# Induced maps on tensor spaces

1. Oct 31, 2007

### 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.

2. Nov 3, 2007

### 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 $$P(e) \in W$$.

Last edited: Nov 3, 2007