Diffeomorphism between the Grassmannian manifolds $\mathbf{Gr}(n,k)$ and $\mathbf{Gr}(n,n-k)$. This seems to be a common exercise question, however I am having trouble with it. The hint is to use a map that associates the k-plane to its orthogonal complement. But I have not been able to show this as a diffeomorphism. 
I was thinking of using the orthonormal basis and extending it to get a basis of $\mathbb{R}^n$, then using the remaining $n-k$ vectors to generate the orthogonal complement. Am I on the right path??How else should I approach this otherwise??
 A: Yes, you're on the right path. The nice thing is that the implementation of your idea is already built into the description of $\text{Gr}(n,k)$ as $O(n) / O(k)\times O(n-k)$ which you seem to have used to turn $\text{Gr}(n,k)$ into a manifold: Representing a $k$-plane $E\subset {\mathbb R}^n$ by some element of $O(n)$ means choosing an orthormal basis $v_1,....,v_n$ of ${\mathbb R}^n$ such that $E = \langle v_1,...,v_k\rangle$, which you can then send to the $(n-k)$-plane $\langle v_{k+1},...,v_n\rangle=E^{\perp}$. In other words, on the level of sets your map is given by the composition $$\text{Gr}(k,n)\xleftarrow{\langle v_1,...,v_k\rangle\leftarrow A=(v_1,...,v_n)} O(n) / O(k)\times O(n-k)\xrightarrow{\alpha:\ A=(v_1,...,v_n)\mapsto \langle v_{k+1},...,v_{n}\rangle} \text{Gr}(n,n-k).$$
If you now want to show that this is even a diffeomorphism, you have to factor $\alpha$ through the bijection $$O(n)/O(n-k)\times O(k)\to \text{Gr}(n,n-k),\ A=(v_1,...,v_n)\mapsto\langle v_1,...,v_{n-k}\rangle$$ through which $\text{Gr}(n,n-k)$ gets its topology and manifold structure. Do you see how this can be done?
After request of details: You have the two sets $$\text{Gr}(n,k) := \{E\subset {\mathbb R}^k \ |\ \text{dim}\ E = k\}\quad\text{ and }\quad \text{Gr}(n,n-k) := \{E\subset {\mathbb R}^k \ |\ \text{dim}\ E = n-k\}$$ which you endow with smooth manifold structures through the bijections 
$$O(n)/O(k)\times O(n-k)\xrightarrow{\varphi_k} \text{Gr}(n,n-k),\quad A=(v_1,...,v_n)\mapsto\langle v_1,...,v_{k}\rangle\quad\text{and}$$
$$O(n)/O(n-k)\times O(k)\xrightarrow{\varphi_{n-k}} \text{Gr}(n,n-k),\quad A=(v_1,...,v_n)\mapsto\langle v_1,...,v_{n-k}\rangle,$$ respectively. Now you want to show that with respect to these smooth structures, the bijection $$(-)^{\perp}: \text{Gr}(n,k)\to\text{Gr}(n,n-k),\quad E\mapsto E^{\perp}$$ is  a diffeomorphism, that is, that the induced bijection $$\tau := \varphi_{n-k}^{-1}\circ (-)^{\perp}\circ \varphi_k: O(n)/O(k)\times O(n-k)\to O(n)/O(n-k)\times O(k)$$
is actually a diffeomorphism. Everything fits together in a commutative diagram:
$$\begin{array}{ccc} O(n)/O(k)\times O(n-k) & \xrightarrow{\varphi_k} & \text{Gr}(n,k)\\{\tiny\tau}\downarrow && \downarrow{\tiny(-)^{\perp}}\\ O(n)/O(n-k)\times O(k) & \xrightarrow{\varphi_{n-k}} & \text{Gr}(n,n-k)\end{array}$$
Now the point is that $\tau$ admits an explicit description on representatives: As an intermediate step, note first that $$(-)^{\perp}\circ\varphi_{k}: O(n)/O(k)\times O(n-k)\to \text{Gr}(n,n-k)\ \text{ is given by }\ A = (v_1,...,v_n)\mapsto \langle v_{k+1},...,v_n\rangle.$$ Hence, you only have to find an explicit lift of this map to a map $$O(n)/O(k)\times O(n-k)\to O(n)/O(n-k)\times O(k)$$ on representatives of which you can then see that it is a diffeomorphism. Can you do that?
