# elementary question regarding differential forms

Is it possible to give a high level explanation why changing the order of differentials will give rise to a minus sign ? I.e. why do we have $$dx\,dt = - dt\,dx$$ (I am going to take a course on manifolds next semester which will hopefully shed some more detailed light on this, but for the moment I'd need some explanatnion to help me along whilst I'm still lacking the proper foundations -if such an explanation is possible)

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I dont know what you mean. – mick Sep 21 '12 at 20:09
@mick where does the minus come from ? – Beltrame Sep 21 '12 at 20:14
Have you had a look at the wikipedia article link ? – Raeder Sep 21 '12 at 20:17
@Raeder I m afraid for the moment I may need something simpler if possible, even though I realise it will need some handwaving :-( ... but I hope to be able to rectify that downside over the next couple of months. – Beltrame Sep 21 '12 at 20:30
I think you ask about the wedge. – mick Sep 21 '12 at 20:30

I think it might be better to leave the wedges in: $dx\wedge dt = - dt\wedge dx$. Otherwise this quickly leads to confusion. Now to you question.

For one thing: If you do it this way, the determinant pops out naturally under change of bases. And this is essential for integration, which is what these things were designed to do (remember the change of variables formula from multi-variable calculus!).

Here's a $2$-dimensional example. Let $$A = \begin{pmatrix} a& b\\ c & d\end{pmatrix}$$ be the change of coordinates matrix from Euclidean coordinates to some other set of coordinates. Then \begin{align} Ae_1 \wedge Ae_2 &= (a e_1 + c e_2)\wedge (be_1 + de_2) \\ &= ab\; e_1 \wedge e_1 + ad\; e_1\wedge e_2 + cb \; e_2\wedge e_1 + cd \; e_2 \wedge e_2 \end{align} Now using the fact you mentioned we see that the first and last terms are zero and obtain $$Ae_1\wedge Ae_2 = (ad - bc) \; e_1 \wedge e_2 = \det(A) \; e_1 \wedge e_2$$ But then if we write $e_i = dx_i$ and interpret them as infinitesimal lengths, their wedge product as infinitesimal area, then this is nothing but the change of variables formula in this special case.

It might be fun to prove that this formula generalizes to higher dimensions (an exercise in linear algebra).

So you see that this is exactly the kind of behaviour one needs in integration. A problem that immediately pops up is that of orientation - what do we do if $\det A < 0$? How might we build a theory of integration on manifolds from this? Can we do so on any manifold?

For the answers to these questions you will have to learn some differential geometry. :)

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Consider the 2-form:

$$\omega = a(\mathbf{x}) \, dx_1 \wedge dx_2$$

It assigns to a 2-surface $\Phi$ the number:

$$\int_{\Phi} \omega = \int_D a(\Phi(\mathbf{u}))\frac{\partial(x_1, x_2)}{\partial(u_1, u_2)}\,d\mathbf{u}$$

Now consider the 2-form:

$$\overline{\omega} = a(\mathbf{x}) \, dx_2 \wedge dx_1$$

It assigns to the same 2-surface $\Phi$ the number:

$$\int_{\Phi} \overline{\omega} = \int_D a(\Phi(\mathbf{u}))\frac{\partial(x_2, x_1)}{\partial(u_1, u_2)}\,d\mathbf{u}$$

The difference is the determinant. If you swap two rows of a determinant, it changes sign. Thus, We have:

$$\frac{\partial(x_2, x_1)}{\partial(u_1, u_2)} = - \frac{\partial(x_1, x_2)}{\partial(u_1, u_2)}$$

Combined this with the fact that:

$$\int_\Phi -\omega = - \int_\Phi \omega$$

And now it should be clear why:

$$dx_2 \wedge dx_1 = - dx_1 \wedge dx_2$$

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Following Tao it's important to have the intuition about orientation. n-dimensional space could have 2 orientations. In 1-dimension we would use dx and -dx. In general we want the exterior product to capture the oriented Vol. (So it should be multilinear.) Also we should have dx$\wedge$dx=0. For any nice algebra A(+,*) that have XX=0 we have (X+Y)(X+Y)=0

(X+Y)(X+Y)=XX+XY+YX+YY= 0+XY+YX+0 So XY=-YX.
Note that on the other hand (dx$\wedge$dy)$\wedge$(dv$\wedge$dw)= dx$\wedge$dy$\wedge$dv$\wedge$dw= -dx$\wedge$dv$\wedge$dy$\wedge$dw= ... =(dv$\wedge$dw)$\wedge$(dx$\wedge$dy)

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"Following Tao"? What do you mean by that? What source are you referring to? – Michael Albanese Sep 29 '15 at 19:36