3.2. The Inequality of Christoffel Symbols of Mixed Subscripts
In Riemannian differential geometry, the equality of Christoffel symbols of mixed subscripts is usually adopt the doctrine. But no forceful researches could provide reliable supports. The truth is that the problem of mixed derivatives of bases in a Riemannian space are far different from the problem of normal mixed derivatives of a 3-dimensional surface in Euclidean geometry.
We could find out the truth that in the deduction of in Eq. (43), has been used instead of so that to gain . But we can easily calculate that and are not equal. It is said that the Eq. (43) could lead to nonvanishing value of , which really relates to time derivatives of bases that must be determined to be zero in rest field. We might have found out the problems.
3.2.1. Coordinate Transformations and Bases
Any points in a Riemannian space of Riemannian manifold of dimension has a neighborhood homeomorphic to a subset of Euclidean space of dimension , so that there must be probable maps between the neighborhoods and the corresponding subsets. It is just to say that the coordinates of any points in Riemannian space could be expressed with the coordinates of corresponding points of Euclidean space, and reversely. If a part or the entire of a Riemannian space are continuous and differentiable, Euclidean coordinate lines could be drawn in the part or the entire of the Riemannian space. On the other side, coordinate lines of Riemannian space could also be drawn in the corresponding Euclidean space. For convenience, the Riemannian space could be called covariant space, and the corresponding Euclidean space could be called contra variant space. A contra variant space is curved in the view of its covariant space, and the covariant space is also curved in the view of contra variant space.
It is obvious that transformations of spaces are actual coordinate transformations. These transformations could happen between covariant space and contra variant space, as well as they could happen among homeomorphic Riemannian spaces. Coordinate transformations may perform in the way with unequal metrics as well as the way with equal metrics.
The more effective method for coordinate transformation is to define bases and distances for spaces. Two examples would be presented firstly for definitions and for following discussions.
Example 1: Bases of Riemannian manifold of super surface
The derivative vectors of 3-dimensional surfaces were usually employed to form bases in classical differential geometry. The curve space has an extra dimension than a plane space that could be called the super surface. The Riemannian manifold of the super surface in the 3-dimensional space
would have a homeomorphic Euclidean space
in the space
. The coordinate lines
and
in contra variant space could be transformed to be
and
in covariant space, while
,
and
in covariant space be transformed to be
,
and
in the space
as shown in
Figure 3 and
Figure 4.
as contra variant space in 3-dimensional space.
The super surface could be determined by a vector function (57)
And the function could also be written as
(58)
or
(59)
At the same time, the contra variant space could be defined by (60)
There will be varieties of available ways to develop the expressions of bases and distances. I prefer to put forward the followings might as well.
The way in super space:
In super space, the differential has 3 components
(61)
That of differential could be simplified to be 2 dimensional because it just locates in the space (62)
To define a set of covariant bases for a position in covariant space by
(63)
It should be pointed out that, in some publications, coordinate and vector symbols have been used reversely like and , which would have brought about confusions.
In covariant space, the differential expressed by with covariant bases
(64)
Differential distance could be defined as
(65)
If the bases are orthogonal, there is
(66)
We have seen that the covariant bases are defined in covariant space to help contra variant coordinates to form covariant distances.
There are more complexities for a transformation between a super surface in 3-dimensional space and that the 2 covariant bases would have 3 components
(67)
Define the contra variant bases for a point in the plane that
(68)
The 3 contra variant bases all have 2 components as
(69)
The differential expressed by in contra variant space
(70)
Of course, one can create transformation matrix to perform the relationship between and directly.
The distance could be defined as
(71)
If the bases are orthogonal
(72)
One could imagine that this condition you cannot give the relationship of metrics that equals to , in that the covariant bases have 3 components and contra variant bases have 2.
The way in tangent space:
Consequently, the issues could be simplified in tangent spaces. At a position
in the covariant space, there is a neighborhood which will be labeled with coordinate lines of
and
, at the same time at the position
, there is a corresponding neighborhood in contra variant space labeled with coordinate lines of
and
, as shown in
Figure 5 and
Figure 6. Generally, coordinate lines could be set orthogonal. In most of publications,
and
were seen as
and
, but one should realize that the difference really matters.
One could define the differential vector in covariant space
(73)
As a result, the bases
(74)
Again, there is the distance
(75)
The differential keep the form as Eq. (62), so that the contra variant bases could be defined as
(76)
Also, there is
(77)
In this case, the relationship of metrics go harmonized that equals to . It should be pointed out that the ways of expressions of bases are all equivalent except that the substitutions of coordinate lines and might have hidden some information of super surface, so that I would like to make analysis within super space in most cases.
Example 2: Bases of Riemannian manifold of equal dimension
As a Riemanian manifold has equal dimension with its contra variant space, it could be called equal dimension manifold. A plane space
maps to a plane space
could be taken for granted, as shown in
Figure 7 and
Figure 8.
A differential vector in covariant space is
(78)
The differential vector in contra variant space is
(79)
Thus, the definition of contra variant bases could be
(80)
To express with (81)
Also, there is the definition of covariant bases
(82)
So that the expression of by should be
(83)
Something different is that a covariant base has 2 components
(84)
And a contra variant base also has 2 components
(85)
In the case that bases are orthogonal, the distance
(86)
and
(87)
3.2.2. Inequalities of Mixed Derivatives of Bases
Now it is the time to carry out the first discussion on the inequality of mixed derivatives of bases. The mixed derivatives of bases are just special defined for bases alternative derivations. As transformation from contra variant space to covariant space is concerned, the covariant bases could be considered to be derivated by the coordinate lines in chain rule
(88)
where, and are the direction derivatives along the coordinate lines and in covariant space, and and are their differential lengths in covariant space, which could be called the covariant lengths. And there will be a setting that Einstein summation convention does not act on double and double .
It should be pointed out that, in most mathematics and physics, mixed derivatives being confirmed to be equal is because in the Eq. (88) and is incorrectly understood to be the differential length in covariant space , but they are really the lengths in contra variant space . That is the reason we have carried out the concept of covariant length and .
Thus, the mixed derivatives will be
(89)
and
(90)
Consequently 3 conditions could be focused on:
Condition 1:
If there is an equality between the first items of the two equations that
(91)
For example, in the super surface, the mixed derivatives of course have the equality just as the equality of normal mixed derivatives of a 3-dimensional surface in a Euclidean space.
In this case and if there is another equality for the last items of the two equations that
(92)
Then that must come to the conclusion
(93)
Otherwise, that depends.
It should be pointed out that and do not equal in general conditions because they have different directions and, in most cases, they are usually set orthogonal, so that if that equality of Eq. (92) happens, it asks for
(94)
We will see that in some cases it is really well satisfied.
Condition 2:
Most special if
(95)
that indicate the first items of the two equations are not equal, but at the same time if the total equations are still equal that
(96)
We will still obtain the equality that
(97)
Condition 3:
This is the condition after the previous two conditions and else to them. Generally if
(98)
No matter the first items of the two equations are equal or not, the mixed derivatives will perform inequality
(99)
Then, turn to the issue of geometrical influence that the inequality of mixed derivatives will cause closure errors [
6,
7]. I prefer to give a brief presentation. Consider a differential in a curve line coordinate system expressed by bases along different coordinate paths as shown in
Figure 9 that
in path 1,
(100)
The irregular expressions of same symbols of integral variables and integral range could be adopted in special cases.
By Taylor’s approximation, it could be written as
(101)
We could also get the differential in path 2,
(102)
Trimming off the 3-order infinite small quantities, the difference of and is
(103)
There will be a closure error in close path if the mixed derivatives of bases do not equal.
3.2.3. Verifications and Discussions
Example 1: Polar coordinate system
Polar coordinate system that we are familiar with is a transformation from its contra variant space
, as shown in
Figure 10 and
Figure 11.
A position in contra variant space could be expressed by vector
(104)
and the differential is
(105)
Corresponding position in covariant space, will be expressed by
(106)
and the differential is
(107)
In the contra variant space, the differential distance between two positions could be defined as
(108)
The system we have used to is the one that has experienced transformation from space with
(109)
The bases
(110)
(111)
so that
(112)
Thus, there is the covariant distance
(113)
The mixed derivatives of bases
(114)
(115)
We have seen the mixed derivatives got equal
(116)
It could also be verified in Eq. (89) and Eq. (90) that
(117)
and
(118)
The vector is
(119)
Because is radius length , and is arc length , then
(120)
(121)
Thus, the first item of Eq. (117) is
(122)
The second item is
(123)
And the first item of Eq. (118)
(124)
The second item
(125)
so that
(126)
and
(127)
Obviously there is
(128)
and
(129)
but the mixed derivatives are equal totally that
(130)
Again, we have got the equality of mixed derivatives. But to our surprise is that this solution really subject to condition 2. It is said that the first items of Eq. (117) and Eq. (118) do not equal. One of the reasons in this case, is that there is no super surface.
Example 2: Spherical surface coordinate system
A spherical surface coordinate system is also a transformation of the corresponding contra variant space as in
Figure 12 and
Figure 13, in which
(131)
Differential distance is
(132)
And the coordinates of covariant space will be expressed with
(133)
The spherical coordinates could be transformed to Cartesian coordinates,
(134)
The bases could be defined as
(135)
and
(136)
Thus, there is the covariant distance
(137)
The derivatives
(138)
so that
(139)
It could also be verified in Eq. (89) and Eq. (90) that
(140)
and
(141)
The vector is
(142)
Because is arc length , and is arc length , then
(143)
(144)
Thus, the first item of Eq. (140) is
(145)
The second item is
(146)
And the first item of Eq. (141)
(147)
The second item
(148)
With Eq. (145) to Eq. (148) we found that
(149)
and
(150)
so that there is
(151)
One can see that this is of condition 1.
Example 3: Spherical coordinate system
A spherical coordinate system is also a transformation of the corresponding contra variant space as in
Figure 14 and
Figure 15, in which
(152)
Differential distance is
(153)
And the coordinates of covariant space will be expressed with
(154)
The spherical coordinates could be transformed to Cartesian coordinates,
(155)
The bases could be defined as
(156)
and
(157)
Thus, there is the covariant distance
(158)
The derivatives
(159)
so that
(160)
It could also be verified in Eq. (89) and Eq. (90) that
(161)
and
(162)
The vector is
(163)
Because is radius length , is arc length , and is arc length , then
(164)
(165)
(166)
Thus, the first item of Eq. (161) is
(167)
The second item is
(168)
And the first item of Eq. (162)
(169)
The second item
(170)
With Eq. (167) and Eq. (170) we found that
(171)
but there is
(172)
One can also calculate that
(173)
At the end we can obtain
(174)
One can see that one of them is of condition 1 and the others of them are of condition 2.
Additional discussion: deformed bases of example 3
If one of the bases in example 3 be set deformed as
(175)
where, is a function of coordinate .
One will find that the derivatives
(176)
and
(177)
while and will still keep the results as in Eq. (159), that will cause
(178)
This result reminds us that the same performance would have happened in gravitational space time that will be put into discussions in next section.