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Suppose one has a $k$-manifold given by $f^{-1}(0)$ for some $C^1$ map $f: U\to \mathbb{R}^{n-k}$ (where $[D f(x)]$ is surjective). How can one construct a form-field $\omega$ that orients the manifold?

In particular I would like to use the fact that $$\Sigma(\vec{v_1},\dots,\vec{v_k}):=\text{sgn}\det[\Delta f_1(x),\dots,\Delta f_{n-k}(x),\vec{v_1},\dots,\vec{v_k}].$$

Using this orientation, how can one construct a form field? In the case of $f(x)=x^2+y^2-1=0$, with $f^{-1}(0)$ the circle, I know one can take $\vec v=[dx,dy]^T$and solve $\det[\Delta f,v]=2xdy+2ydx$, which is a form that orients the circle.

In general, say if one has a 2-manifold, how does this scale?

EDIT: My tentative answer is taking the $v_i$ to be $[dx_1,\dots,dx_n]^t$ and doing the same, but it's not clear to me whether this works in higher dimensions.

EDIT: I'm most interested in the elementary approach, but would be happy to hear how something like the Hodge star operator ties in if there is an explicit relationship.

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2 Answers 2

Do you know about the (Hodge) star operator? This is a great way to get from an ($n-k$)-form to a $k$-form. This will accomplish what you've sketched.

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I don't. Is it related to the construction I am attempting? At the end of the day I have to be able to explain this to people at a rather elementary level. –  Wouter Zeldenthuis Apr 23 '13 at 12:47
    
Let's try this: Since $Df(x)$ has rank $n-k$, locally some $(n-k)\times (n-k)$ submatrix is nonsingular. Let's assume it's the first $n-k$ columns. Then the coefficient of $dx_1\wedge\dots\wedge dx_{n-k}$ in $df_1\wedge\dots\wedge df_{n-k}$ has a constant sign. Using $\omega=dx_{n-k+1}\wedge\dots\wedge dx_n$ should do it. –  Ted Shifrin Apr 23 '13 at 14:18
    
But your construction is local, whereas (assuming the manifold is connected) the form field should orient the entire manifold. So the sign has to be globally consistent. –  Wouter Zeldenthuis Apr 23 '13 at 14:21

Does this in fact trivially follow from what I already showed? If we take the $v_i$ to all be $[dx_1,\dots,dx_n]^T$, then $\det[Δf_1(x),…,Δf_{n−k}(x),v_1,…,v_k]$ is certainly a form-field.

The only remaining question is whether this always has a consistent sign.

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Yes, I believe you're correct. Because the gradient vectors $\nabla f_1$, ..., $\nabla f_{n-k}$ are everywhere linearly independent, you are getting a consistent orientation this way. Be careful, however, when you evaluate that determinant with the $dx_j$ in there ... you need wedges and a consistent ordering. –  Ted Shifrin Apr 23 '13 at 15:29
    
The wedges come in from the interpretation of the determinant, no? Do you agree that all of the $v_i$'s should be of the same form when determining these? –  Wouter Zeldenthuis Apr 23 '13 at 16:30
    
Yes, I agree on the latter. But what does $\left|\begin{array}{c} dx & dy \\ dx & dy \end{array}\right|$ mean? You want it to mean (some constant times) $dx\wedge dy$. Etc. –  Ted Shifrin Apr 23 '13 at 18:13

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