What is the Factor of F in the Differential Forms Problem on Smooth Manifolds?

[Fredrik]

Introduction to smooth manifolds, by John Lee, page 304. The right-hand side of (c) near the top of the page has a factor $\omega_I\circ F$. I've been doing the calculation over and over for hours now and I keep getting just $\omega_I$. Is that F supposed to be there?

Edit: I should add that $$\omega_I=\omega(\partial_{i_1},\dots,\partial_{i_k})$$, so the left-hand side is just $F^*\omega$.

 

[quasar987]

Well, mmh... (3) can be proven from (1) and (2) and the facts that
(a) the pullback commutes with the exterior derivative
(b) the pullback of a 0-form f is f o F.

Granted this,
$$F^*\left(\sum'_I\omega_Idy^I\right)=\sum'F^*\omega_{i_1,...,i_k}F^*dy^{i_1}\wedge ...\wedge F^*dy^{i_k}=\sum'(\omega_{i_1,...,i_k}\circ F) d(y^{i_1}\circ F)\wedge ...\wedge d(y^{i_k}\circ F)$$

 

[Fredrik]

Thank you. That solved the problem for me. After a few minutes of verifying the relavant identities, I understand your solution perfectly. But I still don't see what's wrong with my failed attempt. Experience tells me that I'll see it just before I'm done typing this, but if you're reading this I guess it didn't work out that way.

$$F^*\omega(X_1,\dots,X_k)=\omega(F_*X_1,\dots,F_*X_k)=\sum_I{}^{'} \omega_I\ dy^{i_1}\wedge\dots\wedge dy^{i_k}(F_*X_1,\dots,F_*X_k)$$

$$=\sum_I{}^{'} \omega_I\ \sum_J(F_*X_1)(y^{j_1})\cdots (F_*X_1)(y^{j_k})\ dy^{i_1}\wedge\dots\wedge dy^{i_k}\left(\frac{\partial}{\partial y^{j_1}},\dots,\frac{\partial}{\partial y^{j_k}}\right)$$

Note that

$$(F_*X_1)(y^{j_1})=X_1(y^{j_1}\circ F)$$

and that this is a component of $X_1$ in the coordinate system $y\circ F$. We also have

$$dy^{i_1}\wedge\dots\wedge dy^{i_k}\left(\frac{\partial}{\partial y^{j_1}},\dots,\frac{\partial}{\partial y^{j_k}}\right)=\delta^I_J$$

where the delta is =0 unless there's exactly one permutation P that takes I to J, and =sgn P if there is. Since $y\circ F$ is a coordinate system too, we can also write

$$d(y^{i_1}\circ F)\wedge\dots\wedge d(y^{i_k}\circ F)\left(\frac{\partial}{\partial (y\circ F)^{j_1}},\dots,\frac{\partial}{\partial (y\circ F)^{j_k}}\right)=\delta^I_J$$

So the original expression can be written as

$$\sum_I{}^{'} \omega_I\ \sum_J X_1(y^{j_1}\circ F)\cdots X_1(y^{j_k}\circ F)\ d(y^{i_1}\circ F)\wedge\dots\wedge d(y^{i_k}\circ F)\left(\frac{\partial}{\partial (y\circ F)^{j_1}},\dots,\frac{\partial}{\partial (y\circ F)^{j_k}}\right)$$

$$=\sum_I{}^{'} \omega_I\ \sum_J d(y^{i_1}\circ F)\wedge\dots\wedge d(y^{i_k}\circ F)(X_1,\dots,X_k)$$

which is everything we want, except that $\omega_I$ hasn't changed. I still don't see what I've done wrong, so I'll post this so you can point and laugh.

 

[quasar987]

$$F^*\omega(X_1,\dots,X_k)=\omega(F_*X_1,\dots,F_*X_k)=[...]$$

You can't really write this because if F:M-->N, then F*w is a k-form on M, while w is a k-form on N. So the LHS is a map M-->R and the RHS is a map N-->R. So it is important to keep track of the point p in M at which the map of the LHS is evaluated. If you follow the definition of F*w given by Lee on page 303, you will agree that when evaluated at p in M, the equation you meant to write is actually

$$(F^*\omega)_p(X_1|_p,\dots,X_k|_p)=\omega_{F(p)}(F_*(X_1|_p),\dots,F_*(X_k|_p))$$

and then when you expand the RHS, you will have

$$\sum_I \omega_I(F(p))\ dy^{i_1}|_{F(p))}\wedge\dots\wedge dy^{i_k}|_{F(p))}(F_*(X_1|_p),\dots,F_*(X_k|_p))$$

etc.

 

[Fredrik]

Thank you again. I understand now.