# How to recover calculus integral from tensor and differential forms

Consider the integral on the manifold $\mathbb{R}^2$ defined by the differential form: Levi-Civita tensor:

$$\int \epsilon= \int \epsilon_{\mu\nu}dx^{\mu}\otimes dx^{\nu}$$

Now suppose we want to recover ordinary double integral in calculus in coordinates $x,y$ from the above definition.

The nonvanishing components of the Levi-Civita tensor are $\epsilon_{12}=1; \epsilon_{21}=-1$

We therefore have: $$\int \epsilon_{\mu\nu}dx^{\mu}\otimes dx^{\nu}=\int (dx^1\otimes dx^2-dx^2\otimes dx^1)$$

But I should get $$\int \epsilon_{\mu\nu}dx^{\mu}\otimes dx^{\nu} = \int\int dxdy$$

When descending from tensor to calculus.

For the metric $\eta_{\mu\nu}dx^{\mu}\otimes dx^{\nu}$ we use this convention:

$$dx^idx^k:=(dx^i \otimes dx^k + dx^k \otimes dx^i)/2$$

To switch to calculus in metric calculations we have $$dx^2=dxdx$$$$=(dx \otimes dx + dx \otimes dx)/2=dx\otimes dx$$

In the first line above $dx$ is read as the infinitesimal change in calculus whereas in the second line above $dx$ is a basis one-form or tensor that operates on a tangent vector to point $p$.

My question is that we cannot use the same convention introduced to descend from integral on tensor to a calculus double integral.

Because $$dxdy=(dx \otimes dy + dy \otimes dx)/2 \neq (dx \otimes dy-dy\otimes dx)= \epsilon$$.

Do we have double standards or conventions to descend from forms/tensors to calculus differentials one for the metric calculations and another for the integral calculations?

• Isn't $dx^1 \otimes dx^2 - dx^2 \otimes dx^1 = 2 dx^1 \wedge dx^2$? – Matthew Leingang Sep 20 '16 at 13:02
• Yes it is. But when switching to calculus should we take $dx \otimes dy$ or $dx \wedge dy$ as $dxdy$. You know when writing the metric we take for example $dx \otimes dx$ as $dxdx=dx^2$ – user169903 Sep 20 '16 at 13:23
• If you think about what $dx\wedge dy$ does to two vectors, compared to $dx \otimes dy$, I think you'll see that $dx\wedge dy$ is the volume form on $\mathbb{R}^2$. – Matthew Leingang Sep 20 '16 at 13:34
• Well if we apply $dx \wedge dy$ to $(\partial_{y}, \partial_{x})$ we get a different result compared to $dx \otimes dy$. So the wedge product defines an "oriented" area element. So can we say that for the metric calculations we can consider $dx \otimes dy$ as $dxdy$ whereas for integral calculations we can consider $dx \wedge dy$ as $dxdy$? This is my main question here. – user169903 Sep 20 '16 at 14:10
• I think so. But when you say "for the metric calculations", I'm not recalling how that is defined. What does your reference say? – Matthew Leingang Sep 20 '16 at 14:39

The meaning of things like $dx\,dy$ in any context is always a convention. For example, in Riemann integrals (classical calculus) it is just a part of notation showing the names of variables.

When we consider a differential form, say, $f\,dx\wedge dy$, the integral of the form over a orientable differentiable surface $S$ is defined via the Riemann integral: $$\int_S f\,dx\wedge dy:=\iint_U f(x,y)\,dx\,dy$$ where $U$ is a domain of a chart containing the support of $\omega$.

Only in such a definition (with wedge product of coordinates' differentials) we automatically have symmetry in notation between Riemann integrals and integrals of forms. For example, you can substitute $x$, $y$ by another forms, then expand the expression according the wedge product properties, and you will get the same result as in Riemann integral in the new variables (which is includes the Jacobian).

This is why $\wedge$ used here.

In case of tensors, it's naturally to express it using basis of tensor products of $\partial_i$ and $dx^i$. For example, Riemannian metric $g$ as a $(0,2)$-tensor is naturally expressed in coordinates as: $$g=g_{ik}\,dx^i\otimes dx^k.$$

This is why $\otimes$ used here.

Physicists like to express metric via $ds^2=g_{ik}\,dx^i\,dx^k$ expression. Here $dx^i\,dx^k$ is the symmetrization of $dx^i\otimes dx^k$. The $ds^2$ is compact and intuitive ("element of length"). This is why $dx^i\,dx^k$ used here.