The vector space dimension of the cohomology group of the $2$-plane Grassmannian $\mathrm{Gr}\_{2,n}$ is given by the number of tuples $(\lambda\_1,\lambda\_2)$ satisfying
$$
n - 2 \geq \lambda\_1 \geq \lambda\_2 \geq 0.
$$
Explicitly this is given by
$$
\binom{n}{2}.
$$
This also happens to be the dimension of $V\_{\pi\_2}$ the second fundamental representation of $\frak{sl}\_n$. I am guessing this is not an accident, especially since the $2$-plane Grassmannian corresponds (in the usual way) to $V\_{\pi\_2}$.


Does this extend to the general identity
$$
\mathrm{dim}(H^{\*}(\mathrm{Gr}\_{d,n})) = \mathrm{dim}(V\_{\pi\_d})?
$$
If it does, then what is a conceptual explanation for this?


EDIT: Since $V\_{\pi\_d}$ is isomorphic to the exterior power
$$
\Lambda^d(V\_{\pi\_1})
$$
and $V\_{\pi\_1}$ is of dimension of $n$, we see that the RHS of the claimed identity is the binomial coefficient
$$
\binom{n}{d}.
$$
It follows from the general formula given in this [answer](https://mathoverflow.net/questions/196546/hard-lefschetz-theorem-for-the-flag-manifolds) that the LHS is the same binomial coefficient. Thus the identity does indeed extend from $2$-planes to $d$-planes. So the question is if there is a conceptual reason for this . . .



The dimension of the Grassmannian cohomology ring $H^*(\mathrm{Gr}_{n,d})$ and the fundamental $\frak{sl}_n$-representation $V_{\pi_d}$
This is very standard. For a compact complex variety admitting a cell decomposition, the (co)homology is the free Abelian group generated by the cells (over $\mathbb{C}$ there is no room for the differential in the cellular complex). In the cellular decomposition of the Grassmannian, the (Schubert) cells are indexed by possible reduced row echelon forms of a $d\times n$-matrix, that is by possible positions of pivots. To the cell having pivots in positions $i\_1,\ldots,i\_d$ you can assign the basis element $e\_{i\_1}\wedge\cdots\wedge e\_{i\_d}$ in $\Lambda^d(V\_{\pi\_1})$.


