How Do Induced Maps Affect Higher Rank Tensors?

[quasar_4]

Hi all,

Given a map P: V-->W for vector spaces V and W and the map P*: W* --> V* we have the relationship that many of us are familiar with:

For e in V, f in W, E in V* and F in W*, we can say that

(P*(F))(e)=F(P(e)).

This is nice and fine. So this is kind of the case for a rank 1 tensor. Now can anyone help me generalize this to rank r tensors, namely those of type (0,r) and (r,0)? We can't worry about the case of type (r,s) unless we know that P is an invertible map. But I'm having a REALLY REALLY hard time understanding the case for some higher rank tensor.

I am also hoping that whoever can help with this can introduce it with also explaining which of these induced maps is the pushforward and which is the pullback.

Thanks.

 

[llarsen]

To use the same notation as you have used here, it is convenient to represent a tensor as an r form using the covariant tensor: F_1 \otimes F_2 \otimes ... \otimes F_r. To map this to a real number (as you did with a rank 1 tensor above), we contract this with a contravariant tensor e_1 \otimes e_2 \otimes ... \otimes e_r. Extending this is actually rather simple, since each tensor term is simply mapped individually as follows:

[P^*(F_1 \otimes F_2 \otimes ... \otimes F_r)](e_1\otimes e_2 \otimes ... \otimes e_r) = [F_1 \otimes F_2 \otimes ... \otimes F_r](P(e_1\otimes e_2 \otimes ... \otimes e_r))

which can be rewritten as:

[P^*(F_1) \otimes P^*(F_2) \otimes ... \otimes P^*(F_r)](e_1\otimes e_2 \otimes ... \otimes e_r) = [F_1 \otimes F_2 \otimes ... \otimes F_r](P(e_1)\otimes P(e_2) \otimes ... \otimes P(e_r))

which reduces to the scalar product:

[P^*(F_1)](e_1) \cdot [P^*(F_2)](e_2) \cdot ... \cdot [P^*(F_r)](e_r) = F_1(P(e_1)) \cdot F_2(P(e_2)) \cdot ... \cdot F_r(P(e_r))

Covectors are mapped via pull back P^*(F) \in V^* whereas vectors are mapped via push forward P(e) \in W.