Signature and Orientation It is clear that a reordering of the elements in a chosen basis for an n-dimensional vector space induces a permutation on n elements, and conversely such a permutation corresponds to a re-ordering of the basis.
I am wondering why the signature of a permutation is associated to the orientation induced by a basis. I understand that one can do this technically - but is there an intuitive way to understand why this is so? 
That is, is there a way to understand  why the concept of orientation, as experienced in 1,2 and 3 dimensional Euclidean space, is formalized (and thereby generalized to abstract vector spaces of arbitrary finite dimension) using the signature? 
 A: We can define the signature of a permutation to be the parity of the number of transpositions required to transform the given permutation into the trivial permutation.  With this definition your question reduces to: why does switching exactly two basis vectors reverse the orientation of the basis?
Of course the real answer is that it follows from the definition of orientation.  But it's possible to make intuitive sense of this by going back to the motivation for nailing down the concept of an orientation in the first place.  In $\mathbb{R}^2$ one wants to parametrize rotations using real numbers (at least locally).  But there is an ambiguity: what is the difference between rotating by $+60$ degrees and by $-60$ degrees?  This comes down to deciding whether $+60$ degrees should be a clockwise or a counterclockwise rotation, and this choice can be encoded by specifying an ordering for the standard basis of $\mathbb{R}^2$: to pass from $e_1$ to $e_2$ you have to rotate counterclockwise, while passing from $e_2$ to $e_1$ requires a clockwise orientation.
Something similar happens in $\mathbb{R}^3$ except now you have two angular coordinates and the ambiguity is more severe than just clockwise versus counterclockwise.  The problem can be framed as follows.  Fix an angular coordinate system $(\alpha,\beta)$ and choose a plane orthogonal to the axis of rotation for one of $\alpha$.  Position yourself at the origin of this plane so that, from your point of view,  positive $\alpha$ corresponds to a counterclockwise rotation.  Now $\beta$ is a rotation transverse to the plane that you're standing on, so positive $\beta$ either rotates up or down from your point of view.  A positive orientation on $\mathbb{R}^3$ corresponds to an angular coordinate system in which positive $\beta$ rotates up.  This is consistent with the orientation conventions used by physicists studying electromagnetic fields, for example.
Now we just need to relate this definition of orientation to a choice of basis.  Given a basis $\{v_1, v_2, v_3\}$ of $\mathbb{R}^3$, rescale the vectors so that they are all unit vectors i.e. points on the unit sphere.  Imagine drawing a small geodesic triangle on the sphere with vertices $v_1$, $v_2$, and $v_3$ and orient the edges of the triangle so that $v_1$ points to $v_2$, $v_2$ points to $v_3$, and $v_3$ points back to $v_1$.  Look at the triangle straight on from outside the sphere and ask: does the triangle go clockwise or counterclockwise?  If you define an angular coordinate system corresponding to $\{v_1, v_2, v_3\}$ by allowing $\alpha$ to be a rotation in the direction from $v_1$ to $v_2$ about the axis perpendicular to the plane spanned by $v_1$ and $v_2$ and similarly for $\beta$ except with $v_1$ and $v_3$, then a positively oriented triangle is one which goes counterclockwise.  Finally, in this picture it is clear that if you swap any two vectors (i.e. any two vertices of the triangle) then the orientation goes from positive to negative.
