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Path: blob/main/09Manifold_LinearFormsTensors_onTp.ipynb
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9. Tensors on
This notebook is part of the Introduction to manifolds in SageMath by Andrzej Chrzeszczyk (Jan Kochanowski University of Kielce, Poland).
Warning. In this notebook there are many repetitions with respect to the notebook 9a. Tensors on modules. Although mathematically tensors on modules are generalizations of tensors on the tangent spaces, the SageMath Manifolds code in the present notebook differs significantly from that in notebook 9a.
Differentials of functions, linear forms, covectors
If is a smooth manifold, , is the tangent space to at , and , then the differential of at , denoted by , is defined by
ParseError: KaTeX parse error: Undefined control sequence: \label at position 65: …} X_p\in T_pM. \̲l̲a̲b̲e̲l̲{} \tag{9.1} \e…Let us show some examples of scalar functions and differentials.
Example 9.1
Define a 3-dimensional manifold a point , the tangent space at , a tangent vector and some scalar function on .
Now let us define the differential of at :
and check that the value of at on the tangent vector is equal to :
Linearity of
Recall that for a vector space (or a module) the map is linear iff
To check that is linear let us note that
Cotangent space
The space of linear forms on is by definition the dual space of and is denoted by . Elements of are called covectors or covariant vectors and is called the cotangent space to at .
The vector space operations in are defined in a natural way
for and .
To check what kind of object is we can use the parent
method:
Example 9.2
If are local coordinates on , then differentials are covectors from . From (9.1) it follows that if the tangent vector is expressed as , then
ParseError: KaTeX parse error: Undefined control sequence: \label at position 54: …X_p)=X_p(x^i), \̲l̲a̲b̲e̲l̲{}\tag{9.1'} \e…i.e. is the covector that maps the tangent vector at to its -th component in the basis
Example 9.3
Let us define the scalar functions representing the i-th coordinate of the point.
And next their differentials:
We can check that the just defined differentials, when applied to the tangent vector give its i-th component .
Bases in the cotangent space
Using the definition of the basis tangent vectors and the fact that we see that and consequently . The last relation implies that:
The family is a basis of .
In fact if the linear combination is equal to zero covector, then i.e., the coefficients of this combination vanish, so the family is linearly independent.
If , then for in the form we have
From the definition of the differential it follows , so
consequently
ParseError: KaTeX parse error: Undefined control sequence: \label at position 91: …_p\Big)dx^i_p. \̲l̲a̲b̲e̲l̲{} \tag{9.2} \e…Thus every covector from is a linear combination of
Example 9.4
For example the differential of is a linear combination of with coefficients
In SageMath Manifolds the index denoting the point is always dropped.
The function display
shows the differentials of functions and general linear forms as linear combinations of
Example 9.5
Let us define a general linear form on the tangent space of a 3-dimensional manifold.
The linear form applied to a tangent vector gives a scalar.
The value of a tangent vector on a linear form is by definition equal to .
Tensors in tangent spaces
Recall, that if is a vector space or a module, by a multilinear or more precisely -linear form we mean a function which is linear in each of its arguments i.e., for
Assume that is a smooth manifold.
Space of tensors of mixed type
is the space of multilinear maps:
where is the tangent space at and is the corresponding cotangent space.
Since for linear forms and tangent vectors the assignment defines a linear form , the elements can be considered as elements of . On the other hand as the space of linear forms on is equal to . Thus, the tensor spaces generalize the tangent and cotangent spaces.
Example 9.6
Let us show how to define tensor modules of type (1,0) and (0,1) in SageMath Manifolds.
Space - of covariant tensors of rank
is the space of multilinear maps
where denotes the tangent space at .
For k=1 we obtain the space of covectors at , i.e., linear forms on
In we introduce the vector space structure by
where , and .
For we define the tensor product by
ParseError: KaTeX parse error: Undefined control sequence: \label at position 115: …. , v_{k+m} ), \̲l̲a̲b̲e̲l̲{eq:tensor_prod…for .
In SageMath Manifolds the symbol of tensor product is simply .
Covariant tensors in local components
If the tangent vectors are expressed as , then by the multilinearity of and the definition (9.3) of the tensor product we have for :
so
where ParseError: KaTeX parse error: Undefined control sequence: \label at position 139: …}\big|_p\big). \̲l̲a̲b̲e̲l̲{}\tag{9.5} \en…
Example 9.7
Let us check the formula (9.3) in the case of two general tensors from for a 2-dimensional manifold .
First we define a list of components of four tangent vectors:
and the vectors:
Next we define the components of two (0,2)-type tensors and .
Now we are ready to check the formula (9.3).
Algebraic properties of tensor product of covariant tensors
Tensor product has the following properties
for and for arbitrary covariant tensors (the addition is defined only for tensors of the same rank ).
To check the first formula, let us note that for
and similarly for the second formula.
The associativity follows from
for
Example 9.8
Let us show an example of a tensor . First let us define the corresponding 3-dimensional table of components.
Now we define a manifold , a point , tangent space and the tensor .
Next we show how to define a tensor of type (0,3) on a 2-dimensional manifold with concrete components.
Define the local basis of .
Check that the component of the tensor from the previous example is equal to .
Space of contravariant tensors of rank
is the space of multilinear forms:
where is the cotangent space at .
In we introduce the vector space structure by
where , and .
For we define the tensor product by
for .
Contravariant tensors in local components
Let and let all covectors be expressed as (cf. (9.2)). We have
Since we obtain
Example 9.9
Let us give an example of a tensor .
First we define the corresponding 3-dimensional table of components with upper indices.
Now we define a two-dimensional manifold, the tangent space and the tensor t with component from the table st.
We are ready to display the general tensor of type (3,0) on a 2-dimensional manifold in components.
Here is a tensor of type (3,0) on a 2-dimensional manifold with concrete components:
Let us define the basis of .
Check that the component is equal to .
General tensors from
Since the definition of was given above we start from definition of tensor product.
For we define the tensor product by
for and .
Generalizing the formulas (9.5) and (9.7) we obtain the following expression for the general tensor in local components
ParseError: KaTeX parse error: Undefined control sequence: \label at position 289: …es dx^{j_m}_p. \̲l̲a̲b̲e̲l̲{}\tag{9.9} \en…Very often the notation is used and then
To check that the elements
ParseError: KaTeX parse error: Undefined control sequence: \label at position 167: …mes dx^{j_m}_p \̲l̲a̲b̲e̲l̲{} \tag{9.12} \…are linearly independent, assume that the linear combination
vanishes. If we apply this combination to we get .
Since previously we have checked that elements of the type (9.12) span we have proved, that these elements form a basis for .
Example 9.10
To show an example of a tensor , we first define a 4-dimensional table with lower and upper indices.
Now define a 2-dimensional manifold , tangent space at and .
We can display a general tensor from on a 2-dimensional manifold ( components).
A more concrete example can be defined as follows:
Using the bases of and ,
one can check that the component is equal to :
What's next?
Take a look at the notebook Alternating forms on modules.