I am trying to learn differential forms. I have read some scripts about differential forms and now I am trying to solve some problems.

So the problem is:

given $f: \mathbb{R}^2 \to \mathbb{R}^3, f(x_1,x_2):=(x_1+x_2, -x_1, -x_2)^T$

Now I have to calculate the basis representation of

$f^*(dx_1 \wedge dx_2 \wedge dx_3) $ where $f^*$ is a pullback of $f$

I guess I have to calculate the pullback?

According to the definition $$(f^* \alpha)(v_1,\ldots, v_k)=\sum_{I_k} f^* \alpha (e_{I_1},\ldots, e_{I_k}) dx_{I_1} \wedge \ldots \wedge dx_{I_3}$$

where $\alpha $ is a $k$-Form and $I_k :=\{ \{I_1,..,I_k \} | 1 \leq I_1 \leq I_2 \leq \ldots \leq I_k \leq N \}$

since $N = 2$ we get

$\sum_{I_k} f^* \alpha (e_{I_1},\ldots, e_{I_k}) dx_{I_1} \wedge \ldots \wedge dx_{I_3} = f^* \alpha (e_{1},e_{2}) dx_{1} \wedge dx_{2} + f^* \alpha (e_{1},e_{3}) dx_{1} \wedge dx_{3} + f^* \alpha (e_{2},e_{2}) dx_{2} \wedge dx_{3}$

My $\alpha$ is $dx_1 \wedge dx_2 \wedge dx_3$

but I don't know how to continue. Can someone help me with this ? Or give me an example how to calculate it.

| cite | improve this question | | | | |
  • 2
    $\begingroup$ the answer is zero since the two-dimensional domain will not support a nontrivial three-form. When you expand that definition there will be a repeated $dx_1$ and $dx_1$ or $dx_2$ and $dx_2$ hence it vanishes. $\endgroup$ – James S. Cook Feb 9 '16 at 14:30
  • $\begingroup$ ok, but what do you mean by expand? $\endgroup$ – user312018 Feb 9 '16 at 14:34
  • $\begingroup$ what you write does not follow the form of $\alpha$ (pun partly intended). The given $\alpha$ has $N=3$. $\endgroup$ – James S. Cook Feb 9 '16 at 15:51

To carry out the computation even though we know the answer will be zero:

We have $df_1 = dx_1 + dx_2$, $df_2 = -dx_1$ and $df_3 = -dx_2$. Thus \begin{align*} f^*(dx_1 \wedge dx_2 \wedge dx_3) &= df_1 \wedge df_2 \wedge df_3 \\ &= (dx_1 + dx_2) \wedge (-dx_1) \wedge (-dx_2) \\&= dx_1 \wedge dx_1 \wedge dx_2 + dx_2 \wedge dx_1 \wedge dx_2 \\ &= 0.\end{align*}

| cite | improve this answer | | | | |
  • $\begingroup$ I don't get the first equation. how do you get from f* to the rhs? My problem with the definition above is, that $\alpha $ is a 3-Form so I have to plug in 3 arguments. But I only have 2 e.g. $e_1, e_2$ $\endgroup$ – user312018 Feb 9 '16 at 14:56
  • $\begingroup$ @user312018 This is how the pullback is computed in coordinates. It follows from the definition of the pullback on functions and the fact that the exterior derivative $d$ and pullback operations commute. See e.g. Lee's "Introduction to Smooth Manifolds" Proposition 6.13 or Bott & Tu "Differential Forms in Algebraic Topology" I.2 right above Proposition 2.1. $\endgroup$ – Alex Provost Feb 9 '16 at 15:11
  • $\begingroup$ nicely done. I find myself doing this when faced with explicit problems. Intuitively, just plug in the formulas for the given function and calculate. $\endgroup$ – James S. Cook Feb 9 '16 at 15:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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