of the same orientation, where f* is defined by changing the sign of the first coordinate of f. Call (U,f*) the "modified" chart (U,f).

Thirdly, note that a manifold M is orientable iff given any covering of M by charts, it is always possible to modify them (in the above sense) to obtain a coherently oriented covering of M.
Proof:"<==" O.K.
"==>" Let {(U_i,f_i)} be any covering of M by charts. Take a coherently oriented covering of M by charts {(V_j,g_j)}. For each U_i, chose a V_j that intersects it. Are U_i and V_j of the same orientation? If so, do nothing. If not, modify (U_i,f_i) to (U_i,f*_i). Note that for two charts to have the same orientation is an equivalence relation on the set of all charts of a manifold. Clearly then, this process makes {(U_i,f_i)} into a coherently oriented covering. QED

Now let M be 2 dimensional. Let X=([0,1] x ]-1,1[)/(0,t)~(1,-t) be the Mobius band, and f:X-->M be an embedding. Consider {(U_i,φ_i)} a covering of f(X) by charts lying entirely in f(X) (f(X) is open in M). Since the "pullback charts" (f^{-1}(U_i), φ_i o f) make up a covering of the nonorientable manifold X, it is impossible to modify the covering {(U_i,φ_i)} to make it coherently oriented. Then just extend {(U_i,φ_i)} to a covering of the whole of M. This covering either cannot be modified to be coherently oriented, so M is nonorientable. QED

 

[quasar987]

Well surely if the manifold is orientable, it is orientable on any subset of it, and clearly that would fail if there was an embedded Möbius band.

Nah. As I pointed out, any RP^n (n>1) contains an embedded Möbius band, but only the n even are nonorientable.

And lavina said he\she had an example of a 3-manifold with an embedded klein bottle. What is your example lavina?

 

[lavinia]

Nah. As I pointed out, any RP^n (n>1) contains an embedded Möbius band, but only the n even are nonorientable.

And lavina said he\she had an example of a 3-manifold with an embedded klein bottle. What is your example lavina?

First of all for a 2 manifold it is true that an embedded Mobius band implies non-orientability.
This is because it is not orientable in its own tangent plane and there are no normal directions to allow orientability in the ambient manifold. However in a high dimensional manifold it is possible to have orientability despite embedded Mobius bands. More generally if the embedded non-orientable manifold is of lower dimension that the ambient manifold it is conceivable that the ambient manifold is orientable. This is true even if the embedded non-orientable manifold is not null homologous.

Here is an example.

Let the standard lattice in 3 space act on 3 space by translation. Add one more affine motion, (x,y,z) -> (x+1/2,-y,-z).

The quotient manifold is orientable because this affine motion preserves the orientation of 3 space as does translation by the lattice.

But the image of the xy-plane ( or the xz-plane)is a Klein bottle. Further it is not homologous to zero.( In fact there is a continuum of Klein bottles that all intersect in the same circle.)
If one extends a tangent frame on the embedded Klein bottle to a 3-frame then translation of this frame around the Klein bottle preserves orientation. As the orientation flips on the Klein bottle's tangent plane it also flips in the normal direction.

 

[lavinia]

here is a problem. Suppose a non-orientable manifold is a boundary. We know that the ambient manifold can not be orientable by homology arguments.

How do you show this with a frame translation argument?

BTW: the Klein bottle is the boundary of a 3 manifold.

 

[quasar987]

An orientation on the ambiant manifold induces an orientation on its boundary.

 

[Jamma]

Nah. As I pointed out, any RP^n (n>1) contains an embedded Möbius band, but only the n even are nonorientable.

And lavina said he\she had an example of a 3-manifold with an embedded klein bottle. What is your example lavina?

Sorry, I meant only for 2-manifolds. Then obviously it fails with the reasoning I gave.

 

[Jamma]

I should ellaborate, the orientability of the manifold in question would imply an orientability of the non-orientable embedded manifold if the two are of the same dimension, so clearly an orientable manifold can have no non-orientable manifold of the same dimension embedded in it.

First of all for a 2 manifold it is true that an embedded Mobius band implies non-orientability.
This is because it is not orientable in its own tangent plane and there are no normal directions to allow orientability in the ambient manifold. However in a high dimensional manifold it is possible to have orientability despite embedded Mobius bands. More generally if the embedded non-orientable manifold is of lower dimension that the ambient manifold it is conceivable that the ambient manifold is orientable. This is true even if the embedded non-orientable manifold is not null homologous.

Here is an example.

Let the standard lattice in 3 space act on 3 space by translation. Add one more affine motion, (x,y,z) -> (x+1/2,-y,-z).

The quotient manifold is orientable because this affine motion preserves the orientation of 3 space as does translation by the lattice.

But the image of the xy-plane ( or the xz-plane)is a Klein bottle. Further it is not homologous to zero.( In fact there is a continuum of Klein bottles that all intersect in the same circle.)
If one extends a tangent frame on the embedded Klein bottle to a 3-frame then translation of this frame around the Klein bottle preserves orientation. As the orientation flips on the Klein bottle's tangent plane it also flips in the normal direction.

This is a really nice example, thanks for sharing it!

Oh, and I thought that the boundary of a manifold was always orientable? I must be going mad :/ (I have no idea why I thought this)

 

[lavinia]

An orientation on the ambiant manifold induces an orientation on its boundary.

Right, so are you are saying that if you parallel translate a frame along a curve on the boundary then it can not rotate in the normal direction?
Maybe not.

 

[lavinia]

Oh, and I thought that the boundary of a manifold was always orientable? I must be going mad :/ (I have no idea why I thought this)

It is easy to see that the Klein bottle is a boundary if you think of it as a cylinder whose edge cirlces are identified via a reflection. Fill the cylinder until it is solid and identify the edge discs via the same reflection (the centers of the discs are identified with each other.)

The solid cylinder becomes a 3 manifold whose boundary is the Klein bottle.

 

[Jamma]

It is easy to see that the Klein bottle is a boundary if you think of it as a cylinder whose edge cirlces are identified via a reflection. Fill the cylinder until it is solid and identify the edge discs via the same reflection (the centers of the discs are identified with each other.)

The solid cylinder becomes a 3 manifold whose boundary is the Klein bottle.

Ah, yes, very nice. I must have forgotten that the boundary of an oriented (compact?) manifold with boundary is an oriented (compact?) manifold (which must have no boundary).

 

[quasar987]

Right, so are you are saying that if you parallel translate a frame along a curve on the boundary then it can not rotate in the normal direction?
Maybe not.

I was only saying that it is a well known fact that an orientation on a manifold with boundary induces an orientation on its boundary like so: at any given point p of the boundary ∂M, a basis for the tangent space T_p(∂M) is, by definition, positively oriented if, when you add to it the outward point normal vector at p, then you obtain a positively oriented basis of T_pM.

So I actually did not propose a solution to your problem, I just dismissed it lazily as "a well known fact". :tongue2:

After thinking about it, an argument using frames could be: Assume ∂M is nonorientable and take a loop of frame on ∂M that fails to preserve orientation (i.e. a nowhere vanishing smooth map $[0,1]\rightarrow\bigwedge^{n-1}T(\partial M) : t\mapsto (\gamma(t), v_1(t)\wedge\ldots\wedge v_{n-1}(t))$ where $\gamma(0)=\gamma(1)$). Then make it into a loop of frame on M by adding $v_n(t):=n(t)$, a loop of outward pointing vectors. Then the loop of frame on M $[0,1]\rightarrow\bigwedge^{n}TM : t\mapsto (\gamma(t), v_1(t)\wedge\ldots\wedge v_{n-1}(t)\wedge v_n(t))$ fails to preserve orientation. So M is not orientable either.

The solution assumes that such a loop n(t) exists, but this is easily constructed using, say partitions of unity.

 

[lavinia]

I was only saying that it is a well known fact that an orientation on a manifold with boundary induces an orientation on its boundary like so: at any given point p of the boundary ∂M, a basis for the tangent space T_p(∂M) is, by definition, positively oriented if, when you add to it the outward point normal vector at p, then you obtain a positively oriented basis of T_pM.

So I actually did not propose a solution to your problem, I just dismissed it lazily as "a well known fact". :tongue2:

After thinking about it, an argument using frames could be: Assume ∂M is nonorientable and take a loop of frame on ∂M that fails to preserve orientation (i.e. a nowhere vanishing smooth map $[0,1]\rightarrow\bigwedge^{n-1}T(\partial M) : t\mapsto (\gamma(t), v_1(t)\wedge\ldots\wedge v_{n-1}(t))$ where $\gamma(0)=\gamma(1)$). Then make it into a loop of frame on M by adding $v_n(t):=n(t)$, a loop of outward pointing vectors. Then the loop of frame on M $[0,1]\rightarrow\bigwedge^{n}TM : t\mapsto (\gamma(t), v_1(t)\wedge\ldots\wedge v_{n-1}(t)\wedge v_n(t))$ fails to preserve orientation. So M is not orientable either.

The solution assumes that such a loop n(t) exists, but this is easily constructed using, say partitions of unity.

quasar this seems right. I guess in your way of doing it the point is that the normal does not switch from inward pointing to outward - and I guess something along this line must be true under parallel translation.

 

[quasar987]

Yes, and more sloppily/intuitively, you can see it this way: if v_n(t) is a loop of vector that completes v_1(t),...,v_{n-1}(t) to a loop of frame on M, then v_n(0) and v_n(1) must both point either inward or outward, otherwise it would mean that for some s, v_n(s) lies in T(∂M) and so v_1(s),...,v_{n-1}(s),v_n(s) would not be a frame of M.

 

[mathwonk]

hold a pencil in the middle and twirl the ends around in a small circle keeping the center fixed. note when the eraser goes around in a counterclockwise direction, the point goes in a clockwise direction. hence the antipodal map reverses orientation.

or note that the matrix of the antipodal map on R^3, hence on the two sphere, is minus the 3by3 identity matrix, hence has determinant (-1)^3 = -1.

since RP^2 identifies a small circle on one side of a sphere with small oppositely oriented circle on the other side, orientation is lost.