Question about vectors in the Grassmannian in this example

Question

Consider $f: \mathbb R^2 \to \mathbb R^3$ defined by $(t,s) \mapsto (t^2 + 2s, t^3 + 3ts, t^4 + 4t^2 s)$.

Let $Gr$ denote the Grassmannian and let $Gr(2, T\mathbb R^3) = \bigcup_{x \in T\mathbb R^3} Gr(2, T_x \mathbb R^3)$.

Note that

$$ {\partial f \over \partial t} = {\partial f \over \partial s} + s(0, 3, 8t)$$

Now let $v_1 = {\partial f \over \partial s} = (2,3t, 4t^2)$ and $v_2 = (0,3,8t)$.

I want to give a vector that is in $Gr(2,T\mathbb R^3)$. I have the solution but I don't understand why it's true:

We column-row eliminate the vectors $v_1, v_2$ so that we have

$$ v_1' = (1, 0, -2t^2) \text{ and } v_2' = (0,1, {8\over 3}t)$$

Then the vector I am looking for is given by

$$ (f, -2t^2, {8\over 3}t)$$

Please could someone explain to me why this holds?

Answer 1

How are you (they?) giving local coordinates on $G(2,\Bbb R^3)$? In a large open set of the Grassmannian, namely, the set of $2$-planes that project isomorphically onto the $e_1e_2$-plane, we take a basis for the plane to be $\{(1,0,a),(0,1,b)\}$, and $(a,b)$ gives a chart on that subset.