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More specifically, if $B$ is the stabiliser of the origin, then there are two actions on $ \mathcal O(0)\oplus \mathcal O(0)$. One where $B$ acts in the usual way on $\mathbb C^2$, the other with $B$ acting trivially on the fiber.
However, the ${\rm SL}_2(\mathbb C)$ equivariant structure on a vector bundle is not unique: It corresponds with a $B$ equivariant structure on the fiber over the origin. The action of the Borel group $B$ is not determined by the K-theory class.
The main reason for the terminology is that this is the definition for affine schemes. One would not want a different meaning for affine group schemes.
One may also use that $\mathrm{SL}_n(\mathbb{Z})$ contains representatives of its Weyl group and that an invariant has to be fixed by them. But the Weyl group acts transitively on the standard basis of $\Lambda^i(\mathbb{Z}^n)$, so this means an invariant would be a scalar multiple of the sum of the basis.
One notes that the invariant is killed by matrices with one nonzero entry, off the diagonal. Now just compute explicitly with respect to the standard bases.
I do not find this bizarre. But then I got a position at that time. With much more freedom than is customary nowadays. Yes, we also had to teach. Times do change.
What is $F$ when $V$ is one dimensional? The symmetric algebra is then a polynomial ring $k[t]$. Does $F$ act on $k$? On $t$? By the way, it is much better to define the Reynolds operator on $V$, rather than on a ring.
