I want to know how to prove the theorem:

If M is a k-manifold in $\mathbb{R}^n$, then it is orientable if and only if there is a volume form defined globally on M.

I'm currently stuck at this step:

If I have a nowhere-vanishing k-form on M, how would I proceed and show that M is orientable from this point?

  • how do you define "orientable"? – James S. Cook Dec 6 '12 at 4:31
  • Orientable means that we could cover M by coordinate charts g: U -> $\mathbb{R}^n$, such that $\{\frac{\partial g}{\partial u_1}, \dots, \frac{\partial g}{\partial u_k}\}$ is a positive basis for the tangent space of M – Enzo Dec 6 '12 at 4:43
  • was my answer helpful? – James S. Cook Apr 12 '13 at 12:40
  • Yes it was! I ended up figuring it out myself. I figured out a slightly different proof that is a bit messier. But this is great! Thank you! – Enzo Apr 12 '13 at 19:44
up vote 9 down vote accepted

Suppose there exists a volume form $\gamma$ on an $k$-dimensional submanifold $M$ embedded in $\mathbb{R}^n$ where $k \leq n$. At each point in $p \in M$ there exist coordinates $x^1,x^2, \dots x^k$ for $M$ such that $$ \gamma = h(x)dx^1 \wedge dx^2 \wedge \dots \wedge dx^k $$ for all points in the coordinate chart $x$ containing $p$. If $\gamma_p (\partial_1,\partial_2, \dots , \partial_k) =h(p) >0$ then by continuity of $h$ the form must be nonzero in a open set near $p$. More to the point, since it is a volume form it follows $\gamma(p) \neq 0$ for all $p \in M$. Consequently, $\gamma_p (\partial_1,\partial_2, \dots , \partial_k) >0$ on the whole domain of the coordinate chart $x$. This corresponds to representing, pointwise, the coordinate vector fields $\partial_j$ by $k$-vectors $v_j$ in such a way that $det[v_1|v_2| \cdots |v_k]>0$. In other words, at each point in the coordinate domain the coordinate basis forms a positively oriented $k$-frame. On the other hand suppose $\gamma_p (\partial_1,\partial_2, \dots , \partial_k) =h(p) <0$ then we can simply permute any pair of coordinate functions to create a chart on which $h(p)>0$. Therefore, on a arbitrary coordinate chart with domain $U$ of $M$ we can assume $h(p)>0$ for all $p \in U$ (again, if negative just change coordinates by swapping the first two coordinate charts).

Suppose $V \subseteq M$ with $U \cap V \neq \emptyset$ and consider the coordinate chart $\bar{x}$ on $V$ again choose the chart such that the evaluation on the $k$-tuple of ordered coordinate vectors is positive (this means $\bar{h}>0$ in what follows). We can write the volume form on $V$ by $$ \gamma = \bar{h}(\bar{x})d\bar{x}^1 \wedge d\bar{x}^2 \wedge \dots \wedge d\bar{x}^k $$ Now, for points in the intersection, $$ d\bar{x}^i = \sum_{j=1}^k\frac{\partial \bar{x}^i}{\partial x^j}dx^j$$ Applying the chain-rule for manifolds $k$-fold times yields: $$ \gamma = \sum_{j_1,\dots, j_k=1}^{k} \bar{h}(\bar{x}) \frac{\partial \bar{x}^1}{\partial x^{j_1}} \frac{\partial \bar{x}^2}{\partial x^{j_2}} \cdots \frac{\partial \bar{x}^k}{\partial x^{j_k}}dx^{j_1} \wedge dx^{j_2}\wedge \cdots \wedge dx^{j_k}$$ Observe that the $x$-expression and $\bar{x}$-expression are merely two formulas for the same differential form $\gamma$. Moreover, it is not hard to show: $$ dx^{j_1} \wedge dx^{j_2}\wedge \cdots \wedge dx^{j_k} = \epsilon_{j_1j_2 \dots j_k} dx^1 \wedge dx^2 \wedge \cdots \wedge dx^k$$ where $\epsilon_{12\dots k}=1$ and otherwise the values are given by the complete antisymmetry of the Levi-civita symbol $\epsilon_{j_1j_2 \dots j_k}$. Consequently: $$ \gamma = \sum_{j_1,\dots, j_k=1}^{k} \bar{h}(\bar{x}) \epsilon_{j_1j_2 \dots j_k} \frac{\partial \bar{x}^1}{\partial x^{j_1}} \frac{\partial \bar{x}^2}{\partial x^{j_2}} \cdots \frac{\partial \bar{x}^k}{\partial x^{j_k}}dx^1 \wedge dx^2\wedge \cdots \wedge dx^k$$ But, this simply says $$\gamma = \bar{h}(\bar{x}) det \biggl[ \frac{\partial \bar{x}^1}{\partial x^{j_1}} \bigg| \frac{\partial \bar{x}^2}{\partial x^{j_2}} \bigg| \cdots \bigg| \frac{\partial \bar{x}^k}{\partial x^{j_k}} \biggr] dx^1 \wedge dx^2\wedge \cdots \wedge dx^k $$ Comparing with $\gamma = h(x)dx^1 \wedge dx^2 \wedge \dots \wedge dx^k$ and using the fact that we chose coordinates, by reordering if necessary, such that $h,\bar{h}>0$ it follows $$ det \biggl[ \frac{\partial \bar{x}^1}{\partial x^{j_1}} \bigg| \frac{\partial \bar{x}^2}{\partial x^{j_2}} \bigg| \cdots \bigg| \frac{\partial \bar{x}^k}{\partial x^{j_k}} \biggr] >0$$ Therefore, given the existence of a non-vanishing volume form we can select an atlas of charts for which the Jacobian determinant of any overlap is positive.

Conversely, if we are given a positively oriented atlas in the sense that the above Jacobian determinant is positive we can construct $\gamma$ by setting $h=1$ essentially. Then the top-forms $dx^1 \wedge dx^2 \wedge \cdots \wedge dx^k$ and $d\bar{x}^1 \wedge d\bar{x}^2 \wedge \dots \wedge d\bar{x}^k$ describe the same differential form on an overlap and we may define $\gamma$ globally in this way.

Note: on a non-orientable manifold it is not possible to globally paste top-forms together, negative Jacobian determinants will spoil the construction.

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