I am trying to understand the proof of the following proposition. It is taken from Lee's book "Introduction to smooth manifolds":
I have trouble understanding the 0-dim. case. In partucluar why $\omega>0$ implies $\mathcal{O}_{\omega}$ is +1 and the other case as well.
Since it says this case is immediate I suppose I am missing something obvious, but unfortunately I am completely stuck. I am also not sure what "consistently oriented" means in the case of a 0-dim. vector space. There is only a definition for the case $n\geq 1$. In Jänich's book "Vector Analysis" I found something that would make sense to me regarding the definion of orientaion in the 0-dim. case. It is the following:
Then $\mathcal{O}_{\omega}=[\emptyset]$, since the empty set is the only ordered basis in $V=0$?
Thank you very much in advance!
Best Answer
Note that my book gives a separate definition of orientations of zero-dimensional vector spaces, which doesn't have anything to do with consistently oriented bases. The definition appears on page 379, about a third of the way down the page:
In the statement of Proposition 15.3, the sentence "if $n=0$, then $\mathscr O_\omega$ is $+1$ if $\omega>0$, and $-1$ if $\omega<0$" is not a claim to be proved -- it's the definition of the orientation $\mathscr O_\omega$ determined by a particular choice of $\omega$.
Since $\Lambda^0 V^* = \mathbb R$ (no matter what the dimension of $V$ is), in the case $n=0$, an element $\omega\in \Lambda^n V^*$ is just a real number, and any such number specifies a unique orientation of $V$ by the recipe quoted above.
The only thing to be proved in that case is the claim that two nonzero $0$-covectors (i.e., nonzero real numbers) determine the same orientation if and only if each is a positive multiple of the other, and this is just the statement that two nonzero real numbers have the same sign iff each is a positive multiple of the other.