Tensors and differential geometry
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F. B.
Feb 2, 2014, 7:28:22 AM2/2/14
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Hi all!
I was considering that it would be great to have the diffgeom module and the tensor module work together, as tensors are also part of differential geometry arising on the tangent and cotangent spaces of manifolds.
The main problem I face is that in the tensor module, indices of a tensor can be declared as belonging to different types. For example, gamma matrices can be declared as a (Lorentz, Spinor, Spinor) tensor. The question is, how to characterize such a tensor from a differential geometric perspective?
The Lorentz and Spinor indices are indices carrying two different representation of the symmetry of the universe, they correspond to two representation of a Lie algebra, and have their own transformation laws. The point is, in SymPy there is no such advanced infrastructure which is able to handle principal bundles, so I was wondering if there can be an easier approach to this problem.
When I consider the Riemann tensors, for example, R(a, -b, -c, -d), this is an element of the tensor product space (T, V, V, V), where T is the tangent space, and V is the cotangent space, of the same base manifold, i.e. the space time manifold.
Do you think that the gamma matrices, as their indices do not belong to the same spaces, can be viewed as a tensor in some power of the tangent space of the product space of two manifolds, say the spacetime and something like a Clifford Algebra which represents the spinor space?
It would be useful to be able to declare a link to a manifold in the object TensorIndexType, e.g.:
in such a way, tensors depending only on L would be immediately linked to manifold M, and it would be possible to use the already implemented algorithms in the diffgeom module to perform covariant and Lie derivative, as well as compute the Riemann tensor, Ricci tensor from the metric tensor.
The problem remains in mixed indices tensors. Any ideas on how to overcome this?
I was considering that it would be great to have the diffgeom module and the tensor module work together, as tensors are also part of differential geometry arising on the tangent and cotangent spaces of manifolds.
The main problem I face is that in the tensor module, indices of a tensor can be declared as belonging to different types. For example, gamma matrices can be declared as a (Lorentz, Spinor, Spinor) tensor. The question is, how to characterize such a tensor from a differential geometric perspective?
The Lorentz and Spinor indices are indices carrying two different representation of the symmetry of the universe, they correspond to two representation of a Lie algebra, and have their own transformation laws. The point is, in SymPy there is no such advanced infrastructure which is able to handle principal bundles, so I was wondering if there can be an easier approach to this problem.
When I consider the Riemann tensors, for example, R(a, -b, -c, -d), this is an element of the tensor product space (T, V, V, V), where T is the tangent space, and V is the cotangent space, of the same base manifold, i.e. the space time manifold.
Do you think that the gamma matrices, as their indices do not belong to the same spaces, can be viewed as a tensor in some power of the tangent space of the product space of two manifolds, say the spacetime and something like a Clifford Algebra which represents the spinor space?
It would be useful to be able to declare a link to a manifold in the object TensorIndexType, e.g.:
L = TensorIndexType('L')
M = Manifold('M')
L.manifold = M
in such a way, tensors depending only on L would be immediately linked to manifold M, and it would be possible to use the already implemented algorithms in the diffgeom module to perform covariant and Lie derivative, as well as compute the Riemann tensor, Ricci tensor from the metric tensor.
The problem remains in mixed indices tensors. Any ideas on how to overcome this?
Alan Bromborsky
Feb 2, 2014, 7:58:52 AM2/2/14
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Currently I am rewriting the geometric
algebra (Clifford algebra) and calculus module for sympy again.
Current work is being kept separately at github.com/brombo/GA for
now. You may want to look at tensor sections in the "GA Notes"
and "LaTeX docs" directories.
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F. B.
Feb 2, 2014, 10:23:02 AM2/2/14
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On Sunday, February 2, 2014 1:58:52 PM UTC+1, brombo wrote:
Currently I am rewriting the geometric algebra (Clifford algebra) and calculus module for sympy again. Current work is being kept separately at github.com/brombo/GA for now. You may want to look at tensor sections in the "GA Notes" and "LaTeX docs" directories.
I had a glimpse at it, but that does not solve my problem (did I miss something?).
My idea is to extend the tensor product to vectors and one-forms of different manifolds, so a tensor like the gamma matrices can have a differential geometric interpretation. I'm just not sure that this approach is correct.
Alan Bromborsky
Feb 2, 2014, 12:07:59 PM2/2/14
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I don't think abstract tensors (index
quantities) are a good fit with differential geometry. A better
approach is multlinear algebra as developed in "Multilinear
Algebra" by Werener Greub in which tensor algebra, exterior
algebra, and Clifford algebra are all developed on an equal
footing. In terms of Clifford algebra a spinor is the sum of a
scalar and a bivector (equivalent to antisymmetric rank-2
tensor), a concept which is not in differential forms.
Matthew Rocklin
Feb 2, 2014, 3:11:38 PM2/2/14
to sy...@googlegroups.com, Stefan Krastanov
Pinging @krastanov
F. B.
Feb 3, 2014, 4:35:41 AM2/3/14
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OK, I found out this wikipedia article:
https://en.wikipedia.org/wiki/Spinor_bundle
Apparently besides the tangent and cotangent bundles, there is also need for the spinor bundle to represent gamma matrices in the diffgeom module.
https://en.wikipedia.org/wiki/Spinor_bundle
Apparently besides the tangent and cotangent bundles, there is also need for the spinor bundle to represent gamma matrices in the diffgeom module.
Alok Gahlot
Dec 19, 2014, 1:20:58 AM12/19/14
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Dear member of the groups , i have a problem i basics ., my question is what is the difference in contravariant and covariant tensor
and how can we decide to choose tensor in any problem? what is the rule by which we know that we use covariant or contravariant or mixed tensor ?
Francesco Bonazzi
Dec 19, 2014, 3:06:02 AM12/19/14
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On Friday, December 19, 2014 7:20:58 AM UTC+1, Alok Gahlot wrote:
Dear member of the groups , i have a problem i basics ., my question is what is the difference in contravariant and covariant tensor
http://en.wikipedia.org/wiki/Covariance_and_contravariance_of_vectors
For questions about math itself I would suggest you to use mathoverflow.
and how can we decide to choose tensor in any problem? what is the rule by which we know that we use covariant or contravariant or mixed tensor ?
If you have a metric, covariant and contravariant representations are equivalent (you can raise/lower the index by using the metric), so it doesn't really matter. Anyways, it depends on what you are working with.
Alok Gahlot
Dec 20, 2014, 7:46:33 AM12/20/14
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Thank you dear,I got that in case of working with metric there is no difference between covariant and contravarient tensor but if we are not dealing with metric then ?
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Francesco Bonazzi
Dec 20, 2014, 4:55:33 PM12/20/14
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On Saturday, December 20, 2014 1:46:33 PM UTC+1, Alok Gahlot wrote:
Thank you dear,I got that in case of working with metric there is no difference between covariant and contravarient tensor but if we are not dealing with metric then ?
Well, there is always difference between covariant and contravariant indices, with the metric you just have a correspondence between the two. If there is no metric there is no such correspondence.
Kalevi Suominen
Dec 21, 2014, 9:12:35 AM12/21/14
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On Friday, December 19, 2014 8:20:58 AM UTC+2, Alok Gahlot wrote:
my question is what is the difference in contravariant and covariant tensor
The simple (and stupid) answer is that they are objects of different type.
In more detail, their components behave differently under general coordinate
transformations.
In more detail, their components behave differently under general coordinate
transformations.
and how can we decide to choose tensor in any problem?
There is no freedom of choice in general. Only if an inner product ("metric") is given,
contravariant, covariant, and mixed tensors can be transformed into each other.
Even then there is usually one form that is best suited to the problem in question.
contravariant, covariant, and mixed tensors can be transformed into each other.
Even then there is usually one form that is best suited to the problem in question.
what is the rule by which we know that we use covariant or contravariant or mixed tensor ?
This is a hard question. I cannot write down any general rule valid in all situations.
Instead I try to give some examples in the case of tensors on a (single) vector space V
as in http://en.wikipedia.org/wiki/Tensor_%28intrinsic_definition%29 .
Loosely speaking, a tensor can be regarded as a (linear or multilinear) mapping,
often in more than one way.
A tensor is of type (p,q) if there are q vector arguments and p vector values.
If the value is a scalar, then p = 0, and q = 0, if there are no arguments.
1. A vector x in V is a tensor of type (1,0). It may be thought of as a mapping with no arguments
and the single vector x as its value.
2. A covector x* in V* is a tensor of type (0,1). It is a linear mapping from V to the scalars.
So its argument is a single vector and its value is a scalar.
3. A linear mapping u: V -> V is a tensor of type (1,1). It takes a single vector x as its
argument and has another vector u(x) as its value.
4. An inner product (or metric) in V is a tensor of type (0,2). It has two vector arguments x and y
and their scalar product x.y as the value.
5. A bilinear mapping VxV -> V is a tensor of type (1,2). It takes two vector arguments
and the value is a vector. An example is the cross product in the three-dimensional space.
Finally, if a metric is given, each vector x in V corresponds to a covector x* in V*, namely the
linear form x*: y -> x.y . Then the difference between contravariance and covariance disappears
under transformations that preserve the metric.
Instead I try to give some examples in the case of tensors on a (single) vector space V
as in http://en.wikipedia.org/wiki/Tensor_%28intrinsic_definition%29 .
Loosely speaking, a tensor can be regarded as a (linear or multilinear) mapping,
often in more than one way.
A tensor is of type (p,q) if there are q vector arguments and p vector values.
If the value is a scalar, then p = 0, and q = 0, if there are no arguments.
1. A vector x in V is a tensor of type (1,0). It may be thought of as a mapping with no arguments
and the single vector x as its value.
2. A covector x* in V* is a tensor of type (0,1). It is a linear mapping from V to the scalars.
So its argument is a single vector and its value is a scalar.
3. A linear mapping u: V -> V is a tensor of type (1,1). It takes a single vector x as its
argument and has another vector u(x) as its value.
4. An inner product (or metric) in V is a tensor of type (0,2). It has two vector arguments x and y
and their scalar product x.y as the value.
5. A bilinear mapping VxV -> V is a tensor of type (1,2). It takes two vector arguments
and the value is a vector. An example is the cross product in the three-dimensional space.
Finally, if a metric is given, each vector x in V corresponds to a covector x* in V*, namely the
linear form x*: y -> x.y . Then the difference between contravariance and covariance disappears
under transformations that preserve the metric.
Alan Bromborsky
Dec 21, 2014, 10:13:22 AM12/21/14
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Attached is another view of tensors
where if a dot product of vectors is defined then given a set of
basis vectors 
we
have a metric 
and a set of reciprocal basis
vectors defined by 
. Then 
.
we
have a metric and a set of reciprocal basis
vectors defined by . Then .--
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Alok Gahlot
Dec 21, 2014, 10:21:17 AM12/21/14
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Thanks Alan it clear the shadow of doubts
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