I was exploring what the determinant’s sign means geometrically. For 2-D, you can swap the axes and you’ve flipped orientation, and once you swap them again, you get the original orientation.
Why should this be the case for higher dimensions? Why is it not possible that in 3-D, I can swap i and j, then swap j and k, and I get a “third orientation”?
How is orientation defined anyway?
Here is a geometric reason for the definition of orientation.
The set of all invertible $n \times n$ matrices forms a group denoted $GL(n,\mathbb R)$, and it is identified with the set of all invertible linear transformations $T : \mathbb R^n \to \mathbb R^n$.
This group decomposes into two subsets, which I'll denote $$GL(n,\mathbb R) = {GL}_+(n,\mathbb R) \cup {GL}_-(n,\mathbb R) $$ where the subscript represents the sign of the determinant.
Here are some key algebraic facts (easily derived from the theorem that the determinant of a product equals the product of the determinants):
These three algebraic facts can be summarized in a single statement: there is a group homomorphism $\sigma : GL(n,\mathbb R) \to \{-1,+1\}$ which assigns $+1$ to a matrix of positive determinant and $-1$ to a matrix of negative determinant.
Here are two key geometric facts:
You asked:
Why is it not possible that in 3-D, I can swap i and j, then swap j and k, and I get a “third orientation”?
Because if you swap $i$ and $j$ using the following reflection matrix (which has negative determinant) $$R_1 = \pmatrix{0&1&0\\1&0&0\\0&0&1}$$ and after that if you then swap $j$ and $k$ using the following reflection matrix (which also has negative determinant) $$R_2 = \pmatrix{1&0&0\\0&0&1\\0&1&0} $$ and then you formulate the product matrix $$R_2 R_1 = \pmatrix{0&1&0\\0&0&1\\1&0&0} $$ then $R_2 R_1$ has positive determinant, as you can check algebraically. Therefore, once you have convinced yourself of the algebraic and geometric facts listed earlier, you will be confident that $R_2 R_1$ does not represent a "new orientation" because it is connected by a continuous path of matrices with every other positive determinant matrix, i.e. with every rotation matrix and with the identity itself.
In fact, and you can verify this by calculation, $R_2 R_1$ is itself a rotation matrix: it is a rotation by $2\pi/3$ around the axis which is parameterized by $t \mapsto (t,t,t)$. If you fix this axis, and then vary the angle of rotation continuously, starting from the angle $2\pi/3$, and varying it down to the angle $0$, you will have produced a continuous path which connects the matrix $R_2 R_1$ with the identity matrix.
To summarize, the reason there is not a "third orientation" is because orientation is defined so as to summarize, and to be consistent with, the key algebraic and geometric facts that are listed above. In other words: invertible $n \times n$ matrices are naturally subdivided into two classes, and we choose to describe those two classes using the terminology of "orientations".
One last word: If you've not seen that theorem before, digging into its proof would be an excellent way to understand the geometric reasons why orientation is defined the way that it is.
A reasonable question, indeed, on geometric grounds.
If you track the details of your swap-$i$-and-$j$, and then $j$-and-$k$, you'll see that that composite action is a rotation (preserving orientation).
More generally, the composition of two reflections (the simplest orientation-reversing isometries) is a rotation (orientation-preserving). You can work out the details either geometrically or algebraically.
