# Basis representation of differential-form $f(x_1,x_2):=(x_1+x_2, -x_1, -x_2)^T$

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.

• 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. – James S. Cook Feb 9 '16 at 14:30
• ok, but what do you mean by expand? – user312018 Feb 9 '16 at 14:34
• what you write does not follow the form of $\alpha$ (pun partly intended). The given $\alpha$ has $N=3$. – James S. Cook Feb 9 '16 at 15:51

## 1 Answer

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*}

• 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. $e_1, e_2$ – user312018 Feb 9 '16 at 14:56
• @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. – Alex Provost Feb 9 '16 at 15:11
• nicely done. I find myself doing this when faced with explicit problems. Intuitively, just plug in the formulas for the given function and calculate. – James S. Cook Feb 9 '16 at 15:50