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Computing the k-volume of a k-parallelogram (i.e. a parallelogram spanned by k n-dimensional vectors) in n dimensions is straightforward: Let $P=[\overrightarrow{v_1},...,\overrightarrow{v_k}]$, then the k-volume is equal to $\sqrt{det(P^{T}P)}$. For instance, if n=3 and k=2, then this formula reduces to the norm of the cross-product and thus gives the 2-volume (area) of the parallelogram spanned by two 3-D vectors.

Now this fact is very useful when trying to find the k-volumes of manifolds (e.g. areas of 2-D surfaces in 3-D) by integration.

My question is, is there differential form or a form field that would serve the same purpose? In $\mathbb{R}^3$ for instance, the elementary 2-forms give you the signed area of the "projection" of the 2-parallelogram onto (x,y) or (x,z) or (y,z) planes depending on the form you use, but not the area of 2-parallelogram living in the space.

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No, you need to know what plane $\vec n\cdot\vec x=0$ your parallelogram in $\mathbb R^3$ lies in to write down the $2$-form that gives oriented area of the parallelogram. One of the distinguishing features of complex (differential) geometry is that there is, for each $k=1,\dots,n$, a $2k$-form on $\mathbb C^n$ that gives the $(2k)$-dimensional area form on every $k$-dimensional complex subspace.

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  • $\begingroup$ Thanks. I just realized that these integrands which arise naturally when you are trying to calculate surface areas or arc lengths all involve some kind of quadratic form over standard forms. They are called tensors I found out. $\endgroup$ – firemind Jun 26 '13 at 17:32
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    $\begingroup$ Yes, generally, you're going to apply your Gram determinant formula above. If you parametrize your $k$-dimensional submanifold by $\vec x = \vec f(u_1,\dots,u_k)$, then its surface area is given by integrating $$\int \sqrt{\det\left[\frac{\partial\vec f}{\partial u_i}\cdot\frac{\partial\vec f}{\partial u_j}\right]} \,d\vec u\,.$$ $\endgroup$ – Ted Shifrin Jun 26 '13 at 18:34

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