# Orientation form on manifold cut out by $m$ functions

Let $f: \mathbb{R}^n \to \mathbb{R}^m$ be a smooth function. If $a$ is such that $f$ has surjective derivative at all points in $f^{-1}(a)$ then this is an $n-m$ dimensional manifold $X$. I'm trying to show that the following is an orientation form:

$$\alpha = \textrm{det}(\frac{\partial f_i}{\partial x_j})^{-1} dx_1 \wedge ... \wedge dx_{n-m}$$

(This is on a particular chart where the first $m$ columns of $T_xf$ are linearly independent)

Where the determinant is taken over $n-m+1 \le j \le n$ and $1 \le i \le m$.

Clearly this is non vanishing on this particular chart (namely where the last $m$ columns of the matrix $Tf$ give an invertible matrix). Now I'd like to show that if we change coordinates this is still non-vanishing and then we will have a global non-vanishing n-form.

I'm stuck on how to see how $\alpha$ changes under different coordinates. I've tried using the fact that on $X$ we have that $f$ is constant and so for each $i$ we have $\frac{\partial f_i}{x_1}dx_1 + ... + \frac{\partial f_i}{x_n}dx_n = 0$ but how do we know which $i$ to use here to transform the form?

Thanks for any help

• Are you sure about "the last mm columns of the matrix $Df$ [being] an invertible matrix?" Because being full rank does not mean that any pick of $m$ columns will be invertible. See that you have given preference to the first $n-m$ coordinate variables. How do you justify? – Behnam Esmayli May 28 '16 at 16:32
• @Behnam thanks for your comment. Well the matrix $Tf$ has rank $m$ so there must be $m$ linearly independent columns, I then pick those ones and after relabelling coordinates say they are the last $m$? – Wooster May 28 '16 at 16:38
• Hint: Go through the proof of the regular value theorem. What are the coordinates (charts) that you construct in the proof? – Reznick Apr 17 '18 at 10:07