# Pullback map distributes over wedge product (proof)

To prove that the pullback map distributes with the wedge product is it first best to prove that it distributes over the tensor product and then use the relation $$dx^{\mu_{1}}\wedge\cdots\wedge dx^{\mu_{r}}=\sum_{P\in S_{r}}sgn(P)dx^{\mu_{P(1)}}\otimes\cdots\otimes dx^{\mu_{P(r)}}$$ where $P\in S_{r}$ is a permutation of the ordered tuple $(1,\ldots, r)$ and $S_{r}$ is the set of all such permutations. ($sgn(P)$ is the signature of the permutation $P$ which is $+1$ if $P$ is an even permutation and $-1$ if $P$ is an odd permutation).

If so, is the following proof for distributivity over the tensor product correct?

Consider two $(0,r)$-type tensors $\alpha, \beta$, and their corresponding tensor product $\alpha\otimes\beta$. We have that the pullback of their tensor product $\phi^{\ast}(\alpha\otimes\beta)$ (induced by the map $\phi:M\rightarrow N$) acts on vectors $\mathbf{v},\mathbf{w}\;\in T_{p}M$ is defined as $$\phi^{\ast}(\alpha\otimes\beta)(\mathbf{v},\mathbf{w})= (\alpha\otimes\beta)(\phi_{\ast}\mathbf{v},\phi_{\ast}\mathbf{w})$$ where $\phi_{\ast}\mathbf{v},\phi_{\ast}\mathbf{w}\;\in T_{\phi_{p}}N$.

Next, we note the definition of the tensor product $$(\alpha\otimes\beta)(\mathbf{v},\mathbf{w})=\alpha (\mathbf{v})\beta (\mathbf{w})$$ and so it follows that $$\phi^{\ast}(\alpha\otimes\beta)(\mathbf{v},\mathbf{w})=(\alpha\otimes\beta)(\phi_{\ast}\mathbf{v},\phi_{\ast}\mathbf{w})\\ \qquad\qquad\quad\,=\alpha (\phi_{\ast}\mathbf{v})\beta (\phi_{\ast}\mathbf{w})\\ \qquad\qquad\qquad\;\;\,=(\phi^{\ast}\alpha)(\mathbf{v})(\phi^{\ast}\beta)(\mathbf{w})\\ \qquad\qquad\qquad\quad\;\;\;\;\;\,=\left((\phi^{\ast}\alpha)\otimes(\phi^{\ast}\beta)\right)(\mathbf{v},\mathbf{w})$$ where we have noted that the pullback of a $(0,r)$-type tensor is given by $(\phi^{\ast}\alpha)(\mathbf{v})=\alpha (\phi_{\ast}\mathbf{v})$. Hence, as $\mathbf{v},\mathbf{w}$ were chosen arbitrarily we conclude that $$\phi^{\ast}(\alpha\otimes\beta)=\left((\phi^{\ast}\alpha)\otimes(\phi^{\ast}\beta)\right).$$

• This looks good to me, though I wouldn't say that we observe that $(alpha \otimes \beta)({\bf v}, {\bf w}) = \alpha({\bf v}) \beta({\bf w})$, this is the definition of the tensor product. – Travis May 28 '15 at 14:48
• @Travis Thanks, I've edited it so it should read better now. – Will May 28 '15 at 14:51
• Better fix your definition of wedge product to make it alternating. – Ted Shifrin May 29 '15 at 13:36
• Whoops, yes. Have fixed it now, thanks. – Will May 29 '15 at 13:50