order of parameters in differential forms

Question

I am completely mystified about parameters in differential forms and wherever I looked it's just glossed over.

Take for example the $2-$form $\omega = dx\land dy$ in $\mathbb{R}^n$.

Usually it's just defined like that, but given two vectors $(x,v_1), (x,v_2)$ in the tangent space at $x$, it shold be that $dx \land dy = dy \land dx$, since $$(dx \land dy) (x)((x,v_1), (x,v_2)) = $$ $$- (dy \land dx)(x)((x,v_2)(x,v_1))=$$ $$ (dy \land dx)(x)((x,v_1)(x,v_2))=$$

According to my book, tough, $dx \land dy = - dy \land dx$. Where am i wrong?

Answer 1

First of all, you're writing incorrect equations. Once you evaluate individual terms, there are no more wedges. The correct equation, using your (cumbersome!) notation, is \begin{multline*} (dx\wedge dy)((x,v_1),(x,v_2)) = dx((x,v_1))dy((x,v_2)) - dx((x,v_2))dy((x,v_1)). \end{multline*} And of course from this you see that, in fact, $dx\wedge dy = -dy\wedge dx$.

Answer 2

Answering my own question.

It's a misunderstanding of the anticommutativity of the wedge product of alternating tensors. I previously thought for some reason that, given $f,g$ alternating k-tensor and m-tensor respectively on $V$, with $V$ vector space over $\mathbb{R}$, $$(f \land g) (x_1, ..., x_k, ..., x_{k+m}) = (-1)^{km} (g \land f) (x_{k+1}, ..., x_{k+m}, x_1, ..., x_k)$$ while of course it is: $$(f \land g) (x_1, ..., x_k, ..., x_{k+m}) = (-1)^{km} (g \land f) (x_1, ..., x_k, ..., x_{k+m})$$ so the error is in the first equality.