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August 3

Isomorphism of geometries

Should the Möbius plane (perhaps with the reflections omitted) be considered to be isomorphic to the complex projective line? —Quondum 02:45, 3 August 2015 (UTC)[reply]

If you forget the other structure and consider them both as permutation groups (with the reflections omitted) then they are isomorphic in that sense. But my understanding is that the complex projective line doesn't really have an incidence structure; it's pretty much just a set acted on by a group. You could add an incidence structure to the complex projective line which is consistent with the group action, but then you're essentially just making a model of the Möbius plane. In an analogous manner, the field of complex numbers is isomorphic to the the Euclidean plane if you throw out some of the structure. Is there a formalized concept of an isomorphism of geometries? Certainly the idea of duality in projective geometry would be an example. The thing is, there is a variety of different structures that you can talk about under the heading of 'geometry'. The Erlangen program is worth mentioning in this context. Anyway, sorry for going on a ramble. --RDBury (talk) 19:42, 3 August 2015 (UTC)[reply]

I like the idea of a geometry being defined as a set with a group acting on it. It allows a formalized definition of geometric isomorphism: the amount of structure used is set precisely. So it seems to me one can at least say that the Möbius plane, but with its group of transformations restricted to exclude reflections, is the same in this sense as the complex projective line. Thank you for confirming my thinking on this. —Quondum 02:53, 4 August 2015 (UTC)[reply]

The answer probably depends on what one means by "isomorphism". The group PSL(2,C) acts on the projective line and the Mobius plane in a compatible way, and there is certainly an equivariant function that maps the points of one space to the points of the other space, which is unique up to an overall PSL(2,C) transformation. But, as RDBury points out, in Mobius geometry, one usually also associates an incidence structure with the space. There is a "natural" incidence geometry on the complex projective line, the "circles", which are just all of the real structures on C^2. I'm inclined to think of these circles as being "really there", because they are natural. (Like in Euclidean geometry, circles are really there as well, even though the axioms refer only to lines.) But others may disagree, with good reasons. Sławomir
Biały 13:22, 4 August 2015 (UTC)[reply]

Defining the Orthogonal Compliment of a given Plane in 4-D?

The Orthogonal compliment of a Plane in 4-D is another plane. However, that compliment plane is only unique if the point in common is defined. So the plane defined by (x=0,y=0) is an entire family (z=c, w=d). However presuming that the crossing poing is (0,0,0,0) is there a clean way to get the definition of the orthogonal plane to a given plane? For example, for the plane (x=y+z and w=2z+y) what equations define the orthogonal plane to that?— Preceding unsigned comment added by Naraht (talk • contribs) 20:58, 3 August 2015‎

There is not a unique set (i.e. pair) of equations. The trivial case you give, (x = 0, y = 0) could also be written x + y = 0, x = 0), or in any number of equivalent ways. And this is true for any plane: any two non-equivalent equations that describe points in the plane will do. Formally what you are asking for is the dual, or the Hodge dual, of the plane; Hodge dual#Four dimensions discusses this particular case.--JohnBlackburne^words_deeds 20:44, 3 August 2015 (UTC)[reply]

I agree that the trivial case, or in fact any case could be written in an infinite number of ways, none of which is necessarily simpler, but I thought that the equations for the dual could be made from the way in which the equation was written for the initial plane. The Hodge dual certainly looks like it applies though.Naraht (talk) 21:45, 3 August 2015 (UTC)[reply]

I don't see how. Intuitively you might think you could just exchange coordinates, as you can to get the perpendicular in 2D, but as there's an infinite number of planes that‘s not possible. You can derive the bivector which characterises the plane easily. Each equation, such as x = y + z describes a 3-space. Writing that as x - y - z = 0 you can write down a vector perpendicular to it, e.g. a = (1, -1, -1). The bivector is simply the outer product of the two vectors, a ^ b. As the two perpendiculars lie in the plane the bivector is the one associated with the plane. It is unique up to a scale factor and so a better description than a pair of equations. I don’t know if there’s an easy way to get a pair of equations out of it.--JohnBlackburne^words_deeds 22:37, 3 August 2015 (UTC)[reply]

Okay, I can now see a way to do it. First though I want to reformulate the problem. Each equation has a corresponding vector, as noted above. The vector is the vector perpendicular to the three space the equation describes, and can just be written down from the equation. So you in effect start with two vectors, a and b, which span and so describe a plane. You want to find two vectors c and d which span and describe the orthogonal plane, as their equations will be orthogonal.

First generate the bivector B that is associated with the plane spanned by a and b, using the outer product from Geometric algebra (in fact all this is done with Geometric algebra)

B=a\wedge b

(typo fixed)

To get the bivector C associated with the orthogonal plane take the dual by multiplying by the pseudoscalar I

C=IB

To find c and d pick two arbitrary vectors, x and y, and project each onto C

c=(x\cdot C)C^{-1}

d=(y\cdot C)C^{-1}

You can then write down the equations from these vectors. If x and y are arbitrary the resulting vectors should be non-parallel and so distinct. If you are worried about this, or just want vectors that are less arbitrary, pick more than two (four, one along each axis) and reject parallel results until you have two non-parallel ones.--JohnBlackburne^words_deeds 01:00, 4 August 2015 (UTC)[reply]

How do you make a plane blush? Tell it an orthogonal compliment.

But seriously, a different approach is to write the equation of the original 2-plane P in a suitably symmetrical form. The initial 2-plane P is specified by a pair of linearly independent dual vectors $A_{\alpha }$ , $B_{\beta }$ (indices from 1 to 4), so P is the set of $Z^{\alpha }$ such that $Z^{\alpha }A_{\alpha }=Z^{\alpha }B_{\alpha }=0$ (here using the Einstein summation convention). Wedge these together to get the skew 4x4 matrix $X_{\alpha \beta }=A_{[\alpha }B_{\beta ]}$ satisfying the Plucker relations. Then P is the set of $Z^{\alpha }$ such that $Z^{\alpha }X_{\alpha \beta }=0$ . Let $Y_{\alpha \beta }$ denote the classical adjoint of $X_{\alpha \beta }$ . Then the orthogonal complement of P has equation $Z^{\alpha }Y_{\alpha \beta }=0$ . (Note, Y is also skew-symmetric, and satisfies the Plucker relations, so this actually does define a 2-plane).

The moral of the story is that instead of regarding a 2-plane as specified by a pair $A,B$ , it is actually more natural to regard a 2-plane as specified by a non-zero skew-symmetric $X$ satisfying the Plucker relation $X_{[\alpha \beta }X_{\gamma \delta ]}=0$ .

This can all be done invariantly, using a metric $g_{\alpha \beta }$ , and there the Hodge dual will appear explicitly. The adjoint of $X_{\alpha \beta }$ can be written (up to constants)

Y_{\alpha \beta }=\epsilon _{\alpha \beta \gamma \delta }X_{\rho \sigma }g^{\gamma \rho }g^{\delta \sigma }.

Here ε is the four-dimensional Levi-Civita symbol, and $g^{\alpha \beta }$ is the inverse metric. Sławomir
Biały 14:19, 4 August 2015 (UTC)[reply]