Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $T: \mathbb{R}^n \rightarrow \mathbb{R}^m$ be a smooth transformation. Define the pullback $T^*: C^k (\mathbb{R}^m) \rightarrow C^k (\mathbb{R}^n)$ (With $C^k(\mathbb{R}^n)$ being the set of functionals on $k$-cells on $\mathbb{R}^n$) by $T^*: Y \mapsto Y \circ T$.

Thus, the pullback of $Y \in C^k (\mathbb{R}^m)$ is the functional on $C_k (\mathbb{R}^n)$ $$T^*Y:\phi \mapsto Y(T\circ \phi)$$

Why is the pullback linear, and why does $(T \circ S)^* = S^* \circ T^*$?

share|improve this question
1  
Both facts follow from a direct verification: have you tried to simply use the definitions of everything in sight? –  Mariano Suárez-Alvarez May 2 '12 at 1:23
    
I guess it is just difficult for me to understand how to correctly manipulate $k$-cells. –  Sam Kidd May 2 '12 at 1:26
    
Say you want to check that $(T\circ S)^*=S^*\circ T^*$. The two sides of the equality are functions, so to chck equality you need to apply both sides to an element of their domain and use their definitions to compute: the two results must be equal. –  Mariano Suárez-Alvarez May 2 '12 at 1:30
    
Alright, is it actually this easy then? $$(T \circ S)^* (Y) = Y \circ (T \circ S)$$ and $$S^* \circ T^*(Y) = S^* (Y \circ T) = Y \circ T \circ S$$ and to show linearity: $$T^*(aY + X) = (aY + X)(T) = aY(T) + X(T)$$ –  Sam Kidd May 2 '12 at 1:34
    
Yes :) (Although in the part about linearity you are missing a few $\circ$s.) –  Mariano Suárez-Alvarez May 2 '12 at 1:39
show 2 more comments

1 Answer 1

up vote 0 down vote accepted

Let $X,Y$ be arbitrary functionals (and $a$ an arbitrary constant). Since linearity is preserved on $k$-forms:

\begin{align} T^*(aY + X) &= (aY + X) \circ T \\ &= (aY + X) (T) \\ &= aY(T) + X(T) \\ &= aT^*(Y) + T^*(X). \end{align}

\begin{align} (T \circ S)^* (Y) &= Y \circ (T \circ S) \\ &= Y \circ T \circ S \\ &= S^*(Y \circ T) \\ &= S^* \circ T^* (Y). \end{align}

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.