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 $$\phi^{\ast}(\alpha\otimes\beta)=\left((\phi^{\ast}\alpha)\otimes(\phi^{\ast}\beta)\right).$$

  • $\begingroup$ 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. $\endgroup$ – Travis May 28 '15 at 14:48
  • $\begingroup$ @Travis Thanks, I've edited it so it should read better now. $\endgroup$ – Will May 28 '15 at 14:51
  • $\begingroup$ Better fix your definition of wedge product to make it alternating. $\endgroup$ – Ted Shifrin May 29 '15 at 13:36
  • $\begingroup$ Whoops, yes. Have fixed it now, thanks. $\endgroup$ – Will May 29 '15 at 13:50

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