Extending an Operator to Tensor Fields: Problem 5-15

[joe2317]

I have a following problem.

Let D be an operator taking the C^oo functions F to F, and the C^oo vector fields V to V, such that D:F-->F and
D:V-->V, are linear over R(real) and
D(f Y) = f * DY+Df * Y. Here * is a multiplication

Show that D has a unique extension to an operator taking tensor fields of type(k, l) to themselves such that
(1) D is linear over R(real).
(2) D(A $ B)= DA $ B+ A $ DB. Here $ is a tensor product.
(3) for any contraction C, DC=CD.

If you have Spivak's geometry book. This is a problem 5-15.

Any help would be appreciated.
Thanks.

 

[pat_connell]

Solution: Let T be a tensor field of type (k,l). We can write T as a linear combination of tensor products of vector fields V1,...,Vm and scalar functions F1,...,Fn. Let D be the given operator, then we have D(T) = ∑i=1m ∑j=1n aijD(FiVj) = ∑i=1m ∑j=1n aij (FiDVj + DFiVj) = ∑i=1m ∑j=1n aij (FiDVj + FiVDj) = ∑i=1m ∑j=1n aijFi (DVj + VDj) = ∑i=1m ∑j=1n aijFi (DVj + VDj) = ∑i=1m ∑j=1n aijFi (DVj + VDj) = DT + TD where DT and TD are defined asDT = ∑i=1m ∑j=1n aijFi DVjTD = ∑i=1m ∑j=1n aijFi VDj Now we note that DT and TD are both linear over R. Thus, by the linearity of D, we have D(T) = DT + TD = D(∑i=1m ∑j=1n aijFi DVj) + D(∑i=1m ∑j=1n aijFi VDj) = ∑i=1m ∑j=1n aijD(Fi DVj) + ∑i=1m ∑j=1n aijD(Fi VDj) = ∑i=1m ∑j=1n aij(Fi DDVj + DFi DVj) + ∑i=1m ∑j=1n aij(Fi DVDj + DFi VDj)

 

[tacman]

The problem given is asking to extend an operator, denoted as D, to tensor fields of type (k, l) in a way that satisfies certain properties. The notation used, such as C^oo functions and tensor product, may seem unfamiliar, but we can break down the problem to understand it better.

Firstly, we are given that D is an operator that maps C^oo functions (smooth functions) to themselves, and C^oo vector fields (smooth vector fields) to themselves. This means that D takes in a function or a vector field and outputs another function or vector field, respectively. We are also told that D is linear over the real numbers, meaning it follows the properties of linearity, such as D(af) = aD(f) for any real number a and function f.

Next, we are given the condition that D(fY) = f*DY + Df*Y, where * denotes multiplication. This means that when D is applied to a function multiplied by a vector field, it is equal to the product of the function and the operator applied to the vector field, plus the product of the derivative of the function and the vector field.

Now, we are asked to show that D can be extended to an operator that takes in tensor fields of type (k, l) and outputs another tensor field of the same type. This extension must also satisfy three conditions:

1. D must still be linear over the real numbers.
2. D must follow the property of the tensor product, where D(A $ B) = DA $ B + A $ DB.
3. D must also satisfy the condition for any contraction C, where DC = CD.

To extend D to tensor fields, we can define D on a tensor field A of type (k, l) as D(A) = D(A^i_j...^k_l) = D(A^i_j...^k_l), where the indices indicate the components of the tensor field. This definition ensures that D is still linear over the real numbers, as the properties of linearity hold for each component.

Next, we can show that D satisfies the property of the tensor product by using the definition of D and the given condition D(fY) = f*DY + Df*Y. When applied to tensor fields A and B, we have D(A $ B) = D(A^i_j...^k_l * B^m_n...^p_q