Tangent bundle of Grassmann manifold I have to prove that the tangent bundle of Grassmann manifold $G_n(\mathbb{R}^{n+h})$ is isomorphic to $\operatorname{Hom}(\gamma^n(\mathbb{R}^{n+k}),\gamma^\perp)$, with $\gamma^{\perp}$ is the ortogonal complement of the tautological bundle $\gamma^n(\mathbb{R}^{n+k})$ in the trivial bundle $\epsilon^{n+k} $.
 A: There is a natural smooth action of $\text{GL}(n + k)$ on $\text{Gr}(n, k)$; fix a point $v$ to obtain a map$$f: \text{GL}(n + k) \to \text{Gr}(n, k),\text{ }g \mapsto g \cdot v.$$The derivative of this map at zero yields a map of vector spaces$$df_v : \text{Hom}(U, U) \mapsto T_v \text{Gr}(n, k),$$where we denote the total vector space $U = \mathbb{R}^{n+k}$. Certainly this map is surjective. Let $V$ be the hyperplane corresponding to $v$. There is a natural surjective map$$\text{Hom}(U, U) \to \text{Hom}(V, U/V)$$given by restriction and post-composition; we need to show that the kernel of this map is the same as that of $df_v$, and then we will be done, since that map will then factor through this one and hence induce an isomorphism.
The kernel of our second map is clearly linear maps which map $V$ into itself. The kernel of our first map is the vectors in directions which fix the point $v$; i.e. corresponding to one-parameter subgroups of $\text{GL}(n + k)$ whose action fix $V$. Hence their associated tangent vectors are the endomorphisms fixing $V$. So we have a fiberwise isomorphism of bundles; certainly it is smooth since it was factored through from $df_v$, which arises from the original smooth map (so locally approximates it by the constant rank theorem), so we are done.
A: This is one of Milnor-Stasheff problems. It is the first half of problem $5-B$. My proof at here is more expository than rigorous justification. For an algebraic rigorous proof, see 
Canonic identification of the tangent space of the Grassmannian
We have to show two things, one is the fibre over $V\in Gr(n+k,n)$ is canonically isomorphic to $Hom(V,V^{\perp})$. The other is there is bundle isomorphism
$$
T(Gr(n+k,n))\cong Hom(\gamma^{n}(\mathbb{R}^{n+k}),\gamma^{\perp})
$$
We let $L_{V}\in T_{V}$ be a tangent vector at base point $V$. Identifying $Gr(n+k,n)$ with a $nk$ dimensional manifold give us a map $T_{V}\rightarrow \mathbb{R}^{nk}$. Then $L_{V}$ is the vector along which we move from $V$ to  a nearby $n$-plane $U$ in $Gr(n+k,n)$. To fix this vector we need an element $f$ in $Hom(V,V^{\perp})$ so that $L_{V}$ corresponds to the $n$-plane spanned by $\langle e_{i}+fe_{i},e_{i+1}+fe_{i+1.}\cdots\rangle$. Therefore the set of all such vectors corresponds to $Hom(V,V^{\perp})$ and is isomorphic to $\mathbb{R}^{nk}$. 
To show the other direction we have to demonstrate that this correspondence is bijective and its inverse is also continuous. Here objectivity is clear, and I think both continuity and inverse continuity is also clear if you think about it geometrically. 
A: Let $V_n^O(\mathbb{R}^{n+k})$ denote the subset of $V_n(\mathbb{R}^{n+k})$ consisting of all orthonormal $n-$frames. Then there is a submersion $\pi^O: V_n^O(\mathbb{R}^{n+k}) \twoheadrightarrow Gr_n(\mathbb{R}^{n+k})$ sending an orthonormal basis to the linear n-subspace $X \in Gr_n(\mathbb{R}^{n+k})$ which it spans. Since choosing $x_i$ is the same as choosing $-x_i$ in a basis, and the order of basis no longer matters when we consider $Span<x_1, \dots, x_n>$, this gives a $(2^n)\cdot(n!)$-to-one cover of $Gr_n(\mathbb{R}^{n+k})$. [$n=1$ in the case of projective space].
Since $V_n^O(\mathbb{R}^{n+k}) \simeq_{diffeo} \{ (x_1, \dots, x_n) \in (S^{n+k-1})^n \text{ }|\text{ } x_{j+1} \in <x_1, \dots, x_j>^\perp \}$, we have that
$\tau(V_n^O(\mathbb{R}^{n+k})) = \{ ((x_1, \dots, x_n),(v_1, \dots, v_n)) \text{ }|\text{ } x_i\cdot x_i = 1,  x_i \cdot x_j = x_i \cdot v_i =0\text{ }\forall i \text{ }, \text{ }i\neq j\}$ since $\tau(S^{n+k-1}) = \{(x,v) \text{ }| \text{ }x\cdot x =1, x\cdot v = 0 \}$. Now $\pi^O$ identifies on tangent space all $n-$tuples of pairs
$$ \{((x_1, v_1), (-x_1,-v_1)), ..., ((x_n,v_n),(-x_n,-v_n)) $$
when we send an orthonormal set of $n$ vectors to the linear $n$-subspace which they span. [Again, choosing $x_i$ is the same as choosing $-x_i$ in a span, and order of basis no longer matters in $Gr_n(\mathbb{R}^{n+k})$.] Now each $v_i$ need not be orthogonal to $x_j$ for $j\neq i$, but if we project away $Span<x_1, \dots, x_n>$ from each $v_i$, this gives an element of $Hom(\gamma^n(\mathbb{R}^{n+k}), \gamma^\perp)$.
