Orientability of Stiefel manifold $V_2(\mathbb R^4)$ What is an easy proof of orientability of Stiefel manifold $V_2(\mathbb{R}^4)$ (pairs of orthonormal vectors from $\mathbb{R}^4$ - subset of $\mathbb{R}^8$)? All  proofs I found deal with Lie groups and other complicated for me stuff. I suppose that there is an easy proof because task was given to me by my teacher in university, and our differential geometry course doesn't include Lie groups.
 A: Finally, I solved this problem. 


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*Lemma Let $f_1(x_1, \dots, x_n), \dots, f_k(x_1, \dots, x_n)$ be real functions; smooth manifold $M$ is solution set of system {$f_1(x_1, \dots, x_n) = c_1, \dots, f_k(x_1, \dots, x_n) = c_k$}  and $\mathrm{grad} f_1, \dots, \mathrm{grad} f_k$ are linearly independent in every point of $M$. Then $M$ is orientable.
Firstly, let's give alternative definition of orientability:
Smooth manifold M is orientable if and only if  $\;\forall$ loop $L(t)$ (continuous $L: [0; 1] \to M$ such that L(0) = L(1)) every continuous frame $F(t)$ of tangent space of $L(t)$ orientation of $F(0)$ is equal to orientation of $F(1)$ in tangent space of $L(0) = L(1)$.
Proof by contradiction. Exist $L$, $F$ such that $F(0)$ and $F(1)$ are not equally oriented. Let us consider matrix $Z(t)$ with rows: gradients $\mathrm{grad} f_1, \dots, \mathrm{grad} f_k$ in $L(t)$ and vectors of $F(t)$. Scalar product of $\mathrm{grad} f_i$ and $v_j$ from $F(t)$ is equal to $(f_i(w_j(0)))'$, where $w_j$ is coordinate line corresponding to $v_j$. $w_j(q) \in M \Rightarrow (f_i(w_j(0)))' = (c_i)' = 0 \Rightarrow $ scalar product is equal to 0. Therefore, for all $t$ all rows in $Z(t)$ are linearly independent, and $|Z(t)| \neq 0$.
Let's look on $Z(t)$ from the other side. It's gradient part is the same when $t = 0$ and $t = 1$, because gradient part depends only on point on manifold, and we have $L(0) = L(1)$. $F(0)$ and $F(1)$ have different orientations. Hence rows of $Z(0)$ and rows of $Z(1)$ are not equally oriented in $\mathbb{R}^n$. This means that, without loss of generality, that $|Z(0)| > 0$ and $|Z(1)| < 0$. $|Z|$ is continuous because of gradients and $F$ are continuous. Using these facts , we deduce existence of $\theta \in [0; 1] : |Z(\theta)| = 0$.
Contradiction.
Orientability of Stiefel's manifolds can be easily deduced from lemma, because Stiefel manifold can be represented as solution set of 
$x_1^2 + x_2^2 + x_3^2 + x_4^2 = 1;$
$x_5^2 + x_6^2 + x_7^2 + x_8^2 = 1;$
$x_1 x_5 + x_2 x_6 + x_3 x_7 + x_4 x_8 = 0;$
One can compute gradients and see that lemma conditions take place.
More generally, this lemma can be applied to any Stiefel manifold.
