I'm trying to work out the dimensions of some examples of Grassmannians but I can't seem to do it.
Here is what I understand:
The Grassmanian $G(k,n)$ is the set of all $k$ planes in $\mathbb R^n$. One can construct the Grassmanian as the quotient space of the set of $k$-frames with two $k$-frames $x,y$ equivalent if and only if there is an invertible matrix $m$ such that $y = mx$.
The whole set has dimension $kn$. The cardinality of one equivalence class is $k^2$. So I guess the dimension of the Grassmanian $Gr(k,n)$ should be $nk - k^2$.
Assuming this the dimension of $Gr(n, T\mathbb R^m) = \bigcup_{x \in \mathbb R^m} Gr(n. T_x \mathbb R^m)$ should be $2m n - n^2$ since the tangent bundle has dimension $2m$.
Now I found an example with $n=2, m=3$ that I don't understand:
Consider the map
$$ f(t,s) := (t^2 + 2s, t^3 + 3st, t^4 4 t^2 s)$$
The goal is to find an immersion $\widetilde{f}$ such that $\pi \circ \widetilde{f} = f$ (where $\pi$ is the canonical projection).
In the example, $\widetilde{f}$ turns out to be
$$ \widetilde{f} = (f,-2t^2, {8 \over 3}t)$$
so this is a map into $\mathbb R^5$ (but it's actually supposed to be an immersion into $Gr(n,T\mathbb R^m)$).
But if I use my formula above for the dimension of the Grassmannnian I get $(2m) n - n^2 = 12 - 4 = 8$ since in this example, $n=2$ and $m=3$.
Please could someone help me understand the dimenions of this example of Grassmanian/immersion?