L4a: A word about orientation.

 When you have a vector space \( W \) with a metric and you want to talk about \( \textrm{SO}(W) \) instead of \( \textrm{O}(W) \). You do need an orientation, i.e., an element, up to positive multiples, of \( \wedge^{\dim W} W \). Of course, this orientation is invisible when you look at them as matrices, since you are choosing a basis, say \( \{w_i\} \), and the basis is giving you an orientation \(  w_1\wedge  \ldots \wedge w_n  \). If you choose a different basis, say \( \{-w_1\}\cup \{ w_2,\ldots,w_n \} \), you may get the opposite orientation, as it happens in this example: \( -w_1\wedge \ldots \wedge w_n \).
 
What about \( V\oplus V^* \)? Apart from a canonical pairing, there is a canonical orientation on \( V\oplus V^* \)! We have to give an element of
$$ \wedge^{(2\dim V)} (V\oplus V^*) = \wedge^{\dim V} V \otimes \wedge^{\dim V^*} V^*.$$ The canonical pairing between \( \wedge^{\dim V} V \) and \( \wedge^{\dim V^*} V^*\) gives that element. If you want, it is \( 1 \in \mathbb{R} \equiv \wedge^{\dim V} V \otimes \wedge^{\dim V^*} V^*,\)
where the isomorphism is given by the canonical pairing.

We will talk about \( \textrm{SO}(V\oplus V^*) \). Notice that the Lie algebra \( \mathfrak{so}(V\oplus V^*)\) is exactly \(\mathfrak{o}(V\oplus V^*) \), as it only depends on a neighbourhood of the identity element, which is the same for \( \textrm{SO} \) and \( \textrm{O} \).