A condition for the pullback $f^*\omega$ of top form is zero

Question

Let $f:\Bbb{R}^n\to \Bbb{R}^n$ be a smooth map. Denote by $f_i$ the coordinates of $f$ and suppose $$f_1^2+2f_2^2+3f_3^2+\dots+nf_n^2=1.$$ If $\omega$ is the top form on $\mathbb R^n$, show that its pullback $f^*\omega$ is the zero form.

My initial idea was to use the fact that $$\sum_{k=1}^nk\cdot(\partial_kf_k^2)=0$$ to show that the Jacobian matrix of $f$ has rank zero by reducing each row. But I have given up after futile calculations. I sense that the first relation above tells us something about the preimage set of $1$; maybe $f$ is submersion onto $f(\mathbb R^n)\ni 1$ and with codimension $<n$, and it can't have a non-zero $n$-form. But I failed to show this.

Any help would be appreciated.

Answer 1

From Lee Introduction to smooth manifolds book page 361

Let $F: M \to N$ be a smooth map between $n$-manifolds with or without boundary. If $(x^i)$ and $(y^j)$ are smooth coordinates on open subsets $U \subset M$ and $V \subset N$, respectively, and $u$ is a continuous real-valued function on $V$ , then the following holds on $U \cap F^{-1}(V)$: $$F^{*}\left(u d y^{1} \wedge \cdots \wedge d y^{n}\right)=(u \circ F)(\operatorname{det} D F) d x^{1} \wedge \cdots \wedge d x^{n}$$ where $DF$ represents the Jacobian matrix of $F$ in these coordinates.

In your case we have coordinates $\mathbb{R}^n$. Thus we need to show that $Df$ is singular. We have $$df=\begin{pmatrix} \frac{\partial f_1}{\partial x_1} & \frac{\partial f_2}{\partial x_1} & \cdots & \frac{\partial f_n}{\partial x_1} \\ \frac{\partial f_1}{\partial x_2} & \ddots & \cdots & \\ \ddots & \dots & & \\ \frac{\partial f_1}{\partial x_n} & \dots & \dots & \frac{\partial f_n}{\partial x_n} \end{pmatrix}=(df_1,\dots df_n)$$. This means that $Df$ will be singular if the vectors $(df_1,\dots df_n)$ will be linear dependent. Take the derivative of the function $f_{1}^{2}+2 f_{2}^{2}+3 f_{3}^{2}+\cdots+n f_{n}^{2}:\mathbb{R}^n \to \mathbb{R}$ to get: $$2f_1df_1 + 4f_2df_2+ \dots 2nf_n df_n=0$$ If the vectors $df_i$ are linearly independent then $2if_i=0$ for $1 \le i \le n$. This cant be true as $f_{1}^{2}+2 f_{2}^{2}+3 f_{3}^{2}+\cdots+n f_{n}^{2}=1$. To conclude $Df$ is singular and $\det Df=0$ so $f^* \omega=0$.